rec.sport.unicycling · Public discussion

Re: Tread Mills

Started by maestro8 · · Last activity · 2 posts · 361 views

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rec.sport.unicycling
Published
26 May 2005
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26 May 2005
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maestro8
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  1. This nerdy thread is largely ignoring the dynamics of the human body.
    The formulae and arguments I'm reading are mostly related to point
    masses / rigid bodies, and the human body shouldn't be reduced to a
    point mass in this case. Ferchrissakes, model the problem correctly!

    Granted, kinetic and potential energies are involved, but to a human
    being the quality that is most directly observed is "perceived work"
    (i.e. energy expended) when engaged in an activity such as running /
    riding, on or off a treadmill. Thinking about the center of mass is
    only useful in determining quantities such as ground speed (which is
    zero for one on a treadmill, quite boring!), net acceleration and drag
    forces (again, zero).

    If one is to attempt to calculate the actual quantity of work done, than
    each segment of the human body (i.e. torso, thigh, calf, foot, etc.)
    needs to be considered separately, as well, coupling equations
    (involving driving and frictional forces) must be considered for the
    intersegment connections (a.k.a joints & ligaments). This can get very
    complicated, very quickly: the hip lifts the leg, the thigh extends the
    calf and foot. Work must be done to overcome the inertia of the thigh,
    calf and foot, the friction in the joints, as well as gravity / body
    weight. Etc, etc, etc.

    Some of the earliest posts in this thread had made similar points in
    this regard.

    Sure, more work will need to be done as the grade is increased or
    decreased from the horizonal, but this isn't a surprise. No complicated
    mathematics are required! ***If one thinks about this problem in the
    frame of reference of the treadmill's belt, then the problem becomes
    much easier*** An inclined treadmill = a hill, while a level
    (horizontal) treadmill = flat ground. This is your first-order
    approximation. 'Nuff said.

    This talk of motors and friction doesn't relate to the rider on the
    treadmill! The rider is merely overcoming gravitational forces (again,
    think frame of reference), while the motor is doing the work to overcome
    the load of the rider: to the motor, the load is created in the friction
    between the belt and the belt support, which is increased by the weight
    of the rider. You can put all the friction you want (or don't want)
    between the belt and its support; the rider won't know the difference!

    --
    maestro8 - Mad Scientists for World Domination

    Those are my principles. If you don't like those, I have others. --
    Groucho Marx
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    maestro8's Profile: http://www.unicyclist.com/profile/7871
    View this thread: http://www.unicyclist.com/thread/21413

  2. maestro8 said:

    If one is to attempt to calculate the actual quantity of work done, than
    each segment of the human body (i.e. torso, thigh, calf, foot, etc.)
    needs to be considered separately, as well, coupling equations
    (involving driving and frictional forces) must be considered for the
    intersegment connections (a.k.a joints & ligaments). This can get very
    complicated, very quickly:

    That's not enough! Unliks ideal machines, human muscles require
    energy to remain in tension. You'd have to account for that energy as
    well if you want to know the energy required to walk or ride on a
    treadmill.

    By the way, the muscles in reptiles like lizards do not require energy
    to maintain tension.

    Ken

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