View Full Version : Re: point of application of force

Paul Guy
04-26-1996, 07:36 AM
Hi Kieran,
In your letter (some parts deleted):
> 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.
> 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):
> Ax =3D Mz / Fy (where moments are taken about the plate surface).
...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.
> 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)
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 =


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