Treadmill Physics Questions (Similar to Airplane Question)

dingle wrote:

jtupper-ware wrote:

Contrary to one a previous poster said, you CANNOT simulate wind resistance on a TM by using a fan since you are not moving forward relative to the air flow of the fan...your body doesn't require any extra energy to "run into" the wind created by the fan. In fact, running into a fan almost completely eliminates the added metabolic cost of treadmill running as a result of increased cooling.

I'm pretty sure that if you put a treadmill in wind tunnel with the wind moving at the same speed as the treadmill, the effect of wind resistance would be very close to running outside in still air. The problem with a fan is that it does not move enough air to simulate wind across the entire surface of the body.

No. The resistance that you have to overcome in overground running is to move your mass against the still air.

In treadmill running, the situation is reversed: you are still and the air from the wind tunner is moving against you. The pressure on your chest from the backward-moving air would be the same as over-ground running but your chest is not moving anywhere because your horizontal velocity is zero. Therefore, no force to overcome.

Interesting result. It doesn't jive with my own personal experience of what it feels like to run on a 10% treadmill vs. a 10% hill, but I've not done any scientific study, that's for sure. Could you provide some references please? I'm willing to bet you get a high variability runner to runner; I'll also bet that one could develop a running style suited to the treadmill incline that would reduce the metabolic cost. Whether or not it costs more energy to run on an inclined treadmill compared to an actual hill doesn't change the fact that running on an inclined treadmill is not like running uphill, and your explanations only reinforce that contention. Another thought as to a possible difference is that the treadmill moves at constant velocity, while I'm willing to bet that if one analyzes running up a steep hill, the motion is not constant velocity. Just conjecture.I do object to the use of the term "re-elevate" though. Maybe "maintain elevation of" is what you mean?

jtupper-ware wrote:

No. The resistance that you have to overcome in overground running is to move your mass against the still air.

In treadmill running, the situation is reversed: you are still and the air from the wind tunner is moving against you. The pressure on your chest from the backward-moving air would be the same as over-ground running but your chest is not moving anywhere because your horizontal velocity is zero. Therefore, no force to overcome.

I don't understand this. I would even venture to say that you are just wrong. Your horizontal velocity is zero because you are overcoming the forces on you. These forces are the backwards force of the treadmill on your body and the backwards force of the wind on your body.

Are you suggesting that if I run on a treadmill against hurricane force winds, I don't have to run harder than if I run on a treadmill in still air?

jtupper-ware wrote:

No. The resistance that you have to overcome in overground running is to move your mass against the still air.

No, the resistance you have to overcome is the force of aerodynamic drag, caused by relative velocity between your body and the air. It doesn't matter which is moving, for a given velocity differential you have a given force, period.

In treadmill running, the situation is reversed: you are still and the air from the wind tunner is moving against you. The pressure on your chest from the backward-moving air would be the same as over-ground running but your chest is not moving anywhere because your horizontal velocity is zero. Therefore, no force to overcome.

You're contradicting yourself. If there is pressure on your chest, there is a force to overcome.

If having zero velocity means there is no force to overcome, why do hurricanes blow houses down?

Also, you might want to consider giving lectures to aerospace guys and formula 1 teams who use wind tunnels, as they are under the impression that moving the air acts the same as moving the vehicle.

lucKY2b wrote:

Sorry, but you are absolutely wrong on this. Change in gravitational potential energy IS mgy_f-mgy_i. If y_f = y_i, there is no net change in gravitational potential energy and you've not done work against gravity; end of story. Any additional energy expended while the treadmill is inclined is just because you do need to lift your knees a little higher, and you do cut off your stride on the back end a little, but that's it. If you were taught differently, you were misinformed.

You're confusing the net change in gravitational energy, and the added energy cost of having to run up/down hill.

It is true that, on the street, you can approximate the energy cost by the gravitational energy lost or gained. However, that doesn't mean that there isn't an energy cost on a treadmill with an incline. There is just not a net change because the treadmill motor is busy replacing the energy as you go.

On the street, the system is simple, it's just the runner, so you can get away with E=mgh. However, on the treadmill, there is another energy input into the system, so you also have to look at the energy from the motor.

Here are a few illustrative examples:

Energy (work) of a body moving a straight line is E=FD (force times distance). Since you don't go anywhere on a treadmill, are you going to assert that it requires ZERO energy input to run on a treadmill?

You could make the example more obvious by thinking about a car driving on a (larger) treadmill. Or, even better, a dyno! By your logic, because the car doesn't go anywhere, it requires zero energy. I may just be missing out, but I've never seen a car make a dyno pull without using any fuel. This is because, like running on the treadmill, there are other bodies in the system which need to be part of the energy balance.

The car example also works to illustrate the treadmill incline/decline. If you set the decline high enough, the car (or the runner) would just roll right off the treadmill (decreasing potential energy). Turning the belt on doesn't change this, it just balances out the potential energy loss with the input from the motor to result in a net static energy (but the car still acts, feels, and behaves as though it is going downhill).

Finally, I encourage you to set up a rope over a set of pulleys and a motor, so that you can have it move down at the speed at which you can climb the rope. By your logic, it will take zero energy to climb it forever, since you never go up, and thus there is no energy cost. I suspect you will find this is not the case. ;)

jtupper-ware wrote:

No. The resistance that you have to overcome in overground running is to move your mass against the still air.

In treadmill running, the situation is reversed: you are still and the air from the wind tunner is moving against you. The pressure on your chest from the backward-moving air would be the same as over-ground running but your chest is not moving anywhere because your horizontal velocity is zero. Therefore, no force to overcome.

Unbelievable. Do you have any comprehension of basic physics whatsoever? Any at all?

Or perhaps you are just trolling...in which case you win.

I think one thing that always messes people up in these debates is that they don't consider that the energy that we expend in distance running virtually entirely takes place at constant velocity.

If we assume a flat run, and a constant pace, then if we consider the runner as a simple mass, the only opposing force would be wind resistance, which is fairly negligible in terms of relative energy costs at paces slower than 12mph.

In other words, it is not the velocity per se that causes us to lose energy. Looked at from the perspective of a simple mass as you would in a standard force-body diagram, moving at 12mph would cost no more energy than moving at 4mph.

This tells us that it is the body movements required to maintain a velocity in distance running that cost energy: faster turnover, longer stride, etc. These body movements the same on the treadmill and on the roads, which is why the energy costs are the same.

QED

OP: you are whiny bitch. who complains about not having enough inertia and thus having a harder time turning?? haha ths is pathetic, stop overanalyzing shit and just run.

I donts know any of that physics stuff but I do knows the following about running on treadsmills

running at 0 incline puts stress on quads and eventually knee pain will occur

running indoor at 72° (average temp of fitness places) with no air circulation is similar to running outside at 85°

running at 12% incline is freaking hard

Mea Culpa.

It was an argument I had a long time ago as an undergraduate with my buddy (who is no slouch, as he did go on and win a Presidential Young Investigators award studying Plasma Physics at MIT) and he convinced me it was so. I have let it sit all these years untested. But you are right, and with fresh eyes, it is very simple (embarassingly simple, really). To stand on an incline requires mgsin(theta) of friction force along the incline. This force through the distance d you move along the treadmill is thus mgdsin(theta) which is mgh. As you pointed out, the net work done to your body is zero, but that is because the belt is constantly trying to remove energy from you that you must replace.

Thanks for cajoling me into straightening out this erroneous "idee fixe".

HUGE thumbs-up to you for this post! I think we had a reasonable discussion, and it was nice to think about basic physics outside of work. Have a beer, friend!

lucKY2b wrote:

Mea Culpa.

It was an argument I had a long time ago as an undergraduate with my buddy (who is no slouch, as he did go on and win a Presidential Young Investigators award studying Plasma Physics at MIT) and he convinced me it was so. I have let it sit all these years untested. But you are right, and with fresh eyes, it is very simple (embarassingly simple, really). To stand on an incline requires mgsin(theta) of friction force along the incline. This force through the distance d you move along the treadmill is thus mgdsin(theta) which is mgh. As you pointed out, the net work done to your body is zero, but that is because the belt is constantly trying to remove energy from you that you must replace.

Thanks for cajoling me into straightening out this erroneous "idee fixe".

Ahhh, a sure sign of winter is the annual return of the treadmill physics thread.

It's a holiday tradition.

Last week was finals week so I was not wasting time on Let's Run.

I still have to finish grades but here are some actual data that are relevant to this topic.

http://spot.colorado.edu/~kram/ftm.pdf

I don't have a pdf copy of the an Ingen Schenau paper (ref 23 in our paper) to share. The Bassett paper is also quite relevant to the mtabolic cost discussion.

Thanks for the link Rodger.

Only having skimmed the paper, I can't offer any meaningful substantive comments relating thereto, so I will accept its conclusions arguendo.

The discussion section states that: "Locomotion on a treadmill with inadequate

power or momentum (i.e., no flywheel) does

indeed differ from overground locomotion. However, on

a treadmill with an adequate motor and flywheel,

where the belt speed does not vary, the kinematics (21),

ground-reaction forces (19), and metabolic cost (2) of

locomotion are nearly indistinguishable from overground

locomotion."

Can you provide more information on HOW locomotion on an inadequate treadmill differs from overground locomotion?

I assume that there would be a greater metabolic cost on the inadequate treadmill, having something to do with the belt decelerating on footfall, and having to be accelerated rearward with each stride.

If true, that could account for my slower speed on the treadmill for the same perceived effort. I don't think that cooling effects play any role for me personally: I have no problem running in 90-100 degrees while staying hydrated; and over 20 mins, with good hydration indoors in a 65-degree environment, I don't believe there is sufficient time for any progressive dehydration or overheating to occur.

Also, would the effect be increased with increasing runner mass, all other things being equal? I have a 200-lb friend who has a heck of a time on our treadmills, and who runs very much better outdoors.

Mr Fluid Dynamics PhD wrote: On a treadmill, only non-conservative forces are attributed to energy expenditure, namely friction in the body and the energy to pump your blood at an elevated rate.

If this is true, I fully expect a bill in the next congress to stop it. All non-conservative expenditures most be eliminated.

I'm guessing, but I think in terms of pure force and not efficiency and form, that an inadequate treadmill would reduce metabolic cost. The inadequate treadmill would slow with each step, mimicking a slower pace than intended.

The slippage would certainly mess with your form and in the end make running very inefficient. This might trump the reduced forces you need to apply and make the net effect to have higher metabolic forces.

Just my speculation.

Okay, I might as well try to sound smart like the rest of you. In the block analogy, the force that is acting on the block is moving in the opposite direction than it would if you were pulling it on the ground. You are supplying the force for both you and the block. The only difference is that the surface is acting on the block in the opposite direction with an equal force as that which is holding it in place: you.

If you drop a ball on a train, it doesn't move relative to the train. If you drop a ball on a treadmill, it will bounce away from the treadmill because the belt itself is not moving relative to the ground like a train is. If you were standing on a very long treadmill moving with it and bounced the ball, it would not move relative to the surface. The only difference in this case is the length of the moving path. As for wind resistance, the environment of the train is also moving at the same velocity as the train, but in a treadmill, the path is moving while the air is not, like running on the roof of a train.

What I'm interested in, however, is how some people, like one earlier poster, have noticed that they can run significantly faster outdoors than they can on a treadmill. I have also noticed it seems harder on a treadmill. Why is this?

Hmmm...

These non-laboratory-grade treadmills no doubt measure distance travelled simply as a mathematical function of the programmed speed, and not as a function of the distance actually travelled by the belt.

Does anybody know if that is true?

Going up to the treadmill now, am going to try to run @ 10mph for 20 mins and see what happens...

The treadmill I used for some years, Cardionics, measured distance by a photo cell counting the revolutions of the motor. So the accumulated distance displayed had nothing to do with the set speed.