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Bike on a Treadmill

  1. Oct 2, 2012 #1

    HXB

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    Hey, guys, my first time posting here but I need some educated individuals to definitively solve a problem from another forum. We are having a discussion about a bike on an incline treadmill (one presumably made for a bike). My contention is that it would still be relatively easy to pedal on this since you are not gaining any potential energy by riding it like you are a outdoor incline, and the gravitational force pulling back is equaled by the friction from the tires on the 'mill surface. Once again I'm not saying it will take no energy but the energy will be fairly minimal if the treadmill is self powered, all you really need to do is provide the energy for the rolling resistance basic inertia of the wheels and the wind resistance of the spokes in the wheels. Thanks for your help!
     
    Last edited by a moderator: Oct 2, 2012
  2. jcsd
  3. Oct 2, 2012 #2
    If what you say were true, you should make virtually no effort climbing up a descending escalator if you keep at the same point in space.
    And you should make a double effort climbing up an ascending escalator!
     
  4. Oct 2, 2012 #3
    You need about as much energy on a treadmill as on a road with the same incline.
    In both cases you have the force of gravity mg and the normal force mg cos(phi) acting on you and your bicycle. (where phi is the angle the road/incline makes with the horizontal)
    The sum of them has size mg sin(phi) and willl points backwards along the incline.

    If both the cyclist on the road and the one on the treadmill move with constant velocity, the total of the forces on them must be 0, So there has to be a friction force mg sin(phi) in the forward direction on the wheels, pointing forwards on the incline, and because of newtons third law, your wheels must exert a force in the backwards direction on the incline. The energy needed to exert this force against an incline
    moving with a speed v with respect to the cyclist is mgv sin(phi).

    The only thing that's needed for this to be valid is that the byclicle moves with a constant speed without acceleration, so the net force on it is 0.
    Driving the threadmill with a motor won't help, The driver needs the threadmill to resist movement. If the threadmill has no friction is will accelerate backwards until the cyclist can't keep up and will fall off backwards
     
  5. Oct 2, 2012 #4

    mfb

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    The escalator is a good, intuitive comparison.

    You can even do the reverse: Attach a generator to the treadmill. The bike provides a constant force (trying to accelerate the belt).
     
  6. Oct 2, 2012 #5

    russ_watters

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    Welcome to PF!

    Where you're going wrong is here:
    Sitting on a hill, you can hold on to the brake and neither lose nor gain potential energy, but you can't do that on a treadmill. The treadmill's speed and angle, combined with your weight provides a constant loss of potential energy that must be made-up with your pedaling. So the potential energy change can be calculated like this:

    Road: 200 - 0 = 200
    Treadmill: 200 - 200 = 0

    In each equation, the first number is the potential energy added by you pedaling on the bike, the second is the potential energy added or removed by what you are pedaling on and the third is the net potential energy change of the bike.
     
  7. Oct 2, 2012 #6

    HXB

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    Thanks for the replies so far, here is my thinking, you do have to provide force to overcome the gravitational force (I didn't think so at first). At an incline of 5 degrees which is 9 percent (this is quite steep) the force I calculated with the sin of the angle 5 with a total mass of bike and rider of 85 kg comes out to 72 newtons, or roughly 16 lbs of force. A crank arm is .55 ft long so you need a force of 29 lb feet of torque on the pedals. At a rate of 60 rpm this is 29 lbft/sec. This comes out to 21 watts, a very low number since I can produce 330 watts for a continuous hour. This force is constant and doesn't change if the treadmill speeds up, therefore even on the incline you could go very "fast" by rear wheel speed. I will say that the treadmill is automatic and a motor powers the treadmill, the bike does not contribute at all to the turning. What do you think?
     
  8. Oct 2, 2012 #7

    russ_watters

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    You're mixing terms which may be a source of confusion here: The torque at the wheels stays constant, but since the wheel is rotating faster, the power in watts increases as per the equation you used previously.

    Worse, your legs operate most efficiently in a very small range of speeds. So if you increase your speed, you should upshift, which decreases the torque available at the wheels, making you have to provide more at the pedals.

    [edit] Oh, you have a math error in there too. Power isn't just torque times rpm, you need to multiply by 2pi as well for rotation in terms of radians/sec (or min). So your power there is actually 180 Watts.

    http://en.wikipedia.org/wiki/Torque#Conversion_to_other_units
     
    Last edited: Oct 2, 2012
  9. Oct 3, 2012 #8

    mfb

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    That rate depends on the velocity. You can change the gear, but then the force depends on the velocity.
    There is an easier way: The required power is 72N*v with the bike&treadmill velocity v. 18km/h = 5m/s requires a power of 360W, for example.
     
  10. Oct 4, 2012 #9
    Lots of good comments here.

    I'll take a stab at simplifying/summarizing:

    Looking at the bike in a static setup (not moving), gravity will try to pull it down the ramp with the force of mass*gravity*sin(angle).

    Since the only other force perpendicular to the surface is the friction from bike tires on belt, that force must exactly cancel the gravity force at all times, or you will move forward/backward on the ramp. This is equal to μ*NormalForce = μ*mass*gravity*cos(angle).

    Static friction (b/c the tires are not spinning independantly of their riding surface) governs this setup. You can calculate a maximum angle (by assuming a static coefficient of friction, or just actually do the setup & measure the angle) by equating the gravity forces & friction forces.
    For μ=1 (clean rubber belt, rubber tires, should be a good approximation), this solves to a max angle of 45degrees (regardless of rider weight or belt speed).

    In normal riding, you overcome air resistance, rolling resistance, and potential (height) changes.
    In this case, only rolling resistance applies.
    This is calculated by the product of rolling friction & velocity.

    So, to use your example numbers:
    mass = 85kg
    angle = 5 deg
    → Power = friction * velocity = 85kg * 9.8m/s2 * sin(5°) * [velocity (m/s)]
    = 72.6 * (velocity, m/s) [watts]

    This shows that for even a slow speed of 5 km/h, you would require (5/3.6)*72.6 = 100 Watts of power.

    As a quick reference, 100 Watts is a reasonable amount of energy expendature to maintain for cardio exercise; 200-300 will burn you out in a couple minutes.
     
  11. Oct 6, 2012 #10

    HXB

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    Thanks all for the replies, you certainly know your stuff. I talked with a physics professor and he stated it similarly. I guess the important thing to realize is that potential energy is relative, in the same way someone on the treadmill, not doing work, would move down the treadmill at its speed. In their perception, I'm gaining potential energy, when in fact I'm simply not losing it, the effect is the same.

    Lastly, consider this, to my understanding now, a wheel, even with no friction bearings, receiving a force via a treadmill, still translates that force to the hub in the same linear direction as the treadmill is going right? What if the wheels had no mass? (I know this calls into question friction and other forces.) Since it wouldn't require any force to spin the wheels, no force would be translated to the bike, how would this condition change our problem?
     
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