Physics Question - Hills on Treadmill

Your center of gravity doesn't move forward when you run on a treadmill with no incline, either.

Right, which is why I don't think that treadmill running with no incline is as good as regular running either.

I was wondering specifically about hills though... the center of gravity issue would seem to be more significant here than with flat treadmill running.

Scherbatsky wrote:

Right, which is why I don't think that treadmill running with no incline is as good as regular running either.

I was wondering specifically about hills though... the center of gravity issue would seem to be more significant here than with flat treadmill running.

The treadmill is providing force at an angle of the incline. This can be broken down into a straight backward force and a downward force. In order to not be thrown off the back (not move your center of mass), you need to provide a forward and upward force.

This sounds right to me. Trust me, I'm a doctor.

I think that's what my roommate is thinking as well. But something still seems amiss to me, like this is an oversimplification.

It's been a while since I've had basic physics, but let's think about the forces. On a road hill, I believe that your body (through your foot) is actually applying both a downward and backward force against the pavement. The pavement (through normal forces and static friction) pushes back.

This is complicated when on a treadmill. There is still a normal force pushing up (not down) and static friction "pushing" forward (not backward... you're the one "pushin" backward). But additionally the plane (treadmill) is moving.

This force analysis also doesn't even factor in how your leg muscles work in lifting your body up and propelling it forward. The same forces would apply to simply standing on a hill.

I don't think I completely botched anything there. But it just seems much more complicated than you're making it.

asda wrote:

The treadmill is providing force at an angle of the incline. This can be broken down into a straight backward force and a downward force. In order to not be thrown off the back (not move your center of mass), you need to provide a forward and upward force.

Actually no. The treadmill is providing a NET ZERO horizontal force and an UPWARD force. The only downward force is provided by gravity.

It is true that in order not to be thrown off the back you need to provide an UPWARD force. You also need to provide a BACKWARD force on the treadmill during pushoff but this is exactly balanced by a forward force that you impart upon landing. The equal and opposite forces provided by the treadmill on you are an upward force that exactly counteracts the gravitational force plus a net zero horizontal force that consists of a forward force on you during pushoff and a backward force on you during landing.

And to all those folks who think that treadmill running is magically different than "real" running, please go back to physics class. The only differences are the lack of wind resistance and the cushioning of the treadmill. The fact that you are not going anywhere and that you have no gain in elevation is completely irrelevant.

This has been a public service provided by physicists anonymous and will be completely wasted on the Letsrun crowd.

Typo correction

Scherbatsky's Roommate here.

Let's take a look at two scenarios (while disregarding wind resistance and running surface, as What the heck has already stated):

Scenario #1:

A runner is on a treadmill, running 8:00 pace at 6% grade. A camera situated on the side of the treadmill is filming the side profile of the runner as he/she runs. In order to remain in the camera's frame, the runner must apply forces on the treadmill belt to exactly counteract the treadmill's speed.

Scenario #2:

A runner runs at 8:00 pace up a hill with 6% grade. Adjacent to the runner is a car with a camera that is also moving up the hill at 8:00 pace and filming the side profile of the runner. Once again, in order to remain in the camera's frame, the runner must apply forces on the treadmill belt to exactly counteract the treadmills speed.

I do not see how Scenario #1 and Scenario #2 are any different (besides, once again for the record, that Scenario #1 has no wind resistance and a running surface which most certainly varies from what is seen on a hill). To me, the runner must perform the exact same amount of work to keep pace with the camera.

There is no other explanation.

Damn, the cut and paste got me.

Scenario #2:

A runner runs at 8:00 pace up a hill with 6% grade. Adjacent to the runner is a car with a camera that is also moving up the hill at 8:00 pace and filming the side profile of the runner. Once again, in order to remain in the camera's frame, the runner must apply forces on the HILL to exactly counteract the CAMERA/CAR'S SPEED UP THE HILL.

Scherbatsky's Roommate wrote:

Damn, the cut and paste got me.

Scenario #2:

A runner runs at 8:00 pace up a hill with 6% grade. Adjacent to the runner is a car with a camera that is also moving up the hill at 8:00 pace and filming the side profile of the runner. Once again, in order to remain in the camera's frame, the runner must apply forces on the HILL to exactly counteract the CAMERA/CAR'S SPEED UP THE HILL.

Scherbatsky's Roommate is correct.

OK, I take it back. There are some smart folks here on Letsrun.

Thanks my good man.

Any other takers?

I guess I'm not even 100% certain that I'm right at this point. I just don't see how I'm wrong. Someone please help show me that I'm wrong.

If treadmill running is "magically" the same as real running, why does the treadmill have a motor ?

You drive a car on a flat treadmill at 20 mph and record the car's mpg.

Incline the treadmill to 6 degrees and record the car's mpg.

Does the inclined car get better, worse, or the same mpg than the car on the flat treadmill? Why?

Isn't the simplest way to think of this is a treadmill and road running are the same, just a change of your frame of reference, to an extent (wind resistance, and things like that have already been mentioned). Physics doesn't magically change when you change your reference frame.

Bingo! This man boiled it down.

zip zero nada wrote:

Isn't the simplest way to think of this is a treadmill and road running are the same, just a change of your frame of reference, to an extent (wind resistance, and things like that have already been mentioned). Physics doesn't magically change when you change your reference frame.

No physics doesn't change. But you can't deny that your foot is being "pulled" backward to some extent by the treadmill, whether flat or on a grade. I remain unconvinced (yet) that road and treadmill running (particularly on hills) are the same.

To "What the heck," we described the same force analysis, but just stating conclusively that treadmill running and road running are the same doesn't do it for me.

And regarding whether treadmill = flat ground running, I think the question is obscured/minimal at typical running speeds. It seems though that the transition from walking to running occurs at a far faster speed on the treadmill. Walking on the treadmill at a slow pace requires virtually no effort as your foot is moved backwards by the treadmill, i.e. you don't move your c.g past your plant foot, instead the treadmill moves your plant foot behind your c.g.. I think that ideal treadmill form (if there were treadmill races) would incorporate greater ground contact time and a longer stride length to exploit the treadmill's aid.

no... your foot is not being pulled back... your body is... if you didn't apply a forward force your body would fly off the back

Scherbatsky wrote:

No physics doesn't change. But you can't deny that your foot is being "pulled" backward to some extent by the treadmill, whether flat or on a grade. I remain unconvinced (yet) that road and treadmill running (particularly on hills) are the same.

Roommate, you are exactly correct. The treadmill DOES indeed "pull" your foot back. But in order to stay on the treadmill, you have to exert a certain force on the treadmill. The faster the treadmill is going (and the faster the belt is "pulling" your foot back), the more force you have to exert to "keep up" and stay on the treadmill.

This equalizing force (the force it takes to equalize the backwards force of the treadmill on your foot) is the EXACT same magnitude as the force required go that same speed up an actual hill with the same percent grade (once again, disregarding air resistance and running surface).

LAWYERED.