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: [1] 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. [2] [3] 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)

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)

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

$I=mr^{2}$.

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 $I={\tfrac {1}{2}}mr^{2}$ 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)

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)