Work done running on an inclined treadmill

[jbriggs444]

Slow down. I was expecting to be contradicted, and rightly. Read my sentence again. I am stating, clearly I think, that my zero-work assumption is wrong.

Fair enough. The zero-work assumption is wrong.

@lewando is trying to direct us on a productive course. But rather than replace legs with wheels, I would prefer to idealize things differently and assume for the moment that we are dealing with massless legs. The legs have two external interfaces. One at the hips where they push upward. One at the feet where they push downward. If we wish to compute the net work done by the legs, we should examine both interfaces.

We can make the further simplifying assumption that there are, on average, no horizontal forces. Realistically there would be wind resistance, but we can reasonably neglect this (or put a fan in front of the treadmill to generate a headwind). Realistically, there is energy dissipation in the strike of foot on track. But I do not think that is the effect that that we are trying to concentrate on. So let us ignore that as well. In the absence of horizontal forces, we are concerned entirely with the upward force of leg on hip and the downward force of leg/foot on track. And we are concerned entirely with the vertical motion at those two interfaces.

We have two scenarios to compare.

In one case (climbing a hill), the hips are moving upward at a vertical velocity of +v and the feet, when planted on the ground, are not moving at all. The force at the hip is equal to the weight of the runner's body. Call that mg. The force at the feet is, on average, equal and opposite (recall that we are still under the simplifying assumption of massless legs). Call this scenario A.

In the other case (climbing a treadmill), the hips are stationary and the feet, when planted on the track, are moving downward. Their vertical velocity is -v. The force at the hip is still equal to the weight of the runner's body. The force at the feet is, on average, equal and opposite. Call this scenario B.

Let us tot up the work being done by the legs in scenario A. That's $mg \cdot v + (-mg) \cdot 0 = mgv$
Let us tot up the work being done by the legs in scenario B. That's $mg \cdot 0 + (-mg) \cdot (-v) = mgv$

Edit: A concern had been raised earlier that the motor on the treadmill was somehow contributing to the energy balance. However, that concern is misplaced. The work and energy associated with the motor and any friction in the bearings are not relevant. Those forces act between the treadmill's mechanism and its belt. They do not enter into a calculation of work done by the runner.

 

[PeroK]

Pick a frame of reference and stick with it.

From the ground frame, the treadmill's surface is moving. But it seems that you wish to adopt the belt frame. That's fine. But in that frame, the runner's legs are lifting his body exactly as if he were running up a fixed incline.

No. Because when you are on the treadmill it's gravity that is providing the force. The force on the treadmill is not being generated by the use of your muscles, but by gravity pushing your weight into the treadmill.

When you genuinely move uphill, your weight is supported, but each step you must move your entire mass against gravity by using your muscles.

 

[PeroK]

We can make the further simplifying assumption that there are, on average, no horizontal forces. Realistically there would be wind resistance, but we can reasonably neglect this (or put a fan in front of the treadmill to generate a headwind). Realistically, there is energy dissipation in the strike of foot on track. But I do not think that is the effect that that we are trying to concentrate on. So let us ignore that as well. In the absence of horizontal forces, we are concerned entirely with the upward force of leg on hip and the downward force of leg/foot on track. And we are concerned entirely with the vertical motion at those two interfaces.

We have two scenarios to compare.

In one case (climbing a hill), the hips are moving upward at a vertical velocity of +v and the feet, when planted on the ground, are not moving at all. The force at the hip is equal to the weight of the runner's body. Call that mg. The force at the feet is, on average, equal and opposite (recall that we are still under the simplifying assumption of massless legs). Call this scenario A.

In the other case (climbing a treadmill), the hips are stationary and the feet, when planted on the track, are moving downward. Their vertical velocity is -v. The force at the hip is still equal to the weight of the runner's body. The force at the feet is, on average, equal and opposite. Call this scenario B.

Let us tot up the work being done by the legs in scenario A. That's $mg \cdot v + (-mg) \cdot 0 = mgv$
Let us tot up the work being done by the legs in scenario B. That's $mg \cdot 0 + (-mg) \cdot (-v) = mgv$

Edit: A concern had been raised earlier that the motor on the treadmill was somehow contributing to the energy balance. However, that concern is misplaced. The work and energy associated with the motor and any friction in the bearings are not relevant. Those forces act between the treadmill's mechanism and its belt. They do not enter into a calculation of work done by the runner.

Let's take scenario C: you are standing on a treadmill that is moving down.

The work being done by the legs is still $mgv$. It's the same forces as before. You can see the error in this case. The leg muscles are doing no external work, as it's gravity that is providing the force. Your body is simply transmitting that force as it does when you are standing.

 

[jbriggs444]

No. Because when you are on the treadmill it's gravity that is providing the force. The force on the treadmill is not being generated by the use of your muscles, but by gravity pushing your weight into the treadmill.

Again, this is completely irrelevant. The work performed by your muscles is the work performed by your muscles.

 

[jbriggs444]

Let's take scenario C: you are standing on a treadmill that is moving down.

The work being done by the legs is still $mgv$. It's the same forces as before. You can see the error in this case. The leg muscles are doing no external work, as it's gravity that is providing the force. Your body is simply transmitting that force as it does when you are standing.

Do the math.

 

[PeroK]

Do the math.

Do the exercise!

 

[jbriggs444]

Do the exercise!

Do the math. The work being done by the legs is indeed zero and not mgv. See if you can find out why.

Hint: There are two interfaces.

 

[PeroK]

Again, this is completely irrelevant. The work performed by your muscles is the work performed by your muscles.

Your muscles do little or no work to stand up. Your body can produce a force of $mg$ downwards without any effort. It's the force over and above $mg$ that is provided by your muscles.

On a treadmill (or step machine), the additional force is to keep lifting your legs.

On a real hill the additional force is to lift your whole body.

If you go to the gym and experiment, you will see this immediately. The difference is fundamental.

 

[jbriggs444]

Your muscles do little or no work to stand up. Your body can produce a force of $mg$ downwards without any effort. It's the force over and above $mg$ that is provided by your muscles.

On a treadmill (or step machine), the additional force is to keep lifting your legs.

On a real hill the additional force is to lift your whole body.

If you go to the gym and experiment, you will see this immediately. The difference is fundamental.

Do the math. Here. I'll do it for you. There are two interfaces. Hips and feet. You are standing on a treadmill in motion.

Hips: $(mg) \cdot -v$ [Upward force on downward-moving hips]
Feet: $(-mg) \cdot -v$ [Downward force on downward-moving track]
Total: $0$

 

[russ_watters]

Completely irrelevant. The work done by feet on belt is the same regardless of what other forces act on the belt.

I'll add a +1, but elaborate:

Consider a zero friction inclined plane. It requires a constant force to hold stationary against gravity's tendency to accelerate the object (you). Now move your legs against that force, creating a distance moved(the moving treadmill). That's the work, and it is exactly the same as on a hill (neglecting wind).

You guys are letting the fact that the treadmill moves itself confuse you. In fact, the fact that the treadmill maintains a constant speed against your pushing on it means it is absorbing power from you, not powering itself when you are pushing on it. We recently had a thread about non-powered treadmills, where we discussed the fact that they required brakes.

 

[russ_watters]

The treadmill moves itself. There is no need to do any work to move the treadmill. To make it go faster, you press a button!

Let me try a different way: since the treadmill moves itself, you have to apply a force to it to keep from falling off the back of it. That force is larger the higher the slope. And that force, times the speed of treadmill, is the power you apply, and it is exactly the same as climbing a hill.

 

[PeroK]

Let me try a different way: since the treadmill moves itself, you have to apply a force to it to keep from falling off the back of it. That force is larger the higher the slope. And that force, times the speed of treadmill, is the power you apply, and it is exactly the same as climbing a hill.

Imagine a flat treadmill. You need to walk to stop going off the end of it, but once you have attained the required speed, you only need the biomechanical energy to keep walking.

On an inclined but stationary treadmill, friction keeps you from sliding down.

On an inclined moving treadmill, you need a combination of these: You need to walk. Friction still holds your weight in place, but your legs are pulled down and you need extra energy to keep lifting them up.

The evidence, however, is clear: you can do speeds and "height" gains on a treadmill that are impossible on a real hill.

 

[jbriggs444]

Yes, but most of that force is provided by gravity, not by your muscles. That's the difference.

The muscles go through the same motion either way. The notion that the force is "provided by gravity" is not well posed and is irrelevant.

Anyway, a force is only needed to accelerate; not to stay stationary.

A force is needed to resist gravity. If you prefer we can treat gravity as a fictitious force and adopt a free fall inertial frame, but that seems pointless.

Imagine a flat treadmill. The same applies. You need to walk to stop going off the end of it, but once you have attained the required speed, you only need the biomechanical energy to keep walking.

Yes. Walking or running on a flat treadmill is equivalent to walking or running on a springy sidewalk. However, walking up an incline requires force to maintain one's state of motion in spite of gravity.

On an inclined but stationary treadmill, friction keeps you from sliding down.

Yes, friction provides a force. But only as long as it is resisted.

On an inclined but stationary treadmill I would think that we are all in violent agreement that the leg muscles require no power. One could sit on a chair and achieve a similar effect.

On an inclined treadmill, you need to walk. Friction still holds your weight in place, but your legs are pulled down and you need extra enrgy to keep lifting them up.

Yes indeed. You need to power your legs in order to maintain the strain that allows the friction to exist. That's power required on the downstroke.

The evidence, however, is clear: you can do speeds and "height" gains on a treadmill that are impossible on a real hill.

Unfortunately, the proffered explanation for this "evidence" is just plain wrong.

 

[PeroK]

Unfortunately, the proffered explanation for this "evidence" is just plain wrong.

You dismiss the evidence too easily, by the way. Why not do the experiment yourself or get a friend to do it? You ought to be willing to test your theory by experiment.

a) How fast can you run up a hill?

b) How fast can you run up a moving treadmill of the same gradient?

You may deduce theoretically that a) and b) are equal, but you shouldn't be unwilling to put this to the test. You will, in fact, undoubtedly find that b) is far greater than a).

It may turn out that my evidence is "plain wrong", but you shouldn't dismiss evidence simply because it doesn't concur with your theorectical conclusions.

I can simplify my explanation (in the ground frame):

When the treadmill is stationary, there is a force of $F = mg \sin \theta$ acting down the treadmill. This is balanced by an equal and opposite friction force. Normal forces are of course balanced. There is no motion.

When the treadmill is moving, there is an additonal tangential force generated by the person's leg muscles of $f$ and the friction force increases to balance this. Again, the centre of gravity of the person remains constant. $f$ is required for walking at the required speed and to lift the legs/moving parts against gravity.

The work done/power by the person's muscles is $fv$. Where $v$ is the speed of the treadmill. There is no need for the muscles to assume the additional burden of providing the force $F$ when the treadmill is moving. $F$ continues to be provided by gravity.

 

[jbriggs444]

It may turn out that my evidence is "plain wrong"

I made no judgement about the evidence. It is the explanation that is incorrect.

 

[jbriggs444]

The work done/power by the person's muscles is $fv$. Where $v$ is the speed of the treadmill. There is no need for the muscles to assume the additional burden of providing the force $F$ when the treadmill is moving. $F$ continues to be provided by gravity.

This notion that a force of one object on another being "provided" by some other source is decidedly unphysical.

 

[FactChecker]

CORRECTION: I have been corrected about this. I no longer believe this is a good way to analyse work. Why? At the point of contact between foot and treadmill, work is being done (force times distance). If the man is not walking, gravity is doing the work and his mass loses altitude. If the man is walking, the man is doing the work and his mass altitude remains constant.

Using anything except the official definition of "work" opens a can of worms. But you can still talk about expending energy while no "work" is officially being done. Your zero-energy path on an inclined treadmill is the downward trajectory you would take while standing still without walking on the moving treadmill. By walking, you expend the same amount of energy to stay in place as if you were climbing a hill. No work is being done, but energy is being expended.

The calculation of the difference in potential energy between the zero-energy trajectory and the "walk to stay in place" is easy to do.

 

[A.T.]

Your muscles do little or no work to stand up. Your body can produce a force of $mg$ downwards without any effort. It's the force over and above $mg$ that is provided by your muscles.

On a treadmill (or step machine), the additional force is to keep lifting your legs.

On a real hill the additional force is to lift your whole body.

If you go to the gym and experiment, you will see this immediately. The difference is fundamental.

When the treadmill runs at constant speed, the only difference is air resistance. Just analyse the treadmill from the inertial frame, where the upper belt surface is at rest. Here you continuously move upwards, just as you would on a real hill.

 

[russ_watters]

On an inclined moving treadmill, you need a combination of these: You need to walk. Friction still holds your weight in place, but your legs are pulled down and you need extra energy to keep lifting them up.

Ok...so that's the work, right?

 

[FactChecker]

My two cents:
Stopping something from falling (either free fall or any other downward trajectory) can not be considered work. It would not be possible to say that a person on an inclined treadmill is doing work while at the same time saying that a table that keeps a book from falling is not doing work. The standard definition of work (force times distance) can not be changed without opening a can of worms.

But you can say that the man on the treadmill is expending energy and the table holding up a book is not. That can be done easily without any tricky definitions or controversy.

 

[PeroK]

When the treadmill runs at constant speed, the only difference is air resistance. Just analyse the treadmill from the inertial frame, where the upper belt surface is at rest. Here you continuously move upwards, just as you would on a real hill.

Yes, but so does everything else. The person standing next to the treadmill is also continuously moving upwards in this frame. Are they moving just as they would on a real hill?

 

[PeroK]

Ok...so that's the work, right?

Yes, the work is only the work needed to keep moving the legs up and down. It's not the work needed to move your whole body mass continuously against gravity.

A treadmill, muscularly, feels like fast stepping only - not much harder than a flat treadmill. It doesn't feel like walking fast uphill. In fact, as already mentioned, the speeds I can do comfortably on a treadmill are practically impossible on a real hill.

There is an aspect of being on the treadmill that I don't understand and can't explain in terms of forces.

Is it simply that the operation of the treadmill's internal engine prevents an analysis using free-body diagrams of the person and the treadmill?

 

[A.T.]

When the treadmill runs at constant speed, the only difference is air resistance. Just analyse the treadmill from the inertial frame, where the upper belt surface is at rest. Here you continuously move upwards, just as you would on a real hill.

Yes, but so does everything else. The person standing next to the treadmill is also continuously moving upwards in this frame.

The person standing next to the treadmill is irrelevant, because it is not interacting with you.

Only the upper belt surface interacts with you. Your gain in potential energy in the inertial rest frame of the supporting surface is the same in both scenarios, and so is the work your muscles have to do (ignoring air drag, psychological effects etc.).

 

[PeroK]

Here's a link that backs up what you guys are saying:

https://physics.stackexchange.com/q...ng-up-a-hill-and-running-up-an-inclined-tread

And yet, there is a massive difference between the two - despite the physics concluding that the runners are making the same biomechanical motions, my experience is clear that the biomechanics are very different in the two cases.

Also, there's a new descending step machine at the gym. It's like a flight of stairs that continuously moves downwards. I've used it a couple of times. It seems to me that you can climb these steps simply by an appropriate movement of the legs. The standing leg does nothing except straighten as one step descends and the free leg comes up to the next step. In this way, you can climb the steps without any sensation of moving your body weight.

The raw physics would be the same as the treadmill: just the same as going up a real flight of stairs. But, by "clever" movements of the legs you avoid the configuration of forces that are being analysed. I wonder if something similar is happening on the treadmill, where there is certainly the sensation (perhaps illusion) of really only moving your legs and not having to move your whole mass at any time?

If, on the other hand, you allow the steps to descend, then ascend a step quickly that more accurately mimics going up stairs.

But, again, in any case, I can set these steps to a speed (about 30m of ascent a minute) that would be unsustainable on a real set of stairs.

Perhaps, alternatively, these machines simply allow you to spread the load and the overall work is the same, but it's easier to sustain because it's all evened out.

I must admit I'm still puzzled about why it is so easy to run up a treadmill way faster than I could ever do on a real hill.

 

[A.T.]

And yet, there is a massive difference between the two

Do you know of any objective proof for this massive difference? Like measurements of used oxygen, etc?

Aside from air drag, you have different visual queues on a treadmill. This could explain some difference, but hardly a massive one.

 

[PeroK]

Do you know of any objective proof for this massive difference? Like measurements of used oxygen, etc?

Aside from air drag, you have different visual queues on a treadmill. This could explain some difference, but hardly a massive one.

Just raw speed. I can do over 6 km/h at 15° on a treadmill, which is at least 1,500m of ascent an hour. And this is just a moderate work out. Next time I'll see just how fast I can go (I just walk fast, but I can try running!).

But, sustained 1,500m of ascent an hour is elite mountain athlete, which I'm certainly not, fit though I am. See my previous link about the "vertical km" in 30mins. Or here:

http://www.irunfar.com/2012/04/the-30-minute-kilometer-a-look-at-the-vertical-kilometer-record.html

I could just about (not quite) do a vertical km in 30mins on a treadmill. Even at my normal moderately hard workout pace it's only 40 mins. On a real hill it would take me at least 75 minutes at lung-busting speed. And, to be honest, I've never tried to go that fast. It's hard enough at 750m of ascent an hour, which is the fastest I've ever done; and harder than 1,500m of ascent on a treadmill.

There's also the lack of engagement of my upper legs on the treadmill. It's really all calves on the treadmill. On a real hill you need to power up using the whole leg. I tend to use the bike or a step machine to get the upper legs working.

 

[A.T.]

Perhaps, alternatively, these machines simply allow you to spread the load and the overall work is the same, but it's easier to sustain because it's all evened out.

Yes, if the slope of the real hill varies a lot, that could explain why it's more exhausting.

 

[jbriggs444]

Yes, the work is only the work needed to keep moving the legs up and down. It's not the work needed to move your whole body mass continuously against gravity.

Again, this is incorrect. The effort to pull the legs upward is not the only work being done by the hip and knee extensors.

 

[A.T.]

Again, this is incorrect. The effort to pull the legs upward is not the only work being done by the hip and knee extensors.

I think the simplest way to avoid such confusions is to look at both scenarios from the rest frame of the support surface. Here the work done on the surface is zero, so all the gain in potential energy comes from muscles.

 

[PeroK]

Okay, so I went to the gym and checked this out. First, I have to admit that @jbriggs444 (and the rest of you) were right all along. The raw physical work is the same.

Also, the angle is 15%, not 15°, which was a major blunder on my part. And makes a big difference to my figures.

Nevertheless, I can do speeds on the treadmill that I cannot do on a real hill. To explain this I did note, however, three major biomechanical differences:

The treadmill greatly increases your stride length. By the time your foot moves forward the treadmill has moved and this was increasing my stride length from about 0.9m to 1.2m. This partly explains the extra horizontal speed that can be attained.

As you push with your foot, your foot is being drawn back towards your body. This allows you to push largely from directly below you. On a real hill, a lot of the work is done by the foot in front of you. This also partly explains why the treadmill focuses on your calves. It's impossible to engage your upper legs fully, because your foot is being sucked under you. This has pros and cons I would imagine. Perhaps pros for speed and cons for endurance.

There is a definite and significant spring in the surface. On the treadmill you are definitely getting some spring out of it and not losing all your energy at every stride. It's very different from walking on a road surface.

So, it's clearly easier on the body to run/walk on a treadmill (both on the flat and on an incline) but that is all biomechanical.