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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.
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)))
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)
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)
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)
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)
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)
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)
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)
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)
The translational kinetic energy of an object in motion is:
Where is energy in joules, is mass in kg, and is velocity in meters per second. For a rotating mass (such as a wheel), the rotational kinetic energy is given by
where is the moment of inertia, is the angular velocity in radians per second, 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
The angular velocity is related to the translational velocity and the radius of the tire. As long as there is no slipping,
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:
Substituting for and , we get
The terms cancel, and we finally get
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."
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:
Coreyjbryant ( talk) 03:13, 4 April 2011 (UTC)
References
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)
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:
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?
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)
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.
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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
[1] 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
[2]. 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)
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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.
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)))
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)
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)
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)
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)
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)
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)
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)
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)
The translational kinetic energy of an object in motion is:
Where is energy in joules, is mass in kg, and is velocity in meters per second. For a rotating mass (such as a wheel), the rotational kinetic energy is given by
where is the moment of inertia, is the angular velocity in radians per second, 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
The angular velocity is related to the translational velocity and the radius of the tire. As long as there is no slipping,
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:
Substituting for and , we get
The terms cancel, and we finally get
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."
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:
Coreyjbryant ( talk) 03:13, 4 April 2011 (UTC)
References
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)
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:
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?
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)
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.
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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
[1] 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
[2]. 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)
References
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