Uphill treadmill question

[NetSquirrel]

Does it require more effort/energy to run "uphill" on a treadmill, or to run up an actual slope of the same inclination at the same speed?  (Air resistance can be ignored.)

There was a similar situation to this on a film that was on yesterday.  I'm mainly asking this question to settle a dispute that me and my friends had about which would be the more difficult to run up.  Not meaning to bias the question, but it's harder to run up an actual hill, right?  Along with the effort of running, there is the added force (and associated energy cost) needed to push oneself up against gravity - approximately an extra 736 joules per metre of height, for a 75kg person.  All a treadmill runner does is run horizontally but has to move their legs in an uphill kind of way, as if there was a hill there.  Who is right?  (If you could offer an explanation that goes easy on the scientific aspects of the problem, that would be perfect!)

[Loosehead]

IThe incline of the tread mill is equal to a continuous real hill of the same slope, they are the same, imagine the treadmill loop cut and spread out and multiplied by the number needed for the distance. You ARE tackling gravity in exactly the same way, if you weren't then it would not be any harder than on the flat.

"All a treadmill runner does is run horizontally but has to move their legs in an up hill kind of way"

What?? think about it , do you not move your legs in an "uphill sort of way" on a real hill.

So to test any theory, take it to extremes. Lets put the treadmill up against the wall, and run vertically .... mmm, just running horizontal moveing legs in an uphill sort of way, I don't think so.

[NetSquirrel]

Loosehead, your answer is very good, but there is one point I do not understand.

 

While running on a treadmill, your centre of mass does not move with respect to gravity.  Therefore, you gain no potential energy as a result of your "climb".  So, to not gain any potential energy in the form of height/altitude, you never had to put the extra energy in during your run, (by pushing harder with your legs, and so on).  The only extra energy required for the uphill treadmill run is that which comes from moving your legs around more to keep your place on the treadmill.

 

Assuming you, my friends, my brother and several other people who I asked this question to today are correct (damn, I'm starting to feel stupid now) this would suggest that running for a certain amount of time on the uphill treadmill would require just about the same amount of energy as running up a hill of equal inclination.  Slight problem: running up the hill increases the runner's potential energy, running on the spot does not.  No energy has been created or destroyed of course...so how do you explain the situation where two runners who have just expended the same amount of energy have ended up with obviously different potential energies at the end?

[Loosehead]

I think I see your problem. You talk of creating energy and potential energy. Both these terms are out of place in this argument. What we are talking about is expending energy, in the one case to stay in place in the other case to progress up a hill. The effort is equivalent.

As you say you do not move relative to the treadmill and on a hill you move up it. Now lets use relativity for a moment. The only reason you think you are moving and not the hill is because of the periphery, take that away and imagine if the hill was being move under your feet, forcing you to run to stay where you are. The hill is rasing you up against gravity.

Now I agree the treadmill is a different sensation but it does simulate an equivalent hill. No energy is created, energy is expended in order to stay still and avoid flying off the back. On an incline each step is higher than the preceeding one even if it subsequently drops to the level at which you are running.

Potential energy is applied to objects that are not at that time expending it, it is not appropriate in this argument. I'm not sure what you mean.

[thekraut]

sorry loosehead, but that relativity thing you are talking about is nonsense. if your movement on the treadmill was perfectly smooth (no small changes in "altitude") your chage in kinetic and potential energy would be zero no matter how fast the belt ran in relation to your body. so the only energy needed would be the enrgy to move your legs in an "uphill fashion" (and of course the heat energy produced)
BUT: your movement on the belt will never be perfectly smooth.
with every step you take you will lift your body a little bit up the belt and the go down again with the belt in order to stay centered. so it's these small gains of potential energy that add up to the feeling of walking uphill.
it's the same as the comparison between running and riding a bike. runnig includes all these little jumps and is therefore much more exhausting than riding a bike, where your body is always on the same level.

[Loosehead]

thekraut: Re read my answer. I did not use relativity in connection with the treadmill. Indeed I started my sentence about the treadmill with

"Now I agree the treadmill is a different...." I was

trying induce a different view point in the actual hill.  

Now instead of nit picking, have a go at answering the original question.

[thekraut]

Sorry Loosehead, didn't mean to nit pick, just wanted to explain why you are completely and utterly wrong. and while doing that I might have forgotten to explicitly answer the question.
so here it comes (again, in a way):
No matter what your viewpoint may be, the effort to run up a treadmill is higher than to run un a horizontal one but not as high as running up an actual hill. and thats in no way relative.

[Loosehead]

Absolutely classic, I'm wrong, tell me why. I used relativity to isolate the "on the hill" I wasn't attempting a dissertation. But you ignored that and went and tried merge it somehow with the treadmill mmmm.... GCSE's you can't beat em!

So "it's harder on a real hill", tells me all I need to know, your mis-understanding of basic physics is so vast I cannot begin to correct you.  

[thekraut]

funny that somebody who's talking about the energy that is needed to prevent acceleration can even spell "mis-understanding of basic physics"

[landie]

From personal experience I cxan tell you that the answer is, it does.

[Ozimandius]

You're doing work against the treadmill. When the treadmill is raised at angle theta to the horizontal, the vertical component of your weight acting along the treadmill belt is equal to W.sin(theta). When you multiply this force by the distance travelled along the belt (force times distance gives work done) this is exactly equivalent to the potential energy that you would have gained on a real hill.

[JohnDTC]

the air resistance encountered when running up a real hill would definately mean more energy is required assuming all other factors are constant. i dont know if that was the reasoning behind the argument for i just read.

[JohnDTC]

just realised u said ignore it. it would be significant over a long distance though.

[NetSquirrel]

Ozimandius, you have hit upon the main source of confusion I initially had with this question.  Work done is indeed equal to the magnitude of the resisting force multiplied by the distance travelled against that force.  E.g. raising a ball of weight 5 newtons a distance of 3 metres requires 15 joules of energy, essentially because 15 joules is the exact amount of potential energy that has been given to the ball.  On a hill, you push yourself against a resisting force (gravity).  In other words, your body is constantly gaining potential energy...and the energy must come from somewhere - your muscles, which have to work extra hard to give your body this energy. 

Another way of considering it is to imagine a radio-controlled toy car driving up a long plank of wood (at, say, 40� to horizontal).  The car climbs up the plank until its batteries have run down to the point where its wheels are trying to turn with just enough force to stay still on the slope without sliding down.  If you now start moving the plank in the opposite direction to where the car was going, by your reasoning the car will keep its position on the plank and be carried backwards...but I have a feeling that the car's wheels will turn and it will (in theory at least) stay "still" i.e. not be carried backwards.  I'll see if I can try this in real life...just need to get a toy car or similar.