Did Usain Bolt really sustain 11 horsepower of exerted power during the 100 m sprint?

I've just come across a WP article which says: "The Jamaican sprinter Usain Bolt produced a maximum of 3.5 hp (2.6 kW) 0.89 seconds into his 9.58 second 100-metre (109.4 yd) dash"

Out of curiosity, I decided to figure out the math behind this. While doing so I found this very detailed paper exactly showing how this was all calculated.

The paper states that the effective work Bolt has put into motion was 6.36 kJ and that the peak power was 2619.5 W for a very brief period in the first second of the sprint. The paper also mentions that Bolt average horizontal force during the sprint was 815.7 N which makes the total work done during the 100 meters race equal to 81.58 kJ. The paper concludes that "This means that from the total energy that Bolt develops, only 7.79% is used to achieve the motion, while 92.21% is absorbed by the drag; that is, 75.22 kJ are dissipated by the drag, which is an incredible amount of lost energy"

Now if we take the figure of 81.58 kJ exerted in 9.58 s, that would give us ~8500 W of average power during the sprint which is a little bit more than 11 hp. So my question now is: can we safely say that Bolt average power during the 9.58 s sprint was 11 hp?

In other words, would an 11 hp engine say in a motorcycle of similar physical characteristics (total mass, drag coefficient..etc) to those of Bolt achieve the same 100 meters in 9.58 s?

• Nice question. I guess the problem would be to reproduce the same drag coefficient. I would say that wheels based motion is largely more efficient than legs based motion. Maybe this would be another interesting question actually. – lcv Mar 17 at 20:54

This paper estimates Bolt's peak power using a model. One can also measure human power output directly. There are two common ways that this is done. One is to measure the actual mechanical work performed, e.g., on a stationary bicycle. The other is to measure oxygen consumption. In a sprint, the body is using anaerobic metabolism, so the power output is considerably greater than the aerobic capacity. Some measurements of anaerobic power output are given in Margaria et al. in 1963. The Margaria paper is paywalled, but some of their data are shown in di Prampero 1999, which I was able to find online. The highest rates of energy output they record, for athletes, are about 370 cal/kg.min, which for an 86 kg body mass (=Usain Bolt's) converts to 2.2 kW or about 3 hp. (About half of this seems to be supplied by aerobic metabolism.)

In the Gomez paper, figure 3 is clearly showing the net mechanical power, which doesn't include work done against internal dissipative forces. (This is why it tapers off to zero in the second half of the race.) The peak value of 2.6 kW seems very plausible, since it's on the same order of magnitude as Margaria's biggest values, Bolt is the greatest sprinter who ever lived, and Margaria's data probably extended over longer times.

The figure of 8.5 kW that the OP estimates seems to be an estimate of the total rate of energy consumption, including work done against internal dissipative forces, which are mocked up in Gomez's model using a drag force. The ratio (2.6 kW)/(8.5 kW)=0.3 represents the efficiency of Bolt's muscles.

So if it seems counterintuitive that Bolt is dissipating power at 8.5 kW=11 hp, keep in mind that this comparison is not being done in a way that is fair to the horse. The horsepower unit was defined using the net mechanical work of a typical horse, not the horse's total energy expenditure including energy evolved directly into body heat. (And the horses were tested pulling a plow for hour after hour, not in short bursts of anaerobic metabolism.)

Margaria et al., "Kinetics and mechanism of oxygen debt contraction in man," J. Appl. Physiol. 18 (1963) 371–377.

di Prampero and Ferretti, "The energetics of anaerobic muscle metabolism: a reappraisal of older and recent concepts," Respiration Physiology 118 (1999) 103–115

• +1 Very interesting. I would add that a lot of the dissipation has to do with the mechanics of the legs. I'm pretty sure a scooter with 11hp would be faster than Bolt even on a short 100mt track, which is mostly acceleration. I'd be curious to see if there are data on sprint on bicycles. Of course the bike gears should be optimized for 100mt. – lcv Mar 17 at 20:50
• I agree the paper is based on a model, but models should have some basis in reality to be credible. The paper's model does not: it takes a so-called "drag force" which is the sum of the aerodynamic drag (which is a rational thing to do) plus a hypothetical term proportional to velocity which has no physical interpretation, and is far higher than the realistic aero drag force. At best, this "drag force proportional to velocity" is a weird method of modelling the runner's internal metabolism, and it sort-of-works because for most of the race the velocity is close to constant anyway – alephzero Mar 17 at 21:29
• At best, this "drag force proportional to velocity" is a weird method of modelling the runner's internal metabolism, and it sort-of-works because for most of the race the velocity is close to constant anyway It seems to work very well in the sense that it fits the data well. It also makes a lot of sense, because we expect the drag force to be a smooth function of speed, so that it can be approximated with a Taylor series. But of course it's just a model, and it would be very naive to interpret this paper as anything definitive or precise. – Ben Crowell Mar 18 at 0:45
• "It seems to work very well in the sense that it fits the data well" What data do you mean? Firstly, what does Margaria's data represent? Mechanical or metabolism power? Secondly, According to Gomez's paper, 81.58 kJ were exerted in 100 meters which is double of what you can burn by running 100 meters and using the widely used 100 kcal/kilometer. So in both cases I can't see how the 81.58 kJ "fits the data well". – Abanob Ebrahim Mar 18 at 8:10
• I also found this: codybeals.com/2014/02/running-numbers-how-much-easier-are In the last section it uses a model that equate the force required for running to (1/4)mg which would be 1/4 of what Gomez used. I guess this along with the normal drag losses would be more representative of the actual mechanical power of Bolt and any additional metabolism power would have been used for the acceleration and deceleration of the legs, internal friction and body heat. – Abanob Ebrahim Mar 18 at 12:11

I would question some of the assumptions made here.

• The propulsive force exerted by the runner on the ground is assumed to be constant. If this was correct, a runner on a treadmill (with no aerodynamic drag) should be able to "run" much faster than on a track. This ignores the "internal work" required to accelerate and decelerate the legs - unless that is somehow assumed to be included in the "drag force" term in the equation of motion.

The assumption that the average force exerted on the ground seems very unrealistic. At the start of the race, the runner's starting blocks ensure the feet remain in contact with the ground for as long as possible to give the maximum acceleration, but during the rest of the race both feet are off the ground simultaneously for most of the time.

• The curve fitting to the position and speed graphs looks nice in the plots, until you look closely. The paper says the position was measured at 0.1s intervals, but the "measured data" speed seems to change at a bigger time resolution - more like 1.0s. During the initial acceleration, the steep slope of the curve hides the fact that the measured and fitted data are quite different - e.g. when the fitted curve is 4m/s, the measured data is about 5, which is a 20% error. The position data is equally bad for the first 1 or 2 seconds of the sprint.

• The division of the drag force into linear and quadratic components looks strange and is not given any physical explanation. From Table 1, the asymptotic velocity $$B = 12.2$$m/s. From the coefficients in Table 2, at this velocity the linear drag force is $$728$$N and the quadratic force $$89$$N. There is no physical explanation of what is generating the linear drag and why it is 8 times bigger than the quadratic aerodynamic drag.

To summarize, I don't believe any of this paper, unless the authors convince me it really is correct.

The paper compares the quadratic drag force with the standard formula for aerodynamic drag and concludes that it is the right order of magnitude. I we just ignore the linear drag force since there is no explanation of where it came from, and also ignore the assumption that the total force is constant, we would get a power output of about 1/9 the figure in the paper, which seems more believable compared with the measured power output of athletes on exercise machines, etc.

• The propulsive force exerted by the runner on the ground is assumed to be constant. If this was correct, a runner on a treadmill (with no aerodynamic drag) should be able to "run" much faster than on a track. I don't see any logical connection between the first sentence and the second. – Ben Crowell Mar 17 at 13:47
• I actually have a different approach which somehow confirms what the author of the paper is trying to say. Doing quick search, I found that a figure of roughly 100 kcal burnt per 1 kilometer run is widely used. Therefore, it would be 10 kcal for 100 m or 41.8 kJ for the 100 m sprint and that is at normal running speed around 22 km/h. At this speed a runner can cover 100 m in 16 seconds which equals 2600 W or about 3.5 hp and this is a normal person. I could imagine that Bolt running at double this speed would certainly result in at least double the power which would be 7 hp. – Abanob Ebrahim Mar 17 at 15:02
• This is also a calculator confirming the above calculations: keisan.casio.com/exec/system/1350959101 – Abanob Ebrahim Mar 17 at 15:04
• @BenCrowell If the runner could exert a constant (horizontal) force on the treadmill independent of speed he/she could accelerate (relative to the ground) independent of the speed of the treadmill, and therefore increase the speed of the treadmill indefinitely. This is clearly nonsense, because at some point all the runner's effort is spent accelerating and decelerating his/her legs, and not applying any net horizontal force through his/her feet. – alephzero Mar 17 at 16:24
• @AbanobEbrahim Your OP (and the paper) appears to be about the amount of mechanical work done by the runner against the "drag force". How much chemical energy (i.e. calories) the runner consumes is a completely different question. – alephzero Mar 17 at 16:34