User talk:Ethan Skyler

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Talk:Reactive centrifugal force
I've replied to your comments on Talk:Reactive centrifugal force. It appears that you are confusing the net force on a single object, which clearly does not have to be zero otherwise nothing would ever accelerate according to Newton's 2nd Law, with the action and reaction forces between two different objects, which must equal according to Newton's 3rd Law. You are also confusing the net force on a single object in a multi-object system with the net external force that accelerates the entire system. The net force on just the car, 1 lb in your example numbers, is not the same as the net force on the car and driver, 20 lb in your example numbers, provided by the pavement.

The "net force" formulation of Newton's 2nd Law does not contradict Newton's 3rd Law. The former is concerned only with the sum of the external forces on a single body and seems to be explained by Newton himself in his Corollary I:
 * A body by two forces conjoined will describe the diagonal of a parallelogram, in the same time that it would describe the sides, by those forces apart. -page 21 of volume 1 of the 1729 translation

The latter is concerned only with the equal and opposite pair of forces between two different bodies. -AndrewDressel (talk) 14:36, 25 January 2010 (UTC)


 * Hi Andrew: You very likely are right about my confusion.  Partly I think it is due to the multi-part nature of the event.  I am thinking about describing a simpler event and going on from there.  Duty calls.  More later.  Thanks for responding!  Ethan Skyler (talk) 15:27, 27 January 2010 (UTC)

Jan 28 Newton's explanations of events leave much to be desired in my mind. I would need to see a drawing with description before understanding could begin to set in.


 * I have added to the reactive centrifugal discussion page. I have several more additions in mind as time allows. If you delay your comments until I am mostly done, that is fine by me.  I read today that this article has a B - Low rating.  Maybe we can raise this rating to the A - High rating it deserves.  Ethan Skyler (talk) 19:55, 28 January 2010 (UTC)

Before some other editor objects and kicks us out, under Talk page guidelines, I've moved the discussion to here. If and when we do come up with some more concrete changes for Reactive centrifugal force, we can continue it there. I hope you do not take offense. -AndrewDressel (talk) 21:06, 29 January 2010 (UTC)

=Comments from January 2010=
 * Reactive Centrifugal Force clearly is an unresolved topic both here in Wikipedia and in physics in general. Certain components of mechanics are awaiting its correct recognition including the "net force" theory of acceleration. Determining exactly how an action force impressed against a 1st object, by a 2nd object, can act as the cause of the 1st object's acceleration at the same time that an equal and opposite reaction force is present at every point one chooses to measure, has proven to be a troublesome long-standing problem. In linear acceleration events this equal and opposite reaction force has no specific name.  We experience it when accelerating in a car or on an airplane that is preparing for lift off. Insert a compression scale between your torso and the seat back and the presence of a forward-directed acceleration action force supported by a mutual rearward-directed acceleration reaction force of equal magnitude is revealed.  The same happens in circular events.  Sit in a swivel chair mounted to and some distance from the axis of a rotating turn table.  Turn your chair so you face toward the axis and then spin the turn table. Insert a scale between your torso and the seat back.  Again observe the scale's reading indicating the presence of an inward-directed acceleration action force supported by a mutual outward-directed acceleration reaction force which is always equal in magnitude.  Here the action force is termed centripetal (directed toward center) and the reaction force is termed centrifugal (directed away from center). Notice that in the airplane the reaction force is directed to the rear which is opposite to the forward-directed accelerational activity.  Now notice on the rotating turntable the reactive centrifugal force is outward-directed which is again opposite to the inward-directed accelerational activity. From this fact one can postulate that the reactive centrifugal force is never the cause of any event.  One can also postulate using logic alone that given the ineffectiveness of the reactive centrifugal force in bringing the object's acceleration to a halt predicts that this reaction force is caused by and is therefore a byproduct of the event causing acceleration action force. Think of the reactive centrifugal force, that is measurably present, as a reaction force being reflected back by the accelerating object. Using this byproduct perspective, one could conclude that the reactive centrifugal force requires different rules that take into account its inability to cancel its own action-force cause.  For this reason it can be ignored when performing a vector addition of the action forces present in accelerational events.  "Net action force" is what is already being determined.  Yet its existence cannot be ignored as it is verified by the scale's reading and furthermore predicted by Newton's LAW III to be the equal reaction force that is "always" present while providing support for the event's net action force.


 * Statements within the Wikipedia article on Reactive Centrifugal Force that are in agreement with the above arguments, in my opinion represent a good beginning. Statements that are in conflict, such as writing that the action force in circular events is outward-directed, which is always opposite to the direction of the object's acceleration, in my opinion are in need of reexamination.  I have no interest in changing someone else's work.  I make these suggestions to encourage others to consider making their own changes.  Ethan Skyler (talk) 09:11, 23 November 2009 (UTC)


 * The caption to the turning car with driver drawing caught my eye. I quote:

" Exploded view showing force components. Each object is subject to a net inward force that is the difference between the outward reactive centrifugal force and an inward centripetal force. This net inward force is the centripetal force upon that object necessary for it to make the turn."


 * This caption tells me that it is correct to think that when the total inward active centripetal force (say +20 lb) is added to the lesser outward reactive centrifugal force (say -19 lb) that the difference (+1 lb) is the "net" inward force predicted to be the centripetal force upon that object necessary for it to make the turn.


 * Yes, that is correct, though your example numbers might not be accurate. -AndrewDressel (talk) 18:43, 24 January 2010 (UTC)


 * If this is correct then 1) there will exist an imbalance of forces present (the inward force is predicted in the caption to be larger than the outward force which is not possible (see Newton's LAW III)) and 2) the object is predicted to experience centripetal acceleration at the rate caused by a 1 lb force. I have performed a centripetal acceleration experiment using a whirling lead weight attached by cord to a tension scale.  Based upon the results of my experiment, I assure you that in this centripetal event the turning object will experience a rate of centripetal acceleration caused by the full 20 lb centripetal force.


 * You are confusing the net force on a single object, which clearly does not have to be zero otherwise nothing would ever accelerate according to Newton's 2nd Law, with the action and reaction forces between two different objects, which must equal according to Newton's 3rd Law. You are also confusing the net force on a single object in a multi-object system with the net external force that accelerates the entire system. The net force on just the car, 1 lb in your example numbers, is not the same as the net force on the car and driver, 20 lb in your example numbers, provided by the pavement. -AndrewDressel (talk) 18:43, 24 January 2010 (UTC)


 * I realize that this caption is written to defend and support the "net" force theory of acceleration. But the "net" force theory sprang from a lesser mind than Newton's.  Newton's LAW III rules all forceful events.  Acceleration occurs while equal and opposite action & reaction forces are immediately present.  Hence the contents of this caption are illogical, contrary to my experiment, and generally misleading.

Ethan Skyler (talk) 06:37, 21 January 2010 (UTC)


 * Again, the "net force" formulation of Newton's 2nd Law does not contradict Newton's 3rd Law. The former is concerned only with the sum of the external forces on a single body and seems to be explained by Newton himself in his Corollary I:
 * A body by two forces conjoined will describe the diagonal of a parallelogram, in the same time that it would describe the sides, by those forces apart. -page 21 of volume 1 of the 1729 translation
 * The latter is concerned only with the equal and opposite pair of forces between two different bodies. -AndrewDressel (talk) 18:43, 24 January 2010 (UTC)


 * Today I am looking at the first drawing with 6 panels of a ball whirling about a post to which it is attached by a string. I agree with 5 of the 6 panels and appreciate the use of green vectors depicting the reactive centrifugal force.  It is Panel 3 that I think needs some attention.  I quote:

"Illustration of forces between objects – 1: Ball in uniform circular motion held by string tied to post; 2: Centripetal force on ball; 3: Reactive centrifugal force on string and post;"


 * I am wondering why the ball is missing from Panel 3? If it were truly missing from this event then there would not be such a large reactive centrifugal force present.  What we are concerned with here is the acceleration of the matter of the string and the ball.  This matter is accelerating away from a straight-line path repeatedly drawn tangent to the matter's curved path of travel. Many different rates of centripetal acceleration are occurring depending upon the distance the matter exists from the axis represented by the post.  The majority of the accelerating matter exists in the ball.  Even within the ball, different rates of acceleration for the ball's myriad of components of matter exist again depending upon the component's distance from the axis.  Portions of the ball's matter closer to the post experience less centripetal force and less centripetal acceleration while portions of the ball's matter farther from the post experience more centripetal force and more centripetal acceleration.


 * The ball is missing from Panel 3 for the same reason that the string is missing from panel 2, merely for clarity, though perhaps it fails at this. -AndrewDressel (talk) 18:43, 24 January 2010 (UTC)


 * If Panel 3 is used to indicate the presence of the reactive centrifugal force in this whirling event and further if the total reactive centrifugal force is depicted by the large green force vector then logically all the matter experiencing centripetal acceleration must be present in the drawing.


 * Perhaps the ball has been left out of the drawing again in support of the "net" force theory of acceleration. Or perhaps it is the "forces only affect different objects" directive that is being obeyed.  Whatever the cause, there is no denying that both the centripetal acceleration-causing action force is effecting the ball's matter at differing magnitudes with this action force being supported at every location with the ball by equal and opposite centrifugal acceleration-reaction forces also of differing magnitudes and also present within the same object.


 * If whirled fast enough, the ball will stretch out becoming oval in shape in response to the differing rates of centripetal acceleration supported internally by the real reactive centrifugal forces present. Key here is that both the action and reaction forces are present within the same object.  Again, if the total reactive centrifugal forces present in Panel 3 are to be shown as depicted by the large green arrow, then logically all objects that are accelerating and reactively generating this force need to be included in the drawing.

Ethan Skyler (talk) 18:03, 22 January 2010 (UTC)


 * Of interest next is that overused and poorly understood term "inertia". The offending quote from the article is as follows:

"Viewed from an inertial frame of reference, the passenger's inertia keeps the passenger moving with constant speed and direction as the car begins to turn; "


 * From a non-accelerating frame of reference, the reader is told that prior to the car turning, the car and passenger are both moving at a constant speed and direction (constant velocity). Then when the car enters the turn, (figuring no friction force between the passenger and any portion of the car) the passenger continues moving straight ahead with constant speed and direction caused by the presence of the passenger's "inertia".


 * This "inertia" reference is kind of silly when you think about it since this term is presented as a cause when no cause is required. I will explain.  Here we have an object (the passenger) being carried along by the car with this object ready to respond as accelerational forces become present.  The passenger's linear motion will increase if a forward action force becomes present.  The passenger's linear motion will decrease if a rearward action force is present.  The passenger's linear motion will instantly switch to a curved motion to the left if a leftward action force is applied.  The same thing will happen in the opposite direction if a rightward action force is applied.


 * In each of these four events, we have the action of acceleration being caused by an action force. Now suppose, while at speed with a constant velocity, all such acceleration-action forces become absent.  (Here allow me to assume a friction-free environment)  What do you think will happen to the car's passenger?  Do you think there is anything present to cause the passenger to slow down...  or to speed up... or to turn right... or left?  No?  You are correct.  So what choice does the passenger/object have?  If the passenger's velocity can only change when a forceful cause is present and with all such forceful causes absent, then does the passenger have but one available choice?  Presented this way, it is easy to see that the only available choice is for the passenger/object to continue in a default manner with no change in his/its state of motion?  The action forces cause acceleration for the passenger.  When they are absent no change of the passenger's motion is possible.  To think that this default forceless state of motion needs to be assigned a cause when the passenger/object is experiencing the causeless event known as uniform motion is just plain silly.


 * We all know that once set into motion in a frictionless environment, an object will continue on forever with its motion unchanged as long as accelerational forces are absent. To say that the continuation of this default, causeless, state of uniform motion is caused by the object's "inertia" is just plain backward.  Backward all the way to Aristotle where he taught that the continuation of an object's uniform motion was caused by a "mover".  Aristotle's "mover" cause is as backward as Newton's "inertia" cause.  Instead uniform motion is a default event for which no cause exists.  Just like we knew all along.

Ethan Skyler (talk) 10:01, 23 January 2010 (UTC)


 * Interesting point. Inertia is what causes the reactive centrigual force, but if there is no force between the car and driver, then the driver's inertia is not a factor. I've removed the implication that inertia keeps the passenger moving in a straight line. -AndrewDressel (talk) 18:43, 24 January 2010 (UTC)


 * Next let us look at using the term "act" or "action" when describing the role of the reactive centrifugal force in an accelerational event. I quote:

"If the car is acting upon the passenger, then the passenger must be acting upon the car with an equal and opposite force (Newton's third law of reaction). Being opposite, this reaction force is directed away from the center, as illustrated in the lower figure, therefore centrifugal: this centrifugal force acts upon the car seat, not upon the passenger.[5]"


 * Here the author mixes it up by saying the reactive centrifugal force from the passenger is "acting" upon the car. He concludes that "this centrifugal force "acts" upon the car seat, not upon the passenger."  In this event, the force causing the acceleration is the action force.  The force providing the support for this centripetal action force is the centrifugal reaction force.  Mixing up these terms in the description quoted above is no help.  The author's confusion is transferred on to the reader.


 * The way it is written now seems clear to me, but I cannot follow your analysis. Also, as Haliday and Resnick explain in their discussion of Newton's 3rd Law: "Either force may be considered the action and the other the reaction. Cause and effect is not implied here, but a mutual simultaneous interaction is implied." (emphasis theirs) -AndrewDressel (talk) 18:43, 24 January 2010 (UTC)


 * Then staking the claim that the passenger's centrifugal reaction-to-acceleration force reacts upon the car seat and not upon the passenger seals this description from being any help at all. What does the author think, that the reaction force from the passenger just suddenly appears at the contact point between the passenger and the seat?  How would that be possible?  I look to the matter that is accelerating.  That is the source of the reactive centrifugal force we are all trying to describe.  Each accelerating component of the passenger's matter is the source of that component's reactive centrifugal force.  These myriad of reaction forces stack up through the passenger's matter in the direction opposite to the direction of the centripetal acceleration.  In a linear acceleration event such as heavy braking, this is the reason your eyes feel heavy and your blood rushes forward to your face.  The action and reaction forces are present throughout the matter of every accelerating object, not just at the contact points between accelerating objects.

Ethan Skyler (talk) 19:22, 23 January 2010 (UTC)


 * I don't see the point of subdividing the passenger further. Two pieces, torso and head, seem sufficient to get the point across. -AndrewDressel (talk) 18:43, 24 January 2010 (UTC)


 * Hello Andrew: I am glad to read from your comments that my words were taken as helpful. Overall I appreciate your efforts in providing recognition for the reactive centrifugal force. I really enjoy thinking about such events.  I make mistakes at times but like to think that I get things right in the end.  Perhaps my perspective on these matters will prove worthy of your consideration. I don't know if you authored the entire article.  To keep things simple, I will assume that it is all your work.  It is easy to brush this subject aside as unimportant. Instead I am convinced that the considerations surrounding centripetal acceleration and centrifugal reaction to that acceleration represent an important cornerstone of this science.  I would like to continue offering suggestions on the balance of the article. I also have some additional thoughts regarding points already touched on in this discussion.  Rather than inserting additional comments above which can make for a point/counter point jumble, I will reopen any further discussion of previous items below.  Ethan Skyler (talk) 15:29, 27 January 2010 (UTC)


 * No, it appears that I have made no contribution to this page until now. -AndrewDressel (talk) 20:23, 28 January 2010 (UTC)

=Comments from January 28, 2010=
 * Jan 28 - I would like to have a second look at the beginning of the article including the 6 panel chart of the net centripetal acceleration-causing action force and the mutual centrifugal reaction force. Earlier I objected to the ball missing from Panel 3. Now I realize that my concern will be resolved if the reactive centrifugal force being experienced by the ball is added to Panel 2 in the same way that these mutual forces are shown affecting the post in Panel 6.


 * The way reactive centrifugal force is currently defined in this article, there is none on the ball. The centripital force acting on the ball that causes it to go in a circle is paired, by Newton's 3rd Law, with its equal and opposite reactive centrifugal force acting on the string. -AndrewDressel (talk) 20:23, 28 January 2010 (UTC)


 * Currently Panel 2 indicates that the ball is experiencing only an inward-directed centripetal acceleration-causing action force. While at first it may seems logical that this is the correct assessment, consider the following. Assumming a sufficiently strong post, string and attachments to the post and ball, no matter what the ball's material makeup, whirling the ball faster and faster will always result in the ball becoming oval in shape and ultimately result in the ball experiencing structural failure as it separates into pieces. The ball will be torn apart, possibly in half. Such a dismemberment of the ball can, in my view, only be caused by the presence of equal and opposite forces. These mutual forces are not only both present at the intersection between the ball and the string but also deep within the ball's matter from the side nearest the post all the way to the ball's far side. In its attempt to exceed the ball's reactive centrifugal force, (which it never does) the active centripetal force causes the ball to oval and eventually to separate into pieces. Two force vectors in Panel 2 are, in my opinion, justified.


 * Ah ha. You must be thinking of the Centrifugal force (rotating reference frame) that already has its own article. In the fixed frames used in this article, that centrifugal force does not appear, just as in the fixed frame of a drag strip, there is no force that presses the drivers into their accelerating seats. -AndrewDressel (talk) 20:23, 28 January 2010 (UTC)


 * The way I view this event is that the active centripetal force experienced by the ball's near side becomes so strong that at first the ball takes on an oval shape as its far side begins to lag behind at a slightly reduced rate of acceleration. Eventually the ball experiences structural failure as the rate of centripetal acceleration of its far side components is reduced to zero allowing the pieces to leave this whirling event while following a new tangential path.


 * I realize that if one thinks of the reactive centrifugal force as having the power to cancel the accelerational ability of the centripetal action force from the post and string then one could be led to think that with equal and opposite mutual forces drawn for the ball in Panel 2, no acceleration could occur for the ball. Yet the reactive centrifugal force is caused by the centripetal force and only present during centripetal acceleration so it certainly cannot be expected to serve in any way to cancel its own accelerational cause. Instead it takes on an acceleration-supporting role as it benignly goes along for the ride.  Ethan Skyler (talk) 19:26, 28 January 2010 (UTC)


 * Yes, in the rotating frame attached to the ball, the centripital force of the string is equal and opposite the centrifugal force due to rotation and the ball does not accelerate. -AndrewDressel (talk) 20:23, 28 January 2010 (UTC)

=Comments from the evening of January 28, 2010=
 * Jan 28 Evening - I think of a force as a push or pull being experienced by an object. This force is invisible but its presence may often, but not always, be detectable by a tension or compression scale.  This force is also absolute in that no matter from what frame you choose to observe, the force and the object's experience remain the same. An object experiencing an acceleration/Action force of a given measurable magnitude will experience a given rate of acceleration regardless of whether you choose to view it from a non-accelerating frame or a frame that is itself accelerating.  Newton's F = ma represents the absolute relationship between the quantity of an object's matter and the magnitude of the acceleration/Action force that results in the generation of but one rate of acceleration.


 * Unfortunately, that has not been found to be valid. From the article on the principle of relativity:
 * According to the first postulate of the special theory of relativity:


 * This postulate defines an inertial frame of reference.
 * This postulate defines an inertial frame of reference.


 * The special principle of relativity states that physical laws should be the same in every inertial frame of reference, but that they may vary across non-inertial ones. This principle is used in both Newtonian mechanics and the theory of special relativity. Its influence in the latter is so strong that Max Planck named the theory after the principle.


 * Also, from the article inertial frame of reference:
 * By contrast, in non-inertial reference frames, the laws of physics are dependent upon the particular frame of reference, and the usual physical forces must be supplemented by what are called fictitious forces. All non-inertial frames are accelerating with respect to all inertial frames.


 * Finally, from the article on general relativity, Einstein's elevator reveals that the force of gravity cannot be distinguished from simple acceleration:
 * for an observer in a small enclosed room, it is impossible to decide, whether the room is at rest in a gravitational field, or in free space aboard an accelerated rocket.


 * ES In our whirling ball event, I will insert a tension scale with digital readout to eight decimal places and set the ball to whirling about the post. I will photograph the reading with a high speed camera setting as the ball whirls by my position.  This photo is my observation taken from a nearly non-accelerating frame of reference of Earth.


 * ES Now I will construct a frame of reference in the form of a small, circular-track-mounted, 4-walled room around the position the ball takes while whirling.  There will be a live video camera installed inside the room to monitor this event.  Our observer, located in a distant lab will have no prior knowledge of this event except that he knows the quantity of matter possessed by the ball.  Once again under way, I will ensure that the ball's rate of acceleration is identical to the previous event.  Next our observer will observe the stationary ball within this small frame and note the force display.  Of course it is a perfect match with the display in my earlier photo.  Also he observes that the ball is hovering within his frame while only being attached to a wire that leaves the ball on an angle that appears about 30 degrees above horizontal relative to the floor of the room.  The wire disappears through a black hole in the wall without touching its sides.  At first our observer considers that this event might be a gravitational one with the scientists trying to trick him by tilting the room to disguise the fact that the ball is hanging on a vertical wire like a stationary plumb bob hanging over a fixed point on Earth.  But being a clever observer, he noted that the scale's display registered a reading greater than what the ball's mass rating would generate.  For this to make sense as a gravitational event, this tilted room would also have to be accelerating in a linear manner in the direction of the wire in order for the ball to be registering a greater-than-its-normal-Earth-weight on the scale.  After watching for a time, the observer decides that if this was true the small room would have to be contained within a cleverly designed rocket with variable thrust so that the room's rate of acceleration does not increase while the rocket's mass is being reduced as the fuel is expelled.  Also he notes that the video feed remains constantly excellent which is not what he would expect if his room was zipping away into outer space.  Our observer decides it is very unlikely that his room is inside an active rocket, yet he is intrigued by the thought that his small frame/room is experiencing acceleration causing the scale reading to exceed the force of the ball's normal Earth weight.


 * ES Next he considers that this frame/room might be located on a turntable and the ball is experiencing a combination of two action forces, one gravitational in the direction toward the room's floor and the other centripetal in the general direction of the wire only parallel to the floor.  He then considers that when the vectors of these two action forces are combined, the resultant is a vector that aligns perfectly with the near 30 degree angle above horizontal the wire takes as it departs the ball. At this point with knowledge of the ball's mass rating, the scale's force reading and the wire's angle of departure, our observer calculates the rate of centripetal acceleration being experienced by the ball.  Finally, our observer concludes that this gravitational/centripetal acceleration solution is the most likely description of this event. Accordingly, our observer decides that although his frame of reference appears in the video to be at rest, in truth it is, on average, experiencing the same rate of centripetal acceleration as that of the ball.


 * ES Note that when the same event viewed by me from within a nearly at rest frame, where I snapped the photo of the whirling scale's force reading, is viewed from within an accelerating frame by an impartial observer, we both observe the same magnitude of force causing the same rate of centripetal acceleration for the whirling ball.  Overall, since the ball's centripetal acceleration remains constant, as verified by the scale's force display, there is not reason for one to expect that merely by changing from a nearly-at-rest frame of observation to an accelerating frame of observation one should expect different results.  An accelerational event with constant values retains those values no matter who is observing.  Should the values differ then one of the observations is in error. This conclusion carries with it the recognition that force and acceleration are absolute values, not relative ones such as velocity. Ethan Skyler (talk) 21:53, 31 January 2010 (UTC)


 * Nice try, but just because your camera is stationary does not mean it is observing from a non-accelerating frame. By freezing the motion, it records an image indistinguishable from that available to the rotating frame. A true observer in a truely non-accelerating frame would see the ball accelerating towards the center and so only need a centripital force to explain the situation. The observer in the rotating frame does not see the ball accelerate and so must have a centrifugal force to explain the tension in the cable recorded by the scale. -AndrewDressel (talk) 00:56, 3 February 2010 (UTC)


 * ES Good point.  My purpose with the photo is to make the scale's reading readable to the non-accelerating observer. Perhaps a handheld repeating electronic display would be better.  I think you have placed the observer inside the rotating frame.  Now I am wondering if your rotating frame observer does not know that he is rotating nor does he see that the ball is accelerating then how does he know that the force displayed by the scale is a centripetal force? Perhaps he would be more inclined to think of the force on the ball as being a static force rather than an accelerational one.  I suppose then he would feel the need to invent an action force in some generally opposite direction to explain why the ball hovers in his room as it does.  With no opposing wire present perhaps he might consider magnetism. Of course, we know he will get dizzy and also if he moves around he will dicover that when dropped, objects will not fall straight down but instead will mysteriously fall slightly toward the wall opposite the black hole where the wire exits.  More forces need inventing.  He will soon realize that it takes considerable effort to keep away from that "attractive" wall. Perhaps he will end up inventing a whole new forceful field of influence unknown to the rest of us nearly non-accelerating observers. I have made more of this event than you intended.  I get concerned when observers start inventing forces that do not exist to explain events that they do not understand. This seemed like a good place to insert this thought.  Ethan Skyler (talk) 17:20, 3 February 2010 (UTC)


 * I doubt any observers really grapple with this any more. It is just how the situation is described. More commonly, the observer is an engineer analyzing a situation and knows full well what is going on. It is just convenient sometimes to attach the frame of reference to an accelerating body, such as when analyzing the dynamics of a robotic hand at the end of a moving arm. When analyzing the motion of the fingers with respect to the hand, it may be necessary to introduce "psuedo" forces such as centrifugal and coriolis in order to be accurate. I believe ballistic missile trajectories and weather system movements, when analyzed with respect to the surface of the earth instead of fixed space, also need these corrections. -AndrewDressel (talk) 01:49, 4 February 2010 (UTC)


 * I do not know what distinction you are making between a "static force" and "an accelerational one". Newton does not make this distinction. Although forces may be implemented by different means, there is only one kind of force in Newtonian mechanics, the kind that causes a mass to accelerate when unbalanced. -AndrewDressel (talk) 01:54, 4 February 2010 (UTC)


 * ES Feb04 Apparently then I go a bit deeper into force than did Newton.  For example, I can accelerate an object horizontally at 1g with a Type 3 external (contact) force governed by Newton's F=ma... or I can accelerate the same object vertically at 1g with the Type 1 internal force of Earth gravitation in a weightless state, again governed by Newton's F=ma, by allowing the object to fall from my hand.  These acceleration/Action forces are identical in magnitude but have different causes and different effects upon the object.  In the horizontal event, the contact force causes a compression/distortion of the object while in the vertical event the object expands a bit as it loses its support from my hand as it enters the weightless state of acceleration popularily known as "free fall" perhaps due to the freedom from the stacking of forces effect of weight that one feels when entering such a state. In my mind, different causes + different effects = different forces.  I think the current assessment that there is only one type of force falls short of reality.  I count 4 types so far.  I will agree that as far as the math is concerned, all four types calculate the same.  Yet there is more to physics than math. Perhaps that is the problem.  I actually go outside and work with this stuff on a daily basis. Maybe this reality of mine lends a broader perspective to my work. Ethan Skyler (talk) 19:24, 4 February 2010 (UTC)


 * Well, yes, one is a contact force and the other is a body force, both are external forces, and both are well known and handled by traditional Newtonian mechanics. When you are meerly keeping track of the motion of an object, there is no need to consider what internal stresses an external force may be causing. However, if internal stresses are of interest for some reason, then internal forces and deformations may be considered and all the necessary tools have long been figured out. See for details. -AndrewDressel (talk) 20:08, 4 February 2010 (UTC)


 * The trick here is to thoroughly question the validity of one's observations. If you, the observer, can determine that your adopted frame of reference is accelerating, then it is your responsibility to see to it that your rate and direction of acceleration is determined and factored into the observed behavior of the objects under study.  Also it is your responsibility to not invent forces or accelerations that, in fact, do not exist in the event under observation.  Whenever possible, I think a scale should be used to actually measure the action and reaction forces being experienced by the object.  Relying on appearance or what you feel as the observer is more subjective and less scientific.


 * Before I move on from the whirling ball event, I would like to address the description for Panel 6. I quote: " The net force on the post is zero, so it does not move."  Do you remember how in my opening remarks I addressed the unique nature of the reactive centrifugal force and how it is caused by and never serves to cancel the effect of the active centripetal force?  Well, here the centripetal acceleration/Action force in red is predicted to have its effect on the post canceled by the centrifugal acceleration/Reaction force in green with the result being that the post "does not move."


 * I don't believe I understand your point here. Is it that:
 * the post does move? (The case can be made, of course, that it always does, even if the post is fixed to the massive earth and the ball has very little mass. They both orbit their barycenter.)
 * the force pulling on the post is not the reactive centrifugal force in response, by Newton's 3rd Law, to the centripital force on the string?
 * the force holding the post stationary in response, by Netwon's 3rd Law, to the reactive centrifiugal force is not a centripital force?


 * When I was young, I flew control line combat model planes. They were large with powerful gas motors.  They carried enough fuel to fly at over 100 mph for about 10 minutes. At first flying was fun with lots of stunts.  But then it became one long test as to who could pull the hardest for the longest period of time.  I became the post in the whirling ball event.  I had to lean back at a noticeable angle to get some help from Earth gravitation in generating sufficient centripetal force to cause the plane to travel its curved path about my person. My experience here means to me that the centrifugal force present at the post in our whirling event does not serve to cancel the effect the centripetal action force has upon the post.  Whirl the ball fast enough and the post will bend right over until it touches the ground.  Green arrow forces in this event provide support for red arrow forces, but do not cancel the effect that red arrow forces have upon the objects drawn.  In fact, red arrow action forces cause acceleration all the while that equal and opposite green reaction forces are present.  Clearly, new rules are needed here as the old rules do not apply. Ethan Skyler (talk) 07:12, 29 January 2010 (UTC)


 * I do not understand the conclusion you draw from your experience. Just because the post is not strong enough to remain vertical under the equal and opposite centrifugal and centripetal forces applied to its opposite ends does not mean that the forces are not equal and opposite. Of course, in reality, there is no such thing as a rigid body. Every object experiences deformation, no matter how slight, due to the forces applied to it. Never-the-less, rigid bodies remain useful idealizations in the analysis of many dynamics problems, just as Netwonian physics is a useful idealization when speeds are only a small fraction of the speed of light. Perhaps the guys that design crumple zones in automobiles need to consider the accelerations that occur during deformation, but we hope the guys designing roller coasters do not. -AndrewDressel (talk) 15:28, 29 January 2010 (UTC)

=Comments from January 29, 2010=
 * Jan 29 - Yes, I see the post as moving while providing the centripetal acceleration/Action force responsible for causing centripetal acceleration for the whirling ball. There does exist equal and opposite centripetal action (red arrow) and centrifugal reaction (green arrow) forces at the top of the post, exactly as drawn in Panel 6. It is the "zero net force" and "the post does not move." statements that I think need to be reexamined. From my control line experience I know that if a tension scale is installed between the string and the post, the scale will display the magnitude of centripetal action force (red arrow) the post is providing as the cause of this whirling event. In response to applying this centripetal force, the post will bend in the ever-changing direction of the ball. Here I see the post does move as its top orbits the event's axis.Ethan Skyler (talk) 21:53, 31 January 2010 (UTC)


 * I do not understand why you worry about the post bending. I propose we can do one of thress things to resolve the issue of the bending post:
 * Accept the idealization of rigid objects. They simply do not bend.
 * Let the ball have so little mass that even at high speeds only a small centripetal force is necessary to keep it in a circle.
 * Replace the flexible post with something suitably rigid, a squat cone or a pillar as wide as it is tall.
 * Otherwise the situation gets complicated and obscures the simple fact that the force holding the post in the ground is equal and opposite to the force applied by the string. -AndrewDressel (talk) 20:58, 29 January 2010 (UTC)


 * I don't much care for the term "net force" for it implies that when the forces present are added, the result is that a single unopposed, unbalanced force remains. I am unconvinced that any force can ever exist without being immediately opposed by an equal force. Instead, I think every force is always opposed by an equal force that is immediately present. Thus there is no room in my mind for the existence of an "unbalanced" or "net" force. Now in our whirling ball event, at the top of the post there does exist a "net action force" (red arrow) exactly as drawn in Panel 6. In applying this net action force to the whirling string and ball, the post reacts by bending in the direction of the ball the same way that a fishing pole reacts while providing the action force to tow or accelerate the hook and fish.Ethan Skyler (talk) 21:53, 31 January 2010 (UTC)


 * The concept of net force is really quite established and uncontriversial. As I pointed out above, even Newton mentioned it in his Corollary I.


 * ES - In Newton's Corollary I, he describes the joining of the effects of two motive forces acting on an object that I think is free to accelerate. His point, as I see it, is the recognition that if the object experiences the effect of each motive force separately for a given duration of time, first one and then the other, the object will end up at the same location as it would if it experienced both motive forces together for the same duration of time.  In this way he is describing a net motive force acting on the object. I agree with his analysis, of course.


 * ES - I did not see any mention that he viewed this combined net motive force as being "unopposed" or "unbalanced" or "unsupported" as is so often claimed in modern texts. If you wish I will locate several such quotes.  This is my focus here, that Newton's LAW III always holds true in every event and especially events that follow the predictions of LAW II.  What force do I see as providing the LAW III required support for the "net motive force acting" in circular events?  It is the reactive centrifugal force that is often measurably present during circular events that ensures that the centripetal motive force does not act alone.  Ironically, most authors of physics prefer to call the reactive centrifugal force "fictitious" while actively denying that there is anything whatsoever real about it. They always avoid using scales to prove their point.  Insead they like to tell us how we "feel" during an accelerational event as if our "feeling" is a suitable replacement for a scale.


 * But he also does not limit the angle between his two forces N and M, so they could certainly be 180° opposed. Then the diagonal of the parallelogram (his A to D) will be colinear with the two forces. If they act at the same time, as he allows, then the motion will be the net of the difference between the motion due to M and the motion due to N. Finally, if they are equal and opposite, there will be no motion at all. -AndrewDressel (talk) 23:47, 1 February 2010 (UTC)


 * ES Makes sense to me.  In the "no motion" event above, I see "M"'s motion occuring first followed by "N"' motion with the result that the object is returned to its initial position.Ethan Skyler (talk) 22:23, 2 February 2010 (UTC)


 * ES - When I look at the whirling ball event there is no place where I can find where the centripetal action force is not equally supported by the centrifugal reaction force. If you think such a place exists, let me know and I will have a look at it.


 * No, there is no place where these pairs are not equal and opposite. The point I'm not sure you see, though, is that these opposite pairs of forces always always act on separate bodies, by Newton's 3rd law. For example, they are not the N and M from his Corollary 1. Instead, the reactive centrifugal force acting on the string is the equal and opposite of the centripetal force acting on the ball, and the centripital force acting on the string is the equal and opposite of the reactive centrifugal force acting on the post. Thus, the two forces acting on the string are not equal and the string experiences a net force that causes it to accelerate towards the post. -AndrewDressel (talk) 23:47, 1 February 2010 (UTC)


 * ES We are not together on portions of this one.  Please see my 2D event below.  I see the centripetal force from the post is responsible for accelerating all the attached objects at all their different rates of acceleration present.  This net action force is at its maximum as it departs the post following the string.  It is slightly less at the string's mid point for a small portion of it is spent causing acceleration for the 1st half of the string.  It is again slightly less at the end of the string as it is again reduced by the amount of centripetal force necessary to cause the required rate of centripetal acceleration for the 2nd half of the string.  The major portion of the net action force from the post now goes to work causing acceleration for the ball.  Each portion causing acceleration for the atoms of the ball causes an equal and opposite reaction force to be generated within those atoms just as was caused within the string's atoms.  I explain the generation of this centrifugal reaction force in my 2D event below.  At every point within the whirling ball's matter equal and opposite forces are present according to LAW III. The reaction force is caused by the action force and serves in no manner in reducing the accelerational ability of the action force.  We are in agreement that the reaction force is present throughout the ball's matter. But you begin at the ball's outer-most atom by saying that "there is no reactive centrifugal force on it". In my 2D event below, I show that this outer-most atom, like every other atom in the ball is reactively generating its own internal reactive centrifugal force in support for the last remaining portion of the post's net action force.  Ethan Skyler (talk) 23:25, 2 February 2010 (UTC)


 * ES - If you think I am taking a position opposed to Newton's work, let me know exactly where and I will look at that too. Ethan Skyler (talk) 00:31, 31 January 2010 (UTC)


 * If you believe that the equal and opposite pairs of forces on connected but different bodies, as described by the 3rd law, mean somehow that the net force must be zero on any single body, then that is opposed to Netwon's work. -AndrewDressel (talk) 23:47, 1 February 2010 (UTC)


 * ES Yes, I think when all action and reaction forces are totaled that can possibly be effecting a single body, that total will always equal zero regardless of whether the body is accelerating or in a uniform motion with rotation absent.  Please make this position clear in your mind.  While a net action force is required to cause acceleration for a body, when it is added vectorially to the equal and opposite net reaction forces present, the total will always equal zero.  There exist no difference in the magnitude of these accelerational forces.  Not even the slight difference you are counting on to explain acceleration.  I think my position is correct that acceleration results from a difference in the types of forces present, not a difference in the magnitudes of the forces present. Ethan Skyler (talk) 22:23, 2 February 2010 (UTC)


 * No, no, no. The equal and opposite reaction force does not action on the body accelerating, it acts on some other body that is supplying the force that is causing the acceleration of the first body. If I push a ball, the equal and opposite reaction force is not also on the ball, it is on me. -AndrewDressel (talk) 01:29, 3 February 2010 (UTC)


 * ES You have repeatedly supported the rule that an action force causing acceleration for an object and the reaction force coming from the accelerating object always effect different objects and never the same object.  I appologize for not remembering to discuss that which follows sooner.  There is a different event than the one we are discussing where I think this "never on the same object" rule falls flat.  It comes to light when the acceleration/Action force is gravitation.  I see the acceleration/Reaction force as a force that originates internal to the atom.  It becomes an external force as it departs the atom to bear as a contact force against its neighbors as described in the 3 atom object.  I also see gravitation, magnetism and the electromotive force you mentioned as forces originating internal to the atom.  For this new event, I will reuse our 3 atom object.  Imagine these 3 atoms in a gravitational event here on Earth.  I will pile them one atop the next and stand this pile on a micro trap door like the hangman's trap door.  I suppose this event should be staged inside a vacuum to cancel collisions with air molecules.  On top is Atom 1 whose internal force of Earth gravitation causes it to bear down on Atom 2 with the now external force of the Earth weight of this one atom.  Atom 2 received Atom 1's weight force and adds its own weight force thereby transferring 2 weight forces down to Atom 3.  Atom 3 receives the 2 weight forces from Atom 2 above, adds its own weight force and transfers this 3 atom weight force on down to the surface of the micro trap door.  This is what I refer to as a "stacking of forces" effect.  Here the individually generated action forces of Earth gravitation stack up from atom to atom down in the direction of the door and Earth's core below. I see this stacking of forces effect as always occurring when an internal force is opposed or supported by an external force.  Thus when you stand on your bathroom scale, I see the internal Earth gravitation forces being actively generated within each of your atoms as stacking up as they are transferred as external contact forces from atom to atom all the way to the scale.  Here the total of these gravitational forces equals the force of your weight against Earth. Ethan Skyler (talk) 16:22, 3 February 2010 (UTC)


 * You and I are using "internal force" and "external force" completely differently. I follow the convention of the all the physics and engineering text books I have seen: internal forces are only between the elements or components of a body or system. They do not appear on a free body diagram. External forces are only between a body or system and elements outside that body or system: they do appear on a free body diagram. -AndrewDressel (talk) 02:08, 4 February 2010 (UTC)


 * ES Now, as you have surmised, I will trip the trap door.  Perhaps a better method is for the trap door to simply disappear and thereby not disturb our three atom stack as it suddenly loses its support against Earth.  Each atom is now being accelerated through an air-less space toward Earth's center of matter by its own internally generated acceleration/Action force of Earth gravitation.  Newton's LAW III is quite clear that for every such action there is always opposed an equal reaction.  In this event, the action is the internal force of Earth Gravitation.  The reaction is the internal force of acceleration/Reaction as present in our whirling ball event.  Only here the action and reaction forces are both internal forces.  Here the internal action force is causing its own reactionary support force all within the confines of the single atom.  This means that none of the internal force of gravitation has any need to travel beyond the single atom to seek support as an external force against any atom beyond.  For this reason, there is now no stacking of forces effect going on within our accelerating 3 atom object.  With no external forces being passed down as before the trap door vanished, our three atoms are now relieved of inter-atom forces.  The stack will get a little taller as it accelerates toward Earth with each atom now in a weightless state. Ethan Skyler (talk) 16:22, 3 February 2010 (UTC)


 * "Newton's LAW III is quite clear that for every such action there is always opposed an equal reaction. In this event, the action is the internal force of Earth Gravitation." I do not know what this means. By Newton's 3rd law, when the earth pulls on the the three atoms, the three atoms also pull on the earth. -AndrewDressel (talk) 02:08, 4 February 2010 (UTC)


 * ES Feb 4  Having lost the support of the trap door, the action force of Earth gravitation beings causing acceleration of our three atoms individually with the Type 1 acceleration/Action force of Earth gravitation present within each atom supported by the Type 1 acceleration/Reaction force that is equally present within each individual atom.  As you correctly pointed out, the rest of this event is the other half where Earth's atoms all individually begin their weightless acceleration in the direction of the 3 atoms.  Again whatever acceleration/Action force present within Earth's atoms causes and thererfore finds present the equal acceleration/Reaction support force.  So once this event is complete, overall we have four forces in play.  This brings up a question I would like you to consider.  What do you say is the minimum number of forces present in any event?  Please describe the event you are considering. Ethan Skyler (talk) 19:56, 4 February 2010 (UTC)


 * I do not see how you are counting four forces, unless you are also including the forces between the three atoms in the stack. If we consider subatomic particles, in order to avoid all consideration of internal forces, and at the risk of violating call kinds of rules in quantum mechanics, the minumum number of forces present in any event between two particles is two: the two forces described by Netwon's 3rd law. Now if you want to get fancy, then you may need also to consider the action/reaction pair of gravitational attractions, the pair of electromagnetic attractions or repulsions, and the strong and weak nuclear forces, etc. On the other hand, we can just sum up all these forces and count only the two net forces, one imposed on each particle by the other. -AndrewDressel (talk) 20:55, 4 February 2010 (UTC)


 * ES Feb 4  I already know that you will respond to my statement of the 4 forces at work in the 3 atom - Earth accelerational event described above... by telling me that Earth's acceleration toward the 3 atoms is the "reaction" to the action of the 3 atom's acceleration toward Earth.  But I will ask you to consider that, if so, you are assuming that the internal acceleration/Reaction forces, whose existance we are supporting in every accelerational event, somehow just disappear in this 3 atom-Earth accelerational event.  Instead, now that we are welcoming the reactive centrifugal internal-to-matter force into reality, we might as well do the logical thing by accepting its presence in every accelerational event, even ones caused by other internal-to-matter forces such as gravitation, magnetism, and electo-magnetic forces.  This has been my understanding for the last dozen years.  Welcome aboard! :-)  Ethan Skyler (talk) 19:56, 4 February 2010 (UTC)


 * "Earth's acceleration toward the 3 atoms is the "reaction" to the action of the 3 atom's acceleration toward Earth." Yes, that is correct. The internal forces between the 3 atoms do not "somehow just disappear." They are all in direct response to the force applied to the bottom atom by the trap door. When you remove the trap door, it no longer pushes up on the bottom atom, which in turn no longer pushes down on the trap door. The bottom atom enters free fall and so no longer pushes up on the atom above it, which in turn no longer pushes down on the bottom atom. Without the bottom atom pushing up on it, the atom above it enters free fall and so no longer pushes up on the top atom, which in turn no longer puses down on the atom below it. Without the atom below it pushing up on it, the top atom enters free fall. The contact forces between the three atoms, if that is how we model the force that prevents them from occupying the same space simultaneously, all are due to the trap door preventing their free fall, and so all go away when the trap door no longer does its job. -AndrewDressel (talk) 20:55, 4 February 2010 (UTC)


 * ES Feb 4  I think we agree that whenever matter is forced to accelerate, the matter's atoms reactively generate an acceleration reaction force that is mutual to the accelerating force. Your description above is a perfect description of the unloading of the support force the 3 atoms are/were experiencing from the trap door.  As this unloading occurs, each atom begins accelerating in the downward direction.  My understanding tells me that this acceleration, like every other acceleration, causes each atom to reactively generate its own acceleration/Reaction support force of equal magnitude and opposite in direction to the gravitational acceleration/Action force present within the same atom.  Here I see, as you describe above, that there is no need for any force internal-to-the-matter of each atom to venture beyond the atom in search for support as they were doing prior to the door's release.  Instead the correct support is now being generated within each accelerating atom.  This makes two types of forces, one gravitational and one a/R at the site of the accelerating atoms and two of the same types of forces within Earth's accelerating atoms for a total of 4 types. Ethan Skyler (talk) 22:15, 4 February 2010 (UTC)


 * The "action" and "reaction" forces due to gravity already exist long before any acceleration begin. The acceleration that ensues when the contact force of the trap door is removed does not change anything. Once the door is gone, the only remaining forces are the "action" and "reaction" force pairs between the earth and the atoms due to gravity. There is no additional "support" generated within each atom. Why would there be? -AndrewDressel (talk) 23:03, 4 February 2010 (UTC)


 * ES Feb 4  We could reduce this event to just two atoms a short distance apart attracting each other and then having the massless support between them removed so that this attraction could begin causing their acceleration toward each other.  At this point I see four forces present.  I think it is not correct to describe the gravitational force being experienced by the second atom as a reaction force to the gravitational force being experienced by the first atom.  In my world, reaction forces are never the cause of any event.  Here the reaction force is predicted to be causing acceleration for the 2nd atom.  I see this as a misuse of the term reaction.  This misuse is common but that should not make it correct.  Instead, especially now that we correctly recognize the reaction-to-acceleration force which heretofore has been deemed imaginary, I think it is correct to recognize that two reaction forces also become present, one within each atom the moment it begins accelerating toward the other.  In a large-scale event, I see the Earth & Moon as a 4 force event, 2 internal gravitational action forces causing both the acceleration of matter and the reactive generation of 2 internal-to-matter acceleration/Reaction forces. I never have thought it logical that a gravitational force within the Moon's matter could possibly be finding its reactive support force within Earth's matter 1/4 million miles away. Instead, I think of reaction forces as being "immediately" present within the matter of any accelerating object. When these objects reach each other, the acceleration is over so the a/R forces cease while the action forces continue on only now they are causing each object to freely bear with a static (non-accelerating) action force against the other.  That is pretty much my description for weight - the external (contact) force that one object is freely bearing against another. Ethan Skyler (talk) 22:15, 4 February 2010 (UTC)


 * It is exactly "correct to describe the gravitational force being experienced by the second atom as a reaction force to the gravitational force being experienced by the first atom." That "action/reaction" pair, atom A pulling on atom B and atom B pulling on atom A, is/are the only forces present. There is no need to distinguish between which force is the "action" and which force is the "reaction". There is no difference between them. Certainly in the case of autonomous agents, such as people, we can say that the block only pushes back once I start pushing on it, but the physical outcome does not depend on that sequence. In the case of inanimate objects, such as our atoms, there is absolutely no sequence: they each attract each other, for all time, with a strength that varies inversely with the square of the distance between them. Under Newton's law of universal gravitation, the Earth & Moon is a two force event. These forces exist independent of acceleration. If the Earth and the Moon were suddenly grabbed by two giant hands, of a being capable of doing so, and held still, the "action/reaction" pair of gravitational forces between them would remain completely unchanged. -AndrewDressel (talk) 23:03, 4 February 2010 (UTC)


 * I continue to fail to understand what you mean by "reactive support force". -AndrewDressel (talk) 23:03, 4 February 2010 (UTC)


 * ES Feb 4 Notice how we began discussing the reality of the reactive centrifugal force and are ending up discussing pretty much the whole foundation of physics.  These concepts are all connected.  Change one (centrifugal force for example) and all the ingredients end up back in the pot.  It is like this science is totally linear, like blocks placed on top of each other in one tall stack.  We are discussing some of the early ones at the bottom.  Risky business.  Glad I am not a physicist or anything close. Ethan Skyler (talk) 22:15, 4 February 2010 (UTC)


 * ES If you are standing on the tip of a diving board installed high over a deep pool, your weight forces stack up toward your feet in the same manner as the 3 atom object. Step forward off the board and your body will be relieved of any stacking forces as you accelerate in a weightless manner toward the water below.  You will get a little taller during these weightless moments.  These weightless events are caused by the confinement of two internal forces within their mutual atom of origin.  This is an event where the action and reaction forces exist within the same body.  I would have brought this up sooner only there is so much going on in our now 25 page exchange that frankly I didnt remember that this is the main reason that I do not use this "rules".  This one in particular has no application during weightless accelerational events. Ethan Skyler (talk) 16:22, 3 February 2010 (UTC)


 * When I said above that "the equal and opposite reaction force does not act on the body accelerating", I was refering speciffically to your example in the paragraph preceding mine. I do not mean in general that the equal and opposite reaction force does not act on a body accelerating. Of course, when two bodies push on each other in the absense of any other external forces, in space or while falling from a diving board, they both accelerate. -AndrewDressel (talk) 15:03, 4 February 2010 (UTC)


 * ES As a side note, your should know that I refer to an internal force that causes or finds its opposition or support within the originating atom a Type 1 force.  When the internal force has to leave the atom as an external force to seek opposition or support as does the reactive centifugal force in our whirling ball event, I  refer to it as a Type 2 force. I also see two types of external forces but will spare you until their descriptions are needed.  If you noticed that I stayed up late and got up early it was in preparation for my second colonoscopy in 5 years.  Thanks for all the distractions.  Our exchange has kept my mind off local matters.  What I would give for a good home-cooked meal right now! Ethan Skyler (talk) 16:22, 3 February 2010 (UTC)


 * I hope your precedure went well. -AndrewDressel (talk) 15:03, 4 February 2010 (UTC)


 * ES Feb 5  It did go well.  Thanks for asking.  They are so good at this procedure these days.  The process leading up is demanding but once at the hospital, it is smooth sailing.  Our first "inspection" was 5 years ago.  Due to no family risks, my wife was told to come back in 10 years.  4.5 years later she was back due to faint internal bleeding.  It was a cancerous tumor that was close to blocking passage.  It had not spread. They saved her life.  If ever you suspect anything, the only waiting you should do is waiting for surgery and hopefully that won't be more than a month.


 * Glad to hear it. -AndrewDressel (talk) 17:29, 5 February 2010 (UTC)


 * ES I think we agree that the reaction force is present throughout the myriad of components of the matter of an accelerating body.  I think you understand that I see every component (atom) as reacting in support of the force causing its acceleration.  From my 2D event below, if the body is composed of a single atom then you understand how I view this atom's behavior during acceleration by an external contact force.  The atom is forced to exist in an unbalanced state by the accelerating force.  It responds to this imbalance with the generation of an acceleration/Reaction force all the while that acceleration continues.  Thus I think you understand that I see a balance of a/A and a/R forces all the while that acceleration continues.


 * Yes, and acceleration is not necessary for an object to generate a reaction force. If you push on an atom with 1 unit of force, it pushes back with 1 unit of force whether it accelerates or not. What determines whether it accelerates or not are the possible other external forces applied to it. -AndrewDressel (talk) 14:30, 3 February 2010 (UTC)


 * ES In my work I have determined that reaction forces only occur in accelerational events.  Thus the reactive centrifugal force in circular accelerational events and its twin in linear accelerational events are the only reaction forces to exist.  The primary feature here is that a reaction force is caused by an action force as I demonstrate in my 2D event.   —Preceding unsigned comment added by 216.231.36.211 (talk) 14:51, 3 February 2010 (UTC)


 * But this, "reaction forces only occur in accelerational events", is completely counter to Newton. He makes no distinction between accelerating and non accelerating events. If object A pushes on object B, then object B pushes back on object A with an equal and opposite force. -AndrewDressel (talk) 15:03, 4 February 2010 (UTC)


 * ES Feb 5 I know, Newton's horse pulling on a rope that is dragging a stone does not differentiate between the initial acceleration and the steady dragging thereafter. Curious that he didn't elaborate more.  I think those were contentious times. His critics were powerful and ever-present.  Perhaps I see a bigger picture than expressed by Newton.  For example, if you and I stand and by hand push against each other, I see two action forces equally opposed where our hands meet with no reaction forces present. I think of an action force as any force causing or capable of causing the activity of acceleration. Rather than being stuck having to call one action force or another present in an event a "reaction" force since Newton's work in this area stopped short, I think it is better to recognize that for every action force there is always immediately present an equal and opposite action or reaction force.  Unlike an action force, I think the reaction force label should only be applied to the force caused by and providing support for the accelerating force.  In circular events this would be the reactive centrifugal force herein being discussed.  I refer to it as the acceleration/Reaction force which covers both circular and linear events. Ethan Skyler (talk) 17:21, 5 February 2010 (UTC)


 * ES If the accelerating object is but a single atom, I think it is safe to say that this single atom will become distorted or somewhat squashed, especially if the acceleration is extreme.  Yet I don't see any harm in saying that the accelerating body acts upon this single accelerating atom and that the single atom generated a reaction force that reacts back upon the accelerating object.  Then again, say that the accelerating object is composed of three atoms in a row.  The accelerating object contacts the 1st atom with the net action force needed to accelerate all three atoms at a desired rate.  Atom 1 receives the 3 unit force, is accelerated by one of the units and reacts by internally generating 1 unit of reaction force that is applied back against the accelerating object.  The remaining 2 units of action force are tranferred forward to the 2nd atom.  The 2nd atom accelerates and generates a 1 unit reaction force that it bears back against Atom 1.  It also transfers forward the remaining 1 unit of action force that it impresses against Atom 3.  Atom 3 accelerates and reactively bears back against Atom 2 with 1 unit of reaction force.  This 1 unit reaction force from Atom 3 stacks up or is combined with the 1 unit of reaction force from Atom 2 and the 1 unit of reaction force from Atom 1 to bear with a full 3 unit force back against the accelerating body.


 * Yes, and it is true whether the atoms distort internally or not. Each atom has 1 unit of net external force on it and so they all accelerate in unison. -AndrewDressel (talk) 14:30, 3 February 2010 (UTC)


 * ES Your single body description above fits fairly well with the last and farthest atom #3.  I think my description fits with all the atoms in the object considering that even the far-most Atom 3 is likely to experience some distortion during accelration.  Here you say the reaction forces do not react on the object composed of 3 atoms.  My description above shows that atoms 1 & 2 pass action force forward and reaction forces back to bear against the 2nd body that is causing the acceleration.  In the process, these three atoms change shape as they become compressed against the 2nd body. This type of distortion is commonplace during accelation caused by an external (contact) force.  To me the distortion indicates that the accelerated body's own internally generated reaction forces have a role of support to play within the accelerating object prior to leaving the object as a contact force impressed against the 2nd body that is causing the acceleration.


 * I have never said that "the reaction forces do not react on the object composed of 3 atoms." I have only tried to say that reaction forces, those forces generated by a body in reaction to external forces imposed on it, should not be included when the external forces are summed in order to determine the acceleration of that body. Each body accelerates or not in response only to the net sum of external forces imposed on it. The reaction forces any body generates do not contribute to nor detract from its acceleration. -AndrewDressel (talk) 14:30, 3 February 2010 (UTC)


 * The important point to keep clear, as I mention above, is that a non-zero net force on a single body is necessary for it to accelerate. The equal and opposite forces described by Netwon's 3rd Law are applied to two different bodies: when body A pushes on body B with 1 Netwon of force, body B pushes back with the same 1 Netwon of force. Whether either body moves depends the vector sum of that 1 Netwon force and all the other applied forces.
 * It is not clear from your comments if I am just not making myself clear or you understand what I am saying and disagree. -AndrewDressel (talk) 20:58, 29 January 2010 (UTC)


 * ES - I agree a non-zero net action force on a single body is necessary for it to accelerate. If you leave out the "action" term yet recognize the reaction force as present during acceleration then I think you will have to conclude, as I have, that any total of the forces present will equal zero making a "net" force impossible.  In other words, the net force in every event is zero regardless of whether acceleration is present or absent.  But a net "action" force is always present during every such event where accelerational is present.  This is my point.  Recognizing the reactive centrifugal force supports my position.  I think it supports Newton's work as well.


 * Here again, I believe you are mixing the equal and opposite pair of forces on two different bodies, as described by the 3rd law, with the non-zero net force on a single body that causes it to accelerate, as described by the 2nd law. -23:47, 1 February 2010 (UTC)


 * ES - Regarding Newton's LAW II and LAW III, I push them together in my mind.


 * That is a mistake, I believe. The former applies to a single body, and the latter to two separate bodies. -AndrewDressel (talk) 23:47, 1 February 2010 (UTC)


 * ES Feb 7 I was looking over my copy of PRINCIPIA today.  Among other things, I found the following quote from Corollary III.  "For action and its opposite reaction are equal, by Law III, and therefore by Law II, they produce in the motions equal changes towards opposite parts."  I suspect you are going to tell me that Newton's words in Corollary III do not justify my merging the mutuality of forces in Law III with the proportionality rules governing accelerating objects and forces in LAW II when considering accelerational events.  If so then Newton and I will move to the darkest corner of the room and sulk together.  If not then the three of us will share the light while telling tall physics stories.  Either way, I will be in good company. Ethan Skyler (talk) 04:21, 8 February 2010 (UTC)


 * Yes, of course Laws 2 and 3 both apply to a single event, but notice that Newton writes "they [action and its equal and opposite reaction] produce in the motions equal changes towards opposite parts." (emphasis mine) That is the part I believe you are missing. You seem to insist that Laws 2 and 3 together mean that there is always a zero net force on every object, when that is not what Newton is saying at all. Law 3 explains that when object A pushes on object B, object B pushes back on object A with an equal and opposite force. Law 2 explains that the force on B changes the motion of B in the same way that the force on A changes the motion on A. In the simple case of gravitational attraction between two bodies, say A and B, which is primarily what he was analyzing, there are only two forces, equal and opposite by Law 3, one on A and one on B, and both A and B accelerate according to Law 2. -AndrewDressel (talk) 15:44, 12 February 2010 (UTC)


 * * ES Feb 12 Well put except for the two force event. I think my "problem" is that I prefer to view events whole.


 * I am viewing the entire event, and as far as Newton was concerned, it is only a two force event. See the diagram at the top of Newton's law of universal gravitation. He only considered bodies as particles, with no significant dimension and so no internal forces or deformations. -AndrewDressel (talk) 21:29, 12 February 2010 (UTC)


 * * When doing this, I see every force opposed or supported by an equal force, hence when all these equal and opposite pairs of forces are totalled, the final result is always zero.


 * Yes, as far as the entire universe is concerned, momentum is concerved and so the center of gravity of the universe never accelerates (Newton's Corollary IV), but what use is that conclusion in the real world? -AndrewDressel (talk) 21:29, 12 February 2010 (UTC)


 * * In your two force event above, I accept that the reaction-to-acceleration force (such as the reactive centrifugal force) is present in every event where acceleration is present. I can think of no reason for it to be present in events where an external-to-matter contact force is causing acceleration, but absent in events where the accelerating force is internal-to-matter such as gravitation.


 * I do not know what you are saying here. There is no distinction between action and reaction forces, and if you treat objects as particles as Newton did, there is no distinction between contact and body forces. -AndrewDressel (talk) 21:29, 12 February 2010 (UTC)


 * * Plus when gravitation is the cause of acceleration of two distant objects, each toward the other, I can think of no possibility where the gravitation force generated within A in B's direction finds opposition against the distant gravitation force being generated within B in A's direction.


 * Well, that is exactly what Newton stated. He may have had ideas about how gravity was actually implimented, but he did not write them in the Principia. We still only have theories and don't actually know. And, since we cannot make matter appear or disappear, we never really have to worry about the time it takes for the effect of gravity to propagate at least on the surface of the earth. It decreases with distance, but it is always present. It is simply force at a distance and for all practical purposes acts instantaneously. -AndrewDressel (talk) 21:29, 12 February 2010 (UTC)


 * ES Feb 22  I find it so hard to remember Newton's words on particular subjects.  In re-reading His LAW III, The first half deals with every action having a reaction.  In the second half he writes of "mutual actions" between bodies which I think represents his position on actions at a distance such as Earth's action within the Moon compared to the Moon's action within Earth.


 * I disagree. I take his ": or" to mean "in otherwords" or "another way to say this is". If he were writing about two different situations, don't you think he'd emphasize the difference more? -AndrewDressel (talk) 20:23, 28 February 2010 (UTC)


 * ES Feb 22  Notice that he specifically refers to these action-at-a-distance forces as being composed of two "actions", not one being the "reaction" to the other "action" as I think is your position.


 * On the contrary, my position is that the distinction of "action" and "reaction" is arbitrary. Calling both ends of a single force "action" is just as bad. "Action" as come to mean "motion" in our language, and Newton only meant "force". The reason he didn't just use "force" is because the term was poorly defined at the time. He had already used it to mean two different things in the principia: that which causes acceleration, and that which resists acceleration. -AndrewDressel (talk) 20:23, 28 February 2010 (UTC)


 * ES Mar 2  Nope.  Here Newton joined two seperate sentences with "or" forming a complex statement in logic.  Any two mutual forces are an action/reaction combination "or" they are an action/action combination.  Each action/reaction combination has one force causing accelerated motion and a mutual reaction force caused by and providing support for the action force.  Since the accelerated motion has but one direction, it is not logical to rename the action force a reaction force for to do so would redefine the former reaction force as now being the cause of the acceleration despite it always being directed opposite to the direction of the event's acceleration.  Overall, I think this "renaming" each force in an action/reaction combination certainly does not belong in the science of physics. If you would like to give me an event involving accelerationed motion and show me how the action/reaction force conbination works equally well after such a renaming, I will be happy to have a look at it.  Ethan Skyler (talk) 06:38, 3 March 2010 (UTC)


 * ES Mar 2  After Newton's "or" the mutual forces he was referring to are both action forces with each acting as the cause of acceleration for a portion of an object, say a portion of Earth or a poriton of the Moon.  Of course each action force that is causing accelerated motion for a portion of Earth is also causing its own reaction force which in this case is the reactive centrifugal force present within the same portion of Earth that is experiencing the accelerated motion.  Hoping for Newton to give multiple examples or even one clear example of his position on any subject is a lost cause.  He understood what he was writing about.  It is up to us to play catch up if we are able.  There not only is a difference in his statements before "or" and after, there is also a special need for each.  The first half of LAW III covers the equality of mutual action/reaction forces in common contact force and internal-to-matter force accelerational events. Each action force causing acceleration causes its own immediately present support in the form of a mutual reaction force.  The second half of LAW III is where he explains that the action forces being generated within the matter of one object in the direction of a second object, due to the presence of the matter of that second object, are opposite in direction and equal in magnitude to each other regardless of any difference in the quantity of the matter of the two objects.  Thus he was telling us that despite Earth being more than 80 times as massive as the Moon, the overall mutual action force within each body toward the other are exactly equal in magnitude. This is the "action at a distance" point of the second half of Newton's Law. It certainly is not a simple rewording of the first half. )  Ethan Skyler (talk) 06:38, 3 March 2010 (UTC)


 * ES Feb 22  He had the option to make these two actions as an action/reaction combination especially considering he had just written in the first half of LAW III about action/reaction combinations during events where contact forces are involved.  Instead, in the second half, he specifically chose two "actions" so I think we can acknowledge that he was addressing a different combination of forces than in his earlier action/reaction combination in the first half.  Thus I see my position of recognizing two actions, one within Earth and the other within the Moon as being compatible with his position on such an event. Further, I think his choice of two "actions" in the second half indicates to me that like me, he saw the combination of two actions in the second half as being different from the action/reaction combination he used in the first half. Thus I think his words show that Newton himself thought of an action force as being distinctly different from a reaction force in accelerational events. I take the same position as evident in my 2D Atom event below.  Ethan Skyler (talk) 07:01, 23 February 2010 (UTC)


 * ES Feb 22  I see nothing in his words to prevent me from applying the first half of LAW III to a particle that is experiencing the second half of LAW III. Thus I recognize that for every LAW III 2/2 internal-to-matter centripetal action force (within an Earth particle accelerating in the Moon's direction) there exists a LAW III 1/2 internal-to-matter centrifugal reaction force caused by and providing mutual support for the acceleration-causing centripetal action force.  Here, equal and opposite action and reaction forces are present within the same object all the while acceleration is present.  This understanding is the source of my recognition of 4 forces present in the Earth/Moon system. Ethan Skyler (talk) 07:01, 23 February 2010 (UTC)


 * ES Feb 22  My point here is that I see no reason to turn a blind eye to recognizing that each such "action" within each body causes its own reaction-to-acceleration force to exist in nearly the same way that the centripetal force on the whirling ball causes to exist the reactive centrifugal force within each of the ball's whirling component of matter. Ethan Skyler (talk) 07:01, 23 February 2010 (UTC)


 * ES Feb 22  Isaac Newton does describe the mechanism of gravitation in PRINCIPIA in the SCHOLIUM on page 192 of my copy as follows: "These Propositions naturally lead us to the analogy there is between centripetal forces and the central bodies to which those forces are usually directed; for it is reasonable to suppose that forces which are directed to bodies should depend upon the nature and quantity of those bodies, as we see they do in magnetical experiments. And when such cases occur, we are to compute the attractions of the bodies by assigning to each of their particles its proper force, and then finding the sum of them all.  I here use the word attraction in general for any endeavor whatever, made by bodies to approach to each other, whether that endeavor arise from the action of the bodies themselves, as tending to each other or agitating each other by spirits emitted..." Ethan Skyler (talk) 07:01, 23 February 2010 (UTC)


 * ES Feb 22  My understanding for gravitation, developed around 1996 is quite close to his which I found by accident on 10/11/1999.  If you look at the energy being emitted from Earth's particles as traveling at the speed of light in the Moon's direction and upon reception by the Moon's particles causing an imbalance in the operation of the Moon's particles such that each is continually displacing itself a bit in Earth's direction, you end up with an internal-to-matter force of "attraction" within the Moon's "particles" that results from an imbalance or "agitating" of those "particles" by "emitted" energy or "spirits" that travel the distance of separation at light speed (words in quotes are Newton's). Am I the first one to discover this description for gravitation of Newton's? In the margin of my copy of PRINCIPIA, I wrote a single word, "Wow!" and the date. Ethan Skyler (talk) 07:01, 23 February 2010 (UTC)


 * ES Feb 22 You have referred to Newton as always treating an object as a point meaning to me that you do not see him treating an object as a whole which is the way I often look at objects in our events.  Notice in the quote above how Newton discusses finding the force of attraction for a whole body by tallying up the force of each of its particles.  I realize that shrinking an object down to a point has its advantages, but I think Newton saw the big picture as well.


 * ES Feb 22 Also notice in Newton's quote that the forces he is computing within each of a body's particles is directed toward the other body.  I think it is clear that he saw this single particle force being generated within the particle through agitation of the particle by energy spirits received by the particle that were emitted in the particle's direction by the other body some distance away.  Thus I think Newton saw that energy emissions are exchanged between distant objects and further that the reception process resulted in the generation of gravitational forces within each of a body's particles causing centripetal acceleration of the body toward the other body. This guy was good! Ethan Skyler (talk) 07:01, 23 February 2010 (UTC)


 * ES Feb 23  This morning I decided to read on in Newton's SCHOLUIM beginning on page 192 in my copy of PRINCIPIA.  He made some references to forces that I found most interesting.  One of the key points in our discussion here is the vast difference in our positions on forces.  Your position is that a force is a force meaning to me that there is only one type of force. My position is the opposite that there are by my count 4 types of forces based upon their causes and behaviors in events.  On page 192 Newton weighs in on this force or forces subject with the quote that follows:Ethan Skyler (talk) 16:31, 23 February 2010 (UTC)


 * ES Feb 23  Quote from Newton's SCHOLIUM "In the same general sense I use the word impulse, not defining in this treatise the species or physical qualities of forces, but investigating the quantities and mathematical proportions of them; as I observed before in the Definitions. In mathematics we are to investigate the quantities of forces with their proportions consequent upon any conditions supposed; then, when we enter upon physics, we compare those proportions with the phenomena of Nature, that we may know what conditions of those forces answer to the several kinds of attractive bodies. And this preparation being made, we argue more safely concerning the physical species, causes, and proportions of the forces.  Let us see, then, with what forces spherical bodies consisting of particles endued with attractive powers in the manner above spoken of must act upon one another; and what kind of motions will follow from them." End of Newton's quote. Ethan Skyler (talk) 16:31, 23 February 2010 (UTC)


 * ES Feb 23  After discovering this passage today, I noticed right away that he was writing about force in the plural, not singular as you would hope to strengthen your one-type position. Next I see that twice in his quote he referred to the physical species of forces indicating to me that he and I share the same general understanding that there is more than one type of force in Nature.  He even hinted that there exists different "causes" for these different "species" of forces to which he is referring.  There you have it.  Right from Newton's pen come words in full support for my more-than-one-force-exists-in-Nature position.  I will be the first to point out that Newton was stingy with his thoughts on these important issues.  Perhaps he didn't think of them as issues at the time. I think we all would have been better served if he had explained his understandings more thoroughly.  Instead we are forced into the position of archeologists, going over Newton's words with a feather duster so we can better see without disturbance. Ethan Skyler (talk) 16:31, 23 February 2010 (UTC)


 * * Here I apply LAW III to A and to B which assures me that there is indeed an acceleration/Reaction force present within the matter of each accelerating object. Thus like with every other possible event, this too is a four force event, two acceleration/Action forces reactively generating their own acceleration/Reaction forces of support within the same object. Ethan Skyler (talk) 19:17, 12 February 2010 (UTC)


 * This is completely counter to Newton and perhaps the fundamental flaw in your approach. -AndrewDressel (talk) 21:29, 12 February 2010 (UTC)


 * ES Feb 22 See Feb 22 postings above.  I think my 4 force understanding of the Earth/Moon system for example is fully supported by Newton's work as explained above.  Perhaps you are spending so much time focusing on finding the "fundamental flaw" in my work that you are missing the various logical steps I am taking toward what I see is a greater understanding of accelerational events than is currently enjoyed within conventional physics.  Ethan Skyler (talk) 07:17, 23 February 2010 (UTC)


 * * ES Feb 12  While I accept that you will continue to think that A's action force finds LAW III opposition or support against B's action force, supposing A and B are seperated by a light year of distance, by what mechanism do you think this opposition is possible? Ethan Skyler (talk) 19:17, 12 February 2010 (UTC)


 * As I stated above, Newton did not know, and we do not need to know, in order to calculate correctly the effect of gravity on those two bodies. It turned out, of course, that Newtonian mechanics cannot correctly calculate the precession of the orbit of Mercury, but that is a different issue altogether. -AndrewDressel (talk) 21:29, 12 February 2010 (UTC)


 * ES - In other words when I see a net action force is required to be present as the cause of an object's acceleration, I tell myself that for every such action there is always an equal and opposite reaction. In contact force events such as our whirling ball, I see equal action and reaction forces present at every point one may choose to measure. The swivel at the post is under tension in both directions.  The string is taut and reduced in diameter as it is under tension in both directions.  The ball is slightly oval a low rotation rates as it is under tension in both directions.  I don't see any place where the action force gets past the reaction force to be "unsupported" by the reaction force while acting as the cause of the ball's acceleration as others do.  Show me a point within the ball's matter where you think the reaction force is not present and I will have a look at it.


 * It is not clear why you want to complicate things with deformation, but consider this, once the final rotational speed is reached, no more deformation occurs, unless you want to consider creep. The string stretches no farther, and the ball does not become any more eccentric. Then, there is no relative motion between elements of the system, and we can return to considering it to be composed of rigid bodies.


 * The very tip of the ball, beyond which there is no more ball, experiences a centripital force that causes it to accelerate towards the center, and that is it. There is no reactive centrifugal force on it. In the rotating reference frame of the ball, there is a fictitious or pseudo centrifugal force, but that is something entirely different. The next element, molecule, or atom in, depending on how you want to slice it, does have a reactive centrifugal force on it, the equal and opposite pair to the centripital force on the outermost element. It also has a centripital force, and the two forces on it do not sum to zero, and so it also accelerates toward the center. This chain of forces and accelerations continues all the way to the post at the center. -AndrewDressel (talk) 14:46, 2 February 2010 (UTC)


 * ES - Overall, I think this understanding of acceleration always caused by net action forces supported by equal and opposite net reaction forces is an important position that the readers of Wikipedia should have the option of learning. I don't know of any subject or issue that is truly one-sided. Instead we relish in our freedom to "hear both sides" of an issue.  Well I represent the other side on the subject of "fictitious centrifugal force."  Search this subject on the Internet and you will find lengthy pages with endless speculation on the subject.  The longest one I have found is on a golf forum. It is a popular subject but not properly expressed.  The best article I have found on real centrifugal reaction force is here on Wikipedia.  Yet even it is not altogether correct. As long as I am allowed to finish my work here in my own private box, I will continue to do so.  I don't like to leave any jobs incomplete.  Some good will come of it, I am sure.


 * There is a separate article about the "fictitious centrifugal force" the the controversy surrounding it is delt with there. The force described in this article is very real and the only controversy is due to many authors not bothering to describe it and readers confusing it for the "fictitios" one. -AndrewDressel (talk) 23:47, 1 February 2010 (UTC)


 * ES - I have failed to respond to each of your points as we went along. I tend to really focus on developing my analysis as I go along which leaves me ill-prepared to answer questions until I am finished.  I am sure you have heard a speaker say: "Please, questions will be taken at the end."  My goal was/is to point out areas in the reactive centripetal article that need improvement. Since I only planned on placing these comments in the end of a lengthy discussion page, I thought I would avoid the conflict and bullying that is inherent with Wikipedia.  Also you did not identify yourself as anything but another visitor, such as myself.  I was hopeful you were an author with some built-in support for recognition of the reactive centrifugal force concept.Ethan Skyler (talk) 00:31, 31 January 2010 (UTC)


 * Here at the moving post, we have a net action force (red arrow) opposed or supported by a net reaction force (green arrow) of equal magnitude. Now suppose I end the whirling event and instead stand next to the post and pull as hard as I can on the top.  The post will move or deflect in my direction.  Now suppose you join in and pull in the opposite direction on the top of the post. Now the post will relax its deflection and return to a non-moved upright position.  While maintaining equal and opposite action forces, draw a vector diagram of this event.  It will look like Panel 6 except now there will be two opposed red arrows and the post will not move.  My point here is that with red and green arrows of equal magnitude, the post moves and with opposing red arrows the post does not move.


 * Here it appears that you are simply taking issue with where the forces are applied: at the top or at the bottom of the post. If, in your example of two people pulling on the post in opposite directions, one pulled on the top and the other pulled on the bottom, the post would bend exactly as it does with the whirling ball. -AndrewDressel (talk) 20:58, 29 January 2010 (UTC)


 * By stating that the post does not move during the whirling event, the author is pretending that the old rules apply regarding how equal and opposite forces are supposed to cancel each other's effect resulting in a "net force" of 0. Admitting that the post does move by orbiting the event's axis indicates to me that the presence of a "net action force" >0 is a better indicator to explain why the post moves despite the presence of an equal and opposite "net reaction force". Earlier I admired the use of red and green colors in the 6 Panel Chart. This way one can plainly see there is a difference between a red and a green arrow acting and reacting on the moving post verses two red arrows as when you and I pull in opposite directions on the non-moving post.


 * But the coloration is simply to indicate which forces get the name "reactive centrifugal force". As far as the post is concerned, there is no difference between a reactive centrifugal force generated by a whirling ball on a string and a force generated by someone simply pulling on the post, except that one may be constantly moving while the other remains stationary.
 * The real difference between the two scinarios you describe is that the ground provides the force necessary to keep the post stationary at its base, and the second person provides the force necessary to keep the post stationary at its top. -AndrewDressel (talk) 20:58, 29 January 2010 (UTC)


 * ES Feb 11 I have been considering the turning car event for a few days.  I know this post discussion seems pointless to you.  To me it is very important that you see what I see. In my paragraph above "By stating that the post does not move..." I expressed my understanding effectively.  You think the following: (correct me if I am wrong) Ethan Skyler (talk) 08:41, 12 February 2010 (UTC)


 * 1) A force is a force meaning there is only one kind or type of force.
 * 2) The green force in Panel 6 is in every way, except direction, the same as the red force.
 * 3) When these two forces, present at the top of the post, are added, the net force on the post is zero.
 * 4) The post does not move or accelerate. Ethan Skyler (talk) 08:44, 12 February 2010 (UTC)


 * Yes, that is correct. -AndrewDressel (talk) 16:49, 13 February 2010 (UTC)


 * (I have not tried to misrepresent anything here. Even if you want to point to a much higher red force at the surface of the ground on the post, I think leverage will reduce it to the same magnitude drawn for the red vector in Panel 6 when located at the top of the post.) Ethan Skyler (talk) 08:41, 12 February 2010 (UTC)


 * ES Feb 11  Now I want to make the post a flexible fiberglass fishing pole so we can easily see the effect different force combinations will have on the pole. The goal here is to question the validity of the 4 positions identified above.  Not long after I set the ball to whirling about the fishing pole, I see its tip is no longer vertical.  Instead it is initially accelerated away from vertical in the direction of the whirling ball.  This acceleration of the pole's tip indicates to me that Position 4 is wrong. Ethan Skyler (talk) 08:41, 12 February 2010 (UTC)


 * If the two forces (depicted in panel 6) are equal and opposite and both applied to the top of the fishing pole, then why would it move? Instead, its tip remains vertical, and Position 4 remains correct. -AndrewDressel (talk) 16:49, 13 February 2010 (UTC)


 * ES Feb 11  Position 3 tells me that when the centripetal red vector action force at the top of the pole is added to the centrifugal green vector reaction force, the net force on the pole is zero.  If it is zero then I think the pole should remain vertical.


 * My point exactly. If the two applied forces are indeed equal and opposite, the top of the post could even be replace with a strand of hair it it would not move, though it might sag straight down out of the 2D plane of consideration under its own weight.-AndrewDressel (talk) 16:49, 13 February 2010 (UTC)


 * ES Feb 11  Since it is not remaining vertical I can only conclude that the pole is having to exert a net force on the string to cause centripetal acceleration for the whirling ball.  I also think that even through the centrifugal reaction force from the ball and string are equal in magnitude to the pole's centripetal action force, adding their vectors is pointless for the reason I have stated several times herein: The centrifugal acceleration/Reaction force never serves to cancel the effect of it centripetal acceleration/Action force cause.  Panel 6 makes this addition while the article draws the zero net force conclusion and the post does not move conclusion, neither of which represents the truth in this event.  The post does accelerate in response to the presence of a positive net force.  Position 3 is wrong. Ethan Skyler (talk) 08:41, 12 February 2010 (UTC)


 * ES Feb 11 So is the green reaction force different from the red action force?  I think so.  Let's change the event a bit by replacing the green force at the post by another red force.  I will set two balls to whirling this one post.  Each is directly opposite the other like opposing blades on the main rotor of a small helicopter.  I see that each ball is experiencing centripetal acceleration caused as before by a red vector force the fishing rod is applying to the string.  I see two such red vectors present at the post.  I also see that the fishing rod post is at last vertical.  It is not moving.  Here I see that one centripetal acceleration/Action red vector force truly does cancel the effect an opposing centripetal acceleration/Action red vector force has upon the fishing pole post. To me this means that the effect a red vector acceleration/Action force has upon the fishing pole is not canceled by its own green vector acceleration/Reaction force but is canceled by an opposing red vector acceleration/Action force.  From this analysis I conclude that the green vector acceleration/Reaction force has different characteristics and therefore is a different type of force than red vector acceleration/Action forces.  Position 2 is wrong. Ethan Skyler (talk) 08:41, 12 February 2010 (UTC)


 * ES Feb 11 This leaves Position 1 which contends that a force is a force meaning there is only one type for force.  Since I just finished identifying that green a/R forces are of a different type and behavior from red a/A forces, the answer here is clear.  Position 1 is false as well.  I think that some rules written for red action forces are not suitable for green a/R forces. Ethan Skyler (talk) 08:41, 12 February 2010 (UTC)


 * ES Feb 14 Thank you for asking above the following question. "If the two forces (depicted in panel 6) are equal and opposite and both applied to the top of the fishing pole, then why would it move? Instead, its tip remains vertical, and Position 4 remains correct."  In fact better than that, thank you for persisting in our exchange. Sometimes it takes a long time discussing a subject before the really important points come to light for consideration.  Your question is about a really important point. Before we have another go at answering your question as to why, I want to discuss the fact that the tip of the pole is indeed moving.  First consider my control line experience.  In order to pull on and thereby provide the centripetal acceleration/Action force required to cause the airplane to circle my body, I employed Earth gravitation's help by leaning backward a few degrees from vertical.  Leaning backward in the direction of pull is what we do to apply a significant horizontal force in such an event.


 * Ah, but your analagy does not hold. I thought we decided that the top of the post was held in place by any of several mechanisms: it is very thick and stiff and stuck firmly in the ground, or it extends above the string and is held in place at both ends, etc. You, on the other hand, do not have any of those benefits. Your feet are not long enough, or your ankles are not strong enough to resist the torque generated by pull from your plane and keep you vertical. You must augment them with a little help from gravity. That does not mean that every such central support for a rotating object will suffer the same problem. -AndrewDressel (talk) 23:38, 14 February 2010 (UTC)


 * ES Feb 14 The fishing pole is somewhat different.  When the net action force on the pole is zero, the pole will remain vertical.  In order for the pole to apply an action force on the string and whirling ball, the pole must become bent forward opposite to the direction of the pull.  Thus when a fisherman pulls on a game fish, we know the pole is pulling in on the fish when the pole's tip is bent in the opposite direction which is out toward the fish.  Thus, in order to pull in, the pole's tip must be bent out.  I hope this makes sense to you and matches with your experience.  Ethan Skyler (talk) 09:00, 14 February 2010 (UTC)


 * The poll does not need to bend in order to generate a force. Newton certianly never said so. It bends simply because it is not stiff enough to not. If it were perfectly rigid, it would still be able to generate the force necessary to lift a fish or remain vertical when pulled by the centrifugal force of a rotating ball. -AndrewDressel (talk) 23:38, 14 February 2010 (UTC)


 * ES Feb 14 Now consider our whirling ball event.  We know a centripetal action force is being applied by the pole along the string to end up causing centripetal acceleration for the whirling ball.  In order to apply this inward-directed centripetal acceleration/Action force, according to the analysis above, the pole's tip must be bent out toward the ball.


 * I do not see the analysis, simply the assertion. I know of no reason why an object must flex in order to generate a force. Even if we agree that there is no such thing as a perfectly rigid body in reality and so every body flexes under any load, that still does not make flexing necessary to generate a force. Similarly, every mass generates gravity, but that gravity is not necessary for it to generate a force. -AndrewDressel (talk) 23:38, 14 February 2010 (UTC)


 * ES Feb 14 We know the whirling ball is accelerating.  We know that Law I says an action force is present as the cause.  We know that pole is providing this action force.  Therefore we can accept ahead of the event that in order to apply this action force, the pole's tip must be bent out in the direction of the ball as the tip travels around the center of the ball's curved path.  Ethan Skyler (talk) 09:00, 14 February 2010 (UTC)


 * Nope. I don't see it. If the ball were a puck on an air hockey table, and instead of any kind of post, the string simply went through a hole in the middle of the table, all 6 panels in the article would still be correct. We can also mount the table firmly to the ground, without legs, to further reduce any chance of anything flexing. -AndrewDressel (talk) 23:38, 14 February 2010 (UTC)


 * ES Feb 14 Following Galileo's lead by testing every theory, I suggest you tie a short rope to the handle of a bucket with a half gallon of water inside and whirl it about your person.  You will find a leading force is required to get the bucket whirling.  Once up to speed you will find yourself leaning back to maintain the centripetal acceleration/Action force.  To stop the bucket safely, you will find a trailing force is necessary to slow its circular motion gradually.  If you stop quickly by ceasing your own rotation, the bucket will wrap its rope around you and then crash into you so be careful.  Let me know when you have completed this experiment. Ethan Skyler (talk) 09:00, 14 February 2010 (UTC)


 * Yes, of course, a sufficiently heavy object with sufficient velocity on the end of a sufficiently strong rope would probably be able to bend anything, but I do not see what that proves. Instead, try this experiment: tape a penny to the end of a short string, tie the other end of the string to your belt, and rotate in place just fast enough for the penny to swing slightly away from your body. How much must you lean back to support the horizontal load? -AndrewDressel (talk) 23:38, 14 February 2010 (UTC)


 * ES Feb 14 eve I did try your penny experiment as requested.  I was unable to turn smoothly enough to maintain a stable rate of centripetal acceleration for the penny.  It was hard to determine that centripetal acceleration was occurring for the penny since I only had my eyes to tell me when the string was no longer vertical.  I also tried causing the penny attached to 24" of string to swing in a 1" diameter circle.  It took close to two seconds per orbit.  The centripetal action force applied to the penny would be just slightly above 0 lb so that would be the amount this test would require of my body to cause.  No leaning back required.  I will run a bucket test tomorrow.  I have a tension scale around here somewhere so will attach it between my hand and the rope.  I will try to make the axis to water distance equal to 6' and will attempt to rotate the bucket as fast a possible to have the rope whirl as close to horizontal as I can make it happen.  It will be interesting to measure exactly how hard I will have to pull inward on the rope to cause the centripetal acceleration present.  I will do the test over soft lawn while wearing a motorcycle helmet in case I should fall.  I'll look forward to hearing about your whirling bucket experiment. Then, after we both have this experience, we will draw some conclusions. ([User talk:Ethan Skyler#top|talk]]) 03:25, 15 February 2010 (UTC)


 * ES Feb 11 I think statements within the whirling ball article that are in conflict with the above analysis should be changed or removed.  Thus a net force of the red vector magnitude is present at the post causing the post to initially accelerate away from vertical and then to maintain this bent shape as the tip orbits the center of the ball's curved path.  What do you think? Ethan Skyler (talk) 08:41, 12 February 2010 (UTC)


 * ES Feb 11 I see the failure of the "net force" rule when adding a/A and a/R forces can be avoided simply by adding the net acceleration/Action forces present. For example, the one red a/A force in Panel 6 equals a positive net action force.  As verification, the post bends.  Change to opposing red a/A forces in the twin whirling balls event and the vector total equals a net action force of zero.  As verification, the post does not bend.   Ethan Skyler (talk) 08:41, 12 February 2010 (UTC)


 * ES Feb 15  I am wondering why you object to using a flexible fishing pole as the post.


 * Simply because it adds needless complexity to the problem and reveals nothing new. Yes, a flexible pole will flex when its base is held fixed and a force is applied to its tip. So what? -AndrewDressel (talk) 04:16, 18 February 2010 (UTC)


 * ES Feb 15 Is it because a flexible pole gives us the opportunity to observe that the tip is visibly moving all the time the single ball is whirling about and not moving all the time the twin balls are whirling about as I have described above?  If so then we must share the opinion (me willingly, you unwillingly) that by using the flexible pole, it becomes obvious that "the post is not moving" prediction stated in the article is a false claim.


 * Certainly not. Both situations, one ball or two, are handled completely with a rigid post. In the first case, the post is acted upon by two equal and opposite forces, the string pulling on it and the ground holding it still. In the second case, no force is acting on the post in the horizontal plane. -AndrewDressel (talk) 04:16, 18 February 2010 (UTC)


 * ES Feb 15 You keep wanting to make the post stiffer and stiffer so we will not be able to see this experimental proof as to exactly how false this "not moving" claim is proving to be.  We agree that the net force on any post you care to choose is 0. The "net force" theory tells you that the post cannot move, yet the single ball orbiting the flexible pole experiment proves the opposite that the post does move.


 * Not at all. It is merely more complicated because of the flexible pole. Instead of easily separable bodies and the forces between them, the flexible pole introduces a continuum from the completely flexible and mobile string attached to its tip to the rigid and fixed base. Each element of the pole still obeys Newton's 2nd and 3rd laws as classically understood. -AndrewDressel (talk) 04:16, 18 February 2010 (UTC)


 * ES Feb 15 You know this to be true and wish to obscure this visual clue by wanting to remove the legs from your air hockey table to keep the table from "moving" as the puck orbits the 0 "net force" hole. Ethan Skyler (talk) 20:14, 15 February 2010 (UTC)


 * Go ahead, put wobbly legs under the table. Let them be tugged around by the puck. May they be attached firmly to the ground? Or, will you insist that they also bend slightly the bolts that hold them to the concrete slab? May the concrete slab be rigid and fixed? Or will it too flex and slosh slightly in the hole into which it was poured? Of course, even the crust of the earth is a flimsy sheet floating on a sea of magma. Where will it end? -AndrewDressel (talk) 04:16, 18 February 2010 (UTC)


 * ES Feb 15 So much for the "net force" theory and all of its false conclusions regarding accelerative events.  Like the Bismark, no matter how stout and important it is thought to be, it is still going to end up on the bottom. With a net force of 0 on the pole, how can the pole be moving?


 * Obviously, the net force on the flexible pole is not zero, because it is moving. The net force on the bottom of the pole that is attached to the ground, if you will allow anything not to move, is zero. -AndrewDressel (talk) 04:16, 18 February 2010 (UTC)


 * ES Feb 24 I think I am beginning to better understand your position on the whirling ball event.  While it is unlikely we will reach an agreement, I would not be pleased to learn that I was mistaking some point important to your position. I have a couple of questions that may clear things up for me.  When I think of force, I think of it as being present as the action/reaction forces causing and supporting accelerational events as per Newton's work.  I also recognize force as being experienced by a non-accelerating object as well.  My definition for force is "a push or pull being experienced by an object".  I am not asking you to accept that definition, just to use it as a comparison with your understanding of force.  I think it would help if you explained your view of and definition for force. Ethan Skyler (talk) 08:27, 25 February 2010 (UTC)


 * ES Feb 24 I like thinking about your whirling hockey puck where the string drops down one of the air holes.  I am beginning to see why you prefer a rigid post and bolted down table.  If you see no movement then you feel free to conclude that no acceleration is present for the rigid post or bolted table and with no acceleration present, you see this as evidence that the net force on the post/table is zero. You signify this zero net force conclusion with the statement that the post/table "does not move".  The zero net force significance of your "does not move" statement was lost on me.  Instead, I kept pointing out that a net action force was being experienced by the rigid post which could flex and even break, the flexible pole with its visible evidence, and the legged hockey table. Now that I think I have your position correct, I would like you to explain the forces and their accelerational roles in the whirling hockey puck event at the location where the string plunges down through the hole.  I would like to add a small tension scale below the hole in such a way that the string goes down the hole and is attached to the upper hook of the scale and a second string is attached to the lower hook and from there goes on down to be tied to a non-moving eyebolt cast into a concrete slab.  Please mention the forces on the string, the non-moving hole, and the non-moving eyebolt.  With your description from the accelerating puck and string to the non-accelerating eyebolt, I think I will have a complete picture of your position.  Differences in our positions will remain but at least I will know that I am not missing an important part of your position.


 * ES Feb 15 The reason is the acceleration/Reaction force is caused to exist by the acceleration/Action force present at the top of the pole.  Adding these two equal and opposite forces using the rules of the "net force" theory yields a net force of 0 along with the false prediction that at 0, acceleration will be absent for the pole.  Instead, acceleration is present for the pole due to the simple fact that here the reaction force is of a different type of force than the action force and in no way serves to cancel or balance out the action force.  Instead, it serves to provide support for the action force as it goes along for the accelerational ride.  No cancelling of the accelerational ability of the action force is occurring here.  The pole continues to bend and orbit the axis of the ball's curved path of travel as long as it continues to cause centripetal acceleration for the whirling ball.  Rather than stiffen the post to obscure this fact, I think it is more scientific to view this evidence and consider another solution.  Ethan Skyler (talk) 20:14, 15 February 2010 (UTC)


 * ES - All forces applied to the post are at its top.


 * Why would you make that assertion? In the original example, before I started suggesting modifications in order to prevent the post from flexing and introducing needless complications, the string pulls on the top of the post, and the ground applies a force (push or pull depending on how it is stuck in the ground and what the ground is made of) on the bottom of the post. -AndrewDressel (talk) 14:46, 2 February 2010 (UTC)


 * ES - The article incorrectly predicts that the post does not move by orbiting the event's axis. This position is taken since the green force vector is thought to cancel the effect of the red force vector as drawn in Panel 6. In truth the green force vector supports the red force vector all the while centripetal acceleration is present.  Thus the text here contains two mistakes, one stated and the second one determined by old-style vector addition.  Proof of these two mistakes is the experimental evidence that the post does, in fact, orbit the event's axis and therefore does "move" outward while providing the centripetal net action force that is causing inward-directed acceleration for the ball.Ethan Skyler (talk) 00:31, 31 January 2010 (UTC)


 * By "the event's axis", do you mean its barycenter? For the sake of simplification without loss of generalization, may we consider the post being fixed to an infinite mass so that the barycenter of the system is at the center of the post? Then we can let the post be stationary. Then also, the pair of forces on the post must be equal and opposite so that the post does not move. The force generated by the ground exactly matches the reactive centripital force applied to the post by the string.


 * If you refuse to allow an infinite mass, then let us calculate where the barycenter must be between a post rigidly attached to the earh and a 1 kilogram ball at the end of a 1 meter string. From the barycenter article:
 * In a simple two-body case, r1, the distance from the center of the primary to the barycenter is given by:


 * $$r_1 = a \cdot {m_2 \over m_1 + m_2} = {a \over 1 + m_1/m_2}$$


 * where:
 * a is the distance between the centers of the two bodies;
 * m1 and m2 are the masses of the two bodies.


 * Using 5.9736 kg for the mass of the earth yields a distance of 1.674 m for the distance from the center of the post to the barycenter of the system. That is less than the upper limit estimated for the radius of an electron: 10−22 m. I don't think we'll be measuring that any time soon. Of course, a string pulling on a post stuck in the surface of the earth causes the earth to rotate as well, but the earth also has a pretty big moment of inertia and so I think we're pretty close to the correct order of magnitude. -AndrewDressel (talk) 14:46, 2 February 2010 (UTC)


 * ES By the event's axis I mean its rotational axis which if the post top does not move is at the center of this post. I see the post as being cast in a large concrete block.  As the post is having to provide a stronger and stronger centripetal force on the string to whirl the ball, I see the post bending from the block on up so that its top now orbits the accelerational axis.  Increase the rate of rotation with a large ball on a long wire and the post will bend so much that it will likely suffer metal fatigue and break off at the block.  I describe this event merely to show that the article's position, that when subjected to equal and opposite action and reaction, red and green force vectors, the post "does not move", is in fact incorrect.  My point is that green reaction forces do not ever cancel the accelerational ability of red action forces. Instead they are caused by and end up merely provide support for the red action forces present.  See my 2D event below. Ethan Skyler (talk) 00:40, 3 February 2010 (UTC)


 * The post "does not move" is only incorrect if it is too weak to support the ball. I do not understand why you keep insisting on the post bending and now breaking due to fatigue. Can we simply suppose that the post is strong and the ball is light, or is the flexing of the post somehow central to your analysis of the situation? -AndrewDressel (talk) 01:29, 3 February 2010 (UTC)


 * ES I suspect the post continues to be an issue since you don't see what I see.  The post is claimed to be non-moving by the article's author and further that the net force on the post is zero in support of the belief that the red and green vectors (both are at the top of the post, yes?) cancel each other in the same way that the two red vectors cancel each other when you and I tug in opposite directions on the top of the post.  I think we agree that when the red and green vectors are present as drawn in Panel 6, the post actually does move (admittedly just a tiny amount), then I see this as an indication that this non-moving post claim is an error and further that this error is based upon the rules of the "net force" theory of acceleration that I see is also in error.  At this point I think you do not agree with either conclusion so I will be content to learn that at least you understand my position. 216.231.36.211 (talk) 07:46, 3 February 2010 (UTC)


 * No, I know of no reason that the red and green vectors must both be at the top of the post. The green arrow represents the centrifugal force generated by the string attached to the top of the post. The red arrow represents the force generated by whatever the post is stuck in, the ground or a concrete block, to hold the post still. The "non-moving post claim" is merely a simplification, just as the post, string, and ball being "rigid bodies" is merely a simplification. Including the slight movement of the post does not add to understanding of the situation any more than does including elongation of the string. I am afraid that I still do not understand your position. To me, your insistance on the post moving and the ball ovalizing seems as helpful and necessary as would an insistance on applying a Lorentz transformation to our calculations to correct our measurment according to Special relativity. At low speeds, Newtonian mechanics is plenty good enough, and so are rigid-body assumptions. -AndrewDressel (talk) 15:03, 4 February 2010 (UTC)


 * ES Feb 5 I have spent so much time on this page I missed some of the text on the article's page.  You are right in that the article's 6 panel caption clearly states that "the centripetal force indicated by the red arrow is applied by the post-hole that holds the post."  I think I was confused by two things.  First I remember reading in the text the statement that "The centripetal force of panel 2 is applied by the post to the end of the string." which left me thinking that the red arrow represented the centripetal acceleration-causing action force on the string.  Now I see the caption says otherwise, that the red arrow indicates the centripetal action force at the ground.  I am left wondering two things.  First, where is the red arrow that should be representing the centripetal force of panel 2 applied by the post to the end of the string? Second, I do not know the height of the post but regardless, I think of it as a lever which means to me that any force at ground level in the red arrow centripetal direction will have to be considerably greater than true centripetal force being applied to the string attached to the top of the post.  Since I did remember that the red and green vectors were drawn to be of the same magnitude, I guess I could only imagine them to be at the post's top regardless of the perfectly clear instructions to the contrary written in the 6 panel caption. Ethan Skyler (talk) 16:57, 5 February 2010 (UTC)


 * Panel 2 is just about the ball. The centripetal force applied by the post to the end of the string is shown in panel 4. -AndrewDressel (talk) 17:29, 5 February 2010 (UTC)


 * ES Feb 6 Yes, the article's comment telling us that panel 2's centripetal force on the ball being applied by the post to the inner end of the string and transferred along said string out to cause centripetal acceleration for the whirling ball, is included in the panel 4 description where you thought it belongs. Ethan Skyler (talk) 17:18, 6 February 2010 (UTC)


 * Yes, if the post has a finite height, then a single centripetal force applied at its bottom, as indicated by the red arrow, is not sufficient to hold it in place. How about if the post is supported at its top and bottom and the string is attached to its middle. Then the top and bottom supports would each supply half of the total centripetal force necessary, and when seen from above, these two arrows combine to become the one shown. -AndrewDressel (talk) 17:29, 5 February 2010 (UTC)


 * ES Feb 6 Interesting solution.  It would require an overhead structure based outside the radius of the whirling ball and the whirling room riding on a circular track we added to the event.  I like the simplicity of the post.  I also like the panel 6 drawing where equal and opposite red & green vectors are drawn.  I think we should accept that the location of the post's red vector, as drawn, exists at the post's top in line with both the string and the centripetal vector on the string as drawn in panel 4.  This leaves the two comments about the role of the post hole to reconsider.  I don't suppose we could leave them out altogether.  But if included, then I think Earth's acceleration should also be included to complete this forceful event.  Also the "net force on the post is zero" needs to go.  Instead the article should discuss how the post is experiencing tension on its side farthest from the ball and compression on its side nearest the ball with these forces required to exist in order for the post to be able to apply the event-causing centripetal force to the string's inner end.  Ethan Skyler (talk) 18:08, 6 February 2010 (UTC)


 * In the article about Reactive centrifugal force, we don't have to discuss the post support at all, just as we don't have to discuss deformation or internal forces. Of course they are all there, but they are only diversions from the point of the article. Same goes for the Earth's acceleration. The example as described would be the same in the vacuum of space far from any large mass, changes very little on the surface of the Earth, and so does not benefit from the addition of gravity. The internal forces in the post are better discussed in the article about bending of beams -AndrewDressel (talk) 20:35, 7 February 2010 (UTC)


 * If the post is not accelerating, then "net force on the post is zero". Why would that phrase need to go? -AndrewDressel (talk) 20:35, 7 February 2010 (UTC)


 * ES Feb 6 Remember your earlier suggestion that this whirling event is best considered as rigid or frozen in motion?  I think this is a capital idea.  I think the 6 panel drawing is too confusing.  The article would be better served by one frozen drawing where the centripetal action force, fundamentally caused initially by the difference in the existing motions of Earth and the ball being suddenly connected together by a string tied from Earth's post to the ball's eyebolt with the result that the ball begins its orbital path about the post resulting in the dual actions of centripetal acceleration for the ball and some kind of circular acceleration for the post and Earth below resulting in the post's center orbiting the axis of the whirling ball, that I think you are better equipped to describe.  If this entire event is frozen or becomes rigid, including the bent post and the slightly oval ball, and we are to identify and label all the acceleration/Action and acceleration/Reaction forces present at various points in this "collision" between the former motions of Earth and the ball, then I think we could better understand how Earth ends up causing acceleration for the ball and how with an equal and opposite force the ball ends up causing acceleration for Earth.  Ethan Skyler (talk) 18:08, 6 February 2010 (UTC)


 * Panel 6 is as simple as it can be. Adding gravity or motion would only complicate it. The motion of the Earth is completely irrelevant. -AndrewDressel (talk) 20:35, 7 February 2010 (UTC)


 * Again, I think new rules are needed to explain accelerational events, taking into account the "net action force" present. The old "net force" theory of acceleration is just not up to the task. Ethan Skyler (talk) 19:42, 29 January 2010 (UTC)


 * I still don't see where you have addressed my point that Netwon's 2nd law applies only to the net force on a single body (or system) and his 3rd law describes the equal and opposite pair of forces only between two different bodies. Until you do, I'm afraid that you will not be able to persuade me to see your point. -AndrewDressel (talk) 20:58, 29 January 2010 (UTC)


 * ES - Since you are telling me that Newton's LAW III applies only to mutual action/reaction forces present between two contacting bodies and that LAW II applies only to how an acceleration-causing action force causes acceleration for a single body, I take it that this stance prevents us from considering the likelihood that Newton's LAW III remains fully in effect throughout the body's matter during its acceleration. Since we are dealing with an important part of Newton's work, possibly you know of a part where he tells us that LAW III does not apply within the body during its acceleration. I mean that when a net action force is causing acceleration for the body that Newton tells us that no net reaction force is present within the body and busy providing forceful support for the net action force.  If you know of such a quote then I am suggesting that experiment shows him to be wrong.  If you don't find such a quote, then it is likely that someone added this position to his work at a later date.  Again I will suggest that this later-date person is wrong due to experimental evidence I have already described herein but will be happy to expand upon if requested. Ethan Skyler (talk) 00:31, 31 January 2010 (UTC)


 * I only have time for a quick reply right now, so I'll just address this one point. Apparently Newton only considered what we call today point masses. The application of body dimensions, the significance of the center of mass, and angular momentum balance are all attributed to Leonhard Euler about 50 years later in what are now described as Euler's laws.


 * Yes, Newton's 3rd law does apply to all the "particles" that make up a rigid body, but all those internal forces exactly cancel out, as specified by that same law, and so in the idealized case of rigid bodies do not contribute in any way to the acceleration of the center of gravity by the net external force as described by his 2nd law.


 * That's all I've got for now. -AndrewDressel (talk) 22:00, 31 January 2010 (UTC)


 * ES Thanks for the note explaining the move.  Glad you did.  I will be gone for a few days while I finish up my analysis of the balance of the article.  From your research above I am glad to learn that Newton's LAW III does apply.  Next up for discussion is your saying "all those internal forces exactly cancel out" and more specifically the meaning of the word "cancel".  Until then,  Ethan Skyler (talk) 09:19, 1 February 2010 (UTC)


 * First, according to C. Truesdell in Essays in the History of Mechanics, as cited by McGill and King in Engineering Mechanics, An Introduction to Dynamics, Newton never considered the finite dimensions of the objects to which he applied his laws. That was done about 50 years later by Leonhard Euler.


 * For simplicity, let us assume the rigid body model and disallow deformation. Let us also assume that the material is homogeneous and isotropic. Then we can divide it up into n infinitesimal elements, but not molecules or atoms which introduce many additional complications. Each element perfectly abuts its surrounding neighbors. Finally, if an external force is applied to an element on the exterior, it will push up against its neighbors whether it moves or not, and each neighbor will push back with an exactly equal and opposite force, by Newton's 3rd law. If we allow contact and field forces (gravity), then each element experiences a net force from each of the n-1 other elements. We can label each internal force as fij, the force on element i from element j. If we add up all the internal and external forces, all the internal ones will have exactly one equal and opposite and so will sum perfectly to zero: fij = -fji . The only remaining net force will be the original external force. If all the internal forces do not cancel, it can only be due to an accounting mistake on our part, or the body is deforming which violates our original assumption. -AndrewDressel (talk) 23:47, 1 February 2010 (UTC)

=Comments from February 2, 2010=
 * Feb 2nd, 2010 I generally agree with your "For simplicity" paragraph above.  But....  to fully understand the difference between the acceleration/Reaction or a/R force as I term it and the acceleration/Action or a/A force, we will need to go down to the atomic level.  Nothing fancy, just necessary since the atom is the source of the a/R force.  Also I refer to a force that is generated actively or reactively within an atom as an internal force.  All other forces that come to exist outside the atom are external forces.  An external force is a contact force, like the ones you describe between the elements in your example.  I understand that these terms are currently in use in physics with different meanings.  They serve my purpose well as you will see. Ethan Skyler (talk) 21:18, 2 February 2010 (UTC)


 * SKYLER'S 2D MODEL OF AN ACCELERATING ATOM
 * The following is an atomic model designed to reveal the acceleration-supporting, not acceleration-cancelling role played by the acceleration/Reaction force. This role is the same in linear events as it is in circular ones.  On an overly large air table, I will place a light, plastic hulu hoop to represent a 2D substitute for the distant electrons that encircle the massive atomic nucleus.  I will lay a solid steel disk with a 5 kg mass rating at the hoop's center.  Next I will place a dozen tension springs evenly around the hoop and attach one end of each spring to the hoop and the other end to the edge of the steel disk the way bicycle spokes are attached and tensioned to the hub.  When complete, by moving it around a bit it becomes clear the nucleus of our  2D atom automatically strives to keep itself centered within the electron hoop. Ethan Skyler (talk) 21:18, 2 February 2010 (UTC)


 * Now I will turn on the table's air supply. The steel disk will float just above the table's surface.  It is your job is to cause acceleration for the 2D atom by applying an external acceleration/Action force in one horizontal direction to the electron hoop.  As you push away on the hoop, you will note that the massive steel disk is slow to respond.  It lags behind in its acceleration away from your position as the tension springs on the far side of the disk stretch in length while the near side springs reduce in length thereby providing visual evidence of the effect your acceleration-causing net action force is having on the steel disk.  Ethan Skyler (talk) 21:18, 2 February 2010 (UTC)


 * Before the 2D atom gets away, you tug in the opposite direction on the electron hoop to stop its motion of departure. The hoop stops right away but again the heavy steel disk is slow to respond.  This time the near side springs stretch and the far side springs relax as your net acceleration/Action force begins causing negative acceleration for the 2D atom.  Once the nucleus's motion has been accelerated to a stop, it will rebound perhaps several times until it is once again positioned at the center of the electron hoop. Ethan Skyler (talk) 21:18, 2 February 2010 (UTC)


 * Now let us consider the 2D atom's behavior in a circular event. I will tie a short rope to the hoop, put myself in the middle of the large air table and begin whirling the 2D atom round my person.  You will photograph the atom as it whirls past your location which is next to the table.  The photo shows the steel nucleus constantly off center within the electron hoop during this centripetal acceleration event.  Here the hoop's inside springs are stretched indicating that an inward-directed net action force is being applied to the nucleus.  By adding the several vectors of the near-side springs, a "net action force"  on the nucleus in the direction of the event's axis can be determined.  I refer to this force as the centripetal acceleration/Action force. Ethan Skyler (talk) 21:18, 2 February 2010 (UTC)


 * As this external net action force delivered by the hoop and inside springs is busy causing centripetal acceleration for the steel nucleus, the nucleus presents an external net reaction force to the springs that in total is exactly equal in magnitude and opposite in direction to the previously described net action force. Here while the force of your centripetal pull is causing constant acceleration for the 2D atom with over 99% of this force required to accelerate the massive nucleus, the nucleus is constantly lagging behind its preferred central location within the electron hoop.  The acceleration results in the nucleus becoming off-center within its hoop.  The nucleus reacts by attempting to pull itself back to the center or perhaps more accurately reacts to pull the electron hoop back so that it is once again centered around the nucleus. Ethan Skyler (talk) 21:18, 2 February 2010 (UTC)


 * Next, I will slice the steel nucleus in half with the slice at right angle to a radial line drawn from the axis through the nucleus. I will add two small tension springs between the two halves and then ask you to repeat your whirling event.  A photo will show these central springs to be stretched longer than normal as the nucleus whirls by your location.  This indicates that the net action force being applied to the curved near side of the nucleus  is also present at a little more than 50 % strength at the nucleus center.  It also indicates that the nucleus reactive centrifugal force, (net  reaction force) is equally present at the nucleus center in the opposite direction and at the same 50%+ value.  Here, everywhere a net action force is measured, a net reaction force of the same value is equally present.  The net action force is never greater than the net reaction force regardless of how much we want it to be so.  It is important to dispel this erroneous desire from our minds. Ethan Skyler (talk) 21:18, 2 February 2010 (UTC)


 * Be careful not to confuse some feature of centripetal acceleration with general linear acceleration. Of course the centripital force varies with the radius for the same angular rate:
 * $$F = m r \omega^2 \,$$
 * This will not be the case in linear acceleration. This may also be observed at the scales installed into two strings of different lengths whirling two balls with the same mass at the same angular rate.-AndrewDressel (talk) 01:29, 3 February 2010 (UTC)


 * (Note: I see a force as a push or pull being experienced by an object.  Further, I refer to a force that is generated actively or reactively within an atom as an internal force.  All other forces that come to exist outside the atom are external forces.  In practice, an external force is a contact force.) Ethan Skyler (talk) 21:18, 2 February 2010 (UTC)


 * Or gravitational, or electro-magnetic. -AndrewDressel (talk) 01:29, 3 February 2010 (UTC)


 * Also, your definition internal and external forces is counter the definition used in the physics and engineering texts I have seen. As I imply above, whether a force is internal or external depends on the definition of the "body" or "system" being used at the time. If the body or system is an automobile, then the forces generated in the suspension between the wheels and the chassis are internal. If the body or system is the chassis, then the forces generated in the suspension between the wheels and the chassis are external. External forces are included on free body diagrams and internal forces are not. -AndrewDressel (talk) 15:50, 4 February 2010 (UTC)


 * Inventing and studying this event has helped me to understand that external accelerational forces (contact forces) are applied to the electron hoop, and not directly against the nucleus. Therefore this accelerational force is only indirectly applied to the massive nucleus.  The result is a tug of war where the nucleus lags behind thereby losing its preferred central position and reacts by attempting to re-center itself with this reaction taking the form of the internally-generated acceleration/Reaction force.  Provided that the accelerating force is an external force (not generated within the atom) as this internal a/R force departs its originating atom it take the form of an external contact force as it stacks up by joining with a myriad of identical internal reaction forces from the other accelerating atoms within the object to achieve the total net reaction force providing support for the net action force.  Here we have a net action force, causing acceleration for an object's atoms, causing their nuclei to lag behind becoming off-center, causing the nuclei to forcefully attempt to regain that central position, causing the internal generation of what becomes the net reaction force that provides equal and opposite support for the event-causing net action force.  All in full agreement with Newton's LAW III " To every action there is always opposed an equal reaction,..." Ethan Skyler (talk) 21:18, 2 February 2010 (UTC)


 * If this "sloshing" of the atomic nucleus is necessary for Newton's 3rd law to be true or to generate the reaction force it predicts, how did Newton come up with it, and why did no one point out that it was wrong before the internal structure of the atom was proposed in 1913? Also, does this conclusion mean that Newton's 3rd law does not apply to interactions between subatomic particles? -AndrewDressel (talk) 15:50, 4 February 2010 (UTC)


 * ES Feb 5 I like your term sloshing in reference to a rebounding nucleus.  I think sloshing is a nice explanation for the amazing rebound abilities of a ball bearing when dropped a few feet to impact with an equally dense steel surface.  You would think you had just dropped one of those golf-ball-sized super balls sold at toy stores.  I envision upon impact, the relatively massive nuclei of all the ball's atoms find themselves below center when the negative acceleration during impact is complete and then all rebound at the same time causing the ball bearing to jump back up off the hardened surface with gusto.  Have you ever tried this test? Ethan Skyler (talk) 06:45, 6 February 2010 (UTC)


 * This model of the net reaction force or as I term it the acceleration/Reaction force solves other mysteries in physics as well.  But that is for a later time.  Most importantly, for now, it offers a way out of the "net unbalanced force theory of acceleration" prison.  Once the acceleration/Reaction force is recognized, one becomes free to see that an unbalanced, unopposed force is an impossibility in Nature along with being in direct violation of Newton's LAW III.  His prediction leaves no room for such a false concept for he employs the all-encompassing word in Latin that is translated in PRINCIPIA to "always".  "Always" in Newton's LAW III means that action and reaction forces always remain fully balanced at every point one chooses to examine in every accelerational and non-accelerational event described using the predictions of Newton's LAW I and/or LAW II. (C) Copyright Ethan Skyler - 2010  Permission granted to all Wikipedia readers to freely copy SKYLER'S 2D MODEL OF AN ACCELERATING ATOM provided that my copyright notice is included on each copy. Ethan Skyler (talk) 21:18, 2 February 2010 (UTC)


 * In classical Newtonian mechanics, as it is generally accepted and taught, there is no instance where Newton's 3rd law does not apply. None of the examples presented in the Reactive centrifugal force article suggest that any single force is not accompanied by its equal and opposite. As far as I can tell, the issues you have with the current theory stem from using a different definition of "internal" and "external" and from pushing together Newton's 2nd and 3rd law in your mind.


 * I don't know if any of this will ever get printed on Wikipedia but as you can tell, I think the reactive centrifugal force is a grade A subject of the highest importance to physics.


 * ES I will continue working on suggestions for the reactive centrifugal force article.  I was planning on including the above 2D Model.  Now that it is written I can get your equal and hopefully not opposite reaction.  Thanks Andrew for encouraging me on this adventure.  Ethan Skyler


 * I doubt you will find success with that. At the least, it runs counter to the Wikipedia policy of no original research. Especially in this field, you would need to find a reliable, published secondary source. -AndrewDressel (talk) 15:56, 4 February 2010 (UTC)


 * ES Feb 17  In looking at the car rounding the curve event, I like how the article points out that as the car enters the turn, the passenger sitting on a slippery seat will receive insufficient centripetal force to begin rounding the curve at the same rate as the car.  So for a few moments, the seat slides sideways under the passenger while the passenger continues on generally with his or her straight-line motion only modestly changed.  This would be like when a magician quickly snaps a table cloth out from under a cup and saucer at rest on the table.  Eventually, the car door contacts the passenger and finally delivers sufficient centripetal acceleration-causing action force to cause the passenger to round the curve at the same rate as the car. My only suggested change would be to refer to the seat as sliding under the passenger instead of saying that the passenger slides along the top of the seat.  It is a minor change which I think better describes this event since it is the seat that is trying to accelerate the passenger and not the other way around. Ethan Skyler (talk) 03:54, 18 February 2010 (UTC)


 * ES Feb 17 In the next paragraph, the seat is properly described as acting on the passenger.  But then the passenger's reactive centrifugal force is twice described as "acting" on the seat.  This unique acceleration/Reaction force being reactively generated within the passenger's body is caused by and reacts in support of the centripetal action force from the seat that is causing the passenger's acceleration. So here I object to referring to the reaction force that is always directed opposite to the action of acceleration as being an "action" force.  If it is an action force then it will be aimed in the same direction as is the event's acceleration. Ethan Skyler (talk) 04:10, 18 February 2010 (UTC)


 * There is no law I know of, Newtonian or otherwise, that requires the force opposite of a force that causes an acceleration to be called a "reaction force". Which force of the pair is the "action" force and which is the "reaction" force is completely arbitrary. -AndrewDressel (talk) 00:45, 19 February 2010 (UTC)


 * ES Feb 25 Yes, recognizing that the myriad of mutual forces reactively generated within the myriad of components of matter within an accelerating object, including the reactive centrifugal force in circular events and its unnamed twin in linear events,known by me as the acceleration/Reaction force of matter, is core to my understanding of accelerational events. I think of this internal-to-matter reaction-to-acceleration force as being caused by the acceleration/Action force which in circular events is the centripetal force. Since the a/R force is always directed opposite to the direction of acceleration and recognizing that the acceleration is the action and further that the force causing this action is the accelerating action force, I recognize that the a/R force is never the cause of any event including the work claimed in the article herein.  In truth, I think it best to accept the a/R force as merely a reflection of the a/A or acceleration-causing action force.  Evidence supporting my position here takes the form of the law of mutual pressure in contact force events.  It is not possible to push against an object granting it an accelerated motion without there existing an equal and opposite force from the object that in no way serves to cancel, block, reduce or destroy the accelerational ability of the accelerating action force of your push.  Being so benign proves to me that this reaction-to-acceleration force back from the object is meerly a reflection of the action force.  They are one and the same.  Thus you can see why I think it is an error to add these two force vectors and draw any conclusion from the result.  It is like adding the force on a tensioned horizontal chain link to the force on the next link and so on.  You would be adding the same force to itself and drawing the wrong conclusion about the total force expected at the far end. Ethan Skyler (talk) 17:45, 25 February 2010 (UTC)


 * ES Feb 25 What does Newton say about this? During yesterday's browsing through PRINCIPIA I came up with the following quote regarding the role of the acceleration/Action force in circular events named by Newton as the centripetal force.  This quote is located in Definition V on page 3: "That force which opposes itself to the endeavor, and by which the sling ontinually draws back the stone towards the hand, and retains it in its orbit, because it is directed to the hand as the centre of the orbit, I call the centripetal force."


 * ES Feb 25 Notice in Newton's Definiton 5 quote above,"that force which opposes itself". Here I see that Newton is agreeing with my position that the centripetal force opposes itself by causing to exist the centrifugal force. Never at an earlier time did I appreciate the significance of these words from Newton. I will be using them in the future, for certain.  In my world, every force is immediately opposed or supported by an opposite force.  Immediately means right there at the site of the action force, not millions of miles in the distance across the vacuum of empty space.  Hence the acceleration/Action force of Earth gravitation being generated in the Moon's matter "opposes itself" by generating its own reflection force known herein as the reactive centrifugal force.  Likewise within Earth's matter.  Tally these forces and you find that 4 is the minimum number of forces present in any event you care to describe. Newton indirectly agrees with this 4 force recognition since in his quote he describes that every centripetal force (Earth gravitation with the Moon's matter for example) "opposes itself" by causing its centrifugal force reaction or reflection to exist at the same immediate site. This opposition isn't really an opposition force which in other types of events cancels the action.  Here this "opposition" is really a benign support force that provides LAW III support while serving in no way to reduce of cancel the ongoing acceleration.  The centrifugal force is caused to exist by the centripetal force. It is the same force being reflected back from the accelerating matter.  Newton saw this truth.  I saw this truth.  Now it is your turn to see this truth.  Here I conclude as did Newton that the reaction to acceleration force is a true reaction force with a unique cause and unique behavior.  No matter how much you may want it to be otherwise, the reaction force only exists in accelerational events and is in now way interchangeable with the action force in accelerational events. Ethan Skyler (talk) 17:45, 25 February 2010 (UTC)


 * ES Feb 25 Which begs the question, are there reaction forces in non-accelerational events? The answer is no.  In non-accelerational events, such as you standing in a door frame and pushing against each side with each hand in the horizontal direction, there exists two mutually opposed action forces with no reaction forces present.  Here if you wish to misidentify one of the action forces from one of your hands as a "reaction" force then you can switch these names from one force to the other with no consequence. But in truth there exists two event causing action forces that loop in both directions around the door frame to provide effective opposition and cancellation of each other.  No reaction forces are present.  Thus the only reaction forces that exist, exist only in accelerational events and there they possess unique reflective abilities which are shared by no other force.  In accelerational events "action" is not interchangeable with "reaction" or using my terms, acceleration/Action is always the cause of the acceleration and acceleration/Reaction is always the force providing reflective non-cancelling support for the acceleration. Ethan Skyler (talk) 17:45, 25 February 2010 (UTC)


 * ES Feb 27 I have been looking over my copy of Fundamental Physics 2nd Edition Revised Printing by Halliday & Resnick. I came across a drawing on page 92 of a boy pulling a sled horizontally along snow. Included is a drawing of "the forces acting" on the sled in this non-accelerational event.  "T is the boy's pull, W the sled's weight. F(k) the frictional force, and N the normal force exerted by the surface on the sled." Later on page 107 they refer back to this event with the following quote: "(c) The four forces that act on the body" (of the sled - my words) "are shown in Fig. 6-6b." 216.231.36.211 (talk) 17:20, 27 February 2010 (UTC)


 * ES Feb 27 I call your attention to this event for I think it is a good example showing that like in every accelerational event, the minimum number of forces present in an event where acceleration is absent also remains at 4.  Here the authors referred to these 4 forces as "acting" on the sled with no mention of there being any reaction forces present. As a side note, I see their "Normal" action force drawn upward against the bottom of the sled as Earth's gravitational weight against the sled. They did not make this truth clear in their text which is too bad since every such gravitational event works both ways, not just directed from the small object toward the large one. 216.231.36.211 (talk) 17:20, 27 February 2010 (UTC)


 * ES Feb 17 I take issue with the article's use of the net force theory to explain the centripetal and centrifugal force components of the turning car.  We now know from the whirling ball event that the centripetal action force at the top of the pole in Panel 6 is supported by an equal and opposite centrifugal reaction force yet we are learning that the reaction force does nothing to reduce the effect of the action force as evidenced by the moving pole.


 * I'm afraid we know no such thing. Flexible bodies are complicated to analyze and so usually handled by numerical simulation. Never-the-less, as I describe elsewhere, we can limit out analysis to the stationary base of the pole. The forces on it are equal and opposite and so it does not move. The tip, on the other hand, is subject to unbalanced forces and so it moves. -AndrewDressel (talk) 00:45, 19 February 2010 (UTC)


 * ES Feb 17 So here in the cornering car, it is a mistake to claim that the passenger's internal-to-matter reaction force can be counted on to reduce the centripetal action force from the car against the seat.  Our single whirling ball event shows that the reaction force (green vector) does nothing to reduce the accelerational ability of the centripetal force (red arrow).  Yet in the cornering car event we find the author is now counting on this same type of reaction force to cancel an equal portion of the same type of action force.  You cannot have it both ways. Now that we know that the centrifugal reaction force supports but does not cancel or prevent the centripetal action force from doing its job, we need to find a non-"net force" explanation. Ethan Skyler (talk) 04:35, 18 February 2010 (UTC)


 * ES Feb 17 I will now describe this event using net action force terms.  Let us say that the car, without a seat or passenger, contains 3,200 lb.m (pounds/mass) which is the amount of matter that when weighed against Earth at sea level has an Earth weight of 3,200 lb.f.  I'll set the seat at 60 lb.m and the passenger at 160 lb.m.  The centripetal acceleration rate for these three objects will be, on average, 0.5 g or 16 ft/sec^2.  This means that the centripetal action force required to cause each object the correct rate of acceleration will be, according to Newton's LAW II, half the force of their weight against Earth. Ethan Skyler (talk) 04:50, 18 February 2010 (UTC)


 * ES Feb 17 The centripetal action force responsible for causing the car/seat/passenger combination to accelerate is the centripetal force applied by the pavement against the car. It takes a 1600 lb.f to accelerate the car, a 30 lb.f to accelerate the seat and an 80 lb.f to accelerate the passenger.  This total of 1,710 lb.f is the total centripetal action force being experienced by the car.  Now let's break the car into components.  The car alone is receiving the entire 1,710 lb.f, uses up 1600 lb.f and transfers on a 110 lb action force to the seat. The seat receives a 110 lb action force from the car, uses up a 30 lb action force and transfers on an 80 lb action force to the passenger. The passenger receives just the right magnitude of action force left over at 80 lb so that the passenger rounds the same curve at the same rate as does the seat and the car below. Ethan Skyler (talk) 05:21, 18 February 2010 (UTC)


 * ES Feb 17 Notice here that I am dealing only with the action forces effecting these three objects.  Following the rule that incoming action forces are positive for a given object and outgoing action forces being transferred on to accelerate a different object are negative for the given object, the car is receiving a +1,710 lb action force incoming from the road, which when added to the -110 lb action force outgoing to the seat leaves a net action force of 1,600 lb causing 0.5 g centripetal acceleration for the 3,200 lb.m car. The seat is receiving a +110 lb action force incoming from the car, which when added to the -80 lb action force outgoing to the seat leaves a net action force of 30 lb also causing 0.5 g centripetal acceleration for the 60 lb.m seat.  Finally, the passenger is receiving the last of the action force of 80 lb incoming from the seat.  With no outgoing action force the 160 lb.m passenger ends up experiencing the same average 0.5 g centripetal acceleration causing the passenger to round the curve at the same rate as the seat and car below. Ethan Skyler (talk) 05:38, 18 February 2010 (UTC)


 * ES Feb 17 What then of the centrifugal reaction forces?  Their role is to provide LAW III support for the action forces causing centripetal acceleration.  As we have discussed before, these acceleration/Reaction forces are always generated internal-to-matter in the direction opposite to the direction of the matter's acceleration.  This can make for some interesting effects.  In this curving car event, while the centripetal acceleration/Action forces are applied as exterior-to-matter contact forces at the bottom of each of the three objects, the acceleration of the matter of each object is in the horizontal inward direction while the supporting acceleration/Reaction forces being generated within are directly opposed in the horizon outward direction. This means that, on average, the reactive centrifugal outward-directed force vector is drawn beginning at the object's center of matter which is perhaps at mid-object.  Thus the inward-directed action forces are low on the object and the supporting outward-directed reaction forces are higher up at each object's center of matter.  This misalignment, mentioned in the article, helps to explain why when turning left, the car and its contents roll or lean to the right in response to the misaligned action and reaction forces present. Ethan Skyler (talk) 06:02, 18 February 2010 (UTC)


 * ES Feb 17 The net action force recognition, explained above in both the single whirling ball and the curving car event, provides for a logical understanding of acceleration without resorting to the granting of experimentally unproven capabilities to reactive accelerational forces as counted upon by the "net force" theory of acceleration.  As I have shown, acceleration occurs in the presence of equal and opposite acceleration/Action and acceleration/Reaction forces as drawn in Panel 6 of the single whirling ball event.  Ethan Skyler (talk) 06:23, 18 February 2010 (UTC)


 * ES Feb 17 I take issue with the article's claim that the reactive centrifugal force can do real physical work.  My understanding is that the reactive centrifugal force is always directed opposite to the action of centripetal acceleration and therefore is incapable of "acting" as the cause of any event.


 * There is not such requirement. Any force that moves an object does work. The example given, of a centrifugal governor, is as good as any and can be implemented with springs instead of gravity. In that case, the reactive centrifugal force pulls on the spring, does work on it, and so stores potential energy in the spring. -AndrewDressel (talk) 16:03, 19 February 2010 (UTC)


 * ES Feb 17  The centrifugal governor event is presented as the first of two events where the reactive centrifugal force is thought to be performing physical work stored as energy.  Curiously this first example concludes that since the reactive centrifugal force is horizontal while the work performed by raising the whirling weight is vertical, this example of the storing of potential energy in a gravitational field has nothing to do with the reactive centrifugal force present within the whirling matter.  I agree with this conclusion which makes me wonder why this "non-example" is included in the article. Ethan Skyler (talk) 08:08, 18 February 2010 (UTC)


 * ES Feb 18  In the second "real physical work" event involving expanding springs inside a centrifugal clutch, I am thinking the author is mistakenly attributing the stretching of the clutch springs to the reacton force when this role rightly belongs to the action force responsible for causing the shoe's centripetal acceleration.


 * The distinction is simply a matter of how you slice up the system. If the shoes and springs are a single body, then all the work is being done by the fictitious centrifugal force acting on them. If the shoes and springs are separated, then the bulk of the work on the relatively massless springs is done by the real reactive centrifugal force applied to the springs by the shoes. -AndrewDressel (talk) 16:03, 19 February 2010 (UTC)


 * ES Feb 18  Perhaps this event is easier to understand if I convert the acceleration present into a linear event.  Suppose I have a "rail job" dragster tow a small wheeled trailer from start to finish.  I attach the trailer to the dragster's hitch using a long tension spring, somewhat like the spring used to close an old-fashioned screen door.  This spring is 2' in length when relaxed but will extend in length to 6' when experiencing acceleration caused by the dragster as it departs the start line.  In theory, a part of the dragster's accelerational force is delayed in causing the trailer to displace along the distance of the race track as the trailer lags 4' behind the dragster during the positive accelerational portion of the run.  Then when the dragster crosses the finish line and before applying the drag chute and wheel brakes, for a moment the drager's acceleration rate is 0.  Neglecting air friction by accepting that the small trailer is in a perfect draft of the dragster, the tensioned spring will pull or displace the trailer 4' closer to the dragster.  The additional 4' of displacement is viewed as energy being stored in the extended spring during the acceleration run that is now free to perform work on the small trailer by extending its displacement an additional 4'.  (A side issue here is that this same spring is at the same time reducing the displacement of the dragster).  But focusing on the issue at hand, the question here is what force causes the spring to extend in the first place.  With this entire accelerational event being caused by the dragster's rear tires pushing backward against the ground causing the ground to push forward against the tires causing the tires to push forward against the axle and forward against the chassis and thereby pull forward against the tension spring which now stretches out in length an additional 4' in order to provide the strong forward-directed pull required to cause an extreme rate of acceleration for the small wheeled trailer.  Enlighten me here if you think I am wrong in accepting that the force causing the spring to stretch in length is none other than the forward-directed acceleration/Action force from the accelerating dragster. Ethan Skyler (talk) 19:14, 18 February 2010 (UTC)


 * ES Feb 18  Now that we see that it is the forward-directed acceleration/Action force of the dragster that is causing the tension spring to stretch in length until the required magnitude of action force responsible for accelerating the small trailer is achieved, it time to return to considering the centrifugal clutch event.  Here the inward-directed centripetal force is the action force responsible for causing centripetal acceleration for the whirling clutch shoes.  As the engine speeds up in rpms, a greater and greater centripetal force is required to cause greater and greater centripetal acceleration for the whirling shoes.  The source of this centripetal acceleration/Action force is the post at the center of the clutch. As the engine speeds up, the tension springs can only transfer a greater and greater centripetal force from the post to the shoes if they stretch and stretch in length. Again, correct me if you think I am wrong, but here, as in the linear dragster event, I see the acceleration/Action force present as being responsible for all on-going actions present in this event.  Here the acceleration/Action force is the inward-directed centripetal force from the post.  Since the accelerating shoes are not seen as approaching the post, it is easy to get the wrong idea about the role of the centripetal action force.  I find it is better to think of the shoes being continually accelerated away in the inward-direction from the straight-line path of tangency they will travel if suddenly cut free from being contained within the clutch. So here too I accept that the inward-directed acceleration/Action force, known as the centripetal force, is responsible for causing the clutch springs to transfer a greater and greater inward-directed force to the whirling clutch shoes.  The result is that the clutch springs stretch in length until the shoes contact the clutch drum.  From this point on, the drum takes over any increased requirement for more and more centripetal force that is needed to be present to cause the increased accelerational needs of the whirling shoes. In summary, again I do not see there is any validity to the author's claim that the outward-directed reactive centrifugal force is responsible for causing the clutch springs to extend. Instead, it is the increased acceleration in the centripetal direction that causes the increased tension on the springs as they transfer this centripetal action force sourcesed from the central post on out to the whirling shoes.  Again, the reactive centrifugal force is caused by the centripetal acceleration/Action force.  It serves to provide equal and opposite support for the centripetal acceleration/Action force. It is always directed opposite to the direction of any event and therefore is never the cause of any event. It merely fills a supporting role.  Ethan Skyler (talk) 22:03, 18 February 2010 (UTC)


 * ES Feb 18  Overall this means that there are no examples that show the reactive centrifugal force (my acceleration/Reaction force in circular events) in "action" as the cause of any event. It remains a real reaction-to-acceleration force.  That is enough of a role. There is no need to invent an "action" force role for our newly recognized reactive centrifugal force. It does a fine job of reacting in support of the real active centripetal forces present. Ethan Skyler (talk) 22:03, 18 February 2010 (UTC)


 * The distinctions you stress, between "action" and "reaction" forces and between forces that cause motion and forces that do not, are not necessary. As I have written elsewhere, Haliday and Resnick explain in their discussion of Newton's 3rd Law: "Either force may be considered the action and the other the reaction. Cause and effect is not implied here, but a mutual simultaneous interaction is implied." (emphasis theirs) -AndrewDressel (talk) 16:03, 19 February 2010 (UTC)


 * ES Feb 27  I just finished scanning my copy of Fundamental Physics by Halliday & Resnick.  It does not surprise me to learn they see no distinction between action and reaction forces in accelerational events.  They also see no role for the reactive centrifugal force being presented here in Wikipedia.  I could find no mention of any outward-directed reaction force in circular events nor any backward-directed reaction forces in linear acceleration events. Perhaps your copy is more modern and presents a different take on this subject. Ethan Skyler (talk) 06:33, 28 February 2010 (UTC)


 * ES Feb 28  Gathering The Evidence   This morning I have some thoughts to share that are aimed at gathering together all portions of our discussion. I find there are powerful clues to the truth regarding the physics of Nature embedded in Fundamental Physics by Halliday & Resnick.  As already noted in their boy pulling the sled event as described on pages 92 & 107, Halliday & Resnick identify the four forces as action forces "acting on the sled" which is moving along horizontally at a constant speed. Here acceleration is absent and no reaction forces are identified.  They only identify action forces in this event and in every other event I can find in their book. Why do you think they left out reaction forces that we both agree are present in accelerational events? Ethan Skyler (talk) 19:00, 28 February 2010 (UTC)


 * ES Feb 28  The answer is that there is no mathematical need to include the reaction forces present in accelerational events.  Including them makes the net force theory of acceleration unworkable. By leaving them out, Halliday and Resnick end up adding only the action forces that are present.  Using my net action force technique, I do the same thing - adding only the action forces present. Yet unlike them but like you, I recognize that the reaction force in accelerational events is measurably present.  Further I accept that in accelerational events, the reaction force is not to be added to the action force for it is caused to exist by that action force. I contend that for this reason, it is best to think of the reaction force as a reflection of the action force.  So for different reasons and recognitions, Halliday and Resnick and I end up performing the same math operation in accelerational events.  Their position seems to be that reaction-to-acceleration forces such as the reactive centrifugal force do not exist.  My position is that they do exist but are a byproduct of the acceleration and benignly provide support for the net action force as they go along for the ride. I find Newton's quote regarding how the centripetal action force is "That force which opposes itself" supports my understanding and not that of Halliday & Resnick.  Instead, they choose to ignore the fact that measurable reaction forces are present in accelerational events.  I think they make this choice to keep the net force theory operational.  Instead if the net action force technique is used, the reaction forces in accelerational events can be recognized while the computations remain correct based upon the net action force tally, a tally that is the same as the one employed by Halliday & Resnick.  Ethan Skyler (talk) 19:00, 28 February 2010 (UTC)


 * ES Feb 28  We have discussed to no end the forces acting and reacting on the post in the whirling ball event.  I think your position is that since a rigid post does not move, you see this as evidence that the reaction force balances the action force drawn in Panel 6 leaving the net force on the post at 0.  I, on the other hand, see that the reaction force should not be added to the action force since it is caused by the action force. So I do not see that the action force on the post is canceled.  Instead I see it as fully present and continuing to cause acceleration for the whirling ball.  I have tried to show that to truly cancel the action force requires the addition of a second opposing whirling ball with the action force causing centripetal acceleration for the 2nd ball cancels at the post the action force causing centripetal acceleration for the 1st ball.  Here and only here do I see that the net force on the rigid post is 0. Ethan Skyler (talk) 19:00, 28 February 2010 (UTC)


 * ES Feb 28  What then do I see at the post when only one ball is whirling about?  Hoping we have the same copy of Fundamental Physics, on page 100 in figure 6-26 there is a puck whirling on an air hockey table caused by a tensioned string that plunges down through a hole in the table where it is attached to a weight that is suspended above the floor the table rests upon.  Now when the puck is not whirling and the hanging weight is released, it takes little force to cause the puck's linear acceleration directed toward the hole as the weight's gravitationally-caused action force causes it to take on an accelerated motion toward the floor. Now if the puck is caused to whirl rapidly about the hole, a centripetal acceleration-causing action force is required to be present. This action force is a portion of the action force of the central weight's force of Earth gravitation.  What portion is the problem presented by this event.  Spin the puck faster and the action force of the hanging weight might be insufficient.  If this is the case then the weight will rise until it contacts the underside of the table.  If the puck's rate of spin is just right then the weight's force of Earth gravitation will provide the exact amount of action force needed to cause centripetal acceleration for the whirling puck. Here the weight will continue to hang at an unchanging elevation. Ethan Skyler (talk) 19:00, 28 February 2010 (UTC)


 * ES Feb 28  My point in bringing up their event is that I think you would look at the hole and decide that, like your position of the rigid pole, the net force on the hole also is 0 since it is not moving.  When I look at the hole, I see the net action force as fully present in the tension of the string where the hanging weight's force of Earth gravitation is causing the action of centripetal acceleration for the whirling puck.  Thus I see the determination of a net force of 0 on the hole or the rigid pole is misleading and therefore does not represent the truth of such events.  Again I find the Panel 6 descriptions are in error.  Ethan Skyler (talk) 19:00, 28 February 2010 (UTC)


 * ES Feb 28 What Wikipedia is doing with their reactive centrifugal force article is stepping out of the let's-ignore-the-reaction-forces-present-in-accelerational-events position demonstrated in conventional physics texts such as Fundamental Physics, and instead moving ahead to finally recognizing that there does indeed exist a centrifugal reaction force in every event involving centripetal acceleration.  This bold move points to the flaw in the net force theory as I have tried to demonstrate. Also it is a challenging job getting all the details right in your description for reactive centrifugal force.  It is real but it doesn't act as the cause of anything.  It is a mistake to say it is the cause of work.  Including it in vector additions is also a mistake.  A positive note is that you and I do share in recognizing that the reactive centrifugal force is present throughout the matter of the whirling ball from component to component from its inner surface to its outer surface. Ethan Skyler (talk) 19:00, 28 February 2010 (UTC)


 * ES Feb 28 What then of Halliday & Resnick saying that the action and reaction force labels are interchangeable?  I say that reaction forces are present only in accelerational events, linear and circular.  They ignore these reaction forces and instead only identify action forces in accelerational and non-accelerational events.  In non-accelerational events, I recognize that action forces oppose other action forces with no reaction forces present.  Changing these action vs action force labels to action vs reaction labels changes nothing.  It also proves nothing since here every force is an action force and therefore does truly possess the ability to cancel all of the opposing force or a portion thereof as they are able.  In accelerational events the action force causes the event and also, as Newton pointed out, "opposes itself" by causing to exist its own reaction-to-acceleration support force that serves in no way to cancel the effect of the acceleration-causing action force. Here the reaction force does not cancel the action force.  Thus in accelerational events, the acceleration/Reaction force is quite different when compared to the acceleration/Action force.  They are not interchangeable.  They are also not of the same type.  They are different. Ethan Skyler (talk) 19:00, 28 February 2010 (UTC)


 * ES Feb 28  In the text where Halliday & Resnick discuss Newton's third law on page 71, they explain what they think would happen if their action and reaction forces were to be present together on or within the same body. "If they were to act on the same body, we could never have accelerated motion because the resultant force on every body would always be zero." They reach this conclusion because they falsely think the acceleration/Reaction force is capable of canceling the accelerational abilities of the acceleration/Action force represented by your green and red vectors drawn in Panel 6. Clearly the green vector acceleration/Reaction force does not cancel the acceleration being caused by the red vector acceleration/Action force now does it?  Also out in the whirling ball drawn in Panel 2, both the acceleration/Action force (red vector as drawn) and the acceleration/Reaction forces (green vector that is missing) are present throughout the whirling ball's matter all the while that acceleration continues unabated.  Halliday & Resnick missed the fact that the acceleration/Reaction force (reactive centrifugal force in this event) never serves to cancel or eliminate the acceleration present in any such event.  They were led along this false decision path by the false rules of the net force theory of acceleration.  This theory only takes into account the behavior of one type of force, the event-causing action force which they insist can also be interchangeably termed an event-causing reaction force. Here I add "event-causing" since they mistakenly predict that if present on the same body as the action force, this reaction force of their understanding would "act" to cancel the opposing action force down to a "resultant force" of "zero". These imagined abilities are the abilities of another action force such as the one I offered with the addition of the second whirling ball, and not the abilities of the true acceleration/Reaction force also known in circular events as the reactive centrifugal force being discussed herein. Here Halliday and Resnick have employed false logic in support of their decision to support the net force theory of acceleration by ignoring the reactive centrifugal forces and reactive linear forces entirely. Clearly that decision solved their problem but left the careful reader, who is curious about centrifugal force, less than satisfied. Ethan Skyler (talk) 20:09, 28 February 2010 (UTC)


 * ES Feb 28 As you can see, Halliday and Resnick, brilliant mathematicians that they are, get an F for ignoring the reaction forces present in every accelerational event.  I just don't want you to think that they are authorities on this subject. I have read much better understandings of the acceleration/Action and acceleration/Reaction forces.  The older texts are superior in this regard. Also I have yet to read any offering that gets everything right. To this end we endeavor.  Ethan Skyler (talk) 19:00, 28 February 2010 (UTC)


 * ES Feb 18  The remainder of the article discusses the reality of made-up forces which in itself is an oxymoron.  In truth there is no reality of forces that do not exist.


 * Well, yes and no. It depends on your frame of reference. The "reality" of the so-called fictitious Centrifugal force (rotating reference frame) is explained at length there. -AndrewDressel (talk) 16:03, 19 February 2010 (UTC)


 * ES Feb 18  But we need to take a serious look at the author's position which is to grant reality to "fictitious centrifugal force".  First of all, I object to practice of using the scientific terms centrifugal and force to identify a nonexistent non-force that has no source and is not being experienced by any object yet somehow is seen as an action force capable of causing outward-directed acceleration, which in itself is an impossibility.


 * It most certainly is experienced by any object following a curved path, and from the point of view of that object, is a simple and useful explanation for the force experienced. -AndrewDressel (talk) 16:03, 19 February 2010 (UTC)


 * ES Feb 18  All this misuse of the scientific terms centrifugal and force results clouding the reality of the real reactive centrifugal reaction-to-acceleration support force.  Instead, the author might consider using a made-up name for his made-up non-force such as "outward-directed effect".


 * But that is not the role of Wikipedia or its editors. Instead, our job is to make sense of the published writings of original researchers. The so-called fictitious or pseudo centrifugal force is firmly established in the liturature and so should be, and I think is correctly, described in that article. -AndrewDressel (talk) 16:03, 19 February 2010 (UTC)


 * ES Feb 18  This ties the effect directly to the faulty observations made from within an accelerating frame of reference which seems to be where this whole nonscientific effort began. Ethan Skyler (talk) 04:47, 19 February 2010 (UTC)


 * ES Feb 18 The only event explained in the article in support of the "outward-directed effect" is similar to the curving car event.  Only this time, before entering a curve to the left, the passenger, sitting on the right side within the reference frame of the car, is blindfolded to prevent him from using visual clues to determine the type and direction of the acceleration as it becomes present.  The author then tells us that the passenger will "feel" a "magical force" pushing him away from the center of the bend. Understand that I always question any statement that includes a prediction as to how I will "feel" as the passenger/observer. I know that if I am sitting upright in the passenger seat at the front/right side of the car and the car begins turning left, I will experience an action force pushing my lower back, seat, legs and feet to the left.  If the seat is slippery, I will experience the seat sliding to the left under my legs as previously described above.  While feeling this action force to my left, I will feel my legs flopping to the right and my upper torso as well since I can feel no leftward force being impressed against those parts of my body.  Thus my torso and my legs will lag behind in the rest of my body's acceleration toward my left.  Soon the door will move into position up against my flopping torso and legs.  Then I will finally experience the correct magnitude of leftward-directed force from the door against my torso and legs so that they will be pushed left at the same rate as the rest of my body. Ethan Skyler (talk) 06:05, 19 February 2010 (UTC)


 * ES Feb 18 Notice how all the forces I truly do feel are directed toward my left which in this event is in the centripetal or inward direction.  Also notice that the leftward forces I do experience, as they cause leftward acceleration for my body, all are sourced from other objects in contact with my body. Curiously, the author predicts that I will ignore these left-directed action forces being impressed against my body by the seat and the door and instead concentrate on "feeling" a "magical force" with no explainable source that will magically push my body away from the center of the bend which is to my right.  Now I am wondering exactly how ignorant is the author's passenger to the presence of real action forces bearing against his body in the leftward direction.  It is hard for me to imagine any passenger can be fooled into thinking that in this event, real leftward action forces are nowhere to be found while magical rightward action forces need to be invented to explain the rightward flop the passenger experiences during leftward acceleration. Ethan Skyler (talk) 07:29, 19 February 2010 (UTC)


 * ES Feb 18 I think the author never did find such an ignorant passenger.  Instead, I think this author or someone before him imagined that they were watching a passenger flop slightly to the right while experiencing the leftward action forces responsible for the passenger's leftward or centripetal acceleration.  So instead of asking the passenger to describe the forces he feels during such an event, the author on his own decides that a "magical force" to the right needs to be invented to account for the slight flop the author sees is occurring to the passenger.  This second hand analysis of the event explains the off-target "magical solution" the author invents to "explain" the results.  All without actually interviewing the passenger to learn what he actually feels during this event. Ethan Skyler (talk) 08:18, 19 February 2010 (UTC)


 * ES Feb 18 I have read many times this same fabled story regarding how important scientific decisions are made based upon how the author thinks the passenger is "feeling".  Like global warming, this is a fine example of junk science.  I suggest that the scientific term centrifugal be restricted to events where an object is actually experiencing a curved path of travel resulting in the generation within the object a real reactive centrifugal force.  All uses connecting the term centrifugal with the words fictitious or imaginary are bogus.  Finally, the term force should only be used when a real force is being experienced by an object.  All uses connecting the term force with the words fictitious or imaginary also are bogus. Ethan Skyler (talk) 08:18, 19 February 2010 (UTC)


 * That is all well and good, but Wikipedia is simply not the forum. -AndrewDressel (talk) 16:03, 19 February 2010 (UTC)


 * ES Feb 22 I fully agree. This is the forum, between you and me. I am glad I got to finish my thoughts regarding the article. You have done an excellent job of responding with your thoughts and objections.  I accept that our forum on centrifugal force will be deleted soon due as much to its size as to its content.  I think this exchange of ours is already a classic. Ethan Skyler (talk) 07:36, 23 February 2010 (UTC)


 * ES Feb 19  All things considered, I will rate the current article on reactive centrifugal force as promising yet seriously flawed for its contains many incorrect statements making it impossible for the reader to come away with any useful understanding of the workings of reactive centrifugal force.  Worst of all is the adherence to the rules of the net force theory of acceleration which I have shown do not match with our experience while testing these events.  Overall I think this subject is incorrectly graded within Wikipedia.  Instead, I think it is a grade A subject of the highest importance.  It has been my pleasure to be a participant in this discussion.  Best Wishes to All, Ethan Skyler  1944-      Ethan Skyler (talk) 08:18, 19 February 2010 (UTC)


 * ES I recognize that just as it takes a lot of time on my end to properly address this subject, it takes at least an equal amount of time on your end to consider each portion of my work.  I appreciate your considerable help in this matter.  Ethan Skyler (talk) 21:18, 2 February 2010 (UTC)


 * This has been an fascinating project that has forced me to think clearly about a topic I enjoy. Even if we do not come to agreement, it will not have been time wasted for me. -AndrewDressel (talk) 15:56, 4 February 2010 (UTC)