Error in Scissor Lift Equations at Engineers Edge??(Finally fixed)

Error in Scissors Lift Equations??

7.12.2012 Update:

Ok…. finally Engineers Edge has corrected their equations.  So I have modified this post to take out the obsolete references.

I was looking for some lift equations for our Vex students to use.  I derived some myself but decided to check them against a popular  reference site.

The equations stated in this engineersedge scissor-lift link were incorrect prior to their last update but are now correct except the new authors have not included the weight of the frame in their answer.

For the bottom actuator the force  is given as

F = W /tan(phi)

at engineersedge .  With frame weight Wa  added this becomes:

 F =( W + Wa/2)/tan(phi)  

and

for the center pin actuator  the force with frame weight  is

 F = 2*(W+Wa/2)/tan(phi) 

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

The  solution is obtained by using the moment equation written about the center pin of the lift.  This gives:

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

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

where L is the half length of a scissor leg.

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 Wa)  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 + Wa/2)*dh
or
F = (W+ Wa/2)*dh/dx

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

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

When multiple stages are added the force is simply multiplied by the number of stages.  This follows from the energy formulation since each stage adds another dh but the dx displacement of the actuactor remains the same.

An excellent reference for a more detailed proof is from a paper “Mathematical Analysis of Scissor Lifts” by H. M. Spackman .  He also wrote a paper

“Mathematical Analysis of Actuator Forces in a Scissor Lift”

12/18 Update: Pivot torque required to extend a scissor lift

If one wanted to apply a torque τ to the center pivot pin to extend the lift rather than use a base force F (similar to the “Chinese Vex scissor lift”) the equation is

τ  = (W + Wa/2)*L*cos(theta)  where L is the half length of a scissor leg.

Proof:

Let

theta = interior angle between the scissor legs.

d_theta = a small change in this angle.

          τ = torque applied to the scissor joint

The torque must do the same work that was done by the force F derived above.

τ*d_theta = F*dx =  work done to move the lift a height of dh  against gravity.

So    τ = F*dx/d_theta.

Using half of the base triangle  we know that the base x is  x = 2* L*sin(theta/2)  .So  we can differentiate with respect to theta

dx /d_theta = L*cos(theta/2)

We want this in terms of    phi = 90- theta/2      so

dx/d_theta = L*cos(90 – phi) = L*sin(phi)

Now plugging in for F and dx/d_theta

τ = (W + Wa/2)/tan(phi) * L*sin(phi)  or

τ = (W + Wa/2)*L*cos(phi) 

Again, if there are N stages then the total torque = N*τ