Scissors Lift Equations??

I was looking for some lift equations for our Vex students to use. I derived some myself but decided to check them against your site. The equations stated in this [URL="http://www.engineersedge.com/mechanics_machines/scissor-lift.htm"]engineersedge scissor-lift link[/URL] seem incorrect.

For the bottom actuator the force should be F =( W + wf/2)/tan(phi)

and

for the center pin actuator the force should be F = 2*(W+wf/2)/tan(phi)

Where W is the load weight, wf = the total frame weight and phi is the interior angle between the horizontal and the arm.

The proof used for the bottom force incorrectly assumes that the P force which is the vector sum of F and (W + wf)/2 vertical force is colinear with the arm. This cannot be since there is a moment about the center pin of lift arm that is caused by the force of the load. This requires a component of force at the bottom that is normal to the arm at the point where F is applied. So P cannot be colinear with the arm. If the moment equations are written about the center pin of the lift we get:

F*L*sin(phi) = (W + wf/2)*L*cos(phi)

or F =( W + wf/2)/tan(phi)

This is easily checked by an energy approach. If the load W is lifted by dh then the center of mass of the frame (with weight wf) is lifted by dh/2.

We know that if the actuator moves a distance of dx the work input is F*dx which must equal the potential energy increase of the lift masses moving against gravity.

So F*dx =(W + wf/2)*dh

or

F = (W+ wf/2)*dh/dx

From geometry, dh/dx = 1/tan(phi)

Hence F =( W + wf/2)/tan(phi);

When the force F is applied to the center pin of the lift the same dh is achieved with dx/2 movement so the force F must be twice as large to raise the masses.

So F_center_pin = 2*F_bottom = 2*( W + wf/2)/tan(phi);

Hope you structural engineers set one of us straight.

chris