The Physics of an Inclined Treadmill

A bad day for your ego is a great day for your soul. -Jillian Michaels

One of the most popular exercises at the gym is the treadmill. And why wouldn’t it be? Whether you’re running or walking, it’s a great way to get your heart rate up, get your body moving, and for many people, a great way to burn calories.

But however you use a treadmill, there’s one extremely simple thing you can do to dramatically intensify your workout: incline it!

If you’re an outdoor walker/runner, this is the equivalent of going uphill instead of over level ground. There are many physiological differences in walking along an incline versus on level ground, but what does physics have to say about it?

Normally, if you’re on level ground (or a level treadmill), you stay at the same level in the Earth’s gravitational field.

But if you walk uphill (or on an inclined treadmill), you not only need to move forward at whatever pace you were moving at, you also need to climb — a little with every step — out of the Earth’s gravitational field!

The Earth’s gravitational field is no slouch, either. I’m an 80 kg individual, and for me to raise my elevation by just 5.3 meters (about 17 feet) costs me 4,200 Joules of energy, also known as one food calorie.

Now, if I actually exercise, I burn significantly more than one calorie by raising myself those 5.3 meters. Why? The two most significant reasons are as follows:

  1. I am not a perfect engine. This means, in order for me to do 4,200 Joules of physical work, I need to burn about three times as much in food energy in order to get that much useful energy out. Alas, our bodies are inefficient in that manner.
  2. When you exercise and then stop, your body doesn’t know that it’s okay for your heart to slow down for quite some time. So spending an hour walking uphill will elevate my metabolic rate for a lot longer than an hour!

Ahh, the power of exercising. But I’m not a physiologist; I deal in terms of physical work alone. So, just looking at the extra amount of energy you’d have to spend to climb up an incline rather than level ground, what are we talking about?

Let’s make a helpful table. We’ll just look at the total distance you travel (e.g., if you walk at three miles-per-hour for one hour, you go three miles), put in the incline, and see how much extra physical work you need to do!

Distance (miles) Distance (km) Incline (degrees) Extra Work (Calories)
1.0 mi 1.6 km 1 degree 5.3 Cals
1.0 mi 1.6 km 3 degrees 15.8 Cals
1.0 mi 1.6 km 5 degrees 26.3 Cals
1.0 mi 1.6 km 10 degrees 52.3 Cals
2.0 mi 3.2 km 1 degree 10.6 Cals
2.0 mi 3.2 km 3 degrees 30.6 Cals
2.0 mi 3.2 km 5 degrees 52.6 Cals
2.0 mi 3.2 km 10 degrees 104.6 Cals
3.0 mi 4.8 km 1 degree 15.9 Cals
3.0 mi 4.8 km 3 degrees 47.4 Cals
3.0 mi 4.8 km 5 degrees 78.9 Cals
3.0 mi 4.8 km 10 degrees 156.9 Cals
5.0 mi 8.0 km 1 degree 26.5 Cals
5.0 mi 8.0 km 3 degrees 79.0 Cals
5.0 mi 8.0 km 5 degrees 131.5 Cals
5.0 mi 8.0 km 10 degrees 261.5 Cals
10 mi 16 km 1 degree 53 Cals
10 mi 16 km 3 degrees 158 Cals
10 mi 16 km 5 degrees 263 Cals
10 mi 16 km 10 degrees 523 Cals

This is all for a person with a mass of 80 kg (about 176 pounds). Isn’t that a spectacular difference? In other words, if you make a long-term change from walking on a flat ground (or treadmill) to walking up inclined ground (or an inclined treadmill), you burn extra energy with every step you take!

And what’s with the Jillian Michaels quote? Well, I’m no longer the fittest guy on scienceblogs; say hello to Travis and Peter over at Obesity Panacea, our newest ScienceBlog! But whatever you’re doing, don’t forget to take the time to get out there and do something active; you’ll feel better and you’ll be healthier. And who doesn’t want a higher quality of life?

Comments

  1. #1 Brian Shiro
    March 10, 2010

    And this is why I bought a NordicTrack Incline Trainer X3. It goes up to 40% incline!

  2. #2 Duncan Ivry
    March 10, 2010

    When you exercise on an inclined treadmill you do *not* climb out of the Earth’s gravitational field, but you keep your distance from the center of the earth constant. So, what do you want to tell us here?!

  3. #3 Ethan Siegel
    March 10, 2010

    Duncan, every step you take is a step out of the Earth’s gravitational field. It’s not the Earth’s fault that the treadmill is constantly pulling you back, and deeper into the Earth’s gravitational field! (Stair-steppers work the same way, FYI.)

  4. #4 JDStackpole
    March 10, 2010

    Some physics isn’t quite right here.

    Last time I was on an inclined treadmill, my head (and attached torso) did NOT go “UP” as I hiked along. Most of me stayed at the same gravitational potential.

    I did have to lift my legs more than on a flat treadmill but that, near as I can tell, was the only additional work I had to do. Not nearly as much work as climbing a hill.

  5. #5 NewEnglandBob
    March 10, 2010

    JDStackpole, I would think it is MORE work than climbing a hill because on a hill you get higher up and the gravitational field is weaker. On a treadmill, you stay at the same field level.

  6. #6 JDStackpole
    March 10, 2010

    Nope. Gravity just doesn’t drop off that fast. It’s a minuscule (and undetectable by a climber) difference on any hilltop or mountain top relative to the valley.

  7. #7 bonvito
    March 10, 2010

    am a fan of your blog. will your next post be the physics of manny pacquiao’s power and speed? i love boxing and wants to know a bit of physics too. :) thanks.

  8. #8 george.w
    March 10, 2010

    No, really, I’m not lifting my body up any distance at all. It stays just about level even with a steep incline. When I climb actual stairs, my body goes up. The only extra work being done is lifting my legs a bit more. It’s high-stepping; not a bad thing but not the same as climbing a hill.

  9. #9 Sam Pennella
    March 10, 2010

    the effect of altitude on gravity is relatively negligible. but if you wanted to get real exact… yes, you would be expending less energy climbing at an altitude of 10,000 feet.

  10. #10 geopig
    March 10, 2010

    If you want to put anything to do with a treadmill in terms of work, you get zero because there is zero displacement at the end when you get off. Actually moving up or forward doesn’t get you the right answer in this case. You have to look at this incrementally because you don’t get back the energy you burn as you “fall” back down. That being said, there is a great deal of evidence to suggest that running on a treadmill isn’t as hard as pounding the pavement. Many marathon runners have found this out the hard way on race day.

  11. #11 Sam Pennella
    March 10, 2010

    it definitely isn’t as hard as running on pavement. you have to push off with each step effectively using more muscle fibers in your toes and feet.

  12. #12 Jackal
    March 11, 2010

    “it definitely isn’t as hard as running on pavement.”
    And it definitely isn’t as hard on your joints as running on pavement. Save your knees – use a tread mill – (or run on dirt)!

    When I do the treadmill instead of outdoors exercise, I like to put the incline up to 15% (max on that machine). I feel like I’m using a lot more of the power in my legs – and I am. The treadmill is moving me backwards and downwards every second, and every second, I clime to maintain my position.

  13. #13 Peter Janiszewski
    March 11, 2010

    Hello Ethan!

    Thanks for the blog shout-out and the wonderful post.

    Many of the participants that volunteer to exercise in our weight-loss studies, especially the older individuals, really prefer to keep the speed rather moderate but increase the incline. This ensures their heart-rate stays elevated, and they get a decent workout without having to run too briskly, thus reducing some of the impact on their joints. On the other hand, when doing maximal stress tests on our participants using a protocol which progressively increases the incline of the treadmill until they fatigue – I’ve seen a few cases where people stop the test due to severe cramping and discomfort in the lower legs. Thus, while walking up an incline may help burn a few extra calories – this needs to be considered against the potential reduction in exercise time due to cramping, or discomfort. From my experience, 1-3 degrees incline seems to be very doable for most people.

  14. #14 waynerobinson4
    March 11, 2010

    As part of my daily session at the gym, I do 3 km on the treadmill (if I’m lucky) at 6 degrees incline, so that means I’m doing only an extra 50 kcals of exercise. Damn!

  15. #15 William Paysinger
    March 11, 2010

    I’m agreeing with Duncan and JD on this one. Unfortunately inclining the treadmill isn’t going to increase your calorie consumption as much as actually climbing a hill. The only effect the gravitation will have is on the legs having to be lifted higher, feel the burn :)

  16. #16 John cornish
    March 11, 2010

    running on and incline on a Running machine is not as hard as running on the road. The have to take out the effort afforded by the motor moving your legs backwards,

  17. #17 Ishmael
    March 11, 2010

    According to Sir Isaac if a body has a relative motion in the vertical direction that is not accelerating then the laws of physics act as in every other non-accelerating frame including zero vertical motion.
    While not actually rising, the legs fight the same acceleration of gravity as if the body were.
    The same relative velocity effect acts in the horizontal so while your body remains on the treadmill you are pushing backwards the same as if walking the slope but without boundary of the machine.

  18. #18 Fred Waters
    March 11, 2010

    Another advantage to incline training is that it can be less stressful to your joints. If you walk at a steep incline you can burn as much, if not more calories than if you are running on a level terrain. Running causes extreme impact. In fact, I read where running at a fast pace can cause the impact to be 2.5X your body weight. That’s why many runners have to find alternative forms of exercise as they age. The popularity of elliptical trainers is partially from aging baby boomers who are looking for a form of exercise with less impact.

  19. #19 Hilary PhD
    March 11, 2010

    While not a physicist (I used to be a chemist, but fortunately I have since fully recovered) I was at first surprised how many commenters don’t agree that inclining the treadmill means working harder against gravity, even though the runner stays at the same elevation. But thinking about it, I can see why it’s not immediately obvious.

    By the same token, why does running on a treadmill on the flat require work (once up to constant speed)? After all, Newton’s laws say that no force acts on a body that’s not accelerating, and work done is the (integrated) product of force applied by distance travelled. The reason is because the runner needs to exert a backwards force along the bed of the treadmill to overcome its friction, which would otherwise pull him/her backwards. (Note that a somewhat greater force is needed when running without a treadmill, as then the runner moves relative to the air and has to overcome its resistance as well as that of the floor. Also, there must be some secondary forces needed to stabilise the body and avoid falling over.)

    It’s now much easier to see that if the treadmill bed is inclined, the force acting along its length has a gravitational as well as a frictional component, and the runner has to work harder to overcome both.

    I know this is just a long-winded version of what others have said, but hopefully helps to clarify what’s going on.

  20. #20 Hilary PhD
    March 11, 2010

    Me again. Just to add to what Fred Waters and Peter Janiszewski said, I don’t think it’s just about avoiding impact.

    For a given type of exercise, you can raise your work-rate either by moving faster or by increasing the resistance to be overcome and therefore the force you apply. Some people’s muscles seem to suit increased speed, others increased force. Modern cardio machines allow inclines and variable resistance so that each person can find a balance that suits them personally.

  21. #21 Duncan Ivry
    March 11, 2010

    Ethan: “… It’s not the Earth’s fault that the treadmill is constantly pulling you back, and deeper into the Earth’s gravitational field! …”

    No, it does not work like that. people on a treadmill stay at the same position vertically and horizontally all the time, and the surface of the treadmill roles away under their feet. If you want to have the effect of “pulling you back” you first have to run several steps forward and upward, then slow down, and let the treadmill *then* pull you back — and this again and again. I have never observed people doing it like this.

    It is like JDStackpole said: “I did have to lift my legs more than on a flat treadmill but that, near as I can tell, was the only additional work I had to do. Not nearly as much work as climbing a hill.”

    And above all that, exercising on an inclined treadmill is bad for the joints in the long run — pun intended — and even if Fred Waters above says the opposite — because the feet are in a less natural position all the time. Because of this its also bad for the Achilles string (I hope this is the correct word). I would even say: If you want to ruin your feet, use an inclined treadmill.

    To Hilary PhD:
    A lot of physical work — may be, even most of it — has to be done inside the body. Different muscles have to move the body’s parts in different directions again and again. Think about the heart, the lungs, the brain, etc., and the transport of all the substances the body needs.

  22. #22 Sam Pennella
    March 11, 2010

    when you jog on a treadmill, you’re primarily using your femoral extensors and flexors as well as the gastroc and anterior compartment. you don’t use plantar flexors nearly as much as you do when running on pavement.

    for an incline, you’re adding a stair stepping motion to the run. => more calories.

    that’s the bottom line.

    run for 3 mi on a tredmill one day. take a break. then 2 wks later, run 3 mi on pavement. I guarantee you that you will feel much more soreness in the dorsum of your feet.

    add an incline to THAT… and there’s your difference

    Sam Pennella

  23. #23 Hilary PhD
    March 11, 2010

    @Duncan Ivry

    I agree some internal work has to be done – I mentioned stabilising the body and Ethan mentioned physiological inefficiency (his indented point 1). Neither of that alters the fact that the force the runner has to exert against the treadmill is increased by a (quite significant) gravitational component if the treadmill is inclined.

    Please learn something about the physics of vector forces.

  24. #24 Bernard Leikind
    March 11, 2010

    Please review and update, if necessary, your contact information. I was unable to e-mail some relevant material to you that is too long for the comments.

  25. #25 Duncan Ivry
    March 11, 2010

    Hilary PhD: “… the force the runner has to exert against the treadmill is increased by a … gravitational component if the treadmill is inclined.”

    I do not know, where this gravitational component should be. As far as I can tell, a person on an inclined treadmill has an inclined orientation of his/her feet, but nothing else.

    “Please learn something about the physics of vector forces.”
    Thank you for pointing me into this direction. A poor, old mathematician has difficulties understanding just this ;-)

    But to the point:

    Let us observe a person on an inclined treadmill! Does the person move to a greater distance in the gravitational field of the earth, i.e. in a direction from the floor to the ceiling? No.

    Is there work done by going along a path — going along a path! — up in the gravitational field excerting a force against gravitation pulling down? No.

    We have to consider the “the line integral of F dot ds”, with F = force, ds = infinitesimal path; see e.g. Richard Feynman’s Lectures on Physics, page 14-1. Interestingly, Feynman writes on page 14-2: “… running upstairs is considered as doing work …, but in simply holding an object in a fixed position, no work is done.” Then Feynman goes on explaining how physiological work occurs — muscles, nerves, etc. — and how it differs from physical work.

    The person on the treadmill is the “object”, and the person holds “itself” in a fixed position relative to the gravitational field. Ergo: No physical work is done.

    This is *the* important piece: No path, no work!

    @ Ethan: I think, you have access to the Feynman Lectures. Please read these two pages, and tell what you think.

    Anyway … this is my last comment on this topic.

  26. #26 MadScientist
    March 12, 2010

    But – your height is not changing at all although your footing may be more awkward. I’m with JDStackpole; I can’t imagine the slight increase in work being anything like climbing a mountain.

    @Brian: 40 degrees is extremely steep; I carry rope when I walk up/down terrain with that grade – do you really use it at such a setting?

  27. #27 Emlyn
    March 12, 2010

    @MadScientist he said 40%, not 40 degrees, so it’s more like 22 degrees (I think)

  28. #28 Hilary PhD
    March 12, 2010

    Duncan Ivry asks: “Is there work done by going along a path — going along a path”

    The answer is yes. The path is the motion of the person RELATIVE to the bed of the treadmill.

    I rather wonder how many of the people who talk about “slight increase in work” have any actual experience of walking quickly (I wouldn’t run, but I hate running on the flat as well) on an inclined treadmill.

  29. #29 Anonymous Coward
    March 12, 2010

    Skipping over the “I’m not going up on a stairclimber, therefore I’m not doing any work” silliness, there’s another interesting issue here:

    The “extra work” figure is only meaningful if you have a “baseline work” number to compare it to.

    That makes for an interesting order-of-magnitude physics calculation. Obviously if we had wheels, we could “run” on a horizontal treadmill with (in the limit of perfect wheels) no effort. So where’s the work coming from? I’d guess two main contributors:

    1) The up and down motion involved in running and walking (in theory if our bodies were perfect springs, etc. we could regain the energy expended, but I don’t think that’s the case)

    2) The acceleration of our legs and arms back and forth (again, if our muscles could act as springs, etc. we could regain this energy, but I’m pretty sure that’s not the case).

    Neglecting #2 and just considering #1, I’d guess that at each stride we’re looking at an increase in body center-of-mass height that’s roughly 5 to 10% of the stride length. So your baseline work would be equivalent to the “extra work” associated with the 5 to 10% incline numbers.

  30. #30 Sandy
    March 12, 2010

    Wow – very scientific – enjoyed reading your explanation. I always walk at an incline on my treadmill. I set it up to the maximum incline and work out -

  31. #31 Sven DiMilo
    March 13, 2010

    your body doesn’t know that it’s okay for your heart to slow down for quite some time. So spending an hour walking uphill will elevate my metabolic rate for a lot longer than an hour!

    uh…as a physiologist you’re a pretty good physicist. Your body is not stupidly fooled–metabolic rate (and, essentially, therefore heartrate) remain elevated post-exercise for good reason: there is a lot of chemical work to be done to mop up after exercise.

    I did have to lift my legs more than on a flat treadmill but that, near as I can tell, was the only additional work I had to do.

    The only extra work being done is lifting my legs a bit more. It’s high-stepping

    But the extra work in “high-stepping” is not all (or even mostly, I bet) being done by the leg that is stepping up–your power leg pushes the hip forward. On an incline there is an additional upward vector to that pushing force, and that accounts for most of the additional energy required.

    I think.

  32. #32 Keith C
    March 13, 2010

    When walking horizontally, you place your front foot on the same level as the rear foot. Then walking up a treadmill at 10 degree elevation, and taking steps of 2 ft, the front foot is placed about 4 inches higher than the rear foot.
    As you begin to transfer your weight to the front foot, your body must rise further than for a similar horizontal pace. The net effect is smaller than when walking up a 10 degree ramp because the treadmill surface is moving steadily backwards.

    I suspect that an engineering approximation is that the body rises about 2 inches higher during each step, compared to the trajectory for walking horizontally. Calorie consumption would then perhaps be about the same as for walking up a ramp with half the treadmill slope.

    Are there experimental measurements of oxygen consumption on treadmills of different slopes which could resolve the issue.

  33. #33 Annette
    March 17, 2010

    You guys and your judgements, this is a great article and all I know as an experienced marathon runner, is when I do hill work on my mill, I sweat and feel pain, liek I have run 20 miles. When I run in outdoors and do hills, I sweat and feel pain. Incline running, walking or anything to my body is a workout and a half. Level running or walking isnt as intense as an hour of hills. Just go out or stay in and test it. Weigh yourself before and after on a week of hills and level running.

  34. #34 Richard
    March 22, 2010

    Having a master degree in physics, it is a bit frustrating to read people clearly not understanding the basic principles being so cock-sure. If you run on a treadmill, you are lifting your whole body not just raising your feet or whatever. The easiest way to realize that is to imagine that the treadmill is very long, say 10 miles with an inclination of 10%. Then if you are standing still at the top of the treadmill, the treadmill will carry you down 1 mile vertically after a while. If you on the other hand, run with the same speed as the treadmill your vertical position is unchanged. So who did the work to prevent you from “falling” the 1 mile? Your body and nothing else. QED.

  35. #35 Treadmills Australia
    March 30, 2010

    Valuable information. Incline is pretty much common on treadmills now. It is hard to accurate differences between running on treadmill with incline and really uphill running. However incline is great feature for simulating uphill running.

  36. #36 scan tool
    June 10, 2010

    But the extra work in “high-stepping” is not all (or even mostly, I bet) being done by the leg that is stepping up–your power leg pushes the hip forward. On an incline there is an additional upward vector to that pushing force, and that accounts for most of the additional energy required.

  37. #37 scan tool
    June 10, 2010

    But the extra work in “high-stepping” is not all (or even mostly, I bet) being done by the leg that is stepping up–your power leg pushes the hip forward. On an incline there is an additional upward vector to that pushing force, and that accounts for most of the additional energy required.

  38. #38 scan tool
    June 10, 2010

    But the extra work in “high-stepping” is not all (or even mostly, I bet) being done by the leg that is stepping up–your power leg pushes the hip forward. On an incline there is an additional upward vector to that pushing force, and that accounts for most of the additional energy required.

  39. #39 scan tool
    June 10, 2010

    But the extra work in “high-stepping” is not all (or even mostly, I bet) being done by the leg that is stepping up–your power leg pushes the hip forward. On an incline there is an additional upward vector to that pushing force, and that accounts for most of the additional energy required.

  40. #40 scan tool
    June 10, 2010

    But the extra work in “high-stepping” is not all (or even mostly, I bet) being done by the leg that is stepping up–your power leg pushes the hip forward. On an incline there is an additional upward vector to that pushing force, and that accounts for most of the additional energy required.

  41. #41 London
    June 15, 2010

    you’re a pretty good physicist. Your body is not stupidly fooled–metabolic rate (and, essentially, therefore heartrate) remain elevated post-exercise for good reason: there is a lot of chemical work to be done to mop up after exercise.London

  42. #42 tower defense
    June 16, 2010

    “Individuals undergoing SERE training are obviously in a very different situation from detainees undergoing interrogation; SERE trainees know it is part of a training program,” Bradbury wrote, borrowing from the IG report’s conclusion.

  43. #43 Cameron
    June 20, 2010

    I have a question for you Richard:

    You said this: “So who did the work to prevent you from “falling” the 1 mile? Your body and nothing else.”

    Lets say you are standing next to a 1 mile high cliff. If you jump off, you will fall 1 mile. If you just stand on the ground at the top of the cliff, you don’t fall. So who did the work to prevent you from falling the 1 mile here?

  44. #44 Web Hosting Reseller
    June 21, 2010

    Ms. Yoshida’s still better than I ever was though; my top clocked pitch ever was only 62 and 58 was more typical on a very good day.

  45. #45 Johnny
    June 24, 2010

    One step at a time, here is my theory:

    When you lift your left leg up and forward, you have to lift your legs higher than without the incline. This takes some work, but lifting your leg is just a fraction of your body weight.

    Your right leg now is just sliding downward and your body overall just stays in place. No work. If you were climbing a hill, your right leg would now actually have to lift your entire body weight (against gravity).

    Your body is say 4 times heavier than just your leg, so on a treadmill you only burn a fraction, ~1/4 of the additional calories that you would burn by climbing a real hill.

    For example if running flat burns 1000 calories and running up a real hill burns 2000, you would burn about 1250 calories by using the incline.

  46. #46 Cameron
    June 24, 2010

    Very good points Johnny.

    I have been thinking about this problem alot and have come to identical conclusions. The way I say it is that it depends on how you walk.

    Consider stand the case where you stand still and let the treadmill pull you down, then instantaneously walk up (so the treadmill is basically immobile since you move so fast), then repeat. As you walk up, you have to spend just as much energy as you would on a hill, made obvious since the treadmill is not really moving. As it pulls you down, that energy gets put into the treadmill and motor system. So, if you were able to walk this way, then it would take as much energy as walking up a hill. Note that you have to provide energy both as physiological (or internal, whatever) work and physical work.

    Consider however the other extreme: you somehow manage to stay put without moving your body at all. Say if you had roller blades on). Then you would be “walking” the treadmill without having to expend any energy. I probably need to at this point explain why you wouldn’t roll down the treadmill, so I will. Let’s tie a rope around your waist that fixes you to the frame of the treadmill. This rope stops you from sliding down the treadmill. The wheels stop you from having to move and keep your gravitational potential energy constant. I also need to point out (and I’m sure lots of people will argue this because they don’ understand what work, energy and forces really are. I’m just going to ignore them. They can go read a physics textbook then come back) that the cord doesn’t give you any energy. It doesn’t do any work. Remember that work is force times distance, or more technically the integral of a force over the path parallel to its direction. Either way, the rope exerts a constant force on you, but since neither you nor the rope move at all, it doesn’t have a distance to multiply by, or a path to integrate over. It therefore does no work. So by this mechanism, you “walk” the inclined treadmill without doing any work against gravity. In this example, you do neither physical nor physiological work. Note too that if you wanted an example where you did do physiological work, but no physical work, then just use a skateboard and do jumping jacks on it or something.

    Now we go into a more realistic scenario, like the one you mention Johnny. Doing this, we have to make assumptions about how the person walks. For example: their entire torso remains entirely still, only their legs move. You can then say that they fit somewhere in between the two scenarios above. They don’t have wheels, so don’t do zero work, but their torso also doesn’t move, meaning they don’ do as much work as if they stood while going down, then walked all the way back up. My conclusion at this point is that an inclined treadmill takes more work that just high-stepping, but less work than an actual hill. I believe that should satisfy most people (as long as you agree with my arguments, if not, please let me know where you think I went wrong).

    I would be interested in seeing some actual data that could be used to verify Johnny’s theory and my arguments. I see the biggest challenge in to be how to measure how much work the human being does.

    p.s. in case anyone is interested, I’m a third year electrical engineering student. I believe that even with years of education on physics and mechanics, you could incorrectly answer the question by applying logic incorrectly. I also believe that “common sense” can often produce the incorrect answer. Although not conclusive, or quantifiable, I value people with experience running bringing their opinions forwards. What I believe is truly required to answer questions like this one is a bit of physics understanding, a bit of experience in real life, a lot of thinking and logic, and even more discussion… thank God for the internet.

    p.p.s. please bring forwards your opinions, especially if you disagree

  47. #47 ElmerPhuD
    June 25, 2010

    Cameron and Johnny,

    This horse was dead and buried months ago; Ethan slew it definitively in his follow-up (“Einstein”) post,
    and dozens of people in both threads have added nails to the coffin.

    But still the dissenters dissent. Sigh.
    I think you can appreciate why few of us are willing to continue the conversation.
    That said, suspending my own wariness for the moment, I’ll assume you’re sincere in trying to understand the physics and willing to work.

    So let’s cut to the core of the essential problem.
    You’re making it far too difficult.

    Simplify the scenario as follows:
    1) make the room a vacuum, to eliminate air resistance;
    2) replace the human runner with an (idealized) electric car with 100% efficiency — that is, all energy consumed goes into forward thrust;
    3) the treadmill belt has constant velocity V, sloped at angle A from horizontal;
    4) the car has constant mass M, and drives up the treadmill belt with constant velocity V, maintaining its position exactly.

    How much energy will the car (i.e. its battery) consume in time T?

    Hint: the answer, simple and unambiguous, will be in terms of V,A,M,T, and g, the local acceleration of gravity
    (assume Earth sea level, where g=9.8 m/s2).

    Once you find that answer and understand it, most of the complexities you’ve raised will be apparent as distractions.

    Follow-up questions:
    1) Where does the drained battery energy go?
    2) Put the whole setup in an elevator moving up or down with constant velocity. What happens?
    3) Now move the whole setup to Jupiter, or the moon. What happens?
    4) Why is it appropriate to call this energy transfer “gravitational work”?
    5) Why, when talking about gravitational potential energy, do we need to identify reference frames?

  48. #48 online fortune teller
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  49. #49 Cameron
    July 25, 2010

    I have actually been to the other thread ElmerPhuD. I posted a comment on it in response to something you said.

    I have spent a lot of time thinking about this problem. I also talked to other people I know and listened to their insight. Although I hadn’t come to a definite conclusion before, I think I have settled with one opinion now – in some contradiction to what I thought earlier.

    The approach I took is defining the person as a system and identifying the points where work can be done – i.e. energy transfer in or out of the system. I like simplifying the situation the way you did ElmerPhuD, with the remote controlled car. When you do this, it becomes apparent that the only way work can be done by the car is through the interface of the wheels on the tread. Since the car doesn’t move up or down, no work can be done against gravity. So, the question is how much (if any) work is done here.

    A quick free body diagram shows a force normal to the treadmill and another parallel – both to oppose the force of gravity. This parallel force could be a source of work, as long as we can identify a distance to multiply it by. And that is where I start to become less sure. Once argument could be that since the car doesn’t move with respect to the earth, there is not distance – I don’t like this. Another is that as the treadmill and wheels turn, the treadmill moves a certain distance over a time period with respect to the car. I believe this distance should be use. I would appreciate it if anyone had any insight into this.

    By this assumption, a calculation of the work done by the car can be completed:

    Parallel force = M*g*sin(A)

    Energy = Force * Distance
    Energy = M*g*sin(A)*V*T

    1) All of this energy changes to heat through friction in the treadmill. Likely resulting in less electricity form the wall.
    2) An elevator with constant velocity makes no change.
    3) On a different planet (or accelerating elevator) the value of g (or net m*g with the additional kinetic “force”) changes.
    4) I do not think it is appropriate to call this energy work against gravity since no change in gravitational potential occurs. I can see how some people call it gravitational work since it results from the presence of gravity – either way, not a big deal what we call it.
    5) Gravitational energy results from the interaction between two masses. It doesn’t matter what reference frame we use, the change in distance between the two masses is what counts when determining the change in gravitational energy.

    Someone I talked to brought up this hypothetical situation: assume the treadmill you are on is infinitely large (length and width). If so, to a person standing on it, it would seem like solid, stationary ground. Also assume that a uniform gravitational field is directed through the treadmill at a certain angle (without worrying about what has created this gravitational field). If the person stands still with respect to the treadmill, they loss gravitational energy and expend no energy. If they walk up the treadmill, it can intuitively be seen that they will have to do work. It is interesting to note that we can’t exactly identify a reference point in the field of gravity and it doesn’t matter how fast the person walks up the treadmill.

    I am not sure about a person walking yet, because the situation is more complicated. This may or may not make a difference and I have yet to hear a good argument as to why it does or doesn’t. For example; a person spends some time “in the air”, and exerts force with one foot then the other.

    Let me know what you think ElmerPhuD. I will be 100% convinced about the car once someone brings forwards a good explanation of the distance thing. I’m not yet convinced about a person.

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  51. #51 ElmerPhuD
    August 1, 2010

    Cameron,
    Kudos on the homework; you’re almost there.
    I’m glad to see you’re sincere, so I’ll try to explain in full.
    Forget the runner for now; stick with the car until all is clear.

    Main Question:
    Correct. Energy = gMVT*sin(A).
    From now on, let’s talk about Power = Energy/T = gMV*sin(A).
    That’s the exertion rate for an ideal vehicle.
    And it’s the minimum average exertion rate for ANY vehicle/body maintaining position on an inclined treadmill.

    Question 1) Correct: all such exertion goes into treadmill heat — eventually. But it acts as *lift* first. See full explanation below.

    Q2) Correct: additional components of vertical motion (e.g. elevator) don’t change exertion. There’s no such term in the equation.

    Q3) Correct: moving to a location (e.g. Jupiter) with a different value for g changes exertion rate. I.e., gravity matters.

    Q4-5) You’re still confusing yourself.

    First: accept the reality of Power >= gMV*sin(A).
    Go run on a treadmill; you’ll feel it.
    Every interpretation of the problem, using forces, potential energy, etc, is just a different accounting trick to describe that reality. Any formulation which makes that exertion appear to vanish is wrong somehow.

    Second: stop fixating on the relative height between Car and Earth. Yes, that measure of gravitational potential energy is constant. But it’s irrelevant; it answers the wrong question.

    The chronic dissenter’s argument, often smugly asserted, goes:
    “Vehicle never changes height relative to Earth, so there’s no gravitational work.”
    But there the thinking stops, failing to consider these implications:
    Implication #1) if planet-relative height doesn’t change on Jupiter, gravity still shouldn’t matter.
    Implication #2) if height does change, as with the elevator, now gravity should change something.
    But by now you know those are both WRONG! So don’t trust that argument.

    Third: let’s not put semantics before physics.
    I wouldn’t quibble about whether to call inclined treadmill exertion “gravitational” or not, as long as our terminology motivates the correct predictions.
    But if purists, in refusing to call it gravitational, deny the reality of gMV*sin(A) or make either wrong implication above, their language has failed.

    So I’m calling it gravitational work. Because it’s work, and there’s a “g” in the equation, and it predicts what happens on Jupiter.

    Fourth:
    Potential energy, like distance, can be defined relative to any origin. So let go of Earth as an origin; follow Ethan’s “Einstein” advice and consider relative perspectives.

    Set aside your omniscience about the scenario and consider only what the Car “knows”: it’s on a surface, sloped to a local gravitational field. That surface may be long or short, moving or still, attached to a planet or not. That surface, plus any attached planets, elevators, etc, constitute a Thing.
    Earth, if it exists, is just another moving part within Thing; we don’t know and don’t care how it moves.
    Our concern is only with the energy transfer (i.e. Work) from Car to Thing.

    Assign Thing a reference frame with origin at some arbitrary point P on the treadmill surface. P moves downward at speed V*sin(A) relative to Car (and relative to Earth, if we cared).
    Measure all height and potential energy relative to P.

    There are two concurrent components of motion, one rising and one sinking:

    Motion 1) Car is climbing relative to Thing, converting internal chemical energy to gravitational potential energy, relative to P, at a rate of gMV*sin(A).
    That’s the work we care about, the energy transfer between systems, from Car to Thing.

    Now, if we want to be omniscient and figure out where that energy goes within Thing, consider the second “sinking” process:

    Motion 2) P is sinking relative to Earth, one of Thing’s moving parts. Alternatively, Earth is “sinking” upward relative to P, “falling” within Thing at speed V*sin(A). Its fall is braked by friction, converting potential energy (again, relative to P) to heat at the same rate of gMV*sin(A).
    But that’s all internal to Thing; Car doesn’t care.

    Notice: turning off the treadmill stops 2) but not 1); Car keeps climbing relative to P, exerting the same power. Thing just stores it differently, as internal gravitational potential instead of heat. Car doesn’t care.

    In summary:

    Motion 1: Climbing (Car–>Thing): Chemical –> Grav. Potential
    Motion 2: Sinking (within Thing): Grav. Potential –> Heat

    It’s tempting to combine them, making gravity disappear:
    1+2: Chemical –> Heat, no net motion

    But these two components of gravitational potential, although the same magnitude, operate across different system boundaries: once Car–>Thing, once within Thing.
    Gravity participates TWICE, not NEVER. But only once acts on Car and counts as work between systems.

    The “distance problem”, as you call it, is merely a failure to distinguish system boundaries.

    Finally, as for the runner: he/she is just an inefficient vehicle whose momentary power output wavers around an average of gMV*sin(A)+Waste.

    This is all to say: our host Ethan, Professor of Astrophysics, is indeed correct — something for reflexive dissenters to keep in mind.

  52. #52 ev dekorasyon
    August 2, 2010

    You guys and your judgements, this is a great article and all I know as an experienced marathon runner, is when I do hill work on my mill, I sweat and feel pain, liek I have run 20 miles. When I run in outdoors and do hills, I sweat and feel pain. Incline running, walking or anything to my body is a workout and a half. Level running or walking isnt as intense as an hour of hills. Just go out or stay in and test it. Weigh yourself before and after on a week of hills and level running.

  53. #53 Stu
    August 11, 2010

    I doubt anyone is reading this anymore, but I have a bachelor’s degree in physics, and there is no way that an inclined treadmill requires one to burn as many calories as running up an actual hill. We can get at this without considering any equations. Consider: someone running up a hill with incline A and someone running on a treadmill with incline A (with A being some measure of the angle relative to flat ground: degrees, radians, whatever) should being doing pretty much the same motion. But then the person running on an actual hill gains elevation (gains energy) whereas the person running on the treadmill does not. Assuming that both people are running at the same speed relative to whatever it is they’re running on — treadmill surface or ground — we really have to conclude that the person running up the hill is spending more energy than the person on the treadmill. In denying this, you are more or less claiming that gaining elevation does not require energy, which is a definite violation of the laws of physics. Please shoot me an email if you don’t understand this. The person claiming to be a former chemist is misinformed.

  54. #54 cheap sunglasses
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  55. #55 Bingo Deposit Bonus
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  56. #56 Devo
    August 25, 2010

    The difference between treadmill incline running and hill running is that when “climbing” on a treadmill, your body above your legs does not go anywhere near as much distance (vertical) as if you were to climb a real hill. The reason why running up a real hill takes more energy is because your legs now have to push the rest of your body mass up the hill. On a treadmill, your legs only do work on the force of gravity, not your whole body. Simple as that. I was recently told by a sociology student that I was wrong about this. Apparently engineering physics is stupid.

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  58. #58 Myster Menace
    September 6, 2010

    If you ride a bike on a treadmill neither your body nor your legs rise up or down any more than riding on a road.
    First, ride a bike on a treadmill at a 10% incline for 5 miles.
    Then, ride a bike up a hill at a 10% incline for 5 miles.
    Which will take more effort? Why?

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  60. #60 John
    September 12, 2010

    Ok, we are all currently on a giant treadmill. if we start at the equator and face east and don’t move for 24 hours, we have moved the equivalent of the earth circumference without putting forth work. If we look at the distance it would be considered negative distance which would be due to the lack of ‘keeping up’ with the speed of the earth. Now if we were to be able to move forward at about 1000 miles/hour we are now doing the same thing as being on a treadmill. Would this NOT be work? would it NOT be the same as running around the earth if the earth was standing still? everything we do is in reference to other things and it doesn’t change according to the size of those things.

    and as for those who insist that the difference is in the fact that outside you have to pick your feet up more, next time you get on the treadmill try to stay on it without picking your feet up. ha ha!

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  66. #66 Chris CD
    December 13, 2010

    As a physics teacher who has had to try to grapple with the best strategies for clear exposition of such concepts for most of a lifetime, what always intrigues me most about this sort of interchange is not the mechanics per se, but the never ending confusion about the mechanics. Which is the invariable result, in my view, of how badly and inappropriately physics and maths is taught at school and at university.

    Reading the comments shows in particular how much more difficult it is to apply correctly the widely accepted principles of classical mechanics than most folks realise – even those who have laboured away for three years at an undergrad physics degree and believe that they should now have this down pat.

    First and foremost, here’s the Colenso definition of mechanical work:

    “A wee bit of work is done by a system A on system B when system A pushes or pulls system B a wee displacement in the direction of system A’s push on, or pull of, system B.”

    (Note that I have deliberately eschewed here the word “force” and the word “cause”.)

    In real world systems, contrary to what is taught at school and even in undergrad mechanics courses, few systems move very far in a smooth, rectilinear (straight) line. Nor, cruelly, as if that were not already bad enough, do systems, or parts of systems, even follow the path of a smooth circle. Even relatively uncomplicated systems move along loci that need to be described accurately by the geometry of conics – hence ellipses (of which the circle is a special case), hyperbolae and parabolas.

    To add insult to injury, complex, more “realistic”, systems are even harder to analyse than the aforesaid “uncomplicated” systems because complex systems tend to perturbate: the individual parts follow paths made up of sequences of bits of conics. Yes indeed, real life, complex systems are complex, tricky buggers to analyse.

    So the baby-step mechanics that most folks learn at school and uni needs to be understood as a deliberately oversimplified introduction to what can turn rapidly into some quite tricky 4D vector and tensor maths – and like so much introductory mechanics a pretty misleading intro at that.

    In real world complex systems, forces are not constant and they do not act in a nice, straight forward fashion through the centres of idealised bodies with no extensions, again as shown in the introductory mechanics text books for the sake of pedagogical simplification.

    So, in real world systems, we have to analyse moments of inertia not mere inertias; angular momentum not rectilinear momentum; angular acceleration not rectilinear acceleration; torques not forces; not to mention that the magnitude and direction of the various accelerating parts are themselves constantly changing because moments of inertia and torques have a irritating habit of themselves not staying constant throughout the range of a system’s movements.

    Indeed, the rectilinear classical mechanics we get taught at school is best understood as just a special case of angular mechanics where the displacement of the force from the centre of mass of the system on which the force acts happens to be zero – so there’s no torque, no angular acceleration. Nice and easy intro for the purpose of pedagogy that then leaves almost 100% of physics students confused until the end of their days.

    So, back to the human body. Forget the introductory classical mechanics you struggled with at school or even at uni. The human body is a complex system of rotating systems, of which the moments of inertia and the torques acting thereupon at any moment are constantly changing and need subtle and very painstaking analysis to puzzle out. I’ve spent the last forty years trying to mathematically model the human body in motion and still haven’t got very far. (But then I always was a slow learner and worker.) And that’s after we recognise that all animal systems constantly do work even as they rest; more work as they sit; more work as they stand; more work once they start to move.

    The key of course to all this as always is the diagram. Lots of carefully drawn diagrams in fact. How many times have I railed at my students: but where Perkins is your diagram, you wretched imbecile? Answer of course never, because I’ve never taught anyone called Perkins. But if I had, then the rhetoric would be apt.

    So, as an exercise start by drawing a naked human walker lifting one leg to walk up one step. You will need to draw lots of intermediate steps before your walker has finally stepped up onto the one step. All the parts of the body: the toes, the arch of the foot, the heel, the lower leg, the knee, the thigh, the pelvis, the abdomen, the upper back and upper chest, the neck and the head will need to be drawn. And don’t forget all the parts of the arms. You will need to estimate the masses, moments of inertia, torques, initial average and final speeds of rotation of each part to estimate the work being done by different parts of the body on other parts. Part of the problem is that the parts of the whole body are counterbalancing as they accelerate in order to maintain balance.

    When you have wasted the best part of the next month or so of your life on your detailed drawings, it will gradually dawn upon you that while the overall step-up of the person onto the next step has required of course that work has been done against the person’s overall weight (this involves gravity) this is only one tiny part of the whole work picture. Why? Because to accelerate one part of the body, even in the absence of the earth’s gravitational field, requires a torque to be applied to that part. So mechanical work is done. Because as we move, the mini-systems of our body are accelerated.

    Which is why we do work, lots of work, even when we move quickly or slowly on the flat. Even when we skate on ice apparently hardly lifting our limbs at all. Yes, even when there’s a following wind and we skate on ice! Because we are constantly accelerating parts of our body in order to skate along and this therefore requires work.

    But even when we stand still, or sit, or lie down, our muscles constantly do work. Only when we are dead does this process stop. Hence, only when you have worked out the work that all the parts of the step climber’s body does even when he or she stands still; then worked out the extra work the parts of the stepper’s body do when they move in a horizontal plane; then worked out the extra work that we all do when we climb up one step, will you be intellectually at the place where you can see straightaway that running “in place” up an incline on a moving treadmill will certainly require far more work than running on the flat.

    In fact, of course, exactly the same amount of work, if you ignore wind resistance, as running up a stationary, rubberised mat at the same incline at the same speed.

  67. #67 joe
    April 16, 2011

    i actually started walking on the incline a year ago, only on 5incline. and one day i went in early and saw this girl with the incline on 15, i ask how are you doing that, she said, try it, and it will change yout life. so,for 30 mins. with a quick 60 sec rest between 15min. ive been doing the incline 4days a week incline interval of 15 and 5 switching every 3 minutes, with 30sec breaks at 5degrees incline. ive lost 40pd overall, 20 in the year. im looking to up my time here, maybe 45 min. when ever im done it says iveclimb 775 ft. does anyone know what that is equvalent to with thing, like a building or mountains, is this a great thing im doing, whats next.lol

  68. #68 Rich
    April 22, 2011

    Can it be that you are all reading far too much into this treadmill incline problem? With no incline the motor does all the work driving the belt against friction. The more incline, the less work the motor does, and the more work you do. I might just go get my power meter and get some real energy readings.

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  70. #70 matt
    December 10, 2011

    I wonder if it is a problem of work vs. energy, which I think are measured in the same units. A man flapping his wings and gaining altitude will add energy measured by the ascent into his load, but if he flaps his wings only hard enough to maintain elevation he does no work that is reckoned by PE, but he still might burns lots of energy to fight gravity.

  71. #71 matt
    December 11, 2011

    In the case of inclined treadmill vs. flat, users are expending the same KE, but the incline is undeniably harder, and perhaps the total energy there is the PE neded to recover from a dropped back elevation one would find himself at if he stood still relative to the belt. This would be the same PE one would gain in hill climbing at a forward speed equal to the belt speed of the treadmill climber. Both will use the same amount of energy to stand upright and resist gravity. But if there is an overall difference, it may be accounted for in my health club’s machines, which seem to understate vertical ascent. Perphaps the ascent readout it is the equivlanent ascent of hill climing, which they think requires more energy.

  72. #72 brian
    January 4, 2012

    I too as other posters have commented cannot believe that the education level has eroded to the point people cannot figure out simple physics.

    First of all the whole moving in the gravitational field is confusing and incorrect for the point that is trying to be made. For simplicity’s sake, gravity has the same acceleration on an object everywhere at all times. Toss a ball in the air. It moves upward and slows at the deceleration of gravity. Then stops and falls back with the same acceleration. At all times gravity is the same on the object.

    Next, this is an issue of people getting confused by the relative motion of the climber and the treadmill frame/ground/surroundings being zero. If a row boat in a river is rowed with a force equal to the current pushing against it, it will not move relative to the shore. If the same rowing effort were put forth on still water the boat would move at the same rate the river was moving. Add some trigonometry and vectors and this is true in two dimensions as well.

    Try this experiment if you believe that you do not have to pull your own weight up, against gravity, on an inclined treadmill just as you would a hill.
    Straddle the treadmill belt with a foot on the platform on either side. Turn the incline up and set the belt to an extremely slow speed. Reach up with you foot (either foot) and take a stride as high up as you can on the incline. Put all of your weight onto your stepping foot and release the other foot from the platform and into the air. At this point your knee of your stepping foot should be bent with your foot high on the incline and your back foot is in the air. Now DO NOT straighten your knee, that would be you lifting yourself up against gravity. Just wait for the treadmill to pull you up because you do not need to lift yourself against gravity right? Keep waiting, keep waiting… Have you fallen off yet? Obviously the treadmill did not lift you, you continue to move down with the same speed as the belt. You cannot continue on and take another step until you have lifted your body up, against gravity, with your legs. Until you do that, your a$$ will continue to fall with the treadmill belt.

  73. #73 drRus
    February 28, 2012

    Yet more proof that scientists have no common sense, lol.

    It`s really simple and has very little to do with gravity, you numnuts. When you walk on a flat surface you are mostly switching weight from the foot you are raising to the foot on the ground, burning mostly energy needed to move the foot. For example, if you decide to stand on one foot on a flat surface – you will get tired eventually, but mostly because of the extra weight the foot is not used to, and not because you burnt large amounts of energy.

    But if you incline the treadmill, the foot on the ground will not only support the weight, but will have to push your entire weight upwards to lift your body up and keep balance. I.e. the foot on the ground is doing the actual work of pushing your body`s weight, rather than just supporting it. Duh!

  74. #74 Terry
    April 23, 2012

    What about manual treadmills? You have to push the belt backwards with your leg to keep moving forward and you can increase resistance to make that more difficult and use more calories. You could set the belt resistance low enough to use gravity to push it backwards but then you would begin to approach what the motor of a motorized belt does. One poin that hasn’t been mentioned is that hill climbing involves the body tilted forward with respect to the feet and the corresponding leg muscular orientation. You don’t get that with stair stepping for example becasue you are stepping a horizontal surface.

  75. #75 darseygodwin
    Calgary, AB
    October 17, 2012

    Everyone Go to the mall and get on an empty escalator taking people downstairs. Ride for 5 seconds, run up to where you were. Ride for 4 seconds and run back, then 3 seconds etc etc. You’ll see that you are climbing up the whole time.

  76. #76 Ashley
    London
    October 23, 2012

    My background is computer science and not physics so I am unable to comment one way or another with regard to much of the logic. However, Chris CD made an interesting point in that a human is a complex mechanism whose energy expenditure is very difficult to quantify. Therefore, it may be worth tackling the problem from a different direction.
    The body can be monitored very accurately in a gym using a treadmill and pulse rate monitor. If one wanted a comparative test specific to a hill walking (treadmill) scenario then the subjects pulse rate could be recorded at different workloads. Variations of this could then be conducted by changing the speed or incline as desired. These tests would be comparative and probably sports specific as walking / running is load bearing and swimming is the opposite. In an effort to provide a quantitative value the test could be bench marked against a test conducted on the flat through a full range of PR readings and the calorie output recorded.
    This would actually get round some of the problems associated with how the body assimilates carbohydrates and fat at different workloads and workout durations.
    In short, it may be better not to think of effort as being measured in calories but as a percentage of your VO2 Max. This increases as you get fitter i.e. you can run faster for longer and has the added benefit of taking into account your resting heart rate which is a reflection of your general well being.
    A book that may well be of interest to you is “Training Lactate Pulse-Rate” by Dr Peter Janssen. ISBN 952-90066-8-3

  77. #77 James
    January 16, 2013

    This is a very old topic at this point, but there is quite a bit of misinformation in the article and the follow up comments. The human brain is an incredibly adaptive organ and will hone in on the least metabolic cost gait in any situation. This means seeking a stride pattern that keeps the body’s center of gravity as close to stationary as possible. Watch anyone who runs frequently and you will notice that their torso does not change elevation. Olympic runners are incredibly smooth and efficient with their gait. Their center of gravity hardly moves.

    Maybe initially, as your brain trains a new gait pattern, you will tire very quickly as your center of gravity bounces up and down. In this case, yes, inclined treadmill running’s extra energy can be modeled by considering an elevation change over time.

    However, if you train yourself to run on an inclined treadmill, you will eventually develop a new running/walking gait specific to the task. Efficient gait treadmill running will have your body mass center of gravity held nearly stationary. If it is not nearly stationary, you will rapidly tire. Treadmill running is essentially spinning your wheels.

    It does require more energy to run at an incline, but it does not have to do with your body’s center of gravity changing in elevation. Instead, your legs are required to push off the treadmill surface with greater force. The increase in force is proportional to the incline angle and your own body mass. Each stride requires enough push-off force to move the leg forward, plus the vector component of your weight projected on to the treadmill. At 3 degrees of incline, this is about 5% of your body weight (the sine of the angle). The greater force translates directly into additional caloric burn.

    So in effect, the inclined treadmill is most like wind resistance as though your were actually traveling forward through the air. It is not at all like actually running up a hill. At extreme incline levels, your gait is changing more radically — you are between a run and a lunge at this point, which is less efficient in general as your legs are extending much more than they would if you were walking or running. Small incline levels do a nice job of providing a headwind like force that makes it feel (metabolically/energetically) like you are actually running outdoors.

  78. #78 Wow
    January 16, 2013

    However, you will still have Energy = Force x Distance.

    And an inclined treadmill will still have a greater force than a flat one to keep your body CoG steady and push for a greater distance.

    Not as much as if you were moving your CoG up, but still more than flat.

    Mind you the treadmill isn’t like running either. A bad step on the treadmill has your foot brought under your body. On the planet earth, you don’t get that pull and you lose the energy that your body reacting to the too-far-forward foot being an impediment to forward motion.

  79. #79 mark
    January 19, 2013

    The biggest misinformation in the above comments has been corrected several times already, but most of the corrections are more complex than necessary.

    The big misconception I’m referring to is the notion that climbing the treadmill is unlike a real hill due to not having an elevation change, and the accompanying assertion that the only reason it feels harder is that you’re “raising your leg more” each step.

    Topics above like swinging of arms, bouncing up and down, non-optimality of the human machine etc are all nice things to be aware of but aren’t critical to showing the equivalence of the work done on the treadmill and the hill.

    short story: work = force times distance, where we’re dealing with vectors so direction matters. We’ll only be concerned with the vertical component.

    In the case of a real hill you can deduce a lower bound on the energy used by observing the change in potential energy. You can also directly compute a vertical force (your weight) applied over a vertical distance for each step. Adding these up you end up with the same work done as the change in potential energy you computed from the elevation change.

    Now returning to the treadmill, the vertical force is still your weight, the distance is the sine of the angle times stride length, same as taking the same length stride on a real hill. Result: same force, applied over the same distance, same work.

  80. #80 Wow
    January 20, 2013

    Inclined treadmill: E=g * l * (1+sin a)
    Uphill = E=(g+e) * l * (1+sin) a + m * g * l sin a

    l = length of horizontal travel
    m = mass of you
    e = extra acceleration

    At a first approximation, you can assert that e is so small it is nelgigible. Even so, you need more energy to go up a hill.

  81. #81 mark
    January 20, 2013

    Too many unnecessary terms there. Just m*g*l*sin(a), that’s all you need. Any extra acceleration e would be negligible. People don’t accelerate faster and faster as they travel up the hill. Any positive acceleration at the start of a step would have a corresponding negative acceleration toward the end integrating out to zero. Also note the total change in potential energy due to elevation change in hill climbing: the simple formula for work done per step adds up to the expected total energy without needing any extra term e. This should tell you it is correct, save things like friction, wind resistance, human legs not being perfect springs etc.

  82. #82 Wow
    January 21, 2013

    “Just m*g*l*sin(a), that’s all you need. ”

    1) they were teased out to show why they were there.

    2) your alternative isn’t correct (though I missed out m’s).

    “People don’t accelerate faster and faster as they travel up the hill.”

    They HAVE to.

    The earth is making them accelerate downhill. To go UP hill, they have to move faster up to get their velocity over the time period of one step to a positive value than gravity is accelerating them downhill.

    If you hold out a weight at arms’ length, it will be accelerated down and keeping it level will take a force that would, without the gravity, have it accelerating at 9.81m/s/s.

    To lift that weight higher, you need to use more force, which accelerates it.

    “the simple formula for work done per step adds up to the expected total energy without needing any extra term e”

    No it doesn’t. As explained above.

    And even if it did, it wouldn’t matter, since you still have the vertical displacement term adding to the energy equation.

  83. #83 CB
    January 21, 2013

    To get a weight moving upward at a constant 1 m/s velocity, you can apply a 10.81 m/s^2 acceleration to it for one second, and then apply a constant 9.81 m/s^2 acceleration to it thereafter. Aside from the initial acceleration, the force required is the same as the force required for holding the weight at a constant height.

    Running on a treadmill is the same as running on a hill, only the constant velocity you’re trying to achieve is relative to the moving surface of the treadmill instead of the stationary hill. You still need an initial acceleration that is higher than gravity, but then afterward only need to match it. Ignoring wind resistance, the acceleration needed at any given point in time is the same in the two situations.

  84. #84 Wow
    January 21, 2013

    Then, however, you need to take the next step, CB.

  85. #85 CB
    January 21, 2013

    If you’re stopping after every step then you’re doing it wrong.

  86. #86 CB
    January 21, 2013

    Note that the hill and treadmill would still be equivalent, even if you did it that way.

  87. #87 Wow
    January 21, 2013

    If you’re not stopping after every step, you’re on the flat or running.

  88. #88 Wow
    January 21, 2013

    “Note that the hill and treadmill would still be equivalent”

    Nope.

    The treadmill brings the higher ground down.

    An unpowered one has the maximum gradient defined by the friction of you moving the belt.

    A powered one brings your foot back under your body.

  89. #89 mark
    January 21, 2013

    Is the main issue of disagreement here still the notion that the downward stepping force in real hill climbing is exactly g*m, that instead it should be g*m+e thus making the real hill climb take more energy than the treadmill?

    If so we can skip entirely the calculation of work = force * distance for the real hill climb and simply calculate the change in potential energy from the whole climb: m * g * vertical distance climbed, and deduce that the work done in climbing the hill equals the change in potential energy. This gives us energy to climb real hill = m * g * vertical distance climbed = m * g * L * sin(a), where L is the total distance (diagonally) climbed.

    Now returning to the treadmill do you agree that the work done per step in that situation is exactly m*g*l*sin(a)? The force is the climber’s weight, the distance is the vertical component of the step along the treadmill. After a sequence of steps l+l+l+…+l walking along the treadmill for a total distance of L, this adds up to work done of m * g * L * sin(a).

    Returning to the question of whether the stepping force in real hill climbing needs to be g*m or needs to be higher, I’d agree the initial step needs more to initiate motion but all subsequent steps only require exactly g*m to maintain a constant upward velocity. If you want the climber coming to a full stop every step then the direct “work = force integrated over distance” calculation would be much harder with force being a function rather than a constant (for part of the step you would have a stepping force g*m+e and for part of the step you’d have force g*m-e as you negatively accelerate back to an upward velocity of zero at the end of the step), but the key point is that conservation of energy tells you the end result will be the same either way.

  90. #90 Wow
    January 22, 2013

    “Is the main issue of disagreement here still the notion that the downward stepping force in real hill climbing is exactly g*m, that instead it should be g*m+e”

    No.

    Like I said all along: it doesn’t matter if you count e (times m, since e is the extra acceleration applied to move you up hill) as negligible.

    You still have the linear upward displacement term.

    I left it out so that those who were interested in the facts could see it explicitly.

    Apparently it still wasn’t explicit enough.

  91. #91 Wow
    January 22, 2013

    Now returning to the treadmill do you agree that the work done per step in that situation is exactly m*g*l*sin(a)?

    *(1+sin(a))

    So the short answer is “no”.

    Do you agree that after climbing 1000 steps on an inclined treadmill your body is little if any higher than it was at the start?

    Do you agree that after walking up 1000 steps on a hillside your body is much higher than it was at the beginning?

  92. #92 Wow
    January 22, 2013

    And please remember these:

    “An unpowered [treadmill] has the maximum gradient defined by the friction of you moving the belt.

    A powered one brings your foot back under your body.”

  93. #93 Wow
    January 22, 2013

    I suppose the short version is:

    If you’re using an unpowered treadmill inclined, the incline is usually a lot less than the reading indicates if you wish to relate the incline to going up hill.

    Much like treadmills are not equivalent to running on the actual surface of the earth. 5 miles in a treadmill != 5 mile run.

  94. #94 CB
    January 22, 2013

    A powered treadmill is just like running up a hill of equivalent incline and surface through a trivial application of Galilean Equivalence.

    If the treadmill is moving backwards at 1 m/s, then if you stood still then your body would also be moving backward at 1 m/s. To keep your body motionless relative to the rest of the room, you need to run at 1 m/s relative to the treadmill’s surface. It’s the same as if you were on a bus that was moving down a hill at 1 m/s. Running from the front of the bus to the back of the bus at 1 m/s would be equivalent to running the same distance on the hill the bus is on. The treadmill/bus only “brings your foot back under your body” from the point of view of someone standing next to the treadmill/bus, but that’s irrelevant because what’s actually happening is that you are propelling your body forward. It’s exactly the same as if you were running at 1 m/s up a hill, and someone else was riding a scooter at 1 m/s next to you — from their perspective your body would be stationary, the hill would be moving, and it would appear as though the hill “brings your foot back under your body”.

    Draw the force diagram. Draw an inclined plane, draw the gravity vector, and then draw the force vector required to maintain a constant velocity with respect to the inclined plane.

    Now draw a vector indicating that the inclined plane is moving at an arbitrary constant velocity. Notice how it doesn’t, and can’t, affect the force required.

  95. #95 Wow
    January 22, 2013

    No, it isn’t CB.

    Just isn’t.

    Here’s something to try.

    run up a hillside, then stop.

    You won’t slide down hill.

    Run on an inclined treadmill, then stop.

    You will slide down.

    An unpowered one using friction to make the loop of track roll will require the force you place upon yourself against g plus the force required to move the track against frictional forces.

    And that force cannot be any more than you would require to walk or run up such a slope IRL.

    But however you slice it, your opening statement there is false.

    Indeed, there’s no place where Ethan says that the inclined treadmill will act like a hillside of the same slope.

    Nor does he show such working out that they are equal.

  96. #96 CB
    January 22, 2013

    “Run on an inclined treadmill, then stop.

    You will slide down.”

    Not relative to the surface of the treadmill. You will be motionless relative to it. Just like you’d be motionless relative to the hill if you stopped on it, but the guy on the scooter would say that you’re sliding backwards. And he’d be correct, just like you’d be correct to say that the person standing on the treadmill was sliding downhill relative to you standing next to the treadmill. Where you’re wrong is thinking that your reference frame is special because you’re not moving relative to the surface of the earth.

    “An unpowered one” is not what I’m talking about.

    “Indeed, there’s no place where Ethan says that the inclined treadmill will act like a hillside ”

    He doesn’t actually address the issue directly. He just says “if you walk uphill (or on inclined treadmill)…”. So nice non-point. I guess your implication is that Ethan doesn’t believe in the equivalence principle. Ha.

  97. #97 Wow
    January 22, 2013

    “Not relative to the surface of the treadmill.”

    And sliding down a slide therefore means you don’t slide down???

    OK, I’ll see if I can mail a clue to you.

    There must be SOME way to get Teh Stoopid out of you. This ain’t working.

    Look, here’s something to try.

    Take your treadmill.

    Incline it as much as you can.

    Now put it against a hillside, in the valley.

    Which one is taller? The treadmill or the hill?

    This is a tiny clue that these two things are different.

  98. #98 Wow
    January 22, 2013

    Heck, if “Not relative to the surface of the treadmill” were valid, then there’d be no need to incline the treadmill, since you wouldn’t rise or fall compared to the surface of the treadmill there either.

  99. #99 mark
    January 22, 2013

    The lack of elevation gain in climbing the treadmill is the big red herring that I want to address here. Work is force applied over a distance. Looking at net elevation gain is just a convenient shortcut to deduce via conservation of energy that the hill climber must have done at least x amount of work.

    On the treadmill the net change in potential energy is zero yes, but you can still compute the work done by the climber by looking at force applied over distance. If it’s confusing to have a treadmill climber doing work but not gaining any net potential or kinetic energy, just note that the energy expended by the climber would end up as heat in the treadmill.

    Going back to the details of the formula maybe we’re using “l” differently. I was using it as the diagonal distance of the step, which would give the vertical (upward) displacement per step of l*sin(a). If we define “l” instead as the horizontal displacement of the step, that would make the vertical displacement l*tan(a). I still don’t see any need for (1 + sin(a)). I haven’t been leaving out the upward displacement.

    So you can replace sin(a) by tan(a) in all my formulas if you want, but the results still come out the same.
    work per step on hill = m*g*l*tan(a)
    work per step on treadmill = m*g*l*tan(a)
    change in PE on hill = m*g*L*tan(a) (where L = l+l+l+..+l)
    And the fact that in the real hill climb the work per step adds up to the total change in potential energy is a strong clue that the formula is correct.

    Which formula are you asserting is wrong? Are you saying the real hill climber is working more per step than I’m claiming above, or that the treadmill climber is working less per step? If you’re alleging one of those formulas is wrong, specifically what force applied over what distance (and direction) do you believe is wrong or missing?

  100. #100 mark
    January 22, 2013

    I think I can help provide an explanation concerning why the powered treadmill isn’t really doing any work for you even though it seems like it could be helping by pulling your leg back.

    Lets start by changing the situation into an example where the powered treadmill really would be helping. Suppose you have a rope attached to the treadmill in front of you and extending back horizontally and wrapping around your waist. The force exerted directly by this rope could only be horizontal in direction. But that force would indeed allow the powered treadmill to then truly pull the climber’s leg under them doing the job of lifting for the climber. To be specific lets look at all the forces and directions of movement involved to make that happen.

    – gravity acting on the climber: downward force m*g
    – climber’s foot pushing downward to resist gravity: m*g
    – treadmill pulling the climber’s foot back: horizontal force, as your foot isn’t glued to the treadmill: x
    – rope resisting the horizontal force of the treadmill: x

    It doesn’t really matter what x is, the rope will match that force exactly in the same way a table’s upward force exactly matches the gravitational force of whatever’s sitting on it.

    For calculating the work done by the climber, we need to put the above in terms of what force the climber is applying to the things the climber is in contact with, and note the directions. That comes out to
    1. foot pushing downward: vertical force m*g
    2. foot pushing forward on the treadmill: x
    3. back pushing backward against the rope: x
    The distance vector each of these forces is applied over is
    1. l*tan(a), the vertical component of the step
    2. negative l, since the climber’s force is in the opposite direction
    of the movement
    3. zero
    So the work done per step is now
    m*g*l*tan(a) – l*x
    Item 2 in the lists above gives us negative work done by the climber, ie work done by the treadmill for the climber. It’s as if the climber is a spring and the treadmill is compressing the spring giving the climber energy, doing work for them. So here we’ve successfully constructed an example where the treadmill could pull the leg under the climber and indeed do the climbing work for you, giving less work done on this modified treadmill than on a real hill.

    But notice what was required: a rope to hold the climber in place horizontally. In order for the climber not to zoom off the end of the treadmill the net horizontal force on the climber must be zero. The only way the treadmill does work for the climber is if you get the climber to do negative work by pushing horizontally forward into the treadmill while the treadmill motion is backward.

    In the case of the free-standing climber walking up the treadmill the net horizontal forces must be zero so you can’t push into the treadmill to have it do work, thus same work done on treadmill as real hill.

  101. #101 Wow
    January 22, 2013

    “even though it seems like it could be helping by pulling your leg back”

    Thanks mark, but problem: the flat treadmill where you run on the flat is where it “pulls your leg back”. The inclined element isn’t what is changing there.

    On the inclined powered treadmill, the bit you put your forward foot on will drop because the fabric it is on is being pulled down.

    Thanks for trying, but you’re not reading what’s written.

    Ask for clarification before jumping to an assumption.

  102. #102 Wow
    January 22, 2013

    ” Work is force applied over a distance.”

    And if you want to use that method, then you have to show that the leg on the inclined treadmill is exerting the same force as if it were not on a treadmill but on an inclined plane.

    This, however, is not the case with a treadmill.

  103. #103 mark
    January 23, 2013

    I wasn’t really studying the discussion of inertial frames of reference CB was explaining, and wasn’t referring to those situations. But I guess we’ll have to go there to talk about why the stepping force really is m*g.

    Ultimately we’re still talking about the comparison between these two situations right?
    1. walking up an incline of constant angle “a”
    2. walking the same steps along a powered treadmill of the same angle “a”

    Does the disagreement about the equivalence of situations #1 and #2 above center on the vertical stepping force in situation #2 then, which I claim is m*g? You’re saying that due to the constant downward motion of the treadmill the downward stepping force of the climber is somehow less than m*g right?

    The downward force m*g of gravity on the climber must be matched by the climber stepping with force m*g, anything else would result in vertical motion of the treadmill climber. It really is that simple, and the constant vertical movement of the treadmill downward doesn’t matter. It only influences the distance over which the force is applied.

    But I’ll try to be more specific anyway in arguing why the vertical force really is just m*g and why the downward movement of the treadmill doesn’t matter. Lets consider standing on a scale in a few different situations and figure out what weight the scale would read in each. First suppose you’re standing on a scale and raising a heavy rock upward at a constant rate. For this example lets let “m” bet the mass of you and the rock combined. Do you agree that the weight registering on the scale is just m*g, not greater than m*g? For what it’s worth this is similar to the question of what the stepping force is when climbing a real hill at a constant rate, which I hope we’ve already established is m*g and not greater than m*g. It’s true that during the initial acceleration to start the rock moving the scale would register m*g+e and then when you stopped lifting the rock the scale would show m*g-e briefly, but these would cancel each other out when it comes to calculating the work done, and during the time you’re lifting the rock at a constant rate the scale shows exactly m*g yes?

    Now change the example a little to make you very very tall and light so you can lift the rock a great distance and let the rock be responsible for essentially the entire mass m. Has anything really changed or is the scale still registering m*g the whole time you’re lifting the rock at a constant rate?

    Next situation, suppose you’re standing on a scale in an elevator descending at a constant rate. Do you agree that your weight registering on the scale will be exactly m*g the whole time? Again, during the initial downward acceleration you’ll register m*g-e for a moment and then at the bottom when the elevator accelerates back to a stop your weight would register m*g+e for a moment, but it should be clear these cancel each other out.

    Now combine these two examples. You’re in the elevator descending at a constant rate while lifting the heavy rock at the same constant rate so the rock is vertically motionless. What does the scale read?

    This is the same situation as walking smoothly along the powered incline treadmill. Suppose you have a little scale embedded in each shoe while walking the treadmill. What does that scale read? That is the downward force being applied by the stepper.

    To believe it’s anything other than m*g is similar to believing when you drive down the highway at a constant rate that your arms will be pinned back to the seat. But as long as the car is moving at a constant speed with no change in direction (whether flat or up a hill or down a hill) all movement inside the car is perfectly normal. You could toss balls around, stand on a scale, climb a staircase, whatever, and all the results will be the same as if you were at a stop.

    More specifically the car analogy would be that you’re inside a car driving down a hill at a constant rate, and inside that (big) car you’re walking up a staircase vs walking up the same staircase while parked. The little scale in your shoe will be registering the same weight as you climb in both cases.

  104. #104 Wow
    January 23, 2013

    I wasn’t really studying the discussion of inertial frames of reference CB was explaining, and wasn’t referring to those situations

    AFAICT, nobody was.

    Ultimately we’re still talking about the comparison between these two situations right?
    1. walking up an incline of constant angle “a”
    2. walking the same steps along a powered treadmill of the same angle “a”

    Yes.

    Does the disagreement about the equivalence of situations #1 and #2 above center on the vertical stepping force in situation #2 then, which I claim is m*g?

    If it were only g in #1, then you’d not get up hill.

    The downward force m*g of gravity on the climber must be matched by the climber stepping with force m*g, anything else would result in vertical motion of the treadmill climber.

    I’ve allowed it to be slightly higher, but let’s just say yes here for the sake of brevity.

    It really is that simple

    Only because you’re YET AGAIN making shit up.

    Where on EARTH did you get:

    treadmill the downward stepping force of the climber is somehow less than m*g right?

    ???

    YOU are the only one claiming this.

    You place it every time you call “m * g * sin a”.

    YOU are the one claiming “less than g”.
    Not me.

    It really is that simple.

  105. #105 Wow
    January 23, 2013

    And the other simple point is: you’ve only discussed the treadmill.

    Not the hill.

    To go UP THE HILL you need more force than just g.

  106. #106 Wow
    January 23, 2013

    I can run an average speed of ~9.5mph for ten minutes on a treadmill.

    IRL on the ground, even if flat, I can’t keep anywhere near that speed for more than a couple of minutes. Even with a following wind.

  107. #107 mark
    January 23, 2013

    I’ve discussed both the treadmill and the real hill and argued for a stepping force of m*g in each. You did start off claiming the stepping force in climbing the real hill is greater than m*g but I thought you then agreed the supposed extra force was negligible (zero when climbing at a constant rate). I am glad you agree the stepping force on the treadmill is m*g then as I thought that was the harder one to explain in detail.

    But I have to admit, I’m liking the simplicity of the inertial reference frames argument and finding maybe that’s the easier way to show the equivalence than force applied over distance.

    Consider a long bus driving at a constant rate down a long hill. And consider another bus siting parked along the side of the road on the same hill. Do you agree that all aspects of motion are identical and indistinguishable inside the two buses?

    I like the example of tossing a ball up and down in the bus to solidify the intuition that everything is the same. You see that you don’t have to toss the ball forward in the moving bus to counteract the buses movement, you toss it the same in both buses because everything is the same.

    So if I walked along the aisle from the front to the back of the bus, in both cases I have to apply the same force to take the same steps, so I’m using the same amount of energy and doing the same amount of work.

    Now all that’s left is to observe that we’ve just constructed situations equivalent to real hill climbing (walking up the aisle in the parked bus) and inclined treadmill climbing (walking up the aisle in the bus that’s driving slowly down the hill).

  108. #108 mark
    January 23, 2013

    Also didn’t you note the conservation of energy argument for why the stepping force was just m*g in the real hill climb? Compute the work per step using m*g as the stepping force and add up the work over all the steps. You see that the result is the same as the total change in potential energy for the hill climb right? Doesn’t this suggest m*g is indeed the correct stepping force?

    From another direction are you still thinking the real hill climber begins and ends each step from a standstill rather than maintaining a constant upward motion? If that’s the reason you’re wanting the stepping force to be greater than m*g at the start of each step don’t you see that modeling the problem that way also involves a corresponding stepping force less than m*g toward the end of each step so the climber will come back to a stop, and that these would cancel each other out when you integrate force over distance for the whole step?

  109. #109 Wow
    January 23, 2013

    I’ve discussed both the treadmill and the real hill and argued for a stepping force of m*g in each

    And you’ve not shown that they are both g.

  110. #110 Wow
    January 23, 2013

    You did start off claiming the stepping force in climbing the real hill is greater than m*g but I thought you then agreed the supposed extra force was negligible

    Please learn to read.

    “even if” was completely missed by you.

    Deliberately.

    Mendaciously.

    Tiresomely.

    Please try reading first.

  111. #111 Wow
    January 23, 2013

    - treadmill pulling the climber’s foot back: horizontal force, as your foot isn’t glued to the treadmill: x

    If it;s inclined, sin(a) of that is vertical, reducing the force needed to move the upper foot down.

    Which doesn’t happen on hillsides.

  112. #112 Wow
    January 23, 2013

    But if you walk uphill (or on an inclined treadmill), you not only need to move forward at whatever pace you were moving at, you also need to climb — a little with every step — out of the Earth’s gravitational field!

    This may be where I was going differently from the rest of you.

    I hadn’t really considered this as SOLELY walking, I.e. one foot on the surface at all times.

    That gets (probably negligibly smaller but close) the same scenario on a treadmill as a real incline. The difference being the transfer of weight to the forward foot takes time during which your assertion of the height of the step taken is made invalid by the movement of the surface you’re walking on.

    Running is quite a lot different.

    Don’t believe me?

    Go running on a treadmill. Go as fast as you can go for, say, 5-10 minutes.

    Try and do the same on a flat track.

  113. #113 Wow
    January 23, 2013

    Note: your equivalence can been shown to be false by trying this trick walking up an inclined treadmill.

    When you step up and forward, don’t straighten your upper leg. Let the foot carry down and your leg bend so that your upper body doesn’t rise as your leg comes under your body.

    When it’s done so, raise the other leg and drop it forward.

    You can do this much more simply when using a step/bench for aerobic/CV workouts.

    When jumping on to the step, keep your body at the same spot as when your feet were on the ground.

    Your CoM HAS to rise on the treadmill and the amount that rises is how much of an IRL climb you did.

    A treadmill will let your CoM to lower before the next step, ensuring you don’t rise out of the ceiling.

    By careful treading, you can keep your CoM almost steady.

  114. #114 mark
    January 23, 2013

    For what it’s worth I don’t think anyone is disputing that there’s a vast difference between an indoor perfectly smooth treadmill and the wild world of the outdoors with its uneven and constantly varying terrain. We’re only arguing about the comparison between two equally laboratory-perfect idealized setups.

    You say I haven’t shown that the stepping force is g in both cases (I assume you mean m*g, since g is an acceleration, m*g is a force), but I have several times.

    In post 103 I provided a very detailed explanation for why the downward stepping force on the treadmill is m*g rather than less than m*g and how the downward motion of the treadmill doesn’t change that. In 104 you blew up asking where on earth I got the idea that anyone was claiming the stepping force on the treadmill was less than m*g. Now in 111 notice how you’re claiming exactly that.

    In 89 I asked if you were concerned that the downward stepping force in climbing a hill needed to be greater than m*g, and I showed that it need not be. In 90 you said that no, that’s not what you were saying at all. Now in 105 you’re back to insisting that “to go UP THE HILL you need more force than just g” (where you again mean m*g).

    In 90 you also say I’m leaving out an upward displacement term. But my formulas have been including that the whole time. In stepping up a hill, the climber pushes down with force m*g. The vertical (upward) displacement of the step is l*sin(a). This gives a work done per step of m*g*l*sin(a). (In my formulas I’ve switched back and forth between sin(a) and tan(a) depending on what we meant by “l”, I think it’s easier to let “l” be the diagonal distance, the full length of the stride, so I’d like to stick with that definition and that leaves us with l*sin(a) for the vertical component).

    The only formula you’ve provided for work done per step climbing a hill didn’t make any sense
    “Uphill = E=(g+e) * l * (1+sin) a + m * g * l sin a”
    and you haven’t corrected it or explained what forces are supposedly being applied over what distances to justify it. Please be specific about what direction the force is in and what direction the movement is in. I recommend breaking the vectors into their horizontal and vertical components. I’m guessing you’re trying to say that work is being done by the horizontal movement while climbing the hill as well as by the vertical movement, but there is no horizontal force m*g involved.

    Finally in 107 I tried a different approach to showing equivalence, switching to a “frame of reference” argument involving two buses. This gives the simplest explanation yet for why the stepping force is the same in the two situations.

    Consider a long bus driving at a constant rate down a long hill. And consider another bus siting parked along the side of the road on the same hill. You’ve been in cars and elevators before and know that being in the moving car/elevator and being stationary is indistinguishable from inside the vehicle. It’s only when the vehicle accelerates or turns that you feel any forces other than when stationary.

    Next imagine you have a little scale in your shoe. No matter what you do, stand, walk around, jump, climb, if you compare the scale’s reading between the two buses when you’re doing the same action in each bus the scale will have the same reading. This shows that the downward stepping force (whether it’s m*g or not) is the same inside both buses.

    Completing the frame of reference argument I pointed out that the two buses are equivalent to the real hill climb and the treadmill climb.

  115. #115 Wow
    January 24, 2013

    mark, I don’t think anyone is talking ABOUT “an indoor perfectly smooth treadmill and the wild world of the outdoors with its uneven and constantly varying terrain”

    Look, until you manage to actually learn how to READ, there’s not much point in you trying to pretend that you are explaining anything.

    LEARN TO READ.

  116. #116 Wow
    January 24, 2013

    The only formula you’ve provided for work done per step climbing a hill didn’t make any sense
    “Uphill = E=(g+e) * l * (1+sin) a + m * g * l sin a”
    and you haven’t corrected it or explained what forces are supposedly being applied over what distances to justify it

    And this is YET ANOTHER goalpost whiffed about.

    First time you said that.

    You’re just a Gish Galloping troll, aren’t you.

  117. #117 Wow
    January 24, 2013

    Consider a long bus driving at a constant rate down a long hill.

    “Imagine you’re in a little cave all on your own, where you will meet your spirit animal…”

    Try something that isn’t bollocks, will you, mark.

  118. #118 Wow
    January 24, 2013

    Imagine you’ve inclined the treadmill.

    Imagine that you have taken one stride up the incline.

    Then, with legs one step apart, you leave them there.

    Your lets are pushing down just as much as your assertion of how the treadmill works.

    Do that for 1 hour.

    That is therefore the exact same work done as walking up that incline for 1 hour.

    Except it obviously is not.

    Therefore there must be something different between actually moving your CoM up and the force your legs are putting down to keep you upright.

  119. #119 Wow
    January 24, 2013

    Put a bus on a hill and don’t put the brakes on or put the engine in gear.

    Does it stay still?

    No.

    Put a human on a hillside and get them to stand still.

    Do they stay still?

    Yes.

    This is why you can’t go bleating about “imagine a bus…” in an analogy to a human using a treadmill.

  120. #120 mark
    January 25, 2013

    In reference to post 115, that’s not fair. Do you recall your post 106? In the middle of the discussion where you’re claiming real hills are harder than incline treadmills, you offer that you can run for longer and faster on a treadmill than you can in real life. So in passing I addressed the topic of perfectly smooth treadmills vs real-world terrain. Then accused me of not reading again, claiming that no one was talking about that situation, when I was just addressing a topic you brought up. And I was agreeing with you no less, confirming that nothing in any of the arguments we’ve had is trying to dispute your experience described in 106.

    In reference to post 116, saying my request for a specific formula for work done per step on a hill with details is new. I’ve been providing formulas and details, showing my work, listing all the forces involved and the movements. And I’ve asked for specific details if something I wrote is wrong or I’ve left out some factor. My point is your references to forces needing to be larger is way too vague. You’ve got to provide details.

    I do see a mistake in my writeups though that I’d like to fix. The discussion here has been about energy “used” by a climber. Energy is stored in objects in various forms, such as
    – potential energy
    – kinetic energy
    – heat energy
    – energy in the machinery of your body, I’m not sure what to call this one, I suppose ultimately it boils down to food energy, maybe “metabolic” energy
    – “coiled spring” energy

    When we say energy is “used” by the climber there are two possibilities involved in the bookkeeping:
    1. identify energy transferred between two objects
    2. identify energy converted from one form to another within an object

    #1 above is where work comes into play. Work is usually just described as “force over a distance = energy used” but a more detailed and correct description would be that work is energy transferred from one object to another by a force applied over a distance at the point of contact between the two objects.

    For #2 I’ve been using an imprecise “work-like quantity” involving movement of the climber’s center of mass to describe an energy conversion within the climber in terms of a transfer of energy from itself to itself.

    More specifically in the case of the hill climber if we do our bookkeeping with two object, a climber and a hill, lets look first at the work done by the climber. I had said the solution was to compute force over distance with stepping force = m*g and distance = l*sin(a) since that is the distance the climber’s center of mass moves. But that’s not true work. The distance moved by the point of contact is zero, thus zero work. And with this more specific bookkeeping that’s a good thing: it tells us that no energy was transferred from the hill to the climber. Intuitively that’s obvious, as the climber is one doing the work of climbing. So calculating the energy conversion here is a job for conservation of energy. Since no work is being done on the climber the climber’s total energy is not changing. But their potential energy is increasing by m*g*l*sin(a). This has to be matched by a decrease in one of their other energies, specifically it would come from their metabolic energy. So in the end we have the same metabolic energy being used that I’ve been saying all along, but it’s not actually “work” transferring energy between two objects, it’s an energy conversion inside the climber.

    This is the argument I gave in 89, but since I’ve also been framing the hill climb in terms of “work done per step” I want to correct that to “energy converted per step”.

    Where you could still get by with putting this in terms of work is if we break the bookkeeping down further by splitting the climber into multiple objects. Suppose the climber has a “leg object” and a “torso object” and lets put all the mass in the torso, you’d have the leg object applying a force over a distance to the torso object (and moving it at the point of contact) so now you’ve got a legitimate application of work = force * distance showing a transfer of energy from the leg (which loses metabolic energy) to the torso (which gains potential energy). The leg pushes up on the torso with force m*g therefore the torso pushes down on the leg with force m*g. The fact that the leg doesn’t move implies that the force is being balanced, specifically at the foot with a stepping force of m*g.

    I’ll concede the bookkeeping of forces movements and energy transfers inside the body gets convoluted fast. So overall this splitting the climber into multiple pieces to see the energy transfer between pieces is a bad way to model the situation. It’s way easier to just use conservation of energy as above to deduce the energy conversion inside the climber.

    So I’m modifying my claim then from “real climber and treadmill climber did the same work” to “real climber and treadmill climber used the same metabolic energy”. Specifically
    – real climber converted m*g*L*sin(a) from metabolic to potential energy
    – treadmill climber transferred m*g*L*sin(a) of their metabolic energy to the treadmill

  121. #121 mark
    January 25, 2013

    For 117: the long bus driving down the long hill at a constant speed isn’t a random unrelated example, it’s a situation exactly equivalent to the powered incline treadmill. You don’t see the equivalence there?

    For 118: I think this is a fair point that I haven’t been formal enough in the implications of “work” after identifying work done. Am I understanding this example is a long treadmill so I can stand there an hour without falling off the back?

    So you’re saying that in both this case and the case of regular treadmill climbing my analysis of work = force * distance would result in
    – downward force: m*g
    – downward distance: L*sin(a)
    – work done (by the climber to the treadmill causing heat): m*g*L*sin(a)

    And the complaint here is that the climber standing still on the long treadmill is clearly resting (could be sitting in a chair on the treadmill if we want) and yet my formula shows the climber doing work. And all along I’ve been stopping after showing work done, as if that were the end of the problem.

    That’s fair, the last detail is to account for the energy transfer and observe conservation of energy. In both cases the climber transfers m*g*L*sin(a) to the treadmill, so the climber must lose that amount of energy in some form.

    1. standing still on a long treadmill: the climber did lose m*g*L*sin(a) in potential energy, thus accounting for the required energy loss. No metabolic energy needed to be used to satisfy conservation of energy, so the climber is “resting” metabolically speaking.

    2. climbing a regular treadmill: the climber didn’t lose any potential or kinetic energy or heat etc. Thus they must have used their metabolic energy.

  122. #122 mark
    January 25, 2013

    For 119: I like this example, as I think it brings us to a key misunderstanding. You’ve been saying the downward stepping force on the incline treadmill is less than m*g because the treadmill is moving down out from underneath the climber, and quit saying I’m not reading, you did too say that.

    Consider my bus vs your bus. Mine being the one driving down the hill at a constant rate, yours being the one that doesn’t put on its parking break. There’s a big difference here. My bus rolls down the hill at a constant rate, just like the treadmill rolls along at a constant rate. Your bus is accelerating down the hill.

    This makes it no longer an “inertial frame of reference”. Everyone inside the bus would feel lighter than normal (like a partial free fall) and would feel like they’re being pushed toward the back of the bus. This bus is not equivalent to a stationary bus. But a bus moving at a constant speed with no change in direction is equivalent to a stationary bus.

    Anyway in your bus I agree the downward stepping force would be less than m*g. And if the treadmill were accelerating continuously faster and faster like a free-falling bus I would agree with a reduced stepping force on the treadmill. But the constant-speed bus is the equivalent scenario to the constant-speed treadmill therefore the equivalence argument shows the stepping force to still be m*g despite the fact the treadmill is moving.

  123. #123 Wow
    January 25, 2013

    Nope, you’re still incorrect.

    The hillside is not the treadmill.

    The bus doesn’t help because the bus rolls whilst we walk.

    Look, you’re doing the same as when they “proved” kangeroos couldn’t exist because they were sacks jumping up and down.

    You’re making the same mistake here.

    They are different.

    An inclined treadmill does not give you the same work out as walking up an inclined slope.

  124. #124 mark
    January 26, 2013

    So where’s the problem then?

    1. real hill climber must use at least m*g*L*sin(a) energy to gain elevation
    2. treadmill climber performed m*g*L*sin(a) work (transfer of energy to treadmill)
    3. we’ve identified all forces and movements between touching objects, so there is no other work being done, ie no other transfers of energy happening

    If one were to claim the treadmill climber somehow did less work than listed, the most likely candidate for “something’s missing” would be to claim that the powered treadmill did some of the work for you, say by pulling your leg under you. I addressed this in detail in post 100. Short version: if you’re holding onto the treadmills handles then yes it can do work (transfer energy into you) if you’re free-standing it cannot.

    With no other energy transfers happening, this proves
    – the hill climber used at least the amount of energy listed
    – the treadmill climber also used at least that amount of energy
    – if either used more energy than listed, it was internal conversion, not an energy transfer out of them, ie “non optimality of the human machine” producing waste heat

    All that’s left is the possibility that for some reason there’s a difference in this “non optimality of the human machine” between the two cases. Ie somehow the hill climber is more inefficient, wasting more energy as heat than the treadmill climber.

    Nothing in the equations so far forbids that possibility. But the “frame of reference” argument does.

    If you don’t see my bus as equivalent to a treadmill then I probably didn’t described the scenario well enough. The bus I’m claiming is equivalent to the treadmill is driving down a constant slope hill at a constant (slow) rate, and for easier visualization let it be a fairly long bus. Now have the climber walk down the aisle of the bus at the exact same speed the bus is driving down the hill. Now look at this scenario from outside the bus: the climber is motionless, walking along an inclined surface that’s moving at a constant rate underneath them.

    If you replace the bus with a train, and loop the train around in a big circle now you’ve really just got a giant treadmill going. That is why the bus moving at a constant rate down the hill is an exact replica of the treadmill scenario. And the “inertial frames of reference” argument guarantees that everything you can do inside the buses is equivalent.

  125. #125 Seth
    February 27, 2013

    Ironically, I’ve read much of this thread while walking at my treadmill desk at ~10% slope. I’ve been obsessed about answering the question. I hike in the mountains a lot, and my gut tells me that treadmill walking a slope is less tiring than actually walking a slope.

    I thought that post #77 was informative. The question I think ultimately comes down to biomechanics and gait efficiency.

    Imagine a flat treadmill at a slow speed. You could choose to walk normally and do so comfortably for an hour. Or, you could move forward with deep knee lunges and exhaust yourself in 10 minutes, essentially lifting most of your body weight 2 feet with every step.

    You haven’t raised your potential energy, but with an inefficient gait, you’ve done work similar to raising your body weight hundreds of feet.

    On a real hill, you are forced to use a lunge-like gait to raise your body up the hill with each step. I believe that an inclined, powered treadmill allows you to keep pace with a more efficient motion, even though the theoretical work done at 100% efficiency is the same.

    My guess is that an inclined walking treadmill gait requires less effort, but I would be interested to see metabolic measurements.

    Can anyone point to a study?

  126. #126 Mohammed
    May 11, 2013

    Everybody at Sportsplex in Bethel, Connecticut loves passing judgement on anyone who stares which means looking at women for a long time, as if its some sort of “problem”.

  127. #127 Wow
    May 12, 2013

    “1. real hill climber must use at least m*g*L*sin(a) energy to gain elevation”

    Plus move their body.

  128. #128 Robert
    San Diego
    August 15, 2013

    I’ve run lots of real hills. Running real hills is no comparison to running them on a treadmill. A 10% incline on a treadmill is easy for average mid-pack runners. Such a hill on the road will force most runners to walk.

  129. #129 scottbradley66
    Mount Juliet, TN
    September 15, 2013

    OK, let’s understand something here. When you walk or run on an inclined treadmill you are NOT JUST high-stepping. You are high-stepping to a higher point and then supporting your body weight up to that new point. It is a significantly greater effort required to walk or run inclined than it is to walk or run flat.

    While the % incline on a treadmill may not be as challenging as the same % incline on a real hill, flat is still flat, and inclined is still harder no matter how you try to explain it.

    It is silly to think otherwise. Go do some high steps on flat ground and then run up a hill the same distance and you will see a significant increase in effort required to go up the hill.

    As for the gravitational field being reduced significantly when you go up a hill, forget it. You can’t tell.

  130. #130 Wow
    September 16, 2013

    “. It is a significantly greater effort required to walk or run inclined than it is to walk or run flat. ”
    Scott, this is understood.

    “While the % incline on a treadmill may not be as challenging as the same % incline on a real hill, flat is still flat”

    And inclined is still inclined.

  131. #131 Shakes Head
    November 30, 2013

    I think Wow is a-trolling.

  132. #132 Wow
    December 1, 2013

    I doubt the second word there, shakey.

  133. #133 Michael
    USA
    January 8, 2014

    Simply use a thought model to analyze the problem:

    Imagine a big mountain with a 15% uphill slope. You are running up the slope with 4 mph. Now imagine that the whole mountain is slipping downwards towards the center of the earth with an angle equivalent to the 15% with 4 mph. Now ask yourself whether the work you did before, running uphill, much different from the work needed to maintain your distance from the center of the earth against the sliding mountain? I clearly believe not ( forget the gravitational and wind resistance effects )
    Now imagine the mountain is a gigantic treadmill.

  134. #134 Wow
    January 8, 2014

    Imagine a Kangaroo as a sack of potatoes that you have to lift up to the height of a kangaroo’s jump when running.

    Now add up the number of such jumps per day and multiply that by the energy needed to lift that sack of potatoes.

    Now compare that sack of potatoes to the energy from the food eaten by a kangaroo.

    Therefore kangaroos can’t exist.

  135. #135 Hank Antler
    North
    March 6, 2014

    It is really surprising to me that people think its equivalent (minus the aero drag).

    for a mental image let me do my attempt:
    Consider a stepper as the more vertical movement makes it more obvious. A person with a heavy back pack – lets say 100lb/23kg at a good pace. The back pack does not move much at all if the persons technique is good. Do you guys still think that its the same as carrying it up a real hill (actual work done in sense of physics = potential energy changed).

    In a stepper you carry your weight – support it with your legs . you do this through a range of motion which makes it hard but it is NOT he same as actually lifting yourself up. If they weight is not in a vertical movement no work in the sense of physics is being done.

    The oscillation of human locomotion is such that a lot of the energy is going to just the movement of the system, Think walking in place. What makes incline harder on incline treadmill then – I don’t exactly know – my guess is that its the stride being interrupted. Walking or running is sort of controlled falling forward . When you have an incline your forward step is interrupted at a higher level and you have start it from lower, This is not equal to the work of actually raising the mass (real world hill) but still heavier than flat ground.

    Lets go back to the stepper. Do you think its much easier to if the stepper was going backwards? In s setup where the machine was on reverse and you were walking “down” the stairs.
    Besides the coordination challenge that might make the test awkward I think it should be pretty obvious. Walking backwards on the stepper is roughly as hard as walking forward. You still have to support yourself through the whole range of motion. You are not getting any free work done by the machine.
    Both up and downhill stepper examples assume a technique where the center of mass doesn’t move vertically much at all.
    Naturally if your technique oscillates a lot you can get closer to the same situation as in real hill. Both going up and down.

  136. #136 Wow
    March 6, 2014

    “for a mental image let me do my attempt:
    Consider a stepper as the more vertical movement makes it more obvious.”

    A simpler image is that

    Energy = Force x distance

    And if you’re walking up a staircase that is dropping at the same rate, then you’re STILL pushing with g force at the steps and still stepping up one step whilst doing so.

    In broad terms, this is correct.

    In fine, it’s about as correct as the proof that shows kangeroos can’t exist because it takes more energy to lift a kangaroo up into the air and back down again sufficient times to reach the next bit of food than it can expect to get from that bit of food.

    The simple image is simple, but, to misquote Einstein, too simple.

    “Good enough for government work”, maybe.

  137. #137 Dee Buckland
    May 17, 2014

    I want to to thank you for this very good read!! I certainly loved every little bit of it. I’ve got you book-marked to check out new things you post

  138. #138 TD
    May 28, 2014

    Running and walking outside feel and are much harder than the treadmill, but the treadmill is gentler on the body and joints. In terms of pure exercise, what the treadmill does very well is to keep you at a constant pace, while walking and running outside the pace varies without you noticing. So if you tend to slow down when outside, you would be forced to keep the beat on the treadmill.
    So yes, a treadmill is much different to being outside, but I think that it is objectively not better or worse, just different.

  139. #139 Ronald Nicholson
    California
    July 28, 2014

    Running up an inclined does require as much energy as running uphill, even though the runners absolute altitude does not change.

    If there is no gain in altitude, why is any climbing energy required? The energy goes into the treadmill belt due to the angle of the force vector holding up the runner in relation to the slope of the treadmill belt. Normally, with zero incline, the force required to counteract gravity, and keep the runner from accelerating downward, is perpendicular to the belt, and thus has no effect on the belt speed. With the belt at an incline though, this downward vector force resolves into two components against the belt, one component perpendicular to the belt and one backwards against the belt. The backwards force component would thus accelerate a frictionless belt, requiring the runner to run faster and faster till he or she falls off. However the speed control of a treadmill will instead add braking friction or reduce motor power to keep the belt speed the same. Braking friction will add heat to the treadmill. It turns out this added heat is exactly the same amount of energy that is required to climb and gain altitude on actual hill of the same incline. Or if the treadmill’s motor power is reduced to prevent the belt from speeding up, the energy from the wall plug will be replaced by the energy from the runner, again by the same amount as required to climb an actual hill.

  140. #140 JJ
    Chapel Hill
    September 10, 2014

    I know this is really old but I stumbled upon this and just couldn’t help but comment. While there are many good analogies here, I’m amazed at the number of people who profess knowledge of physics yet don’t understand the basic principles.

    A treadmill belt moving at a constant velocity is just as valid an inertial frame of reference as the floor. When you are walking or running up the treadmill, you are in fact gaining altitude relative to this frame of reference.

    If this is hard to visualize then think of this equivalent analogy. Let’s say you are on top of a mountain and a treadmill extends down the entire slope of the mountain. You and a friend are standing side by side on this treadmill at point t = 0. You decide you are going to stay on top of the mountain but your friend decides he is going to exert absolutely no energy. In time, your friend will ride the treadmill all the way down the mountain. Your friend has done nothing to exert energy. He has applied no force at all. He could be dead for all we care. However, relative to your friend you have in fact gained potential energy.

    We can get into all sorts of discussions about where my energy goes but none of that is necessary to answer the question.

    Some other points:

    1. Chris CD might be the worst physics teacher I have ever encountered. While everything he says may be true, you don’t answer physics questions but trying to complicate the issue as much as possible. In this case we don’t care about the physics of the human anatomy. It’s simply not necessary to answer the problem. The problem can be answered simply be recognizing that a moving belt on a treadmill is a valid frame of reference. It’s as simple as that.

    2. For those who think to jog on a inclined treadmill is simply a matter of lifting your leg higher, it’s obvious they have never actually jogged on an incline treadmill. You are constantly pushing your entire body up. The treadmill is pushing your entire body down (just as in my example above) and you must push it up to maintain a constant elevation. This is painfully obvious to anyone who has actually run on a treadmill.

    I like to do the stairmaster a lot. On the stairmaster you gain no potential energy but you are constantly pushing your body upwards (force through a distance). Would anyone really apply that no work is done because no potential energy is gained?

  141. #141 JJ
    Chapel HIll
    September 10, 2014

    Ugh. apply = imply. be = by. Probably more.

  142. #142 JJ
    Chapel HIll
    September 11, 2014

    Wow wrote:

    “In broad terms, this is correct.

    In fine, it’s about as correct as the proof that shows kangeroos can’t exist because it takes more energy to lift a kangaroo up into the air and back down again sufficient times to reach the next bit of food than it can expect to get from that bit of food.

    The simple image is simple, but, to misquote Einstein, too simple.”

    Wow, you are simply incorrect here. I can’t speak for the Kangaroo example but I suspect is has absolutely nothing to do with this topic.

    There is nothing too simple about this. I think you need to study the concept of Galilean relativity. “The laws of motion are the same in all frames of reference in uniform motion.” This is a absolute truth not a simplification used to get an approximate answer.

    Take a very large escalator that is moving at a constant velocity. That is a perfect Galilean frame of reference. A person walking up that escalator is increasing his potential energy in reference to a point that is stationary with respect to the moving escalator. That’s true even if that person is not moving relative to the surroundings.

    Drop a five pound weight at the top of the escalator and then begin climbing to stay on top. That five pound weight is losing potential energy relative to you as it moves down the escalator. It is certainly not doing any work. You must do work to gain elevation relative to that weight.

    You can complicate the issue by asking why you are not actually gaining elevation relative to the ground and where your expended energy is really going but that is simply not necessary. All that is necessary is to recognize that the escalator is a valid Galilean frame of reference.

  143. #143 JJ
    Chapel Hill
    September 11, 2014

    Here is another example to demonstrate the point of how you can do work even though your potential energy relative to the surrounding has not changed.

    Let’s say you are on an elevator that is moving down at a constant speed. That is just as valid of a Galilean frame of reference as the ground. If a person woke up on this elevator while it was in motion, assuming there are no displays or vibrations/noise, that person would have no way to know that the elevator is in motion. All the laws of physics would be the same in either case.

    Now let’s say the same person grabbed a bar on top of the elevator and did a pull up and pulled his center of mass up at the same speed as the elevator is moving down. Someone from outside the elevator would say that that person gained no potential energy relative to the earth. Yet the person’s pull-up was just as difficult than it would have been if the elevator were stationary.

    The person has gain potential energy relative to the elevator. This can be demonstrated once the person lets go. He would fall back down to the bottom of the elevator (kinetic energy relative to the elevator which then gets dissipated as heat).

    It’s a different scenario but the physics principals are identical.

    I’m hoping this disproves all of the “you have exerted energy against gravity because you haven’t gained potential energy relative to the earth” arguments. Maybe I’m being naive.

  144. #144 Wow
    September 11, 2014

    JJ, your analysis would be correct if we were merely sacks of meal that were moved up and down but never deformed.

    However, this is not the case, Even with sacks of meal.

    I refer you to my earlier comment regarding kangaroos and their incapability to exist due to not being able to eat enough food to power their search for it.

    Simple analysis also “proved” bees could not fly.

  145. #145 JJ
    Chapel Hill
    September 11, 2014

    I’m not going to get into some misguided analysis of kangaroos or bees. I imagine someone did make an invalid assumption or logical mistake.

    However, I have not. One Galilean frame of reference is just as valid as any other. My elevator analogy demonstrates how a person can pull himself up (just as if he is on the ground), do work against gravity, gain no potential energy relative to the ground, yet gain potential energy relative to the elevator, and can do useful work with that potential energy if desired.

    Deformations have absolutely nothing to do with this. It doesn’t have anything to do with a person “walking up” an escalator that is moving down either. That person can perform that activity in a manner that is physiologically identical to a person moving up a stationary escalator. (Why don’t you try it and tell me you’re not lifting your body on every step.) The body deformations would be identical.

    If you don’t understand frames of reference then you don’t understand Newtonian physics. You may know F = ma, W = mgh, etc. (or their integral or infinitesimal forms) and be able to solve some problems but you simply don’t understand the reality behind those concepts.

  146. #146 JJ
    Chapel HIll
    September 11, 2014

    Let’s take the elevator analogy a step further so you can really understand it.

    An elevator can lower itself by releasing its potential energy. Maybe it’s a counter-weight that is rising (so potential energy is shifted from the elevator to the counter-weight) or some other mechanism.

    Now when I do my pull-up, I have slightly slowed down the rate of descent of the system (elevator and me) because I’ve changed the center of mass of the system. The energy I put into doing the pull-up has meant that the counter-weight did not rise as quickly while I was pulling myself up. While I have not moved relative to the ground, I have done work to change the energy distribution in the system and my energy relative to the elevator has certainly increased.

    I can then lower myself to the ground. At this point the center of mass of the system (elevator and me) is dropping at a slightly faster rate than before and the counter-weight moves up slightly faster than normal. My additional energy from doing the pull-up has again been transitioned back to the counter-weight.

    You can analyze a treadmill in just the same way. I don’t happen to know the inner details on how a treadmill motor works. But I really don’t have to because all of this is unnecessary. All I have to do is understand that an elevator moving down or a treadmill belt moving are perfectly fine frames of references.

  147. #147 JJ
    Chapel Hill
    September 11, 2014

    Forget my previous comment. It has issues.

  148. #148 JJ
    Chapel HIll
    September 13, 2014

    Expanding on the treadmill example above….

    The arguments start with a few simple formulas we all know:

    W = f*d
    PE = mgh

    What’s lost here is that (just like velocity), distance, height, work and potential energy are all relative concepts. My height or my distance moved depends on my frame of reference.

    To illustrate. Let’s use an escalator since it’s maybe a little easier to visualize. We have three people. Person A is on the top step of an escalator. Person B is beside person A. Person C is at the same height as these two but is standing off of the escalator. For ease we’ll say they all have a mass of m.

    Person A walks up the treadmill to maintain his position. Person B rides the treadmill down. Has person A changed his potential energy? Relative to the frame of reference of the treadmill he certainly has. Relative to that frame of reference he is moving up through gravity. This can be demonstrated by the fact the he is gaining potential energy relative to person B who we know is exerting no energy whatsoever.

    So, what about person C? He’s not exerting energy yet he’s maintaining the same potential energy as person A. However, the ground it exerting a force of mg on person C. In the frame of reference of the treadmill, the ground is in fact doing work on person C.

    They key is that work and potential energy are relative. In the frame of reference of the ground, A and C are not gaining potential energy. The treadmill is exerting a force of mg on person C which is in the opposite direction of his motion. His change in potential energy is –mgh. Relative to the treadmill’s frame of reference, person B’s potential energy is unchanged and A and C are gaining mgh potential energy.

    The key isn’t the absolute value of energy but the difference between A/C and B. In all cases A/C has mgh more potential energy than B.

  149. #149 JJ
    Chapel Hill
    September 14, 2014

    By the way, in the above example I stated all three have identical mass. If they didn’t then I would need to discuss their relative gravitational potentials (J/kg).

  150. #150 Wow
    September 15, 2014

    JJ, humans are more than just unreactive masses.

    Please include this knowledge in any future attempts to make a point, otherwise you’re wasting magnetic domains.

    TIA

  151. #151 JJ
    Chapel Hill
    September 16, 2014

    Please explain the relevance of the fact that we are just unreactive masses.

    It doesn’t matter what humans are like. Human physiology plays no role here. Humans still must obey the laws of physics.

    One of the most basic laws is that the other laws of physics are the same in any inertial reference frame.

    Other than wind resistance, there is no difference between climbing up a smooth hill of 5% incline at 5 mph and walking up treadmill with a 5% incline at 5 mph.

    You seem to think that the physiology of a human invalidates the equivalency of Galilean frames of reference. You are wrong. They rules of physics hold for a human, an ant, or a rock just the same.

    Running up a hill can be viewed the same as the hill falling below you and you’re trying to maintain your position. In physics terms they are exactly equivalent. The only difference is whether the air around you is moving or not. That’s where wind resistance comes into play.

    If you want to argue that a treadmill belt is softer or grips more or vibrates more then those could be valid objections but that is no different than running or rocks, or asphalt, or concrete. They all pose slight differences.

  152. #152 MtnManJohn
    December 10, 2014

    I just came across this article, and I am curious how the “Extra Work” was calculated based on distance and incline? Thanks!

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