A kinesiologist would put a breathing mask over an exercising subject (horse on a treadmill) and measure the oxygen uptake, which would reveal its rate of caloric production (power). But what portion of the calories is doing mechanical work?
A subject burns carbs at 4 kcal/gm (or fat at 9 kcal/gm, or protein 4 kcal/gm) and oxygen, in yielding calories. Roughly speaking, the energy is divided between: 60% in body heat, 20% in metabolic processes, and 20% in muscular activity. But that last number, 20%, could vary between 15% and 25% amongst individuals. Oh! so some individuals are more energy efficient than others.
But that muscular activity is undertaken to yield or react to various mechanical processes: locomotion (that is what we want), inelastic deformation of the track, aerodynamic drag, and mechanical (inelastic) losses in the joints, tendons and ligaments. The portions of the aforementioned is not constant amongst athletes and their environment. Consider that a mushy track will waste energy, as would a poor running gate. But what is the portion resulting in locomotion? I would consider this analysis to be daunting.
Instead, I would solve this problem with a simplified analogue. I would model this as an object moving through a viscous medium where the only energy loss is through hydrodynamic drag. Consider moving a torpedo through water would be easy enough to model mathematically. Then, the parameters for drag loss needs to be calibrated to have the model match the reality of a horse; ie: to suggest their power relative to their track speed.
http://en.wikipedia.org/wiki/Drag_(physics )
Power:
P = Fd*d/t
Resisting force due to drag:
Fd = -0.5*rho*v^2*A*Cd
Legend:
d = Distance travelled.
t = Time of travel.
rho = Density of fluid.
v = Speed moving through the fluid.
A = Reference area of object.
Cd= Drag coefficient.
Parameters:
v = Real speed of horse.
A = Instead, I would use horse's weight.
Cd= Instead, I would use horse's height.
Calibration:
rho= This needs to an arbitrary value used to calibrate the model.
In the end, this simple model might be able to estimate the relative ratio of power between horses. It could not be used to calculate absolute power of those horses.