Dear Wiggo said:
Merckx index said:
In that situation, a larger rider has a very slight advantage in wind resistance, depending on the gradient.
Could you explain the physics behind this please?
Because wind resistance ~= frontal area ~= size of the rider. You appear to be saying it's the reverse.
If we extrapolate, you're saying the smaller the rider, the more wind resistance they encounter at the same speed as a larger rider.
Maybe I just didn't get enough sleep but there is no law of physics that I can remember where increasing frontal area of a body reduces its wind resistance.
Re: Froome's power: they subsequently showed that the asymmetric rings matched the round rings perfectly, underestimating power in a MTB ride around the same circuit for a day.
You raise a good point, and now I see the confusion. Frontal area increases by a square law, whereas power increases with mass, or by a cube relationship. This means that as a rider increases in size, power increases faster than frontal area and therefore air resistance. In a flat ITT, air resistance is almost all that matters, so the larger rider has an advantage.
The same relationship exists on gradients, it's just that there the effect is usually overwhelmed by gravity, and thus power/weight is the key, rather than power/surface area. You're correct that if a larger rider and a smaller rider are moving at the same speed up a gradient, the larger rider will have more drag or air resistance to overcome. But the smaller rider has to put out a larger proportion of his total power to overcome his air resistance, because, again, his power/frontal area ratio is lower. If the two riders are climbing at the same speed, they are putting out about the same power/weight, but the smaller rider is doing it with less absolute power, and that less absolute power has to deal with proportionally more air resistance.
E.g., suppose the larger rider puts out 385W/70 kg = 5.5W/kg. The smaller rider puts out 330W/60 kg = 5.5 W/kg. So they climb at the same speed. The larger rider has a mass of 1.167 x that of the smaller rider, which means his frontal area is about 11% larger (take the cube root of 1.167 and square it). The larger rider thus has to overcome 11% more air resistance, but he does this with about 17% more power. So the amount of his power needed to overcome air resistance is less.
Let’s say, for the sake of an example, that at the speed the two riders are climbing, the smaller rider must use 10% of his power, or 33 watts, to overcome his air resistance. The larger rider must then use 33 x 1.11 = 36.3 watts to overcome his air resistance. This is only 9.4% of his total power. This means his power/weight ratio, expressed as power not devoted to overcoming air resistance, is actually greater than that of the smaller climber, and he will climb a little faster. Of course, if he does climb faster, the air resistance he needs to overcome will be a little more. But the point is, the larger rider effectively can climb as fast as the smaller rider with a slightly lower power/weight ratio.
Thus the larger rider has a slight advantage in this regard. Though it's not enough to overcome the power/weight relationship on steep climbs, on shallower climbs it may be significant. Thus in stage 19, TD was able to drop Aru on the closing finish, which had a reported gradient of about 4%. A gradient in that range is sort of in between a typical climbing finish and a flat stretch, and thus the larger rider's superior power/frontal area ratio can come into play.
It's an interesting exercise. Take a great ITT rider, say Cancellara in his prime, and a great climber, say Contador in his prime. On a flat surface, Cancellara wins (well, except in the 2009 TDF, it seems). On a usual climbing gradient, Contador is faster. But there is some intermediate gradient where they are about equal, because Contador's power/weight advantage is just equalled by Cancellara's power/frontal area advantage.