Sack Attack Scissor Lift Torque Requirements

8/8/2019 Sack Attack Scissor Lift Torque Requirements

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Sack Attack Scissor Lift Torque Requirements (theory)

Below is a copy of a blog post I made today:

Assume that we have a scissor lift that we use to lift W lb of sacks to a height of 36

inches.

Also assume that the lift is actuated by a center pin torque with gearing ratio GRthat is the output torque/motor torque.

A N stage scissor lift requires a center torque = N*(W_lb + W_lift/2)*L*cos(phi)where phi is the interior angle between the arm and the base (see this scissor theorypost).

The max torque occurs when phi = 0 or when the lift is at its lowest point. At this

point the total torque = M* m_torq where M is the number of motors and m_torq =max motor torque set by the PTC limits.

m_torq*M = N*(W+W_lift/2)*L/GR or

W = GR*m_torq*M/L/N – W_lift/2

Lets put in some typical numbers: L = 6 in, W_lift = 1 lb, N = 4 , GR = 7:1, m_torq= 6 in lbs (the 6 in lbs is set by alloting 3.5 amps per two 393′s driven by half acortex or power expander.)

W = 1.75*M – .5 lbs

M motors => lbs (sacks)

1 => 1.25 lbs (2.5 sacks)

2 => 3 lbs ( 6 sacks)

3 => 4.75 (9.5 sacks)

4 => 6.5 lbs ( 13 sacks)

Focusing on the 4 motor case, lets see what happens when elastic is added thatpulls with an upward force on the center pivot. In this case, when the robot ispicking up sacks, the motors would be holding the lift down against the elastic.When the lift goes up the elastic and motor are working together. If we add justenough elastic that the motor can still hold the lift down then this is equivalent todoubling the motor torque when the lift is down. So

Elastic +4 => 13.5lbs (27 sacks)

Or we could also up the gearing ratio by x3 factor…. the lift would be 3 times slower

http://vamfun.wordpress.com/2012/06/27/sack-attack-scissor-lift-torque-requirements-theory/http://vamfun.wordpress.com/2012/06/27/sack-attack-scissor-lift-torque-requirements-theory/http://vamfun.wordpress.com/2012/06/27/sack-attack-scissor-lift-torque-requirements-theory/http://wp.me/pmcbX-9Bhttp://wp.me/pmcbX-9Bhttp://wp.me/pmcbX-9Bhttp://wp.me/pmcbX-9Bhttp://wp.me/pmcbX-9Bhttp://wp.me/pmcbX-9Bhttp://vamfun.wordpress.com/2012/06/27/sack-attack-scissor-lift-torque-requirements-theory/


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but here you would be making one large dump and speed is not required.

3x gear + 4 => 20.5 lbs(41 sacks)

Ok now we are talking…. 41 sacks !! But in Mythbusters style…. more is better so

how about 6 motors.

3x gear + 6 => 31 lbs (62 sacks)

Ok… enough for me. Not sure how you would stack 62 sacks on the upper goal.These numbers are of course rough cuts and ignore friction etc but elastic can beuse to make the theory come true.

So in summary, looks like we should be able to lift 13 sacks to a 36 in height with 4393′s working through a 7:1 gearing ratio. Adding elastic can increase this to 27

sacks or using a 21:1 gearing we can get to 41 sacks or with a 21:1 gearing and 6motors we can lift 61 sacks.


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A scissor lift (jack) or mechanism is device used to extend or position a platformby mechanical means. The terms "Scissor" comes from the mechanism utilized whichis configured with linked, folding supports in a crisscross 'X' pattern. The extensionor displacement motion is achieved applying of force to one of the supports resultingand an elongation of the crossing pattern.

The force applied to extend the scissor mechanism may be hydraulic, pneumatic ormechanical (via a lead screw or rack and pinion system).

Design Equations for Scissor Lift:

For a scissor lift that has straight, equal-length arms, i.e. the distance from thehorizontal-jack-screw attachment (or horizontal hydraulic-ram attachment) point tothe scissors-joint is the same as the distance from that scissor-joint to the top loadplatform attachment.

Scissor Jack - Loading Applied at Bottom

Open Scissor Lift Jack Force Bottom Load Calculator

http://www.engineersedge.com/calculators/scissor-lift-bottom-loaded.htmhttp://www.engineersedge.com/calculators/scissor-lift-bottom-loaded.htmhttp://www.engineersedge.com/calculators/scissor-lift-bottom-loaded.htmhttp://www.engineersedge.com/calculators/scissor-lift-bottom-loaded.htm


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Scissor Jack Equation With Load Applied At Center Pin:

Open Scissor Lift Jack Force Center Load Calculator

Free Body Diagrams:

http://www.engineersedge.com/calculators/scissor-lift-center-loaded.htmhttp://www.engineersedge.com/calculators/scissor-lift-center-loaded.htmhttp://www.engineersedge.com/calculators/scissor-lift-center-loaded.htmhttp://www.engineersedge.com/calculators/scissor-lift-center-loaded.htm


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Static Equations


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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 engineersedge scissor-lift link 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 momentabout 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);

http://www.engineersedge.com/mechanics_machines/scissor-lift.htmhttp://www.engineersedge.com/mechanics_machines/scissor-lift.htmhttp://www.engineersedge.com/mechanics_machines/scissor-lift.htmhttp://www.engineersedge.com/mechanics_machines/scissor-lift.htm


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Hope you structural engineers set one of us straight.

chris

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Sack Attack Scissor Lift Torque Requirements