# Convert running speed uphill to equivilent speed on flat

Given a certain running pace uphill, I want to be able to determine an equivalent pace running with no elevation change. Assumptions: similar effort in both cases (say for example running at 90% max heart rate), ignore wind, slope is constant for simplicity, ignore physiological and bio-mechanical factors, weight of the runner is 135 lbs if that matters.

Example: Elevation change +236 feet, distance traveled 1 mile, elapsed time 6 minutes 55 seconds. What could I theoretically run for 1 mile with no elevation change given the same effort?

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This is a question in exercise physiology and bio-mechanics, and I doubt that it can be answered by the usual crowd around here. Some of the issues that make biomechanics hard to treat in a physics framework are discussed in Why does holding something up cost energy while no work is being done?. Also What's the difference between running up a hill and running up an inclined treadmill? is also vaguely related. –  dmckee Feb 25 '11 at 16:51
I realize there are an infinite number of physiological and bio-mechanical factors involved here. I'm just interested from a pure physics point-of-view. Perhaps something like quantifying the amount of work being done? P.S. I've edit the question to reflect these assumptions. –  Keith G Feb 25 '11 at 19:46
The problem is that from a physics 101 perspective running on the level at constant speed required no work beyond pushing through the air, but if you've even run with the wind so that it feels still you know that there is still considerable effort required. So the bio-mechanics is a big part of the answer, and as yet we don't seem to have a contributor who's knowledgeable on the topic. –  dmckee Feb 25 '11 at 20:51
I agree completely withe @dmckee. As interesting as this is, the best way to solve it from a pure physics perspective is to ignore biomechanics - running at a constant speed (no acceleration) on a level surface for any distance equals 0 work done. (W = F * d = 0 * 1 mile = 0) –  EAMann Feb 25 '11 at 21:27

The easiest (and roughest) way to to do it would be to convert your running "work" into a VO2 score.

The American College of Sports Med's equation is

VO2= Resting Component + Horizontal Component + Vertical Component or VO2= 3.5 + (0.2 x Speed) + (0.9 x Speed x Elevation Gain)

So, using your example of 8.67 mph (speed in the equation is in meters per min)

3.5 + (.2*232.67) + (0.9*232.67*.045) = 59.5

Thus running on flat ground should give you a speed of 280 m/min or 10.44 mph 5 min 44 sec per mile

(I'm an exercise scientist, not a physicist)

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Let's hear it for practical heuristics. Hip! Hip! Hurrah! –  dmckee Feb 27 '11 at 18:26
BTW-- We have an earlier question that features running and might benefit from some expert comment (as us physics types are kinda stumbling around in the dark). –  dmckee Feb 27 '11 at 18:28

Although this is a biomechanics problem it is of interest to see what kind of physics model would apply here. So I shall take the simplest model of movement against friction and gravity. I shall discuss some more aspects of it after some basics.

I have translated some of the numbers into SI and also will introduce some notation.

D = 1609 meters - the distance

T = 415 seconds - the time

H = 71.93 ~ 72 meters - the height increase

$V_{av}$ = 3.877 meters per second - the average velocity

$\theta$ = 2.56 degrees - the elevation angle

$cos \theta = 0.999$ and $tan \theta = 0.045$ (the number used in the V02 formula)

So the most basic equivalent model is that of pushing a mass M against a friction force $\mu$ for a distance D: level ($\theta=0$) and inclined ($\theta$> 0).

The equation for moving level is:

Work Done = $WD^{level}$ = Friction $\times$ D = $\mu mgD$

For moving at incline $\theta$ is:

Work Done = $WD^{incline}$ = Friction $\times$ D + WD on gravity over H

$= \mu mgD cos \theta + mgD sin \theta$

$= mgD (\mu cos \theta + sin \theta)$

If we note that $cos \theta$ is appoximately 1 we can write:

$WD^{incline} = \mu mgD (1 + (\mu)^{-1} tan \theta) = WD^{level} \times \alpha^2$ say.

Thus moving a fixed distance against a common value of friction the extra work done in going upward is $\alpha^2 = (1 + (\mu)^{-1} tan \theta)$. This is greater than one (as expected) and in our approximation is work against gravity modified by the inverse coefficient of friction.

Now we come to the biomechanics. Can we obtain information about the speed increase in a runner (or any powered object)? The parameter we have available here is the coefficient of friction. We also need to assume some model as to how the extra available energy will be used. Let us assume that all of the extra energy is transfered into kinetic energy: so we assume a linear model (in absence of any more detailed model).

Work Done $\propto$ Kinetic Energy = $1/2 m V^2$

In our case all the extra energy $\alpha^2$ is being added to the energy for the plane giving an increase of velocity:

$V_{new}^2 = V_{av}^2(1 + (\mu)^{-1} tan \theta)$

and thus a proportional decrease of time:

$T_{new} = T (1 + (\mu)^{-1} tan \theta)^{-1/2}$

We know the value of $tan \theta$ but do not know $\mu$. Just to put in a first value for $\mu$ which corresponds to rubber on concrete ie 1 we get:

$T_{new} = T \sqrt(1+0.045)^{-1} = 415 \times 0.978 = 406$ seconds.

This is somewhat slower than the V02 formula and we have not really a valid model of movement because this assumes a "continuous dragging" but it is a first approximation. A different approximation is to consider rubber tyres rolling. The coefficient of friction is defined differently, but could be as low as 0.01. This ignores all the other friction effects in the vehicle however.

Inverting the formula and putting in the result from the V02 answer (344 seconds), to derive a value for $\mu$ we get:

$\mu = 0.045 / ((T/T_{new})^2 -1) = 0.045 / (1.455 - 1) = 0.1$

So this "generalised friction" model predicts a $\mu$ value smaller than most static friction cases and in the direction of a "rolling model".

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Nice approach. A low effective coefficient of friction is to be expected, human running---like most animal locomotion---is pretty efficient, but you should probably also attempt to treat air resistance as that is a big part of the effort if you get much beyond a jog. –  dmckee Feb 27 '11 at 18:31
Love this answer, but @TrainWithThom's answer is a little simpler and both methods yield similar results. –  Keith G Mar 3 '11 at 15:29

Ignoring wind and physiological factors, it requires no energy to run at constant pace with no elevation change. By conservation of energy, your energy has to go somewhere. The only place left (after ignoring wind and physiology) is your kinetic energy.

Under the same assumptions, running up hill does require energy. In this case your energy goes into potential energy.

You can make running up hill equivalent to running with no elevation change if you require that the running with no elevation change be under the condition of constant acceleration. Then you'll be equating the rate of change in potential energy (going up hill) with the rate of change in kinetic energy (accelerating at the same elevation).

But none of this is a very good model of human energy.

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