# Scissors Lift Equations??

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• 11-12-2011, 03:33 PM
Vamfun
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
• 11-13-2011, 07:59 AM
Kelly Bramble
You’re the second person whom has challenged Engineers Edge proof for the scissor lift equations. I’m interested to hear what others think as well.. So, let's get this thing fixed or verified!

The attachment to this post was send in by a Jean-Philippe Major for review.. Engineers Edge (me) has not had a chance to get to it..
• 11-14-2011, 01:04 AM
Vamfun
This looks ok but doesn't deal with lift frame weight. Here are some alternate papers that might be better.

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

[/URL] He also wrote a paper “Mathematical Analysis of Actuator Forces in a Scissor Lift”

[ATTACH]958[/ATTACH]
• 03-26-2012, 09:47 AM
TeamTiki
Our robotics team desires to construct a 9 stage scissor lift to lift a 10lb load utilizing a lead screw to extend/lift the scissor lift by applying upward force to the center pivot point of the first scissor stage, see attached diagrams. Might you assist us with the scissor lift force equations for this configuration which we believe are different from the bottom actuator or center actuator equations? We wish to calculate the force required to lift a 10lb platform at the top of the lift, but we're not sure how to compute the force given the friction form the scissor lift. s[ATTACH=CONFIG]195[/ATTACH][ATTACH=CONFIG]196[/ATTACH]

Once we know the equations to compute the required force, we'll can select the appropriate motor gearing and lead screw lead configuration.

Thank you in advance for your assistance.
• 03-26-2012, 10:10 AM
PinkertonD
Team Tiki, Really?? Have you thought this through?

With a vertical lead screw, I see several problems.
With one end of the Scissor mechanism fixed, The lift point travels laterally so the screw base and drive mechanism will need to track with it.
Next, the vertical lead screw will limit the closed position. You show it mounted to a solid base, so I assume from that the screw and drive is not mounted below the base. If you are lifting from one side as it appears from your sketch, then you are asking for big trouble with "jacking." Look it up.

Finally, why on earth would you want to change a design approach that has been around for two hundred years?

Team Tacky, I would call it? :D
• 03-26-2012, 11:20 AM
TeamTiki
Yes a funny name... but that's fun at high school robotics competitions. We compete in the FIRST Tech Challenge robotics.

In our proposed design, there would be two lead screws. One lifting each of the two stage 1 pivot points. The total travel of the lead screw is 8". The motors and gearing are mounted below the scissor lift and lead screw on the robot's drive frame. The robot is approximately 18"x18"x18".

We envisioned both ends of stage 1 moving along a slotted U channel on rollers thus restricting its movement both vertically and horizontally. The lead screws would be mounted at a fixed point at center point of the scissor lift. We believe this will greatly reduce the required force to lift the load rather than pull or pushing with an actuator horizontally along the bottom which is our current robot/scissor lift design. This methods requires us to generate over 800lbs of force to activate the lift and we've experienced synchronization problems between the left and right sides.

Can you help with the equations?
• 03-26-2012, 01:53 PM
jboggs
I'm trying to understand your diagram. Normally in a scissor lift, one floor support will be pivoted, and one will be rolling, like in attachment. But your diagram appears to have both floor contacts to be rolling, so that the center point can move straight vertically. If you REALLY want to do that you must have some kind of lateral guidance for that center point to keep it on a vertical line. I don't think you really do want to do that.
[ATTACH=CONFIG]197[/ATTACH]

P.S. I attended a FIRST Robotics competition this weekend. Really neat. They were scooping up basketballs, shooting baskets, and balancing on seesaws.
• 03-26-2012, 02:56 PM
PinkertonD
[QUOTE=TeamTiki;2507]We envisioned both ends of stage 1 moving along a slotted U channel on rollers[/QUOTE]

But, but, but... if both ends are free to move then any lateral forces applied to the table or load will be passed directly to the lead screws intent on bending them. It is fine to have both sides moving, but how will you stop them sliding on their rollers, as a unit, when extended? That's why a traditional scissor lift has one end fixed.

Also, dare I suggest, that if you are lifting vertically from the center of the scissors, the lifting (load) force will be roughly equivalent to the effort to lift the scissor mechanism and load. There is no mechanical advantage or disadvantage when lifting vertically form the scissor fulcrum.

There are threads here on scissor lift equations and some of them, a little in contention. :)

If you are getting high start-up loads then try starting at a slightly higher angle of the scissor arms. Essentially a scissor is a toggle. With a Toggle-action, the lifting (load) force increase rapidly and effort required to lift, decreases rapidly as the included angle of the toggle approaches zero.

The wheel as already been invented. :)

p.s. 800-lb would not require much torque when talking of a fine-thread lead screw.
• 03-26-2012, 06:09 PM
TeamTiki
We appreciate the concerns you've raised, but we believe our robot design has already addressed them. The equations you reference on the forum and website do not apply when lifting vertically at the center. So what we seek is help with the scissor lift force equations. Solving the Statistics equations ourselves is a few years above our current capabilities. We received the paper by Jean-Phillipe that provides insight but not enough.

Help anyone? :-)
• 03-27-2012, 07:26 AM
TeamTiki
The competition constraints on the robot design, mainly the 18" height requirement is the reason for our unconventional scissor lift design. By raising the center pivot point, we can raise the scissor with 8" of vertical travel on the lead screw. The lead screw is secured in the support towers such that it does not move. The bottom legs of the scissor have pins that slide along slots in a frame, see the attached diagram.[ATTACH=CONFIG]198[/ATTACH][ATTACH=CONFIG]199[/ATTACH]

Our desire is to further optimize our motor gearing to raise the scissor lift in under 15 seconds. This requires us to be able to compute the required force to raise our 9 stage lift knowing the torqure output of our motors. Do you have any insight? :confused:
• 03-27-2012, 09:44 AM
jboggs
Since all your forces are in the vertical direction, you won't need all those fancy trigonometric equations. Look at the attached free body diagram.[ATTACH=CONFIG]200[/ATTACH]

W = [B]HALF[/B] of the weight of everything above that point.
F = force exerted by actuator
R = force exerted by floor guidance track.

Note that R = F. That means the floor connection points will be pulling upward, not pushing downward. They must be held down firmly.

F will always equal 2W, no matter what the angle of the scissor arm is.

Don't forget that the actuator will be connected to two identical links, so the total force it will see is 4W

Again - since both ground contact points are free to move horizontally, the only thing keeping the whole system in place is the position of the center pivot. It must be firmly guided to prevent any horizontal motion. And I would NOT depend on the stiffness of the actuator rod for that. They are not intended for that purpose, and they will fail.

Hope that helps.
• 04-20-2012, 12:37 AM
Absolutely right Vamfun. I did one mechanism simulation of the above mechanism and wondering why my results are not matching with engineer's edge equation (Actual force will be two times). Further investigation leads to me to this forum.
Now.. if it is proven wrong, why engineers's edge is not removing this page. It will save someone from misleading
• 04-20-2012, 02:11 PM
Kelly Bramble
[QUOTE=nishadp;2702]Absolutely right Vamfun. I did one mechanism simulation of the above mechanism and wondering why my results are not matching with engineer's edge equation (Actual force will be two times). Further investigation leads to me to this forum.
Now.. if it is proven wrong, why engineers's edge is not removing this page. It will save someone from misleading[/QUOTE]

Yes, reviewed the scissor equations and calculators and the results are wrong.

I have republished updated equations and fixed the calculators to match see--->

[url]http://www.engineersedge.com/mechanics_machines/scissor-lift.htm[/url]

Equation proofs provided by :

Jean-Philippe Major
Jeffrey Trembley
Dominic Pellerin
• 05-01-2012, 05:28 PM
Serch
Excuse Vamfun, my english is bad (I'm spanish).

Can you tell me how you came to that formula?? (for the center pin actuator)

Thx, Sergio
• 07-19-2012, 11:12 PM
Vamfun
At last!!
[QUOTE=Kelly Bramble;2716]Yes, reviewed the scissor equations and calculators and the results are wrong.

I have republished updated equations and fixed the calculators to match see--->

[url]http://www.engineersedge.com/mechanics_machines/scissor-lift.htm[/url]

Equation proofs provided by :

Jean-Philippe Major
Jeffrey Trembley
Dominic Pellerin[/QUOTE]

What you have is correct now...but as I mentioned before, Jean-Phlippe et all did not include the weight of the frame in their free body diagrams. No biggy...but it would have been nice to include since a zero weight is not practical for most designs:)

I revised my blog post to nolonger point to your site as having an error.
• 07-19-2012, 11:21 PM
Vamfun
Sorry I didn't get back to the forum until July. But I'll answer your question anyway.

It is really based upon energy required to lift the weight of the frame and the load. This energy is the same no matter where you put your actuator. If your bottom actuator is providing force F in the x direction and it moves a small amount dx during the a small change in lift height then the energy supplied is F*dx. If you now place the actuator at the center, this point only moves 1/2 *dx during the lift. Here we assume that one leg if fixed. So to get the energy you need , F*dx, you must double the force ...or 2*F and the product 2*F*(dx/2)= F*dx.
• 08-19-2012, 11:15 AM
SANDIP
ONE THING MAKES ME STILL CONFUSING IF THE LIFT HEIGHT OF THE SCISSOR IS 3 METERS OR 5 METERS HOW MANY NOS OF CYLINDERS WILL BE NEEDED TO OPEN THE HYDRAULIC SCISSOR LIFT OF CAPACITY 500 KGS ?????

AND HOW THIS EQUATION F (FORCE ON THE CENTRE PIN) = (W + wf/2) / tan phi will play the role in this juncture ???

W = Weight of the load.
wf/2 = Weight of the frame.
Phi = Interior angle between horizontal and the arm.

with b'rgds,
b.sandip
• 10-05-2012, 01:55 PM
Vova
[QUOTE]So F_center_pin = 2*F_bottom = 2*( W + wf/2)/tan(phi);[/QUOTE]
And for vertical cylinders:
F = 2*( W + wf/2)
am I right?
• 02-11-2014, 06:30 AM
bgb321
Is anybody tell me how to claculate Force required to lift the load when cylinder is in Inclined position, Say 10 deg. to lift a load of 900 kg(complete weight).
• 10-30-2014, 06:23 AM
jackieann000
This is very helpful info. This is helpful in industrial and the mechanical application. Thanks for sharing.
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