Wednesday, 18 April 2012

Fun with physics



From Novacheck's 1998 article The Biomechanics of Running 
Here’s how.

Assume that every step you take requires a new source of kinetic energy from a push-off of either foot (don't forget that, by definition, running means both feet leave the ground for some instant). Kinetic energy is defined by

Where m is the mass of the runner (kg) and |v| is the absolute speed (m/s). The distance you travel is related to the speed of the takeoff:

Theta is the takeoff angle and g is the gravitational constant (= 9.81 m/s2). It takes a couple lines to derive the equation but it's not difficult (H.S. physics stuff). Here is a diagram of a one-step takeoff and landing (of the other foot). The straight line is the tangent angle at the moment of launch:

Theta is the takeoff angle of the runner's body with respect to the ground. 

Now, while you run let the number of steps you take per second be R, which is around 3 s-1 for most persons. Assuming the duration of the race is (seconds, s) and the distance raced is D (meters, m), the total number of steps taken in race is Rt. Your running speed is therefore D/t (equal to the cosine of takeoff speed v).

Hence the distance of each step can also be expressed by dividing the total race distance D by the total number of steps taken

The energy used per step is, after combining the previous three equations:

We want to know how much energy is consumed over the race distance, which is the energy used per step multiplied by the number of steps: 

We see that the energy used in a race is then 

Notice the canceling out the Rt factor in the numerator and denominator. If we assume the minimum amount of energy is used for a run, the theta is 45 degrees (which isn't the actual takeoff angle, but we'll ignore that for now) where the sine function is unity.

How much energy does it cost to run a kilometer under these assumptions? 

How did we do? According to a 1977 article by Mayhew.
The caloric cost of 0.97 Kcal kg-1 km-1 was in close agreement with values found in the literature and was independent of running speed.
This is not too bad for a back-of-the-envelope calculation. It also goes some way to explaining why easy mileage is no different than hard mileage in terms of basic fitness; either way you burn the same number of calories (if distance D is the same).

This equation implicitly shows that the rate of energy production (power, or Calories consumed per second) is proportional to the speed a which you run.

Hence doubling your running speed doubles the body's power demands. For non-ultra distances (after which energy stores begin to matter) power production is your limiting factor in running fast. Good middle-to-long distance runners (800m - 21 km) generate power more effectively. Energy storage capabilities are less relevant.

Now to backtrack a little. The original assumption, i.e. that each step is independent of the previous, is mostly untrue. It's the job of the achilles tendon to transfer energy from one step to the next so I've neglected an essential body part. This is why we overestimated the total energy cost per kilometer. The takeoff angle of your foot is also closer to 17-25 degrees, not 45. This is because it is faster to make many small steps than a few big ones (just like relaying a baseball between fielders is faster than throwing a ball directly to home plate). Vertical energy is also more costly for the body to absorb/transfer than horizontal motion and why runners try to minimize their 'bounce'.

What is the actual takeoff angle? If we want to estimate this, we can solve for theta in the original launch-distance equations. Matching definitions of D,

For various turnover rates R and running speeds (= D/t) we have, in degrees:

See how faster turnover leads to smaller takeoff angles?

I'm going to end here. I tried using these angles in the energy equation but the numbers don't quite work. The problem lies in the energy equation, which does not account for many factors such as how R and v might be related, the wind speed (and its relation to v), and the spring-like energy transfer in tendons. Real running math is way more complex that what I've shown here, but it's a start.


  1. Okay, here's your biggest problem. You've calculated how much energy is needed to move a body over a distance given your posited method of locomotion. What you did NOT calculate is how much energy a human body expends moving as above, which is what Mayhew measured by measuring oxygen consumption. Humans are not 100% efficient. The generally accepted figure is 20 to 25% efficiency. 4 to 5 Calories expended for each Calorie converted into motion. Which means caloric expenditure would be 4 to 5 times higher than your calculation, which means your calculation is way off.

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  2. Ok, so let's multiply the value by 5. Then we get ~6 Cal/kg/km. For a 68kg runner like myself running a marathon, that would be about 17,000 calories burned in total. Is that closer to what you were expecting?

  3. No but that's what your formula implies. Your calculations were only for force applied against the ground over a distance. Mayhew's observations were of energy consumed by a human body generating that force against the ground. It takes 5 Calories expenditure to put 1 Calorie against the ground. Mayhew's measurements actually imply that it takes 0.2 Calories to move 1 kg of body weight 1 km while about 0.8 Calories per kg per km are lost to heat mostly.

    0.2 Calories per kg per km is the figure you should be comparing your calculations against.

  4. I fully agree that this a woefully incomplete picture. What's interesting is that the values are anything close to reality. I'm a little surprised a grade-school math equation could capture any of the truth of what our bodies face while running. Essentially what we agree on is that a correction factor would be a nice addition. At some level it feels like a cheat, but then again that's exactly what a fugacity coefficient does in atmospheric gas equations...