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Thread: Skiing: Can I set my own bindings?

  1. #21
    Walt
    Guest Walt's Avatar

    Default Skiing: Can I set my own bindings?

    > Is this replacing the annual percent-slope-versus-angle thread?
    >

    Ya know, I don't recall having this particular argument on RSA before.
    And it's been years since we've had a good percent-slope-versus-angle
    dust up.

    Anyway, if you want to have an argument with me you'll have to pay up
    like the other Richard. Would you like the 5 minute version, or do you
    want to go for the full hour?

    //Walt

  2. #22
    TexasSkiNut
    Guest TexasSkiNut's Avatar

    Default Skiing: Can I set my own bindings?

    On Feb 19, 2:10 pm, Walt <walt_ask...@SHOESyahoo.com> wrote:
    > Richard Henry wrote:
    > > Is this replacing the annual percent-slope-versus-angle thread?

    >
    > Ya know, I don't recall having this particular argument on RSA before.
    > And it's been years since we've had a good percent-slope-versus-angle
    > dust up.
    >
    > Anyway, if you want to have an argument with me you'll have to pay up
    > like the other Richard. Would you like the 5 minute version, or do you
    > want to go for the full hour?

    Did Eugene ever have a winner of his "Physics of Skiing Essay Prize"
    contest?

  3. #23
    VtSkier
    Guest VtSkier's Avatar

    Default Skiing: Can I set my own bindings?

    > down_hill wrote:
    >> Walt wrote:
    >>>
    >>> Think of it this way: I'm testing a binding. I place a boot in the
    >>> binding and apply a torque of, say, 50 Newton Meters. The binding
    >>> doesn't release. I've just described a situation where there is
    >>> torque but no motion. Do you say there is no torque here? If so,
    >>> how does one ever test a binding?

    >>
    >> What are the springs doing? Are they not being compressed <motion>?

    >
    > As you increase the torque from zero to 50 the springs compress, and
    > there is a small amount of motion. Once you reach 50 and stop
    > increasing the torque the system reaches a steady state where there is
    > no motion. At that point there is torque but no motion.
    >
    >> I was trying to think of a torque but no motion example...

    >
    > Here's a more visceral example: pick up a brick and hold it straight
    > out in front of you with your arm horizontal. Hold it still. Then
    > please try to explain, without allowing the brick to move, how there is
    > no torque since there is no motion.


    There is FORCE but no TORQUE
    >
    > //Walt

  4. #24
    VtSkier
    Guest VtSkier's Avatar

    Default Skiing: Can I set my own bindings?

    > Is this replacing the annual percent-slope-versus-angle thread?
    >


    No, but close.

  5. #25
    VtSkier
    Guest VtSkier's Avatar

    Default Skiing: Can I set my own bindings?

    > Is this replacing the annual percent-slope-versus-angle thread?
    >


    Do you want to start that one?

  6. #26
    Walt
    Guest Walt's Avatar

    Default Skiing: Can I set my own bindings?


    >> Here's a more visceral example: pick up a brick and hold it straight
    >> out in front of you with your arm horizontal. Hold it still. Then
    >> please try to explain, without allowing the brick to move, how there
    >> is no torque since there is no motion.

    >
    > There is FORCE but no TORQUE


    Ok. Change the brick out for a ski. Grasp it by the binding and hold it
    vertically. Not that hard, is it?

    Now grasp it by the tail and (try to) hold it horizontally. Much
    harder, right?

    Why? What makes one so much harder than the other? The force hasn't
    increased, since the ski weighs the same as it did before.

    So what makes it so much harder? HINT: torque. I double dog dare you
    to hold a ski like that and tell me that you don't feel the difference.

    //Walt

  7. #27
    Walt
    Guest Walt's Avatar

    Default Skiing: Can I set my own bindings?

    > Are you two not saying the same thing? Both of you mentioned that to have
    > torque something must move am I correct?


    No. VtSkier is saying that in order to have torque something must
    move. I am saying it is incorrect.

    Please, let's not get into an argument over whether we're having an
    argument. We are, the point of contention is clear.

    > I guess the difference in your argument is what must move.


    NOTHING has to move for there to be torque.

    //Walt

  8. #28
    VtSkier
    Guest VtSkier's Avatar

    Default Skiing: Can I set my own bindings?

    > JQ wrote:
    >
    >> Are you two not saying the same thing? Both of you mentioned that to
    >> have torque something must move am I correct?

    >
    > No. VtSkier is saying that in order to have torque something must
    > move. I am saying it is incorrect.


    That's what you are saying.

    > Please, let's not get into an argument over whether we're having an
    > argument. We are, the point of contention is clear.


    Walt is quite correct with his analysis of the disagreement.

    >> I guess the difference in your argument is what must move.

    >
    > NOTHING has to move for there to be torque.


    Yes it does.
    >
    > //Walt

  9. #29
    Walt
    Guest Walt's Avatar

    Default Skiing: Can I set my own bindings?

    > Walt wrote:
    >> VtSkier wrote:


    >>> It goes to the definition of VECTOR. My reading, which I
    >>> posted, it that a vector has magnitude and direction.
    >>> Those are the qualities which create a vector, no?

    >>
    >> Yes, a vector has magnitude and direction.
    >>
    >>>
    >>> Magnitude is usually expressed as a unit of length.

    >>
    >> Um....no. Vectors can have many different units. The electric field
    >> is a vector with units of volt/meter. The magnetic field is vector
    >> with units of ampere/meter. Momentum is a vector with units of
    >> kilogram-meters/second. Acceleration is a vector. Angular momentum
    >> is a vector. Angular acceleration is a vector.

    >
    > Yes, and although I haven't looked up each of those units,
    > each one that you note has a unit of length as a component,


    sigh.

    How about angular velocity, which has units of 1/second?

    Vector != length

    //Walt

  10. #30
    Walt
    Guest Walt's Avatar

    Default Skiing: Can I set my own bindings?

    > The confusion is in the concept of total torque and component
    > torques. You can apply a component torque which does not cause
    > motion. In the case of the screw, the torque wrench applies a torque
    > (indicated by the reading on the wrench) and the screw applies an
    > equal and opposite torque such that no motion occurs until it breaks
    > free of the friction. The total torque until the screw moves is zero,
    > or there would be motion. However, the wrench is still applying a
    > torque, which is given by the reading.
    >


    I think you've got it. I was talking about component torque not net
    (total) torque. VtSkier was talking about total torque (I think). So
    we were talking past each other.

    What VtSkier says is correct regarding total torque: If total torque is
    non-zero, there's going to be motion (or more correctly, a change in
    angular momentum). Conversely, if there is no change in angular
    momentum, the total torque must be zero.

    However, the way to calculate total torque is to add up all of the
    component torques. These may be non-zero and yet still sum to a zero
    result. So, you can have (component) torque without motion.

    //Walt

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