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)  


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
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.



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


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

  1. Great post. I really appreciate your hard work. I was doing a project on scissor lifts for my college and i didn’t realize the error in the equation. I will add these valuable info to my project with the site reference. Thanks for sharing. Thumbs up .

  2. YC says:

    Did you send feedback to engineering edge to verify this problem?

    btw, thanks alot coz i was calculating this also and found a difference in my answer using different theory.

    thanks for the verification 🙂

  3. vamfun says:

    yes… I posted my remarks to their forum. They acknoledged that one other person had written them and would try to resolve it. I think they are just undermanned.

    PS. Glad you agree with my findings.

  4. Dave says:

    Hi! first of all, THANKS! You said “For the bottom actuator the force should be F =( W + wf/2)/tan(phi)”. But in the document you attached namely “Mathematical Analysis of Actuator Forces in a Scissor Lift” by H. Spackman gives F =1/2[( W + wf/2)/tan(phi)] when n=1. I’m not saying you are wrong but why are they different? Thanks!! ~Dave

  5. Dave says:


    For one particular equation in the Spackman formulation, it gives me an infinite value for a sloping angle of 45 degree. I understand that it means the actuator force is a maximum at 45 degree but do you know how can i determine the value of the maximum actuator force? since the equation would only result in infinity. Thanks!

  6. vamfun says:

    If you could be more specific re page and exact equation maybe I could help.

  7. onkar gandhi says:

    how to determine proper location of hydraulic actuator if it is inclined??nd wat advantages it will give??
    your post was very helpful thnx a lot..:-)

  8. Rr says:

    hey can anyone tell me why Wa/2 is used ?and not just Wa ?

  9. Rr says:

    how is dh/dx = 1/tan(phi) ??

    • vamfun says:

      This requires a little calculus and the Pythagorean Theorem. By inspection the large triangle formed by the base x and height h is a right triangle so
      h^2 + x^2 =( 2L)^2

      Differentiating both sides with respect to x gives and dividing by 2 gives
      h*dh/dx + x = 0


      dh/dx = – x/h = -1/tan(phi)

      There is a minus sign that shows up but this only changes the direction of the force which is arbitrary depending on which direction represents a positive change in x.

      • Rr says:

        thanks !!! got that now. Can you also tell me why it is (Wa/2) ?

      • vamfun says:

        This falls out of the proofs. Using moment equations, the center of mass of the lift structure acts with half the lever arm. Or as I mentioned in the virtual work (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. This is simply due to geometric scaling… the top of the lift moves twice as far as the center of the scissor lift where the frame weight force is acting. When the dh is factored out from the summation of the W*dh and Wa*dh/2 this leaves the Wa/2 term.

      • Rr says:

        thanks a lot for helping !!! its clear now.

  10. mohammad says:

    hi . hoe to find the dynamic calculation of scissor lift

  11. hi there…we had our project here..scissor lift..i wonder if we were right to locate and attached the hydraulic jack on the roller and attach it on the arm above on a certain point. help pls. for the equation.

    • vamfun says:

      I will try. Can u send a diagram or picture. Any actuator that moves the scissors open or. Closed is ok. The forces just change pending location. What u describe can work. I can help with the force equation if I know the geometry.

  12. So you’re saying there was an error in the equations in the site, but I check out the site and you’re still modifying the calculations here, how’s that possible?

    • vamfun says:

      Vladimir. Thanks for your comment. If you read my post I say the equations on the site are correct… I simply expanded them to include the mass of the lift its self. Engineers edge only considers the load mass.

  13. Tam Pham says:

    Thanks for your papers. I love it. But in fact, the position of cylinder mounted is not always at mid or in bottom of lift. How can i determine the force of actuator need to raise the lift if the angle of cylinder and the links are difference ?

  14. Anan Ramachandran says:

    Hi Vamfum. I have a query. Is the scissor lift design that you’re analysing is different from Spackman? you are analysing for 2 arm scissor lift while spackman is analysing 2 arm on each side which is Left and right and that is why there is a difference in the equation? Thanks and correct me if I’m wrong 🙂

    • vamfun says:

      Yes the Spackman formulation there are left and right forces which are identical. The equations are shown for only one side. Therefore they are half of mine.

  15. Rory van der Walt says:


    I have been trying to derive the equations for a hydraulic cylinder which is mounted on one side, from the base of the fixed end which extends to the opposing arm some distance above the centre pin.

    I keep getting stuck and was wondering if you could help? I could send you a picture to help your understanding if you would like?


  16. Lucien Emch says:

    We really like your blog, it has engaging information, Thank you!

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