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Time-trials, body weight and allometric scaling in cycling

August 5th, 2011

If you want to be a great sprint cyclist, you need to be able produce enormous bursts of power — so being big and muscular helps. If you want to be able cycle up mountains, on the other hand, you need great relative power — power divided by body weight — since every extra pound is deadweight that you have to haul upwards. But what about the middle ground? How does weight affect your performance in, say, a flat 40-km time trial?

Studies dating back to the 1980s have suggested that you need to use “allometric” scaling of weight to get the best prediction of performance in a 40-km time trial. Start by performing a graded peak power output (PPO) test, which is basically like a VO2max test, and PPO is the average power maintained for the last minute before you reach failure. Your PPO is a great way to predict how you’ll do in a 16-km time trial. Divide PPO by your body weight, and you have a great predictor of how you’ll do in a mountain race. And the interesting part: divide PPO by your weight to the power of 0.32 and you’ll have a great prediction of how you’ll do in a 40-km time trial.

This idea was first proposed by David Swain back in 1987, but hasn’t been tested much — which is why a new study just posted online at the British Journal of Sports Medicine, from Rob Lamberts and his colleagues at the University of Cape Town, put it to the test with 45 trained male cyclists. Here are some of the key results:

It’s pretty clear that the bottom graph (power divided by weight to the power of 0.32) provides a much better fit to the data than power (top) or power divided by weight (middle). So this is a useful piece of data for performance monitoring. But left unanswered is the question: why 0.32? Is this just an empirical number that happens to capture the tradeoffs between having more muscle and carrying more weight in exercise lasting about an hour? Or is there some physical or physiological explanation?

  1. Richard Ayotte
    August 5th, 2011 at 16:49 | #1

    0.32 for a 40km time trial is interesting but now I’m curious to see the curve between a 1km and 300km time trial. Would it be smooth or would we see some kind of bump or crossover somewhere to indicate the point at which the body depends on different energy sources and how that affects the power output in large and small muscles.

  2. Phil Koop
    August 6th, 2011 at 11:51 | #2

    I thought you were going to tell us the answer, not ask us the question! Surely this David Swain must have explained why he chose 0.32?

    Anyway, its an interesting result because intuitively one would expect a number close to 2/3. This 0.32 is suspiciously close to 1/3; would PPO / weight^0.32 provide demonstrably better fit than PPO / height? What about just leg length rather than whole-body height?

  3. Arne Vestrheim
    August 6th, 2011 at 12:16 | #3

    I have read this study myself and was asking myself the same questions. First of all, the number 0.32 is awfully close to 1/3, and I was wondering if this is the real number. Again the power of 3 is related to formulas for volume of 3D objects, as the power of 2 is related to the surface. Of course the best explanatory variable for a flat 40 km tt is power output relative to frontal area. The frontal area is related to the height of the athlete, but the weight is also related to the height (in the power of 1/2 if BMI should be considered a fair measure). Is the height in the power of 2/3 therefore an even better explanatory variable for a time trial? Or will the frontal area increase significantly with higher weight?

    Since the study was done on well trained athletes, I guess the fat percentage of the group was rather low on the study group. However “well trained” is a rather wage concept since it is possible to have a relative VO2max of 65 with over 20 percent body fat (I am myself an example of this). What I am asking, will a low share of body fat help you in a time trial, or is the explanatory variable valid only for well trained athletes with a low precentage of body fat?

    I think this is interesting questions when preparing for a time trial. If the weight is truly an explanatory variable, manipulating the weight can be beneficial for the result. But if it’s just a variable related to the real explanatory variable height, you should focus entirely on your power output. It would therefore be of interest to know both the height and fat percentage of the study examples to draw a better conclusion.

  4. Arne Vestrheim
    August 6th, 2011 at 12:54 | #4

    I see Phil has commented the same answer as me while I was writing. I’ve also read that athletes with long legs relative to height was better in TT due to lower aerodynamic drag of the upper body, http://www.4shared.com/document/ld1-rMKE/Characteristics_of_track_cycli.html.

  5. RH
    August 6th, 2011 at 17:47 | #5

    Wow. If the 0.32 is interpreted as a cubic root of weight, this suggests that you’re somehow held back by your linear dimension. That is your length, of, for hat matter, your shoe size, where yo would expect a correction for surface area, the square of your length, or a 2/3 power of your weight.

    Only thing I can think off is it is a correction for drag, taking into account that humans are probably more like cylinders with a constant circumference than like spheres or cubes, so the surface area scales with length rather than the cube of your length.

    This would mean that the 0.32 would also provide good fit for 16 km trial. Is there any infomation on that?

  6. alex
    August 7th, 2011 at 05:58 | #6

    “I thought you were going to tell us the answer, not ask us the question!”

    I was hoping it might be obvious, and someone would be able to explain it and save me the trouble of digging up the old papers! :)

    Okay, okay, I dug up a great review from 1994 by David Swain that goes into this stuff. It’s actually an excellent paper, and very clearly written — if anyone wants a copy, just send me an e-mail.

    First, some general points about the mass exponent. It will (generally) vary between 0 and 1. A mass exponent of 0 means you pay no penalty for additional mass, so bigger cyclists will be favoured (since they can use their extra mass the generate more power) — e.g. sprinting. A mass exponent of 1 means you pay the “maximal” penalty for mass, so smaller cyclists will be favoured — e.g. mountain climbing. The crossover point appears to be somewhere around 2/3, because that’s the exponent with which VO2max scales. (We traditionally express VO2max values as mL/min/kg, but that actually penalizes big athletes: it should be kg^2/3.)

    So this 0.32 exponent… It’s obviously very close to 1/3, and it’s tempting to interpret it that way — though, as several of you pointed out, we’d naively expect the exponent to be 2/3 in order to scale with surface area (since in a flat time trial, air resistance is the overwhelming factor). To make one thing clear, the origin of 0.32 is purely empirical: Swain took data, fit it logarithmically, and came up with 0.32.

    Swain offers two reasons for why the exponent is lower than 2/3. First, riding in the tuck position “hides” much of your trunk, distorting the geometric relationship between mass and surface area. One study explicitly measured surface area in tuck position as a function of mass, and found it scaled with an exponent of 0.55. That gets us partway to 0.32. The second factor is the mass of the bike: given roughly equal bikes, the bike will make up a smaller percentage of the “bike plus rider” mass for larger riders. In other words, larger riders pay a smaller penalty for the mass of their bikes, pushing the overall mass exponent lower. The result: an empirical exponent around 0.32 — though you’d expect it to shift slightly with changing bike technology and so on.

    While I’m on the topic, what about hills? Is the mass exponent really 1? Not quite. Now gravity rather than air resistance is the key factor, so we start with a naive value of 1. But once again, big cyclists pay less of a proportional penalty for the mass of their bikes, so the exponent drops. Though the data is pretty sparse (or at least it was back in 1995), the studies suggest that the oxygen cost of uphill cycling scales as mass to the 0.79 — which is greater than the 2/3 scaling of VO2max, so smaller cyclists are indeed favoured.

    So how much does all this matter? The estimate in the paper is that mass accounts for 11% of the variability in time trials and 17% in hilly stage racing.

    There are, of course, many assumptions in this analysis. All cyclists are assumed to be geometrically similar (so no accounting for trunk or limb proportional lengths), and we’re limited to aerobic energy systems (so no short sprints).

  7. RH
    August 8th, 2011 at 15:35 | #7

    Given the 0,55 scaling of drag an 2/3 scaling of VO2max and knowing what the outcome should be, appears to 0.32 make sense.

    D´Arcy Thompson discusses in “On growth and form” the relationship between speed and size in nature. He invokes “Froude´s law for steamship comparison”, which states that the square root of the velocity scales with weight divided by surface.

    If we correct this with the given exponents, this becomes the square root of weight^0,66 divided by weight^0,55, which is 0,33.

    D’arcy Thompson discusses the possibility that force production might scale with weight with a smaller exponent, because the surface area of the lung might restrict force production. He concludes that even then the bigger creature might be able to put on an better spurt on the end and that “For an analogous reason wise men know that in the ´Varsity boat race´ it is prudent and judicious to bet in the heavier crew.”

    I’d be very interested in the article.

  8. RH
    August 8th, 2011 at 17:22 | #8

    Sorry, “Froude´s law for steamship comparison”, states that the square of the velocity scales with weight divided by surface.

  1. August 5th, 2011 at 17:55 | #1