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Treadmill Equivalent To Reality ?

Posted on 2012-03-12
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Last Modified: 2012-03-12
Suppose I use a treadmill set to 3 mph, with an incline of 15 degrees for 10 minutes.
I then go to a 15 degree hill and walk for 10 minutes at 3 mph.
Are the two scenarios above roughly equivalent in terms of exercise and energy expended?

What I'm really wondering is, if the ascent (i.e. the height I have to lift my body weight) on the treadmill is the same as on the road. It's supposed to be, but is it really ?
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Question by:Eirman
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6 Comments
 
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Expert Comment

by:d-glitch
ID: 37709844
As far as gravitational potential energy is concerned, yes.

Assume your treadmill is 0.5 miles long and running at 3 mph.
If you start at the top and stand still for 10 minutes you will be at the bottom.
You will have to walk 0.5 miles up the 15 degree incline to get back to where you started.

Wind resistance and cooling are different on a treadmill than a real hill.
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Expert Comment

by:d-glitch
ID: 37710000
In both cases the vertical climb is

    (3 mi/hr) * (sin 15) * (10 min) * (1 hr/60 min) *( 5280 ft/mi)  =  683 feet
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by:Eirman
ID: 37710124
The treadmill always seems somewhat easier than real life to me.

Say you take a step forward with your right leg, then the motor brings your right leg back (doing some of the work ??) while you bring your left foot forward .... etc etc.

Is It an illusion that the treadmill seems to be doing some of the work for me ?
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LVL 27

Expert Comment

by:d-glitch
ID: 37710200
If you are hold on to the rails or console, then the treadmill is supporting some of your weight and helping with your balance.  This is energy that you don't have to expend.

If you don't touch the treadmill with anything but your feet, you are doing all the work.
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Accepted Solution

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d-glitch earned 500 total points
ID: 37710212
The treadmill is really moving you downhill at 3 mph.
You have to climb at the same rate to maintain your position.
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Author Closing Comment

by:Eirman
ID: 37710541
Thanks
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