2
$\begingroup$

So I was at the gym today, running on the treadmill, when the question hit me:

If I run with the treadmill on incline, I burn more calories. Since I run at the same speed, kinetic energy is constant, and since I am not travelling towards the sky, the potential energy stays constant as well.

So I come up with a hypothesis : Running on incline burns the same amount of calories than running without any incline.

I immediately dispose of this since I clearly tire a lot faster when running on incline, so obviously I'm doing more work, hence exerting more power.

So my question remains the same : Why does running on incline burn calories faster?

$\endgroup$
2
$\begingroup$

It's actually quite simple when you think about it.

When you're moving on a treadmill, are say you are "running at the same speed"; but what is your velocity relative to the room?

Obviously it's $0$. This makes sense, because the kinetic energy of moving forwards is countered by the force of the treadmill pushing you backwards; your net energy change (relative to the room again) is $0$.

When it's on an incline, that same thing is happening; except now you are adding energy to move forwards and upwards, and the treadmill is now pushing you backwards and down, removing any potential energy as soon as you develop it, just like it does with your kinetic energy. Essentially, the incline does require you to add potential energy, the treadmill just takes it away at the rate you provide it (if not, you would start going upwards or downwards, depending on which one has more energy; but you naturally try to prevent that when running on a treadmill).

$\endgroup$
  • $\begingroup$ Are you implying that when you run on a tread mill that is not inclined you don't burn calories? Your velocity with respect to the room may be zero but with respect to the treadmill itself it will be what the treadmill reads. Instead of thinking about the belt disappearing, think about it being a long continuous belt. From the reference frame of a person standing (not running) on the belt, the person running on the belt will have a velocity. $\endgroup$ – Bob D Mar 20 at 18:46
  • 1
    $\begingroup$ @BobD That's not at all what I'm implying. I'm saying that going up on a treadmill actually does require you to supply additional energy as if you were trying to increase your potential energy by going upwards, because you are going upwards on the incline as the incline pulls you back down. The same as the energy you are adding to make you travel forwards is not converted into net kinetic energy because the treadmill is pulling you back. You are adding a horizontal and vertical velocity relative to the treadmill; so the reason the incline takes energy is the same reason walking on it does $\endgroup$ – JMac Mar 20 at 18:54
  • $\begingroup$ I made a mistake posted my answer as comment. Will delete. $\endgroup$ – kamran Mar 20 at 19:04
  • $\begingroup$ @kamran I don't see why analyzing beyond average speed is really relevant here though? Is there something you feel is missing here? It really doesn't change why you would fundamentally expect an inclined treadmill to burn more calories, so I don't see a benefit of getting into nuances like that. $\endgroup$ – JMac Mar 20 at 19:09
2
$\begingroup$

Because it is equivalent to running upwards in an sloped path vs running in a horizontal path.

When the treadmill is inclined, your forward feet is placed at a higher point than it was before, thus, the force you make has to overcome not only friction but also gravity.

$\endgroup$
1
$\begingroup$

While it is true that your gravitational potential energy remains constant, you are still doing work against gravity. If you climb up stairs and let an escalator get you back where you started, you still did work despite having the same gravitational potential energy right? This is similar but your climbing and you being brought down are happening simultaneously.

$\endgroup$
1
$\begingroup$

You are correct that your overall gravitational potential energy does not change, but approximating your body as a sphere in this case doesn't capture the whole story. If you focus on the lower half of your body, you can see that with every step, you need to move your rear leg forward and upward, as your front leg moves backward and downward. The net work against gravity is zero, but you do need to do additional work to raise your rear leg, and you as a biological machine can't recoup that energy as your front leg moves backwards and downward.

$\endgroup$
1
$\begingroup$

Work is the dot product of force and displacement. Walking on a level treadmill, there is no theoretical reason why you should use any energy at all. The average force of your feet against the belt is equal to your weight, but the vertical component of the motion of your feet while bearing your weight can come arbitrarily close to zero. You can come arbitrarily close to doing no work at all.*

When the treadmill is set on an incline however, then your feet must experience significant vertical displacement while bearing your weight. Force dot displacement can not approach zero in that case.


* Your body mechanics are inefficient though, which is why you burn energy even when walking on a level surface.

$\endgroup$
1
$\begingroup$

In an inclined treadmill, we have to raise our center of gravity on each stride by lifting our foot and planting it at a higher level as opposed to a flat treadmill which we would put it on same level.

Although the planting point will eventually come down to the same level, but our body moves neither in a uniform speed nor in a straight line parallel to the treadmill's surface.

Each foot helps body temporarily go up and accelerate by rotating the ankle and flexing out the toes, adding to the length of that leg, giving suspension to the other leg to move forward and plant one step ahead. Then we retract the toes in and make the leg short so it could lift off for next step.

This cycle of lifting and landing has a component of mg.sin(angle) multiplied by our gait's disparity with smooth movement of the machine, when the treadmill is sloped up than a level treadmill. Even if we could smoothly average out our motion close to rolling on the treadmill, we would still deal with mg. Sin(angle)

This is why we use more energy on a sloped treadmill.

$\endgroup$
1
$\begingroup$

Relative to someone standing in the room, when you run on the treadmill you will have no velocity (kinetic energy) and no increase in elevation (change in gravitational potential energy).

But imagine you are running on a level treadmill belt that is very long and very wide. Now imagine someone standing still on the treadmill to the side of you. From the frame of reference of that person you will pass that person by with a speed equal to the treadmill reading.

Now consider the treadmill is at an incline. Each time you take a stride your front foot has to elevate your body. That requires more effort than when running level and thus burns more calories. But the treadmill brings that foot back down again so you don’t gain elevation relative to the room.

Although your elevation does not change relative to the room, imagine the treadmill spans several floors of a building and imagine again that a person is standing still on the treadmill initially next to you. That person will see you both pass by and gain height because the person is going "down" with the belt. That is your “effective” change height due to your strides on the inclined belt.

Hope this helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.