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Clearly there will be differences like air resistance; I'm not interested in that. It seems like you're working against gravity when you're actually running in a way that you're not if you're on a treadmill, but on the other hand it seems like one should be able to take a piece of the treadmill's belt as an inertial reference point. What's going on here?

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6 Answers 6

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For me it is axiomatic that machine miles are easier than real miles, but let's analyze the situation.

Assume the runner maintains a constant velocity up the hill, or remains stationary in the frame of the gym on the treadmill. In both cases the runner's acceleration is zero, so we know that her legs must provide a constant force with upward magnitude $mg$, and the they have to do this against a surface passing by at an angle $\theta$ below the horizontal and moving with a velocity $v$.

The kinematics in the runners frame of reference look the same. This is not the cause of the difference in perceived difficulty.

I have always assumed that the difference in difficulty was two fold:

  • Wind resistance is not really negligible.
  • The treadmill presents a very uniform reliable surface and the runner need not lift her legs as high to insure non-tripping progress.

Also modern treadmill are designed to be relatively easy on the knees, and the accomplish this by having a slightly springy feeling which presumably returns some energy to the runner.

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From experience, I agree that treadmill miles are easier. Aside from the possibility of miscalibration, I have a guess that the belt slows down a bit when your foot impacts it, then picks back up the lost speed mostly while you're in the air. Hence, you're not pushed back at quite the speed the treadmill is set to. –  Mark Eichenlaub Dec 4 '10 at 22:54
@Mark: That's a good observation, to prevent it you'd have to really overbuild the whole thing which would drive the price up to the point that no one would buy it. –  dmckee Dec 4 '10 at 22:57
I'm inclined to believe this answer, but how does it take into account the conservation of energy? You'll have to excuse me if this is silly; I haven't taken physics since high school (and it's been a few years). –  aaron Dec 8 '10 at 9:44
@aaron: It is not a silly question at all, and the answer is wrapped up in (1) the difference between "work" meaning $W= \vec{f} \dot{} d\vec{x}$ and "work" meaning "man, this is tiring" (see Pavel's answer) and (2) the question of what happens if you just stop running: on the hill, you just stand there; but on the treadmill you come to a stop and stand on the belt and fly back into the gym, land on your butt (if you're lucky) and look really silly because the platform isn't static---the extra energy is expending to maintain your place in the gym despite the motion of the belt. –  dmckee Dec 8 '10 at 23:58
Couldn't the m of moving my body up a hill be different than the m of moving my legs underneath me on an incline treadmill? –  Dave Feb 2 '12 at 16:15
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The word 'difference' may be ambiguous, but let's look at the situation from several points of view.

Energy balance: Indeed, your potential energy does increase in case 1 and not in case 2. Muscles clearly perform the same work, so the energy must go somewhere? Yes, to the electric grid. The treadmill device's engine, to maintain constant velocity, will consume less electrical power to do so (or might even push energy back into the grid, in case of an efficient motor) because your legs are actually pulling it now downwards. If you do the math, you see it exactly compensates.

Muscle work: work, in thermodynamical sense, is not just F*dx. One has to take a machine and consider all interfaces. For example, a spring or a muscle have two ends, and dx in the formula is actually the difference between two paths. Muscle expansion/contraction will be the same, and so is the force. Therefore, they are doing the same work. This work is, the amount of chemical internal energy stored within the muscle converted to mechanical work.

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Assume that the hill and the treadmill have the same angle of elevation (are inclined identically), and that two identical persons A and B are running on them at the same speed $v$. Here the speed of the person B running on the treadmill is obviously zero with respect to the ground, but we will consider the speed of the treadmill's belt to be $-v$.

Let's assume that B and the treadmill are now in a truck which is running up the hill parallel to A, and with the same speed $v$ as A. The truck should be arranged so that B and the treadmill are not inclined while the truck is running up the hill. By our hypothesis, the speed of the upper part of the belt is zero with respect to the ground. This will not affect the effort done by the runner B, because the truck is moving with constant velocity. That is, there are no extra forces caused by inertia, because the truck doesn't accelerate, decelerate or change direction.

By looking now at both runners A and B we see that they are moving parallel with the same speed. They can even do the same moves in synchrony. The angle elevation is the same for both of them too. But B may even not realize that he is climbing a hill, he may think that he is in a room with no windows, which doesn't move. The conditions are identical. So, there is no difference between them.

If our intuition still saids that the treadmill guy is burning less calories, let's imagine that the road on which the person A is running up the hill is a very long treadmill belt. Imagine that underneath the belt there is a treadmill which is doing two things: is moving with the same speed $v$ towards the top of the hill, and the upper side of the belt is moving backwards with $-v$, so that the belt appear to be fixed with respect to the ground. From the outside, the belt doesn't move (and for runner A too). Now, it should be clear that there is no difference.

In the above I assumed that there is no wind, the treadmill and the runners are moving with uniform speed, there is no difference between the runners. I also assumed that gravity is not weaker towards the top of the hill.

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I think this argument is isomorphic with mine concerning what dosen't account for the perceived difference. I like the truck arrangement: that's clever. –  dmckee Dec 12 '10 at 18:50
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i get the impression this is something that doesn't need complicated physics to explain if you apply a quick common sense test first. if you step up a hill, you have to push the weight of your body upward with each step, or you do not continue moving forward. if you are on a treadmill, you may place your foot forward, but at the same time you would be (on a hill) pushing against the higher point on the ground to move yourself uphill, the treadmill is conveniently lowering your foot back down to the starting point, so you didn't have to actually push yourself uphill very much before you move on to the next step. and it just continues like this, with the treadmill continuously re-lowering your step part way before you ever have the chance to expend that energy you'd need to genuinely push yourself uphill. does anyone else see this?

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Let's estimate some of the contributions discussed:


We can compute the power spent gaining altitude.

$$ W_\text{grav} = mg \dot h = m g v \sin \theta \sim m g v \theta $$ for small angles, with some typical numbers $$ W_\text{grav} = ( 180 \text{ lbs} ) ( 9.8 \text{ m/s}^2 ) ( 1 \text{ mile} / 10 \text{ minutes}) ( 5 \text{ degrees} ) \sim 200 \text{ kcal/hr} $$ if I look up the energy burned running on the web, I get $800 \text{ kcal/hr}$.

So, it would seem, running a 10 minute mile on a 5 degree incline outside takes some 25% more work than on flat ground.

Air resistance

Next let's consider air resistance. We have for its work

$$ W_{\text{air}} = \frac{C_d}{2} \rho A v^3 $$ with typical numbers $$ W_{\text{air}} \sim (0.5) ( 1 \text{ kg/m}^3 ) ( 2 \text{ m} \times 0.5 \text{ m} ) * ( 1 \text{ mile} / 10 \text{ min} )^3 \sim 4 \text{ kcal/hr} $$

which is a much smaller effect. About 1/2 %

Non uniform surface

If we assume that running outside requires us to life our legs an extra 3 inches on average, the power contribution would be $$ W_{\text{rough}} = m_{\text{legs}} g h_{\text{extra}} f_{\text{stride}} $$ with some typical numbers $$ W_{\text{rough}} = ( 0.3 \times 180 \text{ lbs}) ( 9.8 \text{ m/s}^2 ) ( 3 \text{ inches } ) ( 2 / \text{ s}) \sim 30 \text{ kcal/hr} $$ which is larger than wind resistance but only a 4% increase over our base running number.

Springing of landing

What if the coefficient of restitution was different for the treadmill versus land, then everytime you impact you'd need to put in less energy springing off, its power saving should be $$ W_{\text{spring}} = (\Delta r) m g (\Delta h_{\text{center of mass}}) f $$ with some numbers $$ W_{\text{spring}} \sim ( 10 \% )( 180 \text{ lbs} )( 9.8 \text{ m/s}^2 )( 6 \text{ inch} ) \sim 20 \text{ kcal/hr} $$

which is a roughly 3% change.

Not sure what to make of these yet...

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Again, if this analysis held then tilting the bed of the treadmill wouldn't make it harder to run. Tilting the bed does make it harder to run. –  dmckee Jul 18 at 3:35
@dmckee You are probably more right than wrong. I've added estimates for the other points you discuss and removed my assertion. But... it seems like gravity has to make a contribution, because I would say running outside is more than 5% harder. –  alemi Jul 18 at 3:58
@dmckee Tilting the bed does make it harder. I never meant to argue against that, but I still want to believe actually gaining altitude should take more work than not. –  alemi Jul 18 at 4:01
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(Running up a treadmill) = (expend energy to keep feet moving at a constant speed) + (other effects)

(Running up a hill) = (expend energy to keep feet moving at a constant speed) + (energy to lift center of gravity by hill height) + (other effects)

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If this analysis was viable tilting the bed of the treadmill would not make the running any harder. That is not the case (try it), and the reason is you have to support the body while pushing against a surface that falls away from you. –  dmckee Dec 6 '10 at 17:21
Maybe you expend energy lifting your center of gravity and then release it (fall back) without recovering it causing a net expendeture. I can account for this in my balance by adding higher-order-terms to it. –  ja72 Dec 6 '10 at 18:17
Well, yes, but you can take that argument to the infinitesimal limit and recover a inertial frame for both cases. Check my answer for an argument based on the kinematics relative the ground as seen in the runner's frame of reference. –  dmckee Dec 6 '10 at 20:35
To explain this within the framework you proposed: You need to add another term to your first equation (expend energy to push down the treadmill). At first, you might think that you're not pushing down the treadmill (it's supposed to roll back on its own after all), but remember that if you weren't pushing it down than you wouldn't be sustained in place by it. –  Bruce Connor Dec 14 '10 at 4:52
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