Friday, April 2, 2010

running power: part 1

Last time on this subject, I introduced the idea of modeling the power required for running. Time for some equations.

Suppose I "run in place", bouncing up and down. When I'm in the air, my trajectory is determined by gravity, out of my control. So if I want to move my center of mass upwards, the more I move it up and down, the less often I can move it. "Cadence", which is the rate at which each foot contacts the ground, is directly related to the motion of my center of mass. So if I run at a particular cadence, and am in the air a given fraction α of the time, then the amount of energy it takes to raise my center of mass each bounce depends on the cadence and on α.

Consider the bouncing ball example. Flying upward under the influence of gravity with acceleration g, simple integration yields that the relationship between its maximum haight h (the vertical displacement of the center of mass) and the time t is:

h = ½ g t²,

where t is the time taken for the ball to reach its maximum height. The amount of potential energy gained during this trajectory is then:

ΔU = ½ M g² t²,

where M is total mass.

As a simple model, I'll assume the body bounces like a ball twice per stride, once on the left, the other time on the right, and spends a fraction α of that time in the air. Then cadence and t are related:

C = α / (4 t ),


t = α / (4 C ).

Assuming the power to raise the center of mass through potential difference ΔU comes from the muscles, and this must then be done twice per foot-stride, one gets:

Pcom = C M g² t²
= C M g² α² / ( 4 C
= M g² α² / 16 C.

This implies the faster the stride, the less the power required in raising and lowering the center-of-mass.

But then consider the power requirements of raising the feet. Assume the foot is raised to a height hfoot twice per foot stroke. Then assume the effective mass of the foot is Mfoot. This mass includes mass from the rest of the leg, in proportion to the amount each segment of the leg is raised. I assume the leg is raised the same amount independent of the cadence (this superposed on the center of mass displacement). Then the power requirements from raising and lowering the foot are:

Pfeet = 2 C Mfoot g hfoot.

To minimize this, you want as low a cadence as possible. Thus total power associated with bouncing around and raising and lowering your feet becomes:

Ptot = Pfeet + Pcom.

To minimize this is a simple exercise in differential calculus. The result:

optimal C = sqrt[ ( M / Mfoot ) g / 2 hfoot ] α / 4

Kenyan RunnersKenyan runners really kick up their feet on the back-stride

I laid down with my knee on the ground and my foot on the scale and got 6.8 kg. Adding on 320 g for my shoe and I get Mfoot = 7.1 kg. This is actually just what I want, because it calculates the effective weight of my calf and foot if I pivot at the knee. That's not a bad approximation for what happens when I kick while running. The thigh contributes as well, but I'll neglect that. Then I use for my body + clothing M = 57.6 kg. I'll guesstimate hfoot = 0.7 meters. If α = 0.7 (spending 70% of the time flying through the air), then I get an optimal running cadence of 94 strides/minute, which is what I generally run.

At the optimal cadence, with the optimal tradeoff between the two power components, the two power components are equal:
Pcom =
Pfeet =
g sqrt[ 2 Mfoot M hfoot g ] α / 4,

Thus the sum is:
Ptot =
g sqrt[Mfoot M hfoot g / 2] α,

which, using my parameters, yields an optimal power running at that cadence of 94 Hz of 218 watts.

So this simple model is really interesting: it says if I run along, my center of mass going ballistic around 70% of the time, each foot pivoting around its knee to a height of around 50 cm, I end up doing around 218 watts of power, independent of how fast I go. This fails to account for the additional work pushing my feet along the ground, propelling my body forward. So what am I missing?

What I'm missing is I arbitrarily estimated α and hfoot. If I run slower at the same cadence, lifting each foot up the same amount, I'll spend less time ballistic, and α will drop. If I can keep at least one foot on the ground at all times, then in principle I can keep my center-of-mass at a constant height, and I reduce the amount of power required. Additionally, at slower speed I don't need as long a stride, and I can kick up less, reducing power further by reducing hfoot. So really this model applies to only a single speed.

More next time.

No comments: