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)

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.

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)