Earlier this week, I wrote about an inclined treadmill, and talked about physical work. Physically, the amount of work that you do is equal to the amount of force that you exert in a certain direction multiplied by the distance you move in that direction.

If you walk up an incline, as opposed to moving on a level surface, you have to also fight the force of gravity to get up that hill, hence you have to do extra work.

I contended that, in order to walk up an inclined treadmill, you also have to do extra physical work, the same way you have to do extra physical work to walk up a real hill.

“Not so fast,” said many of the commenters! Unlike a real hill, where you actually wind up at a higher elevation than you started at, you don’t travel any real distance on a treadmill, and you certainly don’t end at a different elevation than you started at! Therefore, the argument goes, you don’t do any extra physical work.

So, I ask you, what would Einstein have to say about this? Einstein — mastermind of the equivalence principle — realized that as far as forces go, the only thing that you feel is acceleration. Your environment doesn’t matter at all.

What’s a good, analogous thought experiment for our treadmill/hill problem here? Imagine an escalator. Now, you’re going to walk up this escalator, and you’re going to walk up 50 steps on this escalator for three different cases, like Fred here.

If you walk up 50 steps, and each step is 0.2 meters high, then you will raise your elevation by 10 meters relative to not taking those 50 steps. In other words, it doesn’t matter whether that escalator is moving up while you do it, in which case you might wind up raising your elevation by 20 meters total, or the escalator is moving down, in which case you might wind up not changing your elevation at all.

In either case, you have to do work against the force of gravity, regardless of what your velocity is. This even works in the case of a broken escalator, a.k.a. stairs.

And that’s why, regardless of whether your elevation changes or not, you do work whenever you walk up an incline! Hope you’ve had a great week, and I wish you all a great weekend!

Comments

  1. #1 Kurt
    March 12, 2010

    I’m pretty sure Einstein would *not* have said that.

    (I’m not saying that running on an inclined treadmill isn’t harder than running on a level treadmill. But trying running on a 10% inclined treadmill for 10 minutes, and then running up a road with a 10% grade for 10 minutes, and tell us if you still think those two situations are roughly equivalent.)

  2. #2 elspi
    March 12, 2010

    “But trying running on a 10% inclined treadmill for 10 minutes, and then running up a road with a 10% grade for 10 minutes, and tell us if you still think those two situations are roughly equivalent.”

    You are in luck (or rather out of it) I run up a grade every other day. Just this last weekend I was away from home and so I used a treadmill instead (at 15%). It was identical in terms of workout. (ok some parts of my route are >15% but on the grades where it is about 15% it was the same on the treadmill)

  3. #3 John Armstrong
    March 12, 2010

    It’s a lot easier if you stop ignoring the simple fact that the treadmill has a belt running backwards.

    That is, you do work in going up the incline, and then the belt drags you back down. Net effect: you’ve done work.

  4. #4 Roger Migently
    March 12, 2010

    IANAS but … if you walk up an actual incline and reach a greater height, have you not increased your Potential Energy, whereas if you use a treadmill (or a down-escalator) your PE has not changed? If the energy exerted is the same, what has happened to the PE? Does this matter? Am I stupid?

    PS I must use that O-O-O escalator poster in motivational sessions…

  5. #5 Duncan Ivry
    March 12, 2010

    @ Ethan:
    Again it’s remarkable how good you are in explaining something. Because English is not my first language, I have some difficulties, but even in my own language I don’t like teaching, I prefer learning and doing.
    By the way, usually I don’t read your weekend diversion, because I’m not interested in reading something non-astronomical here. It just happened, that some … er … strange statements caught my eyes.

    But: “… analogous thought experiment for our treadmill/hill problem here? Imagine an escalator.”

    The escalator is *not* equivalent to a treadmill “here”, because the escalator *does* move the body on a path up through the gravitational field and the treadmill *does* *not*. This is the point for doing *physical* work. This should really be clear now.

    Don’t get this wrong — I like your blog very much, and it is *your* blog, and all this — but you should talk about astronomy and astrophysics, if you understand what I mean.

    @ Kurt:
    The comparison of running on a inclined treadmill and the same up a road, and saying “tell us if … situations are equivalent” is going more into the direction of what I said before, and what Richard Feynman (who? see below) wrote: Do we talk about physical work — where it is relevant whether the body moves along an appropriate path relative to the gravitational field, and only this! — or about physiological work? May be, you talk about the latter. And, may be, Ethan is partly talking about this and partly about that. Several other commenters definitely were not able to make the difference and to understand the point: Doing *physical* work on a treadmill by utilizing *gravity*.

    @ All:
    For those how don’t know: Richard Feynman was an American physicist (1918-1988) who received the Nobel Prize in Physics in 1965. His Feynman Lectures on Physics is a legendary physics textbook. I used it when studying mathematics and physics at the university, and I still use it sometimes today for my work as a developer of software for the simulation of physical-technical systems. It’s useful, when I want to be sure about the basics again. That’s why I used him here.

    Talking about basics: The problem of work in a gravity field on a treadmill is something undergraduates should be able to solve. Asking what Einstein would say is — sorry — a little bit ridiculous.

  6. #6 kevin
    March 12, 2010

    I’m with Ethan, although at first I wasn’t. The “no change in elevation therefore no work” argument seems convincing. Ethan has explained a different way to view things (by taking the environment out of the picture, Einstein and all…). How to reconcile?

    On the treadmill, if you expend no work, say, by stopping walking, you will *lose* elevation. The work you do by walking isn’t gaining you elevation, it is balancing this loss.

    Now what I can’t figure is whether this is the same as a helicopter hovering (does it do work, ignoring turbulence and other losses?) or holding a weight above your head (a favorite no-work example for high school teachers).

  7. #7 Colin
    March 12, 2010

    First, the closed escalators becoming stairs is a bit by Mitch Hedberg. The picture doesn’t cite.

    Second, I think the hardest part about the inclined treadmill or a helicopter is even if the whole doesn’t move but part does then it can still do work. Just because your reference frame aligns with the helicopter it doesn’t mean the helicopter is not exerting a force on moving air.

    If a boat is moving at 10 knots through still water it unquestionably does work. If you’re in a boat going 10 knots then the other boat appears to not move but it still unquestionably is still doing work. This is no different than being still and having the boat drive 10 knots into a -10 knot current: the boat still exerts a force on the water through the distance of the propeller.

    If your reference frame was on the treadmiller’s foot then you wouldn’t question if the foot was doing work.

  8. #8 David
    March 12, 2010

    if you walk up a real hill, and then fall off a cliff at the end to return to your starting height, you will have done the same amount of work as if you walked up the hill and didn’t fall off the cliff. The treadmill is like falling off the cliff at the end, only a little bit at a time.

    If you think that the “work” of a workout comes from the difference in potential energy after minus before, then why is a flat treadmill (or a run on level ground) any work at all?

  9. #9 John Armstrong
    March 12, 2010

    @David: Same as I explained above. You expend energy (doing work) lifting yourself with each stride, but you don’t get back that energy as you back down. Only on level ground it’s all gravity and no treadmill bringing you back down.

  10. #10 george.w
    March 12, 2010

    Nope, still not buying it, and I have plenty of experience with treadmills, hills, Stairmasters and real stairs. Sure, you work harder on an inclined treadmill because you are lifting your legs a bit more with each step. On level ground this would be called high-stepping, and the weight of your legs is not inconsiderable. But you’re not lifting your body more with each step; you can hold onto the railings while you walk. Your body stays at the same elevation (other than normal vertical movement that occurs while walking). The treadmill and hill are not equivalent.

    And for reasons mentioned above, neither are the escalator and treadmill.

  11. #11 Anonymous
    March 12, 2010

    Put a yo-yo on the treadmill so it can roll up and down, but hold its string so it stays in place.

    Now as the treadmill runs, the yo-yo is doing work (spending energy) against gravity to stay in place?

    Run the treadmill backwards. Now the yo-yo would be GAINING energy. You should be able to store as much as you need if you let it run long enough. Eventually your yo-yo-battery can even store up more energy than a nuclear bomb! Sweet!

  12. #12 Jason A.
    March 12, 2010

    I think people are having a problem with this because they want it to be discrete – i.e. walk faster than the treadmill so you get ahead of it to the front, then stop and let the treadmill drop you to the back, then repeat. Then it’s obvious you do work to run up, then let the treadmill pull you to the back, then again. You get these nice discrete chunks of ‘work’ (walking up the incline faster than the treadmill so your whole body moves up) and ‘no work’ (standing still so your whole body goes down). They can’t see this is equivalent to when you’re walking the same speed as the treadmill so your body never moves up or down (in the reference frame of the room). I don’t know why. Think of an extreme case, when the incline is something like 45 degrees but you’re still walking at the right speed to keep your body ‘still’. If you lift your leg to get on the incline, and your leg ends up under your body again, isn’t it obvious your body has lifted itself (in the reference frame of the walking surface)? How else could your leg get back under your body and extended?

    Consider the discrete case, where you run to the front of the incline, then let the treadmill pull you all the way back to the bottom, and repeat. Everybody agrees you’re doing work in this case. Let the distance you travel up and down each time to be dy. Now look at the limit where dy approaches zero. You’re still moving up and down, but the amounts are so small you don’t notice. You’re still doing work against gravity, even though your body appears to be standing still (with respect to the room).

    Or heck, just consider your average velocity (with respect to the room) in the discrete case. If you’re moving dy uphill then dy downhill, then the velocity up * time up = velocity down * time down, and the average velocity (again, with respect to the room) is zero, which is exactly what was confounding people who thought their bodies weren’t moving.

    The problem is people are wanting to measure their bodies movement with respect to the room rather than the walking surface. That’s why they can get the discrete case (your body does move with respect to the room) but not the continuous one.

    I’m sure that explanation made little sense to some people, but this is for the people who are getting themselves too caught up in the math ( W = F * dy ) to see what’s going on. They’re trying to plug through the formula, and getting zero work because they’re using zero for dy, because they’re not thinking about what dy is measured with respect to.

  13. #13 bob
    March 12, 2010

    Since treadmill testing for cardiovascular fitness shows that O2 consumption increases with increasing inclines, it seems that work must increase when the incline increases. Data, not theory.

  14. #14 Jason A.
    March 12, 2010

    george.w

    But you’re not lifting your body more with each step; you can hold onto the railings while you walk. Your body stays at the same elevation

    Your body is definitely not staying at the same elevation. Stop thinking about what your body is doing with respect to the room and think about what matters, what it’s doing with respect to the walking surface. If you stand still (doing zero work), you end up several hundred feet below (assuming infinite treadmill). You must work to prevent that.

    What you are suggesting is equivalent to saying that climbing a mountain is doing zero work against gravity because your body never climbed above your friend who climbed with you. There it’s obvious what the mistake is, measuring your climb with respect to your friend instead of the walking surface (the mountain). It’s the same mistake you’re making here. The room is like your friend, and it doesn’t matter that you never climb with respect to it, what matters is if you climb with respect to the walking surface.

    You probably want to reply that my analogy is flawed because the room is stationary in the gravitational field but the mountain climbing friend isn’t, and you’d be right about that, but it’s irrelevant. Consider a uniform gravitational field, then the analogy holds.

  15. #15 Anonymous 34
    March 12, 2010

    bob, physiological work would naturally increase as the incline goes up, due to the angles of the forces. But this is not representative of work done against the gravitational potential, as implied by the original article.

  16. #16 Jason A.
    March 12, 2010

    Anonymous:

    Put a yo-yo on the treadmill so it can roll up and down, but hold its string so it stays in place.

    Now as the treadmill runs, the yo-yo is doing work (spending energy) against gravity to stay in place?

    Run the treadmill backwards. Now the yo-yo would be GAINING energy. You should be able to store as much as you need if you let it run long enough. Eventually your yo-yo-battery can even store up more energy than a nuclear bomb! Sweet!

    Yes, if friction with the walking surface does rotational work by rolling the yo yo, you can gain energy and charge a battery like that. This is not controversial, this is how a turbine in a dam works, for example.

    But no, the yo yo does not need to expend work to fight gravity, no matter which way you run the treadmill. By tying it off to something ‘at rest’ with respect to the room, you have changed the relevant reference frame.

  17. #17 John Armstrong
    March 12, 2010

    @Jason: that may be a better way of phrasing it. Let’s combine it with my approach!

    So, you’re exerting a force to move forward. This is the one the equivalence principle bears on. The treadmill’s belt is also exerting a force to move you backward. The net force (and net work) is zero. But you are doing work, and the treadmill does negative work, and these two cancel out in the reference frame of the room.

  18. #18 Jason A.
    March 12, 2010

    Anonymous 34:

    this is not representative of work done against the gravitational potential, as implied by the original article.

    Yes it is. I fail to see how you can justify that it isn’t.

    Consider the treadmill to be stopped and you walking up the incline. You’re doing work against gravity, right? Now, how is this different? The only thing that’s changed is which frame you consider to be ‘still’ and ‘moving’. And the laws of physics are the same in any non-accelerating frame, so if one frame gives work against gravity then the other must as well.

    You’re still trying to measure movement with respect to the room rather than the walking surface, so you say your body doesn’t move up. This is incorrect.

  19. #19 Jason A.
    March 12, 2010

    John, I like that explanation.

  20. #20 bob
    March 12, 2010

    | But this is not representative of work done against the gravitational potential, as implied by the
    | original article.

    Ok, i didn’t read the original article. But I don’t understand your point. If you were on a treadmill in zero gravity, just changing the angle of your body to the treadmill wouldn’t change the physiologic work involved. Doing that work in a gravitational field is what increases the work required.

  21. #21 Anonymous
    March 12, 2010

    Yes, if friction with the walking surface does rotational work by rolling the yo yo, you can gain energy and charge a battery like that. This is not controversial, this is how a turbine in a dam works, for example.

    I agree about this potential for rotational work, but I was talking about whether or not the yo-yo / human does work against the gravitational potential. If they do, then the yo-yo would have to be able to store energy even if it was held on a frictionless string.

    Consider the treadmill to be stopped and you walking up the incline. You’re doing work against gravity, right?

    Correct.

    Now, how is this different? The only thing that’s changed is which frame you consider to be ‘still’ and ‘moving’. And the laws of physics are the same in any non-accelerating frame, so if one frame gives work against gravity then the other must as well.

    No, your relative position in the gravitational field has also changed, and that is important! If you think it doesn’t matter, try considering these two scenarios: one guy walks on an inclined treadmill for an hour. Another guy walks up a hill for an hour, then rides down a turbine to produce some energy, and ends up where he started. To me, the extra turbine energy equates to the extra work spent walking up the hill instead of a treadmill. If you think that the walking work is equal, how do you explain the energy balance? One situation results in more final energy than the other?

    If you guys are going to keep using “Einstein’s relative frames” to argue, then please also explain what is the correct calculation for the rest frame of the room. My take is that if Einstein found gravitational work done in the treadmill frame, he would still calculate zero gravitational work in the room frame, and then realize the discrepancy where the gravitational field is moving in the treadmill frame actually makes it zero gravitational work done.

    The argument about comparing it to you moving discretely back and forth along the treadmill is valid for your feet and your legs to an extent. You do actually do the gravitational work on your feet, and then that stored energy is lost to kinetic energy. But this doesn’t apply to your torso if it’s not actually moving up and down.

  22. #22 Anonymous
    March 12, 2010

    Another two situations to consider:

    Put on roller skates (this is to take the physiological work of maintaining muscle force out of the equation). Tie a rope around your waist and hitch it to a car going up a hill. There is extra gravitational work done here (corresponding to the extra load on the car).

    Put on roller skates, get on a treadmill, tie a rope from yourself to a bar in front of you so that you maintain position while the treadmill is running. There is no gravitational work done here.

    Similarly, walking up a hill takes more energy than walking an inclined treadmill (at the same speed/angle).

  23. #23 george.w
    March 12, 2010

    Here’s an extremely simple way to test this in isolation of the action of your legs on a moving incline. Equip an RC model car to measure voltage and amperage during operation. Set it on the treadmill running level at 3mph and figure the wattage. Incline the treadmill at the same speed and see if the wattage changes. Then run it up an actual hill of the same incline and the same speed and check the wattage there.

  24. #24 Sphere Coupler
    March 13, 2010

    If you break these analogies down to the fundemental interactions, the disparity would be the decrease in atmospheric pressure at higher elevations,and the decrease of gravity from 1g.
    Assuming no temperature,humidity,windage change and of course a static solar system.
    Oh and of course the incoming solar and cosmic winds.
    And all this is assuming your climbing up a down running escalater and your not on the bottem two steps!ha ha!
    (versus maintaining elevation in the center)
    .
    In a twenty minute work out the disparity would be insignificant for a human.

  25. #25 Karl Lembke
    March 13, 2010

    I think the phrase that clinched it for me was

    If you walk up 50 steps, and each step is 0.2 meters high, then you will raise your elevation by 10 meters relative to not taking those 50 steps. In other words, it doesn’t matter whether that escalator is moving up while you do it, in which case you might wind up raising your elevation by 20 meters total, or the escalator is moving down, in which case you might wind up not changing your elevation at all.

    Indeed, in any given reference frame, the work you do against gravity is the same, whether the reference frame is moving up, down, or staying still — so long as the velocity is constant. Constant velocity means acceleration equals zero. That means the acceleration you’re working against is the acceleration of gravity, plus zero.

    Instead of an inclined treadmill, imagine a ramp in a large room. Let the room move downward at a constant speed. Because acceleration is zero, any movement away from the floor of the room works against the acceleration due to gravity.
    Let a person walk up the ramp. Adjust the speeds of the walker and the room so an external observer sees the walker staying at the same height. The walker is still doing work against gravity, despite the fact that he’s not changing level with respect to an outside observer.

  26. #26 Kurt
    March 13, 2010

    Okay, I’m going to have to admit I was wrong about this. Sorry about the snarky comment at #1.

  27. #27 ElmerPhuD
    March 13, 2010

    Here’s a thought experiment which clarifies things perfectly for me: imagine getting rid of the treadmill motor and running on a belt looped over a frictionless surface.
    Now it’s obvious: the steeper the incline, the more energy you’d have to exert to maintain your position. Where does that energy go? Into accelerating the belt ever faster.

    Now put friction back in: instead of endlessly accelerating, the belt reaches some constant speed and additional energy is “burned off” as friction.
    In short, you’re always “climbing”: adding what would become potential energy over solid ground but is immediately converted to heat on the moving belt.

    The treadmill motor just adds a little energy to raise the equilibrium belt velocity.

    The whole system is, almost literally, a giant hamster wheel. Remaining at any position above the horizontal means running harder.

  28. #28 Waydude
    March 13, 2010

    OMG, you are all morons and missing the greater point that the inspirational poster was a joke by Mitch Hedberg!!!

    Mitch Hedburg!!!

    RIP.

  29. #29 mrcreosote
    March 13, 2010

    Walking up an escalator in the opposite direction to which the escalator is going is exactly analogous to the treadmill. For each step you must raise your body by the height of the step just to maintain your absolute elevation. The same is the case with the treadmill. If the treadmill is at 10%, and you take a 1 meter step each time, then you are raising your body 10cm with each step. It is just that on the treadmill, you fall 10cm by the time your foot reaches the back of the stride, whereas on the mountain, it doesn’t.

  30. #30 Hilary PhD
    March 13, 2010

    It staggers me that something so simple raised so much debate.

    Not sure why you wonder what Einstein would have said though. Newton had already cracked this one.

  31. #31 Anonymous
    March 13, 2010

    And the falling 10cm makes all the difference, since it makes it so that you don’t have to spend that extra energy to fight gravity for 10cm, unlike climbing the mountain. It’s that simple.

  32. #32 Art
    March 13, 2010

    As an aside it has to be noted that escalators breaking and becoming stairs is dependent on certain portions of the drive system failing. It is possible, I’ve seen it, where the belt of stairs becomes free-wheeling and the stairs fall back as you climb.

    A lot depends on the design of the mechanism and how good the bearings are but it is possible if the system to very loose to see a healthy person really working and making very slow progress. Arriving at the top winded but well exercised.

    Going down the effect is a bit more dangerous. Stepping on the chain of steps rapidly gains speed and you hit the bottom moving at several times the normal pace but your feet aren’t moving. It is like stepping off an escalator that is running too fast. If your somewhat less than graceful in your footwork a face plant is the likely outcome.

  33. #33 Chris Kuklewicz
    March 13, 2010

    Here are the comparisons that explain that the inclined treadmill is exactly equivalent to a hill:

    1) Einstein liked comparing elevators moving at constant velocities up and down. Let us agree with Einstein that constant up or down velocity of “you plus floor” does not change the physics.

    2) Imagine an infinite planar incline as a hill. You have Alice and Bob and Charlie at the same point at the start of the clock and then Alice climbs the hill. Charlie walks sideways (neither up nor down) and Bob stands still.

    3) Alice is clearly doing more work than Charlie from Bob’s viewpoint.

    4) Now move the viewpoint to one that tracks (at constant velocity) Charlie and you get a level treadmill. Charlie’s walking is doing more work than Bob’s standing still.

    5) Now move the viewpoint to one that tracks (at constant velocity) Alice and you get an inclined treadmill. Alice’s walking plus climbing is doing more work than Bob or Charlie.

  34. #34 coldfrontin
    March 13, 2010

    I’m sad about the guy who explained why broken elevators cannot be used as stairs. He totally ruined the mitch’s joke for me….

    @ #5, the escalator and the treadmill are EXACTLY the same: imagine an escalator with a more subtle incline and with infinitely small steps. The ‘chunks of work’ are just easier to visualize on an escalator.

  35. #35 delosgatos
    March 13, 2010

    22 wrote:

    Put on roller skates [...]. Tie a rope around your waist and hitch it to a car going up a hill. There is extra gravitational work done here (corresponding to the extra load on the car).

    Put on roller skates, get on a treadmill, tie a rope from yourself to a bar in front of you so that you maintain position while the treadmill is running. There is no gravitational work done here.

    For this comparison to be valid, in the treadmill case you would have to tie yourself to a car on the treadmill.

  36. #36 delosgatos
    March 13, 2010

    A good way to approach this might be to consider, where does the energy go?

    If I walk up an incline and then jump off the top, the potential energy I stored by doing work to walk up is converted to kinetic energy. True, once I hit bottom (assuming “bottom” is the same altitude as at which I started) my “net work” is zero, but that’s because I “gave it back”.

    On the treadmill, where is the energy stored by the extra work due to the incline going? I think it’s partly going into moving the treadmill, so that the treadmill engine doesn’t have to work quite as hard. Where else?

  37. #37 CherryBomb
    March 13, 2010

    Now if you REALLY want to see an argument, imagine an airplane on a level treadmill that is set to travel backwards exactly as fast as the wheels are turning. Can the airplane take off? heheheh

  38. #38 Duncan Ivry
    March 13, 2010

    One more time … sigh

    If you want to do *physical* work — “physical” as in “belonging to the science of physics” –, you have to *actually* move along a path *up* through a gravitational field. When you hold your position against pulling gravity — as on an inclined treadmill — you do *not* move up against gravitation. Gravitation usually remains constant in locations with treadmills. Ergo: In the line integral of F dot ds (I mentioned this before) there is no path of length greater than zero, and because of this the integral is equal to zero. The amount of *physical* work is equal to the value of this integral.

    It is pointless trying to argue with the moving surface of the treadmill or a belt around the body (by the way, the woman on the picture above does not use a belt). Only the body’s movement relative to the gravitational field is relevant for calculating the *physical* work utilizing gravitation. The movement of the treadmill’s surface under the person’s body does *not* move the body relative to the gravitational field.

    It is nice, but it is not necessary trying to argue with certain absurd consequences of the “doing physical work on the inclined treadmill against gravitation” … er … hypothesis. it is *definitely” enough to look at the force-times-path-integral with a path relative to a gravitational field, because by this we go back to the definition of physical work done utilizing gravity. This kind of physics is simple, undergraduate stuff.

    It is pointless trying to argue with a hill or an escalator, because in these cases you *actually* *do* move up against gravitation. These cases are not equivalent to the treadmill.

    Several commenters here confound *physiological* work with *physical* work, I think. Yes, you do physiological work when running on the treadmill, and when holding a weight off the ground for a while, and when walking up a hill. You do physiological work when you are standing still on a point doing “nothing”. Yes, you feel it in your muscles, you begin to sweat and breathe harder. But: *physiological* work and no *physical* work.

    It is pointless trying to argue with experiences and feelings when having exercised in different environments. This is about *physiological* work, but Ethan started talking about *physical* work on a treadmill.

    @ Ethan:
    I offer scanning a little bit of what Richard Feynman wrote about physiological work in order to publish it here as a comment. Feynman new a lot about biology. What he wrote would be useful, and he has been a great teacher and communicator, as far as I know. Please, tell me, if you agree.

  39. #39 Left Coast Bernard
    March 13, 2010

    In a thermodynamic analysis, it is important to consider what is the system and what is the surroundings. Work and heat are flows of different forms of energy from one to the other.

    Consider walking up an operating down elevator. If you stand on a step, the gravitational force does work on you, the system, and your potential energy goes down. If you then step onto the next step above the one you are standing on, you are doing work on the escalator.

    If you step smoothly at just the correct rate, you will be doing work on the escalator while your center of mass remains at the same height relative to the escalator’s framework. Thus the gravitational force will not do any work on you.

    It is worthwhile to remember your first week of your first physics class, and to draw the force vector diagrams. In these diagrams, you would carefully keep track of the forces on you from elsewhere, and you’d have another diagram to illustrate the forces you exert on elsewhere.

    Moving one part of your body with respect to another against resistance involves work that is internal to the system, if we consider you to be the system. All of this work eventually ends up as energy transferred to the surroundings, mostly from evaporation of sweat. That energy transfer is heating the environment, but is not work on the environment. Jogging on a level treadmill, or, what is the same thing, an escalator that has steps of zero rise, produces lots of this internal work, but does not involve work by you on the environment.

    Left Coast Bernard

  40. #40 Anonymous
    March 13, 2010

    The other day, my wife complained that I was getting fat and needed to exercise more.

    Me: “But honey, I AM exercising right now. Very hard, in fact”

    Wife: “No you’re not! You’re just laying on the couch watching TV!”

    Me: “Let me show you. Go to the stairs and start walking down at a steady pace. In your downward frame of reference, I will be rising up 10 feet against gravity! 10 feet times 300 lbs is a lot of energy! That is how much work I am doing constantly pushing downward against the couch.”

    Wife: “But in the frame of reference of the couch, you’re not moving against grav-”

    Me: “Tut tut! Einstein said physics is the same in any frame of reference, so it must be the same result as the stair-descending frame. Let me break it down another way: imagine that every second I move myself up 10 cm and then down 10 cm, as if I was doing push-ups. You would agree that pushing myself up 10 cm is work done against gravity, right?”

    Wife: “Yes…so you’re going to start doing push-ups now?”

    Me: “Ahh, but I don’t need to! The overall effect on my position is the same from the 100 push-ups as it is just lying here. Push-ups are just the discrete picture to help you understand. I’m actually doing the exact same amount of work yet keeping my hands free for eating these tasty pork rinds, a win-win!”

    Wife: “…”

  41. #41 delosgatos
    March 13, 2010

    Duncan, along the lines of what LCBernard writes in #39: If you stand still on the inclined treadmill, the system {you + treadmill} loses potential energy as you move down. If you walk “up” the treadmill, staying in place relative to the ground outside the treadmill, the system maintains its potential energy. Do you agree that some work must be done to compensate for what would otherwise be a loss of potential energy?

    Denote the potential energy of the system at initial time T_0 and later time T_1 as E_0 and E_1, respectively.

    In the system where you are on a hill,

    dE/dt = 0 and E_1 = E_0 if you are standing still
    dE/dt > 0 and E_1 > E_0 if you are walking up

    whereas if you are on the inclined treadmill

    dE/dt < 0 and E_1 < E_0 if you are standing still
    dE/dt = 0 and E_1 = E_0 if you are “walking up”

    So if you are walking up (which on the treadmill you might prefer to describe as “walking in place” or something) you end up with more energy in the system than if you don’t walk up. So some work (in the physics, not just physiologic, sense) is being done.

  42. #42 Hilary PhD
    March 13, 2010

    CherryBomb said: “Now if you REALLY want to see an argument, imagine an airplane on a level treadmill that is set to travel backwards exactly as fast as the wheels are turning. Can the airplane take off? heheheh”

    Not unless there’s a hurricane blowing.

  43. #43 Kurt
    March 13, 2010

    For CherryBomb at #37: For another example of physics on a treadmill that generated a lot of argument, take a look at the “sailing downwind faster than the wind” phenomenon. (See here for a summary.)

  44. #44 Sphere Coupler
    March 13, 2010

    To reiterate what some have already stated but perhaps as AE might have said;

    Physically regarding only the standard gravity field as Distance from Mass (DM)in an assumed spherical center. Greater the distance from the assumed geometric spherical center, less joules are depleted as your reference frame gravitational acceleration (g)progressively weakens. Gravitational force is proportional to the square of the distance from the center of mass.

    To receive an equal amount of exorcise from two identical forty foot escalators.(one running, one not)
    Assume moving escalator matches exactly the person’s consistent steps maintaining half way point.
    Assume equal energy usage for both people.
    Assume distance traveled from bottom to top for the person on the stationary escalator.
    The stationary escalator’s relative position would be very slightly lower in the adjoining reference frame and highly insignificant.(A quick guess would be in the centi-meter range)
    OR
    Assume moving escalator matches exactly the person’s consistent steps maintaining half way point.
    Assume distance traveled from bottom to top for the person on an equally placed stationary escalator.
    The joule depletion would be very slightly lower for the person on the stationary escalator and highly insignificant.(A quick guess would be in the centi-joule range)
    OR
    Assume moving escalator matches exactly the person’s consistent steps maintaining a calculated point above half way yet not at the top.
    Assume distance traveled from bottom to top for the person on the stationary escalator.
    The joule/caloric depletion is equal.

    The stationary person works harder to maintain position as the moving person’s work gets easier as he moves thru the geometric gravitational field that falls off at a consistent yet ever greater amount. The stationary person moving to maintain steady position is still actively moving in relation to the gravity field. The length of forty feet produces a highly insignificant result. The above is not to be confused with the G-constant or the g-force (acceleration force)which is equal and opposite to standard gravity, the above is standard gravity only(g).

    gh = go (re/re+h)2

  45. #45 Sphere Coupler
    March 14, 2010

    CherryBomb said:”Can the airplane take off? heheheh”

    The answer to that one is easy…the answer is Yes, but ONLY if the stewardess is serving drinks.

  46. #46 Duncan Ivry
    March 14, 2010

    @ delosgatos
    Sorry, I’m not quite with you.

    Going again back to the definitions for doing work against gravity there are the following preconditions:
    (1) A gravity field is there.
    (2) A body is in the field.
    (3) A force is excerted on the body.
    (4) The body moves along a path relative to the gravity field.
    (5) The line integral of the scalar product of force vector times infinitesimal path vector (usually written as F dot ds; simple example below) along the path is greater than zero.

    Then — only if *all* preconditions are true — the body does *physical* work (“physical” as “belonging to the science of physics”).

    To (4): The path is relative to the field, not anything else, e.g. the surface of the treadmill.

    To (5): This condition is true in the special case, where the body moves driven by a constant force on a straight line to a position in a greater height relative to earth — or simply “up” — this is because the scalar product from above is greater than zero all along the way; the value can be computed as the following product: (scalar value of the force) x (path length) x (cosine of the angle between force vector and path vector); e.g. if the angle is zero, i.e. the body is lifted vertically, then the cosine is equal to one.

    Let us assume a body with mass 70 kg, and a constant force accelerating the body with 1 m/(s^2) (meter per second squared; i.e. velocity is increased by 1 m/s each second), then we have a force equal to mass times acceleration of 70 kg*m/(s^2) or 70 N (N = Newton). If the force lifts the body 1 m, then a physical work of 70 N*m = 70 J (J = Joule) is done.

    As an example, one of the chocolate crackers I like so much — sized about 6 cm x 5 cm x 0.5 cm, half chocolate, half paste — “contains” about 69 kcal = 289.110 J (1 kcal = 4.190 J). So lifting *your* body a couple of steps up the stair is not so much, is it?

    If only one of the preconditions above is not true, e.g. if the integral in (5) is zero, because the path has length zero — there is no movement relative to the field –, then the body does *no* *physical* work.

    Let me emphasize: This is how work utilizing gravity is defined in physics.

    What many commenters here are talking about on and on is *physiological* work done inside the human body. And some — sorry — simply do not understand basic physics. I say “sorry”, and I mean it, because not everybody has to understand this, and because Richard Feynman said it’s difficult, and we have to be careful.

  47. #47 ElmerPhuD
    March 14, 2010

    Duncan,

    The line integral, your condition 5), IS non-zero.

    Look at just one foot: it traces a cyclic path, first downward with the moving belt, then up and back to the starting point.
    There are no forces between the foot and treadmill on the return path, but there are on the downward path.
    Specifically, the foot exerts downward force which supports the body. Meanwhile, the foot is moving downward, through the gravity field, at a rate proportional to the sine of the treadmill angle. The force times displacement scalar is positive. Since there are no forces on the return path, the sum of these scalars around the entire path is likewise positive.
    Same for the opposite foot.

    Therefore, by your own rules, the body does physical work, against gravity, on the treadmill.

    You’re right to defend the mathematical definitions; the problem is that they can be elusive with non-rigid bodies as we have here. That’s why the thermodynamic arguments, equally defensible, are a good check where there’s something wrong.

  48. #48 William Paysinger
    March 14, 2010

    I’m glad you revisited this subject as there are many opposing views on the matter.

    I suggest setting up an actual test, not the “go run and tell us what you FEEL to be true”

    Sounds like a good myth buster episode.

  49. #49 delosgatos
    March 14, 2010

    (In all my posts I am considering the treadmill to be 3 miles long, and I start at the midpoint.)

    Duncan, you’re focusing only on the body on the treadmill. But the system {treadmill, body, earth} is what is storing or dissipating energy. If you stand still on the treadmill, the system dissipates energy. If you walk “up” (relative to the treadmill) you can offset that dissipation and maintain the potential energy of the system.

    Another try, not really different in kind from my previous comment. If I put 500 pounds of rocks in a wagon and by “walking up the treadmill” hold it at a steady elevation on the treadmill, after some amount of time I can use that mass with a pulley to raise some other mass from the base of the treadmill to my level. This is work in the physics sense. If I let the treadmill carry the rocks to the base of the treadmill I can’t use them to do this work. So in the “hold them steady” case something in this system is adding potential energy that is later available to do work. Where is this energy coming from?

    An earlier post asked about putting on roller skates and tying yourself to the bar of the treadmill so that you stay in place, compared to tying yourself to a car driving up a hill. I pointed out that to make the analogy correct you should be tied to a car running its wheels on the treadmill. It’s also worth pointing out that by tying yourself to the treadmill bar you’ve just created a new transmission system. The energy of the treadmill is going into the kinetic energy of the wheels, which could be connected to a winch to pull rocks up the hill. Inefficient transmission to be sure, but still work is being done.

  50. #50 delosgatos
    March 14, 2010

    Duncan wrote:

    Going again back to the definitions for doing work against gravity there are the following preconditions:
    (1) A gravity field is there.
    (2) A body is in the field.
    (3) A force is excerted on the body.
    (4) The body moves along a path relative to the gravity field.
    (5) The line integral of the scalar product of force vector times infinitesimal path vector (usually written as F dot ds; simple example below) along the path is greater than zero.

    Can you cite a source for this definition? I don’t dispute that a body meeting these conditions is doing work against gravity, but this isn’t the only definition of how a system can do work.

    A bicycle on level ground loses kinetic energy to friction if you don’t pedal. If you pedal to maintain speed work is done (in the physics sense; you keep pointing out that some posters are blurring physiological work and physical work, I hope you don’t think I’m doing so). Yet you don’t end up with more kinetic energy that you started with, if you just pedal enough to maintain speed. This is analogous to walking on the treadmill, with kinetic energy analagous to potential energy.

  51. #51 Anonymous
    March 14, 2010

    delosgatos, you (and probably a lot of other posters) are making the classic mistake of treating work as force*time. Often this is because of the physiological connection (it takes up chemical energy to keep your muscles taut over time, but this is just because of the way the body works).

    You obviously aren’t adding any energy to the rocks in a wagon (except for maybe a little rotational kinetic energy to the wagon wheels) by keeping it steady on the treadmill. The gravitational potential energy it has at the end is simply the same as what you gave it when you lifted it onto the treadmill. If you were adding energy to it, then it should be able to do more and more work on a pulley after a longer and longer time on a treadmill…but it cannot.

    Force*time doesn’t even have the right units to be considered energy. What force*time really is (and what you really have in mind), is change to momentum. You ARE making a greater change to momentum with your feet on an inclined treadmill, equivalent to the greater effect on momentum from the treadmill’s normal force + gravity, which perfectly cancels.

    But this is not energy, and not “work” in the physics sense.

  52. #52 Duncan Ivry
    March 14, 2010

    @ ElmerPhuD: “The line integral, your condition 5), IS non-zero. Look at just one foot: it traces a cyclic path, first downward with the moving belt, then up and back to the starting point. …”

    Sorry, this is not correct. Only the center of mass of the body of the person on the treadmill is relevant — especially because all parts are connected together ;-)

    Cyclic pathes of bodies are just cases, where the pertinent line integral is calculated along a closed path and because of this it is equal to zero. By the way, fields for which this is true, are called “conservative”. The gravitational field *is* conservative. Roughly speaking, up “against” the gravity field the body does physical work, and down “with” the field work is “given back” to the body. You may have heard about the exchange between potential energy and kinetic energy. I hope you know what I mean.

    Another thing: “mathematical definitions”.

    No, these are physical definitions. The definition of work done utilizing gravity is something in the science of physics, and has been invented by physicists.

    By the way, now I realize: the “belt” is the moving surface of the treadmill — language barrier ;-)

  53. #53 Duncan Ivry
    March 14, 2010

    @ delosgatos: “But the system {treadmill, body, earth} is what is storing or dissipating energy.”

    Yes, I’m completely with you. But Ethan started with something different, and this is what I talk about: Doing work utilizing gravity. The part “utilizing gravity” is important.

    There are other things going on in and around a treadmill and the human body on it. Heat is produced in the treadmill’s motor, between the belt and what is below (I don’t know how to call it), and inside the human body. Chemical energy is bound in certain substances in the body, and it needs energy to build, transport, and destroy those substances. These are only some examples. The person has to work *physiologically* against this.

  54. #54 Duncan Ivry
    March 14, 2010

    @ delosgatos: “Can you cite a source for this definition? I don’t dispute that a body meeting these conditions is doing work against gravity, but this isn’t the only definition of how a system can do work.”

    Well, it’s the one and only definition of “doing work against gravity”. Of course, physical bodies can do work against electro-magnetic fields, and the fields of the strong and the weak nuclear forces too. Friction, as an example, has to do with electro-magnetic interaction.

    A source for the definition is what I mentioned already:
    Richard Feynman, Robert B. Leighton, Matthew Sands: Lectures on Physics, 3 Volumes.

    For the topic under discussion chapters 13 and 14 of the first volume are especially relevant.

    From the foreword of my copy of the first volume: “This book is based upon a course of lectures in introductory physics given by Prof. R. P. Feynman … ; it covers the first year of a two-year introductory course taken by all Caltech freshmen and sophomores …”

    Since Feynman is a Nobel laureate, it’s a must have at universities, I think. About colleges, highschools, or what you may have in your country, I don’t know. I live in a non English speaking country in continental Europe, you know.

  55. #55 ElmerPhuD
    March 14, 2010

    Duncan@52:

    Consider this variation: a one-legged “runner” rides down the treadmill for some distance, then hops in one leap back the top, and repeats. Now his center of mass follows the trajectory I described.
    On the downward ride, his body exerts a downward force (i.e. weight) on the treadmill and concurrently a downward displacement.
    On the leap up, once airborne, his body exerts NO force on the treadmill.
    Would you now agree that the line integral is positive?

    Would you not also agree that the body must repeatedly exert energy to maintain that average position? So where does that energy go?

    Answer: it goes into heating the treadmill, due to friction between the belt (moving surface) and the platform underneath.

    And yes, I’ve heard of the exchange between potential and kinetic energy; I addressed it earlier in #27, as have several others.

    I’ll try again. Suppose the treadmill belt had no friction against the platform. Then the potential energy lost while riding down would turn into kinetic energy, continuously accelerating the rider and belt. The fact that you don’t accelerate means that the energy is going somewhere else: friction. That is work being done: energy transferred from the body to the treadmill.

    As delosgatos has also explained, keeping the body at the same (average) position requires energy input, which is dissipated as heat in the treadmill.

    What say you?

  56. #56 Anonymous
    March 14, 2010

    ElmerPhuD, if one repeatedly hops like that, then the extra work towards gravitational potential energy is done (and repeatedly lost into kinetic energy of the treadmill).

    But that doesn’t apply in the case where the runner is stationary (only for his feet and half his legs, but not his torso). It is a different case, as I illustrated by comparing it to the difference between doing repeated push-ups and laying on a couch.

    No movement within the gravity well = no work done against gravitational potential.

  57. #57 ElmerPhuD
    March 15, 2010

    Duncan, Anon, and other doubters:

    I’ll try this again, differently.

    What you’re saying would be true standing, lying on the couch, doing pushups, whatever… on solid ground.
    Or on the treadmill, if you support all your weight with your arms on the handrails.
    But that’s not the case when you’re walking/running/standing with full weight on the treadmill.
    The key point is that your floor is sinking.
    Leave out all horizontal components of force and motion and keep just the vertical: in the absence of exertion, standing still on the treadmill, you would be sinking.

    It’s equivalent to standing on a dampened piston which compresses very slowly.
    No input of energy into the system means you would be moving within the gravity well, as potential energy is converted to heat.

    An inclined treadmill is a “perpetual sinking machine”, capable of slowly converting any amount of potential energy to heat.

    Once things are moving, you’re not adding anything to kinetic energy, since the treadmill isn’t accelerating; all lost potential energy is going to heat and must be replaced.

    Delosgatos has explained this well already.
    Zero work means negative displacement (sinking). Therefore zero displacement (not sinking) requires positive work.
    The reason you’re not sinking on the treadmill is that you’re constantly compensating by adding energy — doing work against gravity. All of that energy becomes heat in the treadmill.
    That’s in addition to any “physiological” work from muscle ineffeciency, lateral exertion, etc., which would be needed on a level treadmill. When the treadmill is inclined, with your weight on it, you exert energy just to stay put, constantly re-lifting yourself.

    Do we agree yet?

  58. #58 ElmerPhuD
    March 15, 2010

    Also, Anon@51, with respect to Force*Time:

    You’re putting words in others’ mouths;
    nobody’s talking about F*T as work.
    We’re talking about Work=Force*Distance.
    Specifically, Force = body weight and Dist = vertical displacement.
    Standing on the ground, Work=F*D =F*0 =0, as expected.

    Standing on a treadmill for time T, with belt velocity V:
    D (distance fallen) = VT sin Angle, so
    Work = FVT sin Angle.
    For Angle=0 (level treadmill), V=0 (stopped treadmill), T=0 (instantaneous), or F=0 (you’re weightless),
    Work = 0, as expected.

    But for F>0, V>0, T>0, and Angle>0 (i.e. real body on moving, inclined treadmill for some time), then D>0 (i.e. you’re sinking), so Work>0.
    That’s the energy lost to treadmill friction which must be replaced to maintain height.

  59. #59 Brett
    March 15, 2010

    I’m amazed that anyone could argue this! Obviously there is a group of people that have never used a treadmill. Forget about physics for a moment. The simple fact of having been on a treadmill on an incline makes it a fact that more work is done on an incline regardless of relative motion with the earth. For crying out loud.

  60. #60 Hilary PhD
    March 15, 2010

    It seems to me that there’s confusion between 4 (or possibly 3) sorts of work*. In all cases, the system is the person’s body, since the question we’re interested in is how much work the person is doing:

    a. Work which raises the gravitational potential energy of a system. I agree that a person on an inclined treadmill maintains constant gravitational potential energy relative to the earth’s surface, so might be said to do none of this kind of work.

    b. Mechanical work which involves an energy transfer across a system boundary. In order to maintain constant gravitational potential energy relative to the earth’s surface, the person has to push backwards and downwards against the treadmill to propel herself upwards and forwards relative to its moving bed/belt. By any definition this is real work in which the person exerts force and moves along a path relative to the treadmill, transferring energy to it. This work offsets the gravitational potential that the person would have lost, had she moved down with the treadmill.

    For those who are understand the equivalence of different reference frames, this mechanical work is in fact gravitational work, viewed within the reference frame of the moving treadmill bed/belt. For those who don’t it seems to be a different sort of mechanical work, but it is still real work done by the person on her surroundings and the amount of it that is required is exactly the same as the gravitational work that would be done in running up a hill at the same speed and incline.

    c. Mechanical work which is purely internal to a system (and therefore not actual work done by or on that system). Imagine if instead of running, the person was lying on her back and cycling her legs in the air. Regardless of whether she’s on a treadmill or not, she’s exerting muscular forces to move her legs, but this energy is all dissipated as heat within her body and there is no transfer.

    d. Non-mechanical work which is purely internal to a system (and therefore not actual work done by or on that system). This is all the energy-converting physiological processes going on inside the person’s body except the mechanical work of major skeletal muscle contractions. It is dissipated as heat within the body. Since muscles are inefficient, any internal or external mechanical work demands an increase in this non-mechanical work as well.

    *Incidentally, these are all at least arguably forms of physical work since all physiological processes are in accord with the laws of physics, so I think the use of the term “physical work” is unhelpful. Mechanical/non-mechanical and internal/external (to the person’s body) are more useful concepts.

  61. #61 Duncan Ivry
    March 15, 2010

    @ ElmerPhuD 55: “I’ll try again. … That is work being done: energy transferred from the body to the treadmill. … What say you?”

    Resistence is futile ;-) I always say the same: No work done utilizing gravity. Other work: yes.

  62. #62 selfification
    March 15, 2010

    Agree with Hilary PhD. There is a lot of confusion here.

    The formal definition of work is not invariant under a reference frame transformation. The question of “does this person do work” is *MEANINGLESS* without specifying the reference frame – just as meaningless as “does this person travel distance”. The everyday notion of “work” is in fact not the physics definition of work and, in this case, references the internal energy required to overcome biomechanical inefficiencies and friction required to keep your legs moving at the same speed.

  63. #63 Duncan Ivry
    March 15, 2010

    Brett: “… The simple fact of having been on a treadmill on an incline makes it a fact that more work is done on an incline regardless of relative motion with the earth. For crying out loud.”

    Very true, indeed.

  64. #64 Duncan Ivry
    March 15, 2010

    Well, people, it has really been fun discussing this with you. But there are no new arguments by now, I think, and I have some work to do. I’m out. Thank you.

  65. #65 Sphere Coupler
    March 20, 2010
  66. #66 Mitch
    May 3, 2010

    Why the question? You didn’t read what he said? OK, I cite what he said;
    “Only two things are infinite, the universe and human stupidity, and I’m not sure about the former.”

  67. #67 Cameron
    June 20, 2010

    ElmerPhuD; Are you saying that in order to stop you “sinking” on a treadmill, you need to continuously put energy into it as work?

    Just because you would lose potential energy if you didn’t run up the treadmill doesn’t mean it takes energy to maintain your position. The couch example shows this well. Although I like to think of it as standing beside a cliff. If you were to jump off the cliff, you would loss your potential energy. The cliff is therefore a perpetual sinking machine as well. If you don’t jump off and just stand, then you don’t lose your potential energy. By your argument, that means that you must be doing work somehow. But it is obvious to most of us that you are not doing work by just standing.

    Many people have proved that no work is done against gravity since you don’t increase your gravitational potential energy. This is definitely correct and indisputable at this point. It has however not yet been shown that gravity doesn’t act as a part of a mechanism to make you do more work and subsequently transfer the energy to the treadmill. Throughout the cycle of one step, you might do work against gravity, then that potential energy might be returned to the treadmill through the interface between your feet and the belt. I don’t believe this is the case, but I’m not yet 100% convinced and it is more difficult to prove. Either way, you can’t come to a conclusion without first analyzing how someone runs in more depth. This has come up when people talk about running up, then letting the treadmill pull you down, then running up again.

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  69. #69 clickbank
    July 25, 2010

    Not so fast,” said many of the commenters! Unlike a real hill, where you actually wind up at a higher elevation than you started at, you don’t travel any real distance on a treadmill, and you certainly don’t end at a different elevation than you started at! Therefore, the argument goes, you don’t do any extra physical work.

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  71. #71 Share Messages and Notes
    August 18, 2010

    I used a treadmill instead (at 15%). It was identical in terms of workout. (ok some parts of my route are >15% but on the grades where it is about 15% it was the same on the treadmill)

  72. Throughout the cycle of one step, you might do work against gravity, then that potential energy might be returned to the treadmill through the interface between your feet and the belt.

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    Of course you do extra work. You have to use more force to move. It certaintly takes more energy which will burn more calories.

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  81. #81 Chris CD
    December 13, 2010

    Hmm, just realised I should probably have posted here in the follow-up, rather than in part one:

    So, again -

    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.

  82. #82 Katie~Run for the Bikini
    December 27, 2010

    I just found this awesome post through a google search on treadmill inclines.

    All I can add is that when I run intervals on my treadmill at 12%, I want to die after a minute. I go at a pace of 7 mph usually. I don’t have a lot of hills around here so I can’t compare too much (plus I don’t know what grade a 12% incline would be equal too).

    I do know that on the treadmill, at 1% incline I could hold 7.0 for over an hour and at 12% incline I can hold 7.0 for just just over a minute maybe 2. So I conclude I am definitely doing more work, or at least putting in a whole lot more effort on a higher treadmill incline.

    For anyone who disagrees, try running on a treadmill at 12% then comment.