Talk:Bicycle performance

Poor Science
I realise that many people stil use imperial units in cycling, but surely when calculating phsysical quantities such as kinetic energy S.I. units can be used. As a physics student imperial units seem to me to be from the dark age and incredible difficult to work to. Also could someone tidy up some of the equations so they fit in with standard wikipedia practise. Sorry to rant, I'd do it myself if I had the time.


 * Ok, I agree, and the units were not only old fashioned, but the calculations were wrong.


 * Here is a python program to compute the various numbers:

import scipy.optimize.zeros k1 = 0.0053 k2 = 0.185 # kg/m # (745.69987/375)*0.0083*2.2369363**3 K = 9.8 # m/s^2 #2.2369363 *2.2*745.69987/375 # W print "k1 = %g, k2 = %g, K = %g\n" % (k1, k2, K) speed_str = lambda v:"%.2gm/s (%.2g mph or %.2g km/hr)" % (v,v*2.2369363, v*3.6) psi = lambda w,vg,va,g=0:K*vg*w*(k1 + g) + k2*va**3 rel_drag = lambda w,vg,va,g=0: (k2*va**3)/psi(w,vg,va,g) pim = lambda w,vg,va:745.69987/375*(vg*w*(k1 + g) + 0.0083*va**3) print "power,  relative drag,     equivalent up 7% grade,   rel drag" for w in [90,65]: for vg in [9,11]: P = psi(w, vg, vg) f = lambda x:psi(w,x,x,0.07)-P vh = scipy.optimize.zeros.brentq(f, 0, vg) print ("%.2fW    %.1f%%         %s %.2g%% \n" %     (P, 100*rel_drag(w,vg,vg), speed_str(vh), 100*rel_drag(w,vh,vh)))


 * I left the anecdotes in old units to give people another perspective.

Mile-a-minute Murphy
It might be interesting to mention "mile-a-minute Murphy" here, who went 62 mph (100 kph) in 1899: I'm not sure how notable this is though. --NE2 08:24, 22 January 2007 (UTC)


 * Not very notable, at least in the context of this article. Motor-pacing virtually eliminates aerodynamic drag. Mr Larrington (talk) 12:18, 23 July 2009 (UTC)

Calories vs kilocalories
The "These are actually kilocalories" statement seems unnecessarily confusing. It would be better to use either "calories" or "kilocalories" correctly or not at all. -AndrewDressel 20:29, 22 January 2007 (UTC)
 * Done, finally, 11 months later. -AndrewDressel (talk) 20:44, 28 December 2007 (UTC)

References needed?
Fram: As you said, this isn't your area of expertise. The type of calculations performed in this article are completely routine back-of-the-envelope analysis that physics students are taught to perform in their first year of college. They're not original research. As for whether this subject could be referenced. Ahem. etc.

This material does need to be shortened, revised for tone and moved though. -- pde 09:33, 3 February 2007 (UTC)


 * As I said, I don't mean referencing the formulas (which are indeed routine ones), but the article, the subject matter as an independent field of research. I think you have misread my statement there. But thanks for the references for the subject, they can probably be added to the article (if they are WP:RS soutces). Fram 20:28, 3 February 2007 (UTC)

Here's a great little online calculator for all kinds of configurations by Walter Zorn: http://www.kreuzotter.de/english/espeed.htm Scroll down a bit if you want to see the formulas & explanations. Should be added to the main article's external links, imho. Merctio 17:57, 5 May 2007 (UTC)

There are many assumptions that are made for the calculations that go un-cited or even unjustified. The 24% efficiency figure on human body efficiency, for instance. This article needs citation attention, imo. Trevorzink (talk) 18:50, 6 May 2011 (UTC)

Rotating kinetic energy, need to do some calculus
The claim that the KE for the rotational component is equal to the moving componentis not correct. The claim is only true if ALL of the wheels mass is at the perimeter. The rotational velocity of the wheel at the hub is much smaller, and the hub/spokes are a significant portion of the total mass of the wheel. The rotational velocity is a function of the radius -- v(r). To correctly calculate the rotational KE, you would have to perform an integration over the entire radius of the wheel. In any case the rotational component will be a fraction of the moving component...so the notion that "A pound off the wheels = 2 pounds off the frame." is (at least mathematically) false unless you took that pound entirely from the tires. 13:08, 27 April 2007 (UTC)
 * Yes, it is an approximation, but not a bad one. A quick search online finds a Dura-Ace front hub with 129g, a DT Swiss rim with 415g, a Continental GP4000 tire with 205g, and a tube with 55g. 32 spokes are abaout 300g. So, of the total mass, over 60% is close to the bead diameter. The spokes are less than 30% and the hub less than 12%. Low-spoke-count wheels make the approximation even better. -AndrewDressel 19:18, 27 April 2007 (UTC)

1.4 kcal, even if an upper limit, is only 1.4 Cal. Insignificant. This part of the discussion is called a tempest in a teapot. Acceleration from standstill to speed for any ride of decent length is essentially immaterial. — Preceding unsigned comment added by Ppetrel (talk • contribs) 16:49, 7 October 2012 (UTC)

Inconsistency in the energy efficiency calculations
The comparison of the walking energy expenditure rate of 100 W at 5 km/hr with the cycling rate of 100 W at 25 km/hr is inconsistent with the values in the bulleted list of 3.78 kJ/(km∙kg) and 1.62 kJ/(km∙kg), respectively. The first pair of numbers has a ratio of 5, while the second pair's ratio is 2.3. The absolute numbers can also be related to one another. For example, the 100 W for a 70 kg person at 5 km/hr becomes 1.03 kJ/(km∙kg). The analogous cycling number of 100 W at 25 km/hr becomes 0.21 kJ/(km∙kg). Clearly, none of this hangs together. Typical bike-to-walking ratios I've seen quoted are closer to 2.5 or 3, never 5. Worse, the absolute numbers are off by a large factor: 3.78 vs. 1.03 and 1.62 vs. 0.21.

A more consistent, and perhaps better treatment is given at Cycling Performance Tips and is based on a seemingly reputable book. Using the formulas from this site, the cyclist requires 0.74 kJ/(km∙kg), somewhere in between the 0.21 and 1.62 given in the article. I assumed the values for a road bike and included the mass of the bicycle in the calculation. Note that this page has formulas for mechanical energy at the pedals as well as chemical energy required to produce this mechanical energy. The quoted efficiency of 25% is commonly used, but is only approximate. It is among the many uncertainties that arise in such a calculation. I suspect the larger numbers in the article (bulleted list) are food energy amounts, the smaller ones that derive from the 100 W number are the mechanical energies. In any case, this needs clarification. None of this changes the inconsistency of their ratios. Only one set of numbers needs to be listed since they are related to each other by multiplying or dividing by 4. I think people are generally interested in food energy, hence the use of calories.

In summary, the data in this article are internally inconsistent and not consonant with published formulas. I suggest revising this section to reflect the currently accepted understanding and to reduce the precision of the numbers listed. As the author of the Cycling Tips page noted, "...biking is NOT an exact science." Drphysics 23:52, 21 June 2007 (UTC)

The "Aerodynamics vs power" section cites an equation similar to this one Cycling Performance Tips, but there is no reference. The values for K1 and K2 are most comparable to the Road Bike case. The values from the Performance Tips site translate into K1=0.0042 (cf. 0.0053) and K2=0.012 (cf. 0.0083). These are not too far off, but given the absence of references for the formulas in the article, the Performance Tips values should be preferred. It is also noteworthy that the Performance Tips site has dramatically different values for road bikes vs. mountain bikes, which is glossed over in the "Aerodynamics vs power" section. Drphysics 01:44, 22 June 2007 (UTC)


 * I wanted to signal the very same problem, but I see it's already been noted, though still not fixed after such a long time... :-/ The inconsistency is there, and is real. However the amount of inconsistency isn't trivial, because the original figure said 5 km/h walking vs. 15-25 km/h cycling. So the ratio could be anywhere from 3 to 5. Also note, that the bullets below listed numbers that are said to be used in calculations, and did not mention the variable of speed. It might as well be measured at 30 km/h. I've read in some wiki article, that treadmill manufacturers count with a 12.5% efficiency on the human part, and that anywhere between 10-24% is possible. Also note the variance caused by wind, clothes worn, posture, individual factors, aerobic vs. anaerobic muscle work, fittness, etc. I bet a factor of 2 wouldn't be way off when specifying these constants.
 * To give proper answer, one would need to measure the energy usage of a few representative humans in a few controlled sets of walking and cycling. One method is to measure oxygen usage or CO2 exhalation rate. Or radiated body heat perhaps? Well, basically if you connect an electic motor with a known efficiency to the bike with a driver, you could trivially arrive at the needed power output (not counting the aerodynamics of the motor and the battery in your backpack). Obtaining human efficiency at different power outputs could be solved as a separate problem.
 * I'll keep you updated if I encounter a reliable source of measurement results or scientific paper. bkil (talk) 20:24, 27 January 2008 (UTC)

Having cycled for 10 minutes at around 23 km/h yesterday, I totally disagree about the fact that cycling at 25 km/h takes as much power as a 5 km/h walk! This seems totally impossible to me, unless we are talking about some record-breaking bike with an aerodynamic shell around the rider... —Preceding unsigned comment added by 159.153.144.23 (talk) 00:22, 29 April 2008 (UTC)

I'd like to call attention to the K2 value in the energy formula. It says it's derived from an area of .4 m^2 at a drag coefficient of .7 (.185 kg/m ~= .7 * .4 m^2 * .5 * 1.204 kg/m^3). I'm a big guy, but my estimate for frontal area of my bike + rider is .73 m^2 (I multiplied my hip width by height of helmet while seated non-aerodynamically, which I feel is reasonable given the extra protrusions on the bike). Moreover, the drag coefficient for a bike + rider is cited as .9 over at the drag coefficient wiki article. This combines to make my K2 .396 kg/m (.73*.9*.5*1.204). This is over twice the K2 cited in this article, and means that power required at 20 mph (no wind, flat ground) is 283 watts for the aerodynamic component ALONE, compared to the 181 watts from the equation in this article which encompasses aero AND mechanical losses. I'd say the K2 value on this article is low, but this is original research. I'd like more discussion on the matter. —Preceding unsigned comment added by 67.183.2.174 (talk) 20:43, 11 August 2009 (UTC)

I have a problem with these calculations as well. I know when I ride bicycle I produce about 200W continuously. This is common knowledge and also verified on a stationary bicycle with power display. I weigh 75 kg, and ride 30 km/hr. In one hour I have produced 200 Watts * 3600 seconds = 720,000 Watt seconds = 720 kJ. In the same period of time I have moved 75 kg over a distance of 30 km, which is 2250 kg km. This is 720 kJ / 2250 kg km. Dividing by 2250 yields: 0.32 kJ / (kg km). I can't imagine why it is possible to get to a value of 1.6 kJ/(kg km). This seems to by highly inaccurate. Jlinkels (talk) 13:33, 22 April 2012 (UTC)


 * I've used the example already given: cycling at 15 km/h using 60 W. Thus 1 km is covered in 240 s which equates to 14400 J/km or 205 J/km*kg for a 70 kg person. Assuming a muscular efficiency of 20%, this gives about 1 kJ/km*kg. But if 60 W is assumed additionally as the basic metabolic rate, we get 2 kJ/km*kg. Therefore 1.6 kJ/(kg km) seems realistic. But I have replaced the uncited examples with this calculation. Is it comprehensible enough?
 * Your example above is not valid because the 720 kJ have no significance for this calculation.Theosch (talk) 14:25, 26 March 2015 (UTC)

Kinetic Energy
The translational kinetic energy of an object in motion is:


 * $$E = \tfrac{1}{2}mv^2$$,

Where $$E$$ is energy in joules, $$m$$ is mass in kg, and $$v$$ is velocity in meters per second. For a rotating mass (such as a wheel), the rotational kinetic energy is given by


 * $$E = \tfrac{1}{2}I \omega^2$$,

where $$I$$ is the moment of inertia, $$\omega$$ is the angular velocity in radians per second, $$r$$ is the radius in meters. For a wheel with all its mass at the radius (a fair approximation for a bicycle wheel), the moment of inertia is


 * $$I = m r^2$$.

The angular velocity is related to the translational velocity and the radius of the tire. As long as there is no slipping,


 * $$\omega = \frac{v}{r}$$.

When a rotating mass is moving down the road, its total kinetic energy is the sum of its translational kinetic energy and its rotational kinetic energy:


 * $$E = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2$$

Substituting for $$I$$ and $$\omega$$, we get
 * $$E = \tfrac{1}{2}mv^2 + \tfrac{1}{2}mr^2 \cdot \frac{v^2}{r^2}$$

The $$r^2$$ terms cancel, and we finally get
 * $$E = \tfrac{1}{2}mv^2 + \tfrac{1}{2}mv^2 = mv^2$$.

In other words, a mass on the tire has twice the kinetic energy of a non-rotating mass on the bike. There is a kernel of truth in the old saying that "A pound off the wheels = 2 pounds off the frame."


 * This math is in error! the r in the moment of enertia formula is for the radius of an approximating point mass, not the outside of the wheel. thus the velocity of that mass is not the same as the velocity of the ground speed. They do not cancel! -pvd (130.212.215.215)


 * The formula is correct for a ring. For a ball it would be 2/5 times that, so I think the maths is correct and, more importantly, so is the physics -ejf (194.30.186.250)


 * I agree with ejf. The formula is correct. Its usefulness however depends on how well a thin hoop approximates a particular bicycle wheel. This, I believe is pvd's objection. In reality, all the mass cannot be at the radius. For comparison, the opposite extreme might be a disk wheel where the mass is distributed evenly throughout the interior. In this case $$I = \tfrac{1}{2}m r^2$$ and so the resulting total kinetic energy becomes $$E = \tfrac{1}{2}mv^2 + \tfrac{1}{4}mv^2 = \tfrac{3}{4}mv^2$$. A pound off the disk wheels = only 1.5 points off the frame. Most real bicycle wheels will be somewhere between these two extremes. -AndrewDressel (talk) 22:51, 18 January 2008 (UTC)

Force effectiveness index
The performance section of the bicycle article says "from a mechanical viewpoint, up to 99% of the energy delivered by the rider into the pedals is transmitted to the wheels." I checked the source and it is only referring to the chain drive (chain ring, rear sprocket and chain), the testing never involved a crank arm or pedal, so it seems hard to make the claim stated above.

Now, I've found 4 different published studies that did look into the energy delivered by the rider into the pedals. All 4 refer to this analysis or study as Force effectiveness index, effectiveness index or index of effectiveness. From what these studies show is that the avg effectiveness ranges anywhere from 22%-64% depending on the rpm/cadence, load or watt output and lastly the experience of the rider.

I'm not sure if your familiar with the concept, but to me, when the crank arm is at the 12 or 1 oclock position and I push on the pedal, I sincerely cant see ...logically or physically that 99% of that energy is being transmitted to the wheel, I think a large amount is not transfered. (at the 3 oclock position of course I can see the efficiency or transfer to the rear wheel being much higher).

Sources:
 * 1) J Sports Sci. 1991 Summer;9(2):191-203. Sanderson DJ. The influence of cadence and power output on the biomechanics of force application during steady-rate cycling in competitive and recreational cyclists. Look up
 * 2) ISBS'98 –Proceedings II : THE USE OF FORCE PEDALS FOR ANALYSIS OF CYCLING SPRINT PERFORMANCE by Simon G. S. Coleman, University of Edinburgh, U.K.
 * 3) The influence of flywheel weight and pedalling frequency on the biomechanics and physiological responses to bicycle exercise† Ergonomics Volume 26, Issue 7, 1983, Pages 659 - 668
 * 4) The engineering of sport 7,vol 1: Page 263 Development of multi-platform instrumented force pedals for track cycling(p49) Jean-marc Drouet, Yvan champoux, Sylvain dorel

Coreyjbryant (talk) 03:13, 4 April 2011 (UTC)


 * It appears that you are not contesting the claim in that article, but rather are pointing out an additional consideration: that of transfering power from the legs to the pedals. Correct? -AndrewDressel (talk) 20:00, 4 April 2011 (UTC)


 * I think I would contest: "from a mechanical viewpoint, up to 99% of the energy delivered by the rider into the pedals is transmitted to the wheels" from the article. My point is that the 99% energy efficiency claimed seems to be false or misleading, and the study related to the chain and sproket system, the quoted source never incorporated the bicycle crank arm and pedal. I think an equivalent example would be a claim that a study found a car engine 99% efficient, but only tested the automatic transmission and never included the main crank shaft and piston. (Further more the way the quote is read, someone could easily assume that "the bicycle" is perfected and there is only 1% room for efficiency improvement.)


 * I'm afraid I disagree. It sounds as though you are claiming that energy is lost somewhere between the pedal and the chainring, while the sources you cite have to do with biomechanics, the delivery of torque and power to the pedals. I concede that the 99% claim does not take into account the efficiency of pedals and cranks in collecting torque and power from a rider's legs, but it doesn't claim to either. The biomechanics of pedalling is a fascinating area that could be addressed in this article, but will not change the fact that once the power is delivered to the pedals, a chain and sprocket bicycle drive train can transmit it at up to 99% efficiency to the rear wheel. The equivalent example would be to claim that the manual transmission in some car is 99% efficient without mentioning the efficiency of the power plant. -AndrewDressel (talk) 02:45, 5 April 2011 (UTC)


 * What I'm looking for is to see if someone knows of any concensus as to what is the average bicycle efficiency ( avg total Force applied to the pedal...transfered to the rear wheel). I'd like to see as well if "Index of effectiveness" or "effectiveness index" would be the appropriate number for determining the true bicycle efficiency. I think it would be appropriate to keep the source and the 99% percent efficiency but rewrite it in such away that it refers to only the parts that the study look at.


 * There are probably a few studies now that cover that, as your four examples indicate, but I doubt there is a concensus yet. Problems are that the issue is very dependent on training and that force applied at to the pedals is only a surrogate for metabolic cost, which is the real concern. -AndrewDressel (talk) 02:45, 5 April 2011 (UTC)


 * I've tried searching google "how efficient is a bicycle" and found the link the wiki and a few other sites. ...It is extremely difficult to find any good articles.Wiki uses the 99% quote and many other site use a claim of 99% but they do not provide any reference..maybe they are using wiki? Thanks. Coreyjbryant (talk) 00:44, 5 April 2011 (UTC)


 * I don't believe Google is the best tool for finding academic papers. Have you tried an academic library? -AndrewDressel (talk) 02:45, 5 April 2011 (UTC)


 * 2 things:
 * 1st reply to: andrewDresselcomment 02:45, 5 April 2011: The idea of metabolic cost,: Metabolic cost needs to be discussed/included as well, as it would have a direct relationship to the efficiency. I think metabolic cost and force initially applied to the pedal could be considered equivalent terms/substituted terms.


 * Unfortunately, it cannot, and that is a huge part of the problem. Simply measuring the force applied to the pedal ignores the forces that muscles must generate in different parts of the leg in order to produce that resulting force at the bottom of the foot. Some muscles will be contracting as they generate force and so doing positive work, while others will be extending as they generate force and so be doing negative work. The body pays for approximately the sum of the absolute values, but the pedal receives approximately just the sum. Plus, this little analysis is ignoring moment arms and the rates at which forces are applied. -AndrewDressel (talk) 18:17, 6 April 2011 (UTC)


 * 2nd to the comment 02:45, 5 April 2011 I agree Google is not the best place to find Academic papers...I was trying to make a point that the answer to a basic question "how efficient is a bicycle?" should be easy to find. It shouldn't require someone to go to an Academic Library. Hence Why I'm here...to create discussion and hopefully find consensus as to what an appropriate answer could be and make  the answer easy to find! :o)(wikipedea tends to be the first place people go to find easy answers).


 * An excellent goal. I applaud you for coming here looking for a detail, not finding, and now trying find it elsewhere so that it can be added here. -AndrewDressel (talk) 18:17, 6 April 2011 (UTC)


 * Could we both at least agree that wiki stating 99% efficiency of a bicycle may be misleading/incorrect, considering there is only one source mentioned on this claim and that the study was only looking at the chain drive? I will attempt to get authors of some of the academic studies that have been done to participate or shed incite on this topic. If anyone else knows of anyone that could offer insightful/useful insight into the topic please refer them here. Coreyjbryant (talk) 17:27, 6 April 2011 (UTC)


 * I believe the 99% efficiency number is correct as stated. If you feel it is misleading, you might want to add a caveat about what it ignores. -AndrewDressel (talk) 18:17, 6 April 2011 (UTC)


 * Question: If a new bicycle were designed and the crank arms were removed/changed to a different mechanism (a more efficient design) and if all cyclist were able to produce 20% more Watt's to the rear wheel on the new design compared to what all cyclist produce on the typical design with a crank arm...would this be enough to discredit a 99% efficiency? (stating a 99% efficiency, incontrovertibly states that there will only ever be a 1.01% improvement of bicycle mechanical design) Coreyjbryant (talk) 19:37, 6 April 2011 (UTC)


 * No, because this hypothetical new design would be capturing more energy from the rider, not transmitting a higher percentage of what it receives from the rider to the rear wheel. In fact, the new design may actually require a less efficient drive-train if it involves motions that cannot be efficiently translated into the rotary motion of the rear wheel. The 99% efficiency number concerns only transmitting power from the pedals to the rear wheel and addresses only one aspect of the entire bicycle mechanical design. -AndrewDressel (talk) 22:04, 6 April 2011 (UTC)


 * The last sentence above included "99% efficiency" and the word pedal....my entire point is that the sole source quoted never included/tested a pedal or crank, if there are studies that claims 99% efficiency and included the use of crank arm and pedal in the testing please make reference to it.


 * Well, the article states that they tested a "bicycle drive train in a campus lab", so I think it is safe to say that they applied power to the crank. The second source in the first paragraph of this article, Bicycling Science, is quoted as stating "when new, clean, and well-lubricated, and when sprockets with a minimum of 21 teeth are used, a chain transmission is highly efficient (at a level of maybe 98.5 percent or even higher)." So, unless you are contending that power is lost somewhere between the pedals and the chainring, I believe these two sources confirm that the transmission efficiency from the pedals to the rear wheel is about 99%. -AndrewDressel (talk) 15:59, 7 April 2011 (UTC)


 * This is going to be an unproductive disagreement if we are debating something that you have not read the original source. My point has always been the single source is not credible in stating bicycle efficiency at 99%, if you're basing your entire argument on the university news paper article, (which is funny they never actually referenced the study that SPENCER did..which I think shows to the credibility of the source), the true source of the article is: Effects of Frictional Loss on Bicycle Chain Drive Efficiency Journal of Mechanical Design DECEMBER 2001, Vol. 123 Õ 605 James B SPICER). Coreyjbryant (talk) 14:21, 9 April 2011 (UTC)


 * I am not basing any argument on a single source. The 99% number is well established, if rounded up. Wilson uses 98.5% in is book, which is quoted in the references at the end of this very article. Berto uses 97% in The Dancing Chain. -AndrewDressel (talk) 22:33, 9 April 2011 (UTC)


 * I have the PDF if you would like to read it (my email is my user name @ hot mail), I can assure you they did not use a crank arm or pedal. If you want to see what they used, please look at the picture in the news paper article, its just above the words, "Using electric motors and a computer", in the picture you'll see only a chain ring attached to an electric motor. The second reference above states something similar to the first, that they are only looking at the chain drive mechanism, my contention is that "chain drive mechanism" do not include the crank arm or pedal, I think once the original source is reviewed maybe we could come to some type of agreement. Lastly, it was said" So, unless you are contending that power is lost somewhere between the pedals and the chainring" This is exactly what I have been saying should be looked at, power and energy is lost between the pedal and chain ring. Coreyjbryant (talk) 14:21, 9 April 2011 (UTC)


 * I'm afraid I cannot accept your assertion. Where do you propose that the lost energy is going, since it is also pretty well established that energy cannot be created nor destroyed? -AndrewDressel (talk) 22:33, 9 April 2011 (UTC)


 * When looking at lever arms (crank arms) energy or force can ONLY be effectively transfered when Force is applied at a right angle to the lever. Coreyjbryant (talk) 14:21, 9 April 2011 (UTC)


 * That is simply not true. -AndrewDressel (talk) 22:33, 9 April 2011 (UTC)


 * So when the crank arm is at 12 oclock or TDC, when 10N of force are applied on the pedal of course not all that energy is transfered to the chain ring. Coreyjbryant (talk) 14:21, 9 April 2011 (UTC)


 * I cannot tell if you are using force, energy, and power interchangebly or not. They are related but not equivalent, of course. Since the change in kinetic energy of a particle equals the dot product of the applied force vector and the displacement vector it causes, a 10N force applied to a pedal at TDC does not change its kinetic energy at all if it causes no displacement: no energy it transfered from the foot to the pedal. The fact that the leg used energy to produce that force is an issue of biomechanics and a complicated one at that. It does not alter, however, the efficiency of a bicycle's drive train. -AndrewDressel (talk) 22:33, 9 April 2011 (UTC)


 * I'm unsure now if you have read the sources I have quoted at the beginning, that looked/measured specifically at the forces applied to the pedals, and then calculated how much of that force was effectively transfered to the chain ring. Please email me, and I could send you the pdf's. I'll leave it to you to see if this continues, this is what I interpret, and if we are not on the same page it's not that we are wrong, it's just that we have different perspectives and will never come to any consensus. 1) Chain drive mechanism, the studies looked at chain ring, chain and sprocket and did not include crank arm and pedal. 2) A pedal is a attached to a crank arm, To transfer all energy across a lever arm or crank arm in the form of torque, the force must be perpendicular to the lever..if it is not perpendicular then not all the power that was applied to the pedal is transfered as torque.(effectiveness index looks at this aspect of the bicycle crank arm and pedal) Coreyjbryant (talk) 14:21, 9 April 2011 (UTC)


 * All the sources you cite are about biomechanics, which is an entirely different animal. As I said in one of my first replies above "the biomechanics of pedalling is a fascinating area that could be addressed in this article, but will not change the fact that once the power is delivered to the pedals, a chain and sprocket bicycle drive train can transmit it at up to 99% efficiency to the rear wheel." As I have also stated elsewhere, you are very welcome to start a section that addresses the issue of how efficiently the pedal and crank system captures energy from a rider's legs. -AndrewDressel (talk) 22:33, 9 April 2011 (UTC)


 * Another strong argument would be a new terrible bicycle design(completely impractical) was made... and used a chain ring, chain and rear sprocket in the design and then automatically made the claim that it too was 99% efficient and referenced the above mentioned article. The above mentioned article could be used for any mechanism that uses a chain ring, chain and sprocket. Search youtube for the following: Elliptigo, H-zontal bike, Streetstepper model 09, 4-bar bicycle drive mechanism.


 * Marketeers will claim whatever they want, no matter what is written here. The only way to prevent that is to say nothing, and even that will only send them elsewhere for details to misuse. -AndrewDressel (talk) 15:59, 7 April 2011 (UTC)


 * The above could use the same quote that they are mechanically 99% efficient and reference, "Johns Hopkins Gazette", 30 August 1999" but this would be wrong. When discussing a machines efficiency there should be more parameters then chain drive, and there should be more sources available if there is a claim that something is absolute.!99%! I'm afraid this debate will shortly end as it is draining. I think we need new blood in this discussion to maybe offer new insight into both are arguments.(Thank you for your intelligent, well expressed arguments,too often you see sarcastic replies and discussions) Coreyjbryant (talk) 23:12, 6 April 2011 (UTC)

effect of bicycle tyres
This article doesn't seem to have any mention of drag/rolling resistance due to the tyres; underinflated tyres are widely beleived to cause a lot of drag, but are there any useful references for this effect? Murray Langton (talk) 12:25, 18 July 2011 (UTC)

In fact, anyone who has ridden a road bike and a mountain bike on a road can easily tell that the tyres make a huge difference in the efficiency of the bike. --188.62.165.43 (talk) 09:53, 1 June 2014 (UTC)

Power required to overcome Gravity
In relation to the following extract:

''Power required ''There is a well known equation that gives the power required to push a bike/rider through the air and to overcome the friction of the drive train:


 * P = g m V_g (K_1+s) + K_2 \times V_a^2 V_g

''Where P is in watts, g is Earth's gravity, Vg is ground speed (m/s), m is bike/rider mass in kg, s is the grade (m/m), and Va is the rider's speed through the air (m/s). K1 is a lumped constant for all frictional losses (tires, bearings, chain), and is generally reported with a value of 0.0053. K2 is a lumped constant for aerodynamic drag and is generally reported with a value of 0.185 kg/m.[16]

''Note that the power required to overcome friction and gravity is proportional only to rider weight and ground speed...

I know that this formula appears to come from a credible source (e.g. http://www.sportsci.org/jour/9804/dps.html#ref ), but it doesn't make sense to me that the power required to overcome gravity is proportional to speed. (Perhaps the original author has confused 'work done' with 'force required'?)

This formula suggests that the power required to overcome gravity at zero speed is zero, when in fact isn't gravity a constant downward force accelerating an object by 9.81m/s? On a flat, natural force counteracts gravity, but on a slope, this net force is a function of massive, gravity, and slope angle (specifically m.g.sin(angle) - see http://www.studyphysics.ca/newnotes/20/unit01_kinematicsdynamics/chp06_vectors/lesson25.htm )

I suggest the formula be amended accordingly. Anyone care to comment on this?
 * Since mechanical power is defined to be the dot product of the force vector and the velocity vector, the power is zero if the veloity is zero. On a slope, the force required to remain stationary may be non-zero, but the mechanical power required is zero. -AndrewDressel (talk) 16:17, 20 August 2011 (UTC)

Weight, or Mass?
The section on kinetic energy refers to the mass of the wheel, and seems to be using the weight in the equations. Could someone take a look and comment. If this error has been made it will not change the conclusion that a kg or wheel is equivalent to 1.5-2 kg of frame, but the numbers themselves will be much smaller. — Preceding unsigned comment added by M610 (talk • contribs) 17:45, 19 November 2011 (UTC)

This is embarrassing
x^2 + x != x^3

in fact,

x^2 * x = x^3

So the resistance in fact NEVER increasse with the cube, but, quite simply, with the square. plus some. just like it's written there. Doing such a basic mistake is, sorry if i sound harsh, but it really is embarassing. considering that the rest of the article reads perfectly sound. apart from the wheel k.e. mumbo jumbo.

please refrain from further attempts at mathematizing.


 * It would be helpful if you would indicate to what and to whom you are referring. The article currently states that the force of air drag "increases roughly with the square of speed" and "the power required to overcome drag increases with the cube of the speed." Is there somewhere else in the article that contradicts this? What do you mean by "there"? Who should "refrain from further attempts at mathematizing?" -AndrewDressel (talk) 20:18, 23 May 2012 (UTC)


 * "...increases with the cube of the speed..." turned into a formal statement this becomes
 * P_overcomedrag = P_drag
 * P_overcomedrag=v^3 P_drag=v^2 => P_drag = v^3 = v^2 which is obviously nonsense. Now, if im overly generous i'll say P_overcomedrag = P_drag + P_ridiculouslytinyamount, however, it still makes for an embarassing result.


 * Sorry, but I still cannot find where the article makes this mistake. Nowhere do I see "P_drag=v^2". Instead, I see:
 * "Air drag, which increases roughly with the square of speed, requires increasingly higher power outputs relative to speed, power increasing with the cube of speed as power equals force times velocity."
 * "Even at moderate speeds, most power is spent in overcoming the aerodynamic drag force, which increases with the square of speed. Thus, the power required to overcome drag increases with the cube of the speed."
 * "The aerodynamic drag is roughly proportional to the square of the relative velocity of the air and the bike."
 * All of these correctly assert that the force of air drag "increases roughly with the square of speed" and "the power required to overcome drag increases with the cube of the speed." -AndrewDressel (talk) 22:33, 23 May 2012 (UTC)

oh goddamnit. you are right. i guess it's all the water in my eyes making me blind. all the insults are on me now then. ah, whatever. ill go drown myself in booze. — Preceding unsigned comment added by 87.162.56.150 (talk) 22:35, 23 May 2012 (UTC)

disclaimer: kids, this is what happens when your math professors pump you full of functional analysis instead of doing something practical! — Preceding unsigned comment added by 87.162.56.150 (talk) 22:40, 23 May 2012 (UTC)

Energy/power needed for cycling???
I seem to be profoundly confused - perhaps someone can help me out here. Say I weigh 100 kg and go for a 1-hr ride in which I travel 30 km, hence 30 km/h. There are no hills and no wind. I'd like to know how much energy I need.

Section 1: 1.62 kJ/(km∙kg) = 4860 kJ = 1161 kcal, compatible with what I find on many websites (on the order of 1000 kcal for ~1h).

Section 4.1: using the formula, I get P = 43 + 107 = 150 W, which is compatible with the 175 W example below the formula and commentators on the Tour de France claiming thesr guys do ~250 W. Biking for 1h, I need 150 W ∙ 3600 s = 540 kJ = 129 kcal.

How can the difference of about an order of magnitude(!) between the rules of thumb "a few 100 W in power" on the one hand and "on the order of 1000 kcal in 1h" on the other be explained?

AstroFloyd (talk) 22:20, 8 July 2012 (UTC)


 * OK, I guess there's a factor of four in the efficiency of the human body, as stated further down in Sect. 4.1: "The human body runs at about 24% efficiency for a relatively fit athlete, so for every kilojoule delivered to the pedals the body consumes 1 kcal (4.2 kJ) of food energy." I suppose that brings the difference down to a factor of two... AstroFloyd (talk) 07:46, 9 July 2012 (UTC)

Wheel k.e. mumbo jumbo
The article should start with the "Power Required" section which is relevant and very useful.

The whole thing about the KE of the wheels be put later or in a footnote or another page, because is actual relevance to the efficiency of a bike is very small. That is, all that massive discourse is just about the energy which you put in to get the bike going once when you start, which for most rides and most races is a very very small part of the total energy, and which you in fact can get back if you rest as you let the bike come to a halt at the end of the ride. I have found people (who may have read this) who believe it is important to get the weight of the rims down, compared to the fixed parts of the bike, as though it helped you go faster up hills or steadily along the flat. So the whole KE section could do with some sort of disclaimer on it. The KE of the wheels is irrelevant to the performance a bike being ridden steadily once it is going. TimBL (talk) —Preceding undated comment added 18:59, 9 July 2012 (UTC)


 * I've tried to quantify this and removed the excessively long derivation. Theosch (talk) 10:34, 30 March 2015 (UTC)

'explanations' trimmed off
With sentences like "possible technical explanations"... "placebo effect" and so forth which shows it was guesses and invented contents by editors. Cantaloupe2 (talk) 10:12, 7 December 2012 (UTC)

flex
anyone up on frame flex and wheel hub lateral and torsional flex ? I would love to see a reference link to a slow motion shoot of a hub shifting around on the spokes - this must be out there somewhere--— ⦿⨦⨀Tumadoireacht Talk/Stalk 09:29, 31 December 2013 (UTC)

Reduction of weight vs reduction of rotating mass/Kinetic energy of a rotating wheel
Should this line

"In other words, a mass on the tire has twice the kinetic energy of a non-rotating mass on the bike. This all depends, of course, on how well a thin hoop approximates the bicycle wheel. In reality, all the mass cannot be at the radius."

correctly read

In other words, a mass on the tire has twice the kinetic energy of a non-rotating mass on the bike. This all depends, of course, on how well a thin hoop approximates the bicycle wheel. In reality, all the mass cannot be at the rim/outer edge

--— ⦿⨦⨀Tumadoireacht Talk/Stalk 09:15, 18 March 2015 (UTC)


 * I think this is what is meant, but I agree that it is confusing. Theosch (talk) 16:30, 29 March 2015 (UTC)

Also does anyone have access to high speed film of the lateral flexing of the spoke cluster as the hub drifts three dimensionally under load at speed or calculations for this wobble and its effects on efficiency ?--— ⦿⨦⨀Tumadoireacht Talk/Stalk 09:17, 18 March 2015 (UTC)

Potential energy regained?
The section "Climbing Power" contains this claim: "As this power is used to increase the potential energy of bike and rider, it is returned when going downhill and not lost unless the rider brakes or travels faster than desired."

Ahm, I don't think so. The potential energy is converted to kinetic energy of the bicycle and rider, plus kinetic energy of the air displaced as the rider speeds down the hill (and eventually into heat as the vortices dissipate), plus friction in the bearings and tires and other things that move. If the downhill leg is a very shallow grade, and the downhill leg consequently slow, more kinetic energy will go into the bike-and-rider, and less into turbulence (because turbulence is non-linear). Only as his downhill speed goes to zero will all the potential energy at the top be converted to the bike-and-rider's kinetic energy. Nature just isn't fair, is she?

This means that a hilly course takes more of the rider's energy, for a given speed, than a flat course: the energy put into climbing is never fully recovered. I ride with someone who is entirely convinced the speeds he attains on the downhill portions more than make up for the work required to climb. I don't believe it, and am pretty sure he has no corroborative data. (I should ask him to coast up the next equally high hill.)

But I have made no edit, in case this turns out to be contested. Captain Puget (talk) 06:00, 24 November 2015 (UTC)


 * The potential energy is fully recovered when the original altitude is reached but you are correct that it is converted to heat except for any surplus bit of kinetic energy at the bottom. I've reworded slightly in order to hopefully clarify. Theosch (talk) 12:42, 12 July 2016 (UTC)

Usain Bolt
Under 'Typical Speeds' the article states that a reasonably fit man on a racing bicycle can hit 40kmh-1 over a short distance. Since Usain Bolt, and others, can run at 36kmh-1 over a short distance, I find 40 a bit unamazing. Could someone review that, please? — Preceding unsigned comment added by 27.33.152.79 (talk) 23:42, 22 April 2016 (UTC)
 * Work it out yourself using the cited reference, Walter Zorn's calculator. It takes about 300 W to maintain 40 km/h on a racing bike and most people can't maintain this power for very long when pedalling. Although more power is available briefly, this is used for accelerating. Theosch (talk) 12:53, 12 July 2016 (UTC)


 * Sorry in advance if my comment doesn't follow proper protocol. This is my second attempt to actively participate in Wikipedia conversations. Let me know if I need to modify it.


 * I disagree with the statement based on personal experience, but the statement is vague, so unprovable. How far is a short distance? 300 feet? One mile? I doubt many riders could hold that speed for several miles (several miles is vague--just seeing if you are paying attention). Some people think reasonably fit means able to climb two flights of stairs if the elevator is broken and there are donuts on the third floor. I'm male, almost 57, and ride most days. I would say I'm in better condition than reasonably fit but that is my opinion, not a fact. Yesterday I rode about 0.25 miles up a shallow hill, nominally into the wind (the hill, surrounding trees, etc, more or less blocked the wind), without getting particularly out of breath. A female about my age followed me up the hill, not drafting, at about the same speed. A year ago, I held 32 mph for about the same distance, up a very shallow hill, on a 40 mph road, because I was scared of the traffic coming up from behind. The predominant wind direction in my area makes it likely that it was 45 degree headwind, but assume a tailwind of 7 mph (not the same as zero wind at 25 mph, BTW. Someone tell me how to get the attention of the author on the Power Required / Air Drag section). I had to stop to catch my breath when I was finally able to turn off that road.


 * The anecdotes I cited are useless for supporting or refuting the statement, of course, since 1) I can't provide proof of any of it, and 2) I didn't actually measure distances, road grades, or wind speeds. That being said, real world tests trump theoretical calculations. I'll look for opportunities to test the statement, using myself and my riding buddies as test subjects, and update this with actual distance measurements. I'll use weather.com for wind speed. I don't know how to quantify fitness level. How about, below the fitness level of category 5 riders? (Category 5 is the slowest, least experienced racing category.) RiderBill (talk) 18:12, 2 November 2016 (UTC)

Lead section wording
The last sentence of the lead seems to be ambiguous: "In terms of the ratio of cargo weight a bicycle can carry to total weight, it is also a most efficient means of cargo transportation." If it's only "a" most efficient means, the sentence would read better as "...it is also an efficient means...", but if we're claiming it's "the" most efficient means, the sentence should be written that way (and be cited). — Preceding unsigned comment added by Kajabla (talk • contribs) 20:41, 11 July 2016 (UTC)

The Power to Overcome Aerodynamic Force Equation is Wrong
Does anyone read this? Tell me I'm wrong if you disagree. I don't want to make the change to the main article myself, because I'm admittedly inexperienced with editing Wikipedia pages. I'll come back and change it myself eventually, if no one else will. In the meantime, readers are being misinformed!

This is my first experience actively participating with Wikipedia. Sorry if I don't follow proper conventions or etiquette. Let me know.

First a nit:

Technically, the longitudinal force due to wind resistance is the Axial force, not drag. A lot of sources define Drag as the force opposing motion, which is correct for an airplane, but confusing when applied to bicycles. Aircraft and boats move relative to a free stream fluid. We usually think of bicycles as moving relative to the surface they are rolling on. A more useful definition of drag is, the fluid force acting on a body in the direction of the fluid free stream I'm citing a wikipedia article that cites a book. I probably have a book with that definition. If someone wants the reference, ask me. In other words, Drag is the force in the direction of the relative wind. Axial force is the force in the longitudinal direction. The difference is minor for airplanes under normal conditions, where the angle of attack is typically less than 15 degrees (cos(15°) = 0.966, i.e. 3.4% difference). It can be very big difference for a slow moving bicycle in a strong cross wind. In the rare case that the relative wind is 90 degrees from the direction of travel, the drag force is exactly countered by side force friction between the tires and the road, resulting in zero energy loss (other than a very small increase in rolling resistance due to increased tire deformation), since there is zero relative motion in that direction. That being said, I recommend using the term Drag anyway, to avoid confusing non-rocket scientist types. I'll use the term Drag below.

The important part:

The Drag is correctly identified as 0.5 ρ Va^2. But Energy is the dot product of force and displacement. The force is applied by the tires to the ground, and the displacement is along ground. Power is the rate of displacement in direction the force is applied, i.e. in the direction of travel. More simply put, power is the dot product of force and velocity relative to what the force is applied to, in this case the ground. So,

Power = 0.5 (ρ Va^2)(V), where Va = the component of relative wind speed in the direction of travel, and V = speed relative to the ground.

A simple thought experiment will help: sit on a bike, or stand in a wind without moving. It's the same as holding a weight but not raising it; force, but no energy is required, hence no power applied. If you are a rider, try going 40 mph on level ground, with a 10 mph tail wind. If you can do it at all, you can't do it for long. It's much easier to ride at 20 mph in a 10 mph head wind. In both cases, the relative wind speed is 30 mph, so the drag force is the same. But the tailwind case requires twice as much power, because the ground speed is twice as much.

Metric Unit Whinogram

I say this only partly tongue in cheek:

As in most cases, I find metric units to be user unfriendly: 20 mph is a typical bicycle speed, 10 mph is pretty slow, 5 mph is where stability (balance) and reasonable pedal cadence become easy for most riders to maintain (assuming moderate or lower wind and grade, and typical gearing). 32.2 Km/h? 16.1? 8.05? Really? For power, 0 Hp is standing still, 1 Hp is "Omg! Someone call 911." I can produce 0.3 Hp for a long time, 0.4 Hp for a while, more than 1 Hp in a short sprint. Not many riders can push 1 Kw long enough to say "1 Kilowatt!" One Hp = 768 watts. Quick, what's 40% of 748 W? Can you calculate it in your head before you can no longer produce it with your legs?

However, I live in a world of metric weenies. Feel free to convert units if you are one of them: 64.4 Km/h, 16.1 Km/h and 32.2 Km/h. Use seconds if you are feeling particularly pure of heart: 17.9, 4.5, and 8.9 m/s. Check those conversions; I have no intuition in metric units, and I calculated them in my head at 30 mph. Quick, how fast is that in Km/s?

For you American metric weenies:

1) Shame on you!

2) Don't convert units. If I said I was going 11 Km/s in a 1.1 Km/s tail wind, and breathing hard. Would you assume I was going uphill, downhill, or lying? I'll make it easier: 40 Km/h, and 4 Km/h? I don't know either; I'd have to convert to mph, just like you.

If you live in France:

1) I'm sorry; there's nothing I can do about that. ;^J

2) No credit for getting the above question correct if you used the mixed units (Km/h) to figure it out. Km/h is not SI, after all.

RiderBill (talk) 17:44, 24 August 2016 (UTC)


 * Welcome to Wikipedia and thanks for pointing this out. I've put it in using the term apparent wind, which has its own article.


 * You could change air drag to air resistance, if you like.


 * Your comment on metric units not having round numbers is only valid if they are converted from round numbers in other units. In most places in the world one would for example use 30 km/h and not 18.6 mph. km/h is of course not SI, but very widely used, as are also degrees. Here are Wikipedia's recommendations: Manual_of_Style/Dates_and_numbers
 * Theosch (talk) 18:29, 18 March 2017 (UTC)

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