Wikipedia:Reference desk/Archives/Mathematics/2008 May 27

= May 27 =

How to make a pattern board for a cone shape?
I am doing a project pertaining to a cone shape with specific measurements. I don't know how to make a pattern to do it. The cone measurements are; base openning diameter=17", top openning diameter=4 1/2", height=8".

Your help will be highly appreciated. —Preceding unsigned comment added by Whitegold1925 (talk • contribs) 04:39, 27 May 2008 (UTC)


 * Simple ratio work will determine the "height" at which the top opening would be 0. Then you can draw a circle of radius this height and another (same centre) of this height less 8". Measure 17" around the circumference of the bigger circle. Cut out your shape, bend and tape it up. -- SGBailey (talk) 11:22, 27 May 2008 (UTC)


 * Before you start cutting, you will need to adjust your cut radius to account got the slope of the cone. If you have the height of the pointed cone calculated as described above, you can use it and the radius of the base to calculate the sloped distance from the tip to the base.  This will be your outer cut radius for the pattern.  The inner cut radius for the pattern can be similarly calculated.  Also, the perimeter of the base will be the arc length of the outer cut.  The ratio of the arc length to the outer cut radius will give you the angle of your pie slice in radians.  Be sure to leave extra material to serve as a flap when joining the edges together.  -- Tcncv (talk) 20:45, 27 May 2008 (UTC)


 * I had been interpreting the 8" as the height along the slope. On re-reading it I see that interpreattion is probably wrong. Converting a vertical 8" to a hypotenuse with a horizontal of "(17 - 4.5)/2" is fairly straightforward. -- SGBailey (talk) 21:45, 27 May 2008 (UTC)

Maths in 21st century
How have new developments in mathematics in the 21st century changed our way of living? Including topics remotely linked to maths like computing, etc.

Thanks in advance, 220.244.78.150 (talk) 05:52, 27 May 2008 (UTC)
 * Maybe I'm just jaded, but that sounds a lot like an essay topic. :P -- Kinu t /c  06:34, 27 May 2008 (UTC)
 * They give us something to take classes on when we need more credits. « Aaron Rotenberg « Talk « 17:23, 27 May 2008 (UTC)
 * Mathematical discoveries take time to be assimilated into the industry, and even more time into everyday life. The 21st century has only been around for less than 8 years, so my answer is "they haven't". -- Meni Rosenfeld (talk) 20:06, 27 May 2008 (UTC)
 * Pure maths, certainly. Some applied maths might get used quicker. Computer science probably has a pretty fast turnaround - the time from inventing a new algorithm to someone implementing it and it being used online, say, could easily be less than 8 years. --Tango (talk) 21:42, 27 May 2008 (UTC)


 * Given the importance of primes to cryptography, the discovery of a polynomial-time algorithm for checking primality (in 2000, I think) might have had an effect. Black Carrot (talk) 05:42, 30 May 2008 (UTC)
 * Shor's algorithm (which I think is what you're referring to) can't be used yet because it requires a working quantum computer. Hut 8.5 14:42, 1 June 2008 (UTC)
 * No, Black Carrot is referring to the AKS primality test, which runs in polynomial time on a classical computer. However, what's needed in cryptography is a factoring algorithm, which Shor's algorithm is and the AKS algorithm is not. AKS's practical importance is diminished by the very efficient randomized primality tests which were known before AKS. --196.210.152.31 (talk) 10:06, 2 June 2008 (UTC)

Latex Typesetting
I was trying to type something in Latex and I can't seem to figure out something. What is the command to draw a line through an entire expression such as for canceling? I am in Math mode and I have a fraction. I want to cross out factors in the numerator and in the denominator. For a single character, the \not command can be used. But what if I want to draw a single line across an entire expression.A Real Kaiser (talk) 06:39, 27 May 2008 (UTC)
 * \sout{ } perhaps. I don’t know how to make a diagonal line through the text though. GromXXVII (talk) 10:56, 27 May 2008 (UTC)


 * Perhaps not. At least not in the one implemented in Wikipedia: see \sout{4\pi x}:
 * $$\sout{4\pi x}$$
 * For more info see Help:Displaying a formula and LaTeX (not Latex). --CiaPan (talk) 15:28, 27 May 2008 (UTC)
 * Ahh yes, \sout requires the ulem package. GromXXVII (talk) 21:46, 27 May 2008 (UTC)

Thanks guys, I installed the ulem package and used the \sout command. It works perfect with standard text but if I am using the Math Display environment (with $$) and canceling out a term in the numerator in a fraction, it looks just like an underline. Is there any other way to cross out or strike through text that would work well with a Mathematical expression?A Real Kaiser (talk) 00:09, 28 May 2008 (UTC)

I think I got it. I discovered a Latex package called cancel.sty which simply has the \cancel command and it works beautifully in Math and text mode across any expression (even a long fraction).A Real Kaiser (talk) 01:18, 28 May 2008 (UTC)
 * I went and found that, indeed cancel.sty works great in math. I also tried it in plain text and it gave quite a few (read: 4) errors for me, but still managed to give the correct output. GromXXVII (talk) 10:59, 28 May 2008 (UTC)

Defining outliers
how do i investigate the use of mean and standard deviation to find a general rule to identify outliers? —Preceding unsigned comment added by 124.185.173.208 (talk) 07:31, 27 May 2008 (UTC)
 * Interesting question. I once had the situation of measuring the same quantity four times and reporting the mean. I checked whether the mean was also a median. If not, I suspected an outlier to have occured. Assuming that the four normally distributed measurements are x1<x2<x3<x4, what is the probability that the mean m=(x1+x2+x3+x4)/4 is a median, x2<m<x3? Bo Jacoby (talk) 09:40, 27 May 2008 (UTC)


 * See Outlier, in particular Identifying outliers. -- SGBailey (talk) 11:16, 27 May 2008 (UTC)
 * Also you may want to use some common sense. For example if you ask people what would be the ideal number of hours per day they would like to sleep and someone answers with any number above 24 then it is automatically an outlier given the constraints of the scale you're using.--droptone (talk) 18:43, 27 May 2008 (UTC)


 * If you are working with an approximately normal distribution, then you can standardise your data points to give Z-scores. Then all you need to do is define a "cut-off" a so that a point is an outlier if |Z| > a. A choice of a = 3, for example, will mean that roughly the outermost 1% of points are outliers. Confusing Manifestation (Say hi!) 23:42, 27 May 2008 (UTC)

Polynomial-time reduction from SAT to CNF-SAT
I know that CNF-SAT is NP-complete, but I can't seem to find a source which actually gives a reduction from SAT to CNF-SAT. Is there a polynomial-time algorithm that takes a boolean formula and gives an equisatisfiable one in CNF? --196.210.152.31 (talk) 11:29, 27 May 2008 (UTC)
 * Just wanted to add that I know an algorithm that does this, but it requires a truth table and so it takes exponential time. --196.210.152.31 (talk) 11:30, 27 May 2008 (UTC)


 * Find a Boolean logic circuit that computes the value of the formula and in which each gate has at most three wires (two in, one out, for most gates, but you can also view a branching wire as a gate with one in and two out). Make a 3CNF formula that has one variable for the value on each wire and clauses guaranteeing that each gate behaves the way it's supposed to. —David Eppstein (talk) 14:46, 27 May 2008 (UTC)
 * Thanks. --196.210.152.31 (talk) 16:50, 27 May 2008 (UTC)

Pentagon, golden ratio
Hello. I had an exam just today, and was wondering if I did something right. A pentagon with sides 5cm was presented, with a pentagram built from pentagon (angles forming outside, not inside the pentagon). The task asked me to find a distance AB, which ran from one angle to another (see this, overlapping a side of the pentagon. There, another line of the pentagram intersected the line AB in the point E (arrow in that picture). The picture shows only parts of the pentagram, but it should suffice. Now then, with a side of the pentagram being 5cm, what is AB? I'll tell my own answer after I find out whether or not it was right, although this really is elementary. =) Thank you. Scaller (talk) 13:23, 27 May 2008 (UTC)
 * I'm sorry, I can't figure out what's going on - could you provide the entire picture? --Tango (talk) 19:32, 27 May 2008 (UTC)
 * I didn't get it either. If you have the entire picture (complete with vertex labels) that would be great, otherwise add more details to the verbal description. -- Meni Rosenfeld (talk) 19:41, 27 May 2008 (UTC)
 * By the way, is the pentagon regular? -- Meni Rosenfeld (talk) 20:03, 27 May 2008 (UTC)
 * Assuming a regular pentagon and pentagram, it appears that each point is an isosceles triangle has angles of 72, 72, and 36 degrees. If you know the base (5cm), you can calculate the other sides.  Then its a simple matter of adding up the line segments.  -- Tcncv (talk) 20:56, 27 May 2008 (UTC)


 * I apologize for my lack of clarity, I had hoped the partial picture would give the necessary information away. :) Yes, we here have a regular pentagon, inside a pentagram, the very same we see here, when we look at the shape that the lines of the pentagram create. I was, in my exam, asked to find the length of the line AB (from the top of the star in the pentagram, to the bottom right leg of the star as it stands). This on the basis of knowing that the side of the pentagon had sides of s=5, and knowing that the line AB was cut in a golden ratio (you get my point (of intersection, HAH)), in the point E - which is thus one of the outer angles of the pentagon.
 * Now, I proceeded to find EB. The outer angle of a pentagon is 72 degrees (from 180-108, the latter its inside angle), and I'd use this in a trigonometic approach where tan(72)*(5/2) yields the height of the height of the triangle. We have 180-90-72=18, meaning, for calculation's sake, that the angles of the pentagram are angles of 18*2=36. Pythagoras gets us to EB within a few decimals' precision, and we may simply multiply that with 1.618 to acquire the length of AB. The question is now if I did this at all right, because I seem to remember my approach here being to find EBø, then add EB. I remember that the end result was 21.2, most likely. Oh dear, that must've been wrong. It must be 13.1, isn't it. Scaller (talk) 21:06, 27 May 2008 (UTC)
 * What do you mean by "within a few decimals' precision"? There shouldn't be anything approximate involved in a calculation like this (other than rounding your final answer off to something practical to write down). Your method looks right, however I don't see any need to multiply by the golden ratio - just multiply by 2 and add 5, giving you the length of the 2 parts of the line outside the pentagon, plus the length of the side of the pentagon. That may well get you the same answer (I'd have to find pen and paper to prove it), but it's certainly the easier method. --Tango (talk) 21:33, 27 May 2008 (UTC)


 * Might I add, I later found that really, the shortcut seemed to be 5*1.618 + 5 = 13.09. Scaller (talk) 21:23, 27 May 2008 (UTC)
 * Where did that come from? --Tango (talk) 21:34, 27 May 2008 (UTC)


 * The way I see it, segments on either side of the points are $$\tfrac{2.5 cm}{\cos(72^\circ)}$$, or about 8.09017 cm. (The 2.5 cm is half the base of the isosceles triangle that makes up the point.)  The entire segment from point to point is $$2 * \tfrac{2.5 cm}{\cos(72^\circ)} + 5 cm$$, or about 21.18034 cm.  The difference between this and the other answer given above is the inclusion of a second outer line segment.  I'm not sure where the golden ration comes into play, but must be some relationship with the angles involved.   -- Tcncv (talk) 02:29, 28 May 2008 (UTC)
 * Thank you Tcncv, that seems fabulous, and would indicate I wrote down the correct answer. =) I was rather disoriented (and have remained so) with regards to just what needed adding, and where. As for where the golden ratio comes into play, it's everywhere! In response to Tango, the shortcut to finding AB from the information given is to perform 5*1.618 + 5*1.618 + 5 = 21.2ish, and to back this up by stating the golden-ratio qualities of pentagrams. Thank you, this should be regarded as resolved. :) Scaller (talk) 08:28, 28 May 2008 (UTC)

Missing-block triangle
There was this really awesome paradox-ish thing I saw a long time ago. The triangle was broken into four or five pieces, and put together, they formed...well, the triangle. The pieces were rearranged and formed another triangle, but this one was missing a block. I was wondering if anyone could point me to the page where this is located. Thank you for your time. Nolarboot (talk) 14:42, 27 May 2008 (UTC)
 * Check out Missing square puzzle. Pallida  Mors  15:13, 27 May 2008 (UTC)

Angles


When I have a chance to be a passenger in a light airplane, I often take pictures to illustrate geography-related articles. I usually attach a note about the direction in which the picture is taken by finding the compass direction of the road on a map, measuring the direction that the road would be if the picture were oriented north, and subtracting the picture's direction from the actual direction. For example, in Image:Ohio Valley Career and Technical Center.jpg (attached here), I calculated the actual direction of the picture to be 96º because the road that crosses the picture from left to right seems to go at an angle of 89º, and the road's actual direction is 185º. It seems to work well with this picture, but I've tonight realised that there's an error in some pictures: a new picture that I took is plainly oriented more toward the east than to the north, but my calculation (which I've checked multiple times) has it oriented at 16º! What's more, when I calculate the same picture with a different road as my base, I get an angle of 84º repeatedly. Obviously my method is wrong...any ideas how I correctly calculate these angles? Nyttend (talk) 22:13, 27 May 2008 (UTC)


 * The method sounds right. You'll have to add 360 to any answers that come out negative, of course. Are you sure you didn't just measure the angle of the road wrong (either on the photo, or the map)? --Tango (talk) 22:23, 27 May 2008 (UTC)
 * I disagree. The orientation of a line in a picture does not depend only on the real direction of the depicted object and the orientation of the camera. See for example Vanishing point - in the image there, the lines are all parallel and yet they appear in the picture with different orientation. Conversely, knowing only the geographical direction of a road and the pictorial direction of the road you cannot determine the direction of the picture. Even if you have some additional information, the actual calculation will be more complicated - Perspective (graphical) and the more technical Projective transformation contain some information about the machinery involved. -- Meni Rosenfeld (talk) 23:19, 27 May 2008 (UTC)
 * True, to get a precise answer, you're going to need to be more careful, but you should generally get a pretty decent approximation. Being off by around 90 degrees sounds more like a simple mistake. As long as the road in question is in the foreground so you can assume the camera was roughly overhead, the method should work. If you're measuring things on the horizon, it won't work so well. The best method would just be to take a compass with you on the plane... --Tango (talk) 23:58, 27 May 2008 (UTC)
 * The problem with the other picture (maybe I'll post it soon; it's a useful picture) is that there are two roads, each of which appear on the sides of the photo: in the picture, they appear meet at a 160º angle, although they really meet at a 90º angle. I've realised that I had considered these pictures as if they were overhead, not considering the factors of foreshortening.  At the moment, the only way I can imagine to do this is to make a line from the middle of the bottom of the picture and compare its angle to something higher up in the very middle of the picture, and then to find the angle on the map of a line between the two points: because it's in the very middle of the picture, it shouldn't look mess it up.  And by the way, I asked the pilot, who says that even in light airplanes, there's so much metal that a handheld compass would quickly be messed up; although the plane has a compass, I have no way to know at what relative bearing my picture is taken.  Nyttend backup (talk) 00:14, 28 May 2008 (UTC)
 * Yes, looking at things that are in a vertical line in the middle of the picture and using them to plot that line on a map is the best choice. It's reliable and requires no calculations. --Anonymous, 06:40 UTC, May 28, 2008.
 * Yes, but only as long, as your camera is quasi-vertical, that is it is tilted forward, but not rolled (I mean not turned around the optical axis). If you roll it, the vertical line on the image is not what you expect – it does not correspond to a vertical plane at the point where you took a picture. In that case you need to have a part of the horizon on the picture, and take a line perpendicular to it. Of course that will work if the horizon is horizontal ;) so you can't apply this method to mountains pictures. [Sorry for my poor English language level, hope you can understand me.] --CiaPan (talk) 07:17, 28 May 2008 (UTC)


 * Actually the apparent angle between roads you know to be at right angles should give you exactly what you need. First, pretend you are looking straight down at the ground, and mark two points A and B on a road.  Make point A the origin, call the actual vertical distance between the points y, and call the actual horizontal distance between the points x.  The road will appear to be an an angle $$\tan{\alpha} = y/x$$.  Now pitch the plane that the road is on back by an angle $$\theta$$, and call the distance to point A d.  The perspective projection will give you new projected distances (to a viewing plane a unit distance from the eye point):
 * $$y' = \frac{y \cos{\theta}}{d + y \sin{\theta}}$$
 * $$x' = \frac{x}{d + y \sin{\theta}}$$
 * and the road will thus make a visual angle on the screen of:
 * $$\tan{\beta} = \frac{y'}{x'} = \cos{\theta} \frac{y}{x} = \cos{\theta} \tan{\alpha}$$
 * Now say you are looking at two roads:
 * $$\tan{\beta_1} = \cos{\theta} \tan{\alpha_1}$$
 * $$\tan{\beta_2} = \cos{\theta} \tan{\alpha_2}$$
 * $$\cos{\theta} = \frac{\tan{\beta_1}}{\tan{\alpha_1}} = \frac{\tan{\beta_2}}{\tan{\alpha_2}}$$
 * If you know that the roads are at right angles ($$| \alpha_2 - \alpha_1 | = \pi/2$$) then:
 * $$\tan{\alpha_2} = -1/\tan{\alpha_1}$$
 * $$\cos{\theta} = \frac{\tan{\beta_1}}{\tan{\alpha_1}} = \frac{\tan{\beta_2}}{\tan{\alpha_2}} = -\tan{\beta_2} \tan{\alpha_1}$$
 * $$\tan^2{\alpha_1} = -\frac{\tan{\beta_1}}{\tan{\beta_2}}$$
 * $$\alpha_1 = \tan^{-1}{\sqrt{-\frac{\tan{\beta_1}}{\tan{\beta_2}}}}$$
 * This isn't needed if one of the roads appears horizontal ($$\beta_1 = 0$$) or vertical ($$\beta_1 = \pi/2$$) obviously. --Prestidigitator (talk) 13:57, 29 May 2008 (UTC)