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Re: point of application of force

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  • Re: point of application of force

    Hi Kieran,
    In your letter (some parts deleted):
    >=20
    > I have what may be a very simple question, but the answer to which is
    > very important to my studies. My PhD studies partly involve the
    > evaluation of Bi-lateral symmetry of the lower extremities during
    > vertical jumps. The results are being evaluated at the joint moment
    > level and, therefore, I require the determination of the point of
    > application of force.
    >=20
    > I have access to two platforms which are located parallel to each
    > other ( and seperated by 5cm in the Z direction). With my subjects
    > facing in a positive X direction the point of application of force in
    > the X direction for each plate is equal to the moment of force about
    > the Z axis (Mz) divided by the total vertical force (Fy):
    >=20
    > Ax =3D Mz / Fy (where moments are taken about the plate surface).
    >=20
    ...Make sure also to remove any components in that moment that could =
    be caused by shear force and the effective depth of the gauges. Not all =
    force plates have gauges at zero height. Torgue about the vertical axis =
    can cause a similiar problem if the mechanical and 'electrical' centres =
    aren't coincident. Tables of correction values should be supplied by the =
    manufacturer to fix these problems.

    > >is Ax =3D the sum of moment of force about the Z axis for plates one
    > >and two devided by the sum of the vertical forces for plates one and
    > >two.
    > >
    > >is Ax =3D [ Mz(1) + Mz(2) ] / [ Fy(1) + Fy(2) ]
    >

    Naaah..... determine the effective point of application for each plate. =
    Multiply the coordinates (in global coordinates) by the vertical force =
    on that plate. Do this for the second plate. you now have two 'moments'. =
    Add them up, and divide by the subjects weight (or the sum of the forces =
    on the two plates).
    ie, Zeff=3D((COPz1 * fy1)+(COPz2 * fy2))/(fy1 + fy2)
    where Zeff is the effective z coordinate of the centre of pressure =
    (or application), COPz1,COPz2 are the centres of pressure for each =
    plate, and fy1,fy2 are the vert. reaction forces for each plate. The =
    COPz's are calculated in the normal way.=20
    Your kinematics needs to know where the force plate centres are. If =
    you only measured one side of the body, you need to correct the =
    measurements to get rid of the asymetry, so things like total body =
    centre of gravity line up (assuming that you are using or 'simulating' =
    3D measurements.
    =20
    > P.s.
    > (The coordinate system is the ISB system which is reactionary and
    > follows the left hand rule. Positive Y is vertically normal to the
    > force plate and positive X coincides with the direction the subjects
    > are facing)
    >=20
    It follows the left or right hand rule depending on which way you =
    establish Z+. If z+ is to the subjects right, then the right hand rule =
    applies. The XYZ axes seem silly until you realize that many labs =
    started with a 2D system, often digitized from video or film. Then =
    height is Y, forwards is X+. When you can afford 3D, the obvious =
    coordinate system you'd choose to remain compatible with your old data, =
    is the ISB system. Maybe it's not mathematically kosher, but it's =
    practical.
    >=20

    -Paul

    --------------------------------------------------------------------
    Paul J Guy work phone:519-885-1211 ext 6371 =20
    paul@gaitlab1.uwaterloo.ca home/FAX/:519-576-3090
    pguy@healthy.uwaterloo.ca 64 Mt.Hope St.,Kitchener,Ontario,Canada
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