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Thread: Scissor Lift Design Help

  1. #1
    Associate Engineer
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    Scissor Lift Design Help

    Yes, i know this topic has been hit alot. I really just need help ensuring my math is correct. (its been a while since college) lol. anyways. I'm trying to figure out the optimal set up required to lift a load of approximately 25lbs from completely collapsed scissors to a height of 96" using 2 levels. first, solving for L:

    L =(H/4)/sin(theta)
    where H = height of platform when lift is extended
    theta = angle of L to horizontal

    plugging in:
    L= 24/sin(45) = ~34"
    which means each bar would have to be ~68"


    Now to calculate force required
    F = n((W +(Wa)/2)/tan(theta))

    where n=number of levels = 2
    w = weight of load + platform = 25lbs
    wa = weight of members = 15lbs
    F = force,

    1.When lift is collapsed (lets say angle is 5 degrees):
    F = 2((25 + 7.5)/tan(5)) = ~149 lbs of force

    2. When lift is extended (lets say angle is 45degrees):
    F= 2(25 + 7.5)/tan(45) = ~65 lbs of force

    Then, the distance, D, the base member moves inward should be equal to the distance the top member moves inward as well and proportional to the height each member travels from the horizontal, correct?
    So if im moving the base member via linear actuataor i would need to have it rated for ~150lbs of force with a stroke of 12"?

    Also, i did a 3 level lift, the force required would be higher, but the stroke of the actuator and the members would be shorter?

    This seems very inefficient when opening the lift from collapse, so am I missing something or are my calculations correct?

    Thanks in advance!

    update:
    see attached image for visualscissor lift diagram.jpg
    Last edited by SirAlexander; 01-16-2013 at 07:42 PM.

  2. #2
    Principle Engineer
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    I found a few problems:

    F = n((W +(Wa)/2)/tan(theta))

    I get F=2 n((W +(Wa)/2)/tan(theta))


    1.When lift is collapsed (lets say angle is 5 degrees):
    F = 2((25 + 7.5)/tan(5)) = ~149 lbs of force

    I get 1400+ lbs.
    5 degrees is too small;the price you pay for compactness.

    Yes,scissor jacks are very inefficient at the start.

  3. #3
    Associate Engineer
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    Thanks Zeke, part of my error was i put ".8749" for tan(5) as opposed to ".08749" which threw off my calculations. but why did you put "2n" for the number of levels? based on that, you would need 4 times the force to lift 2 levels. that seems really excessive. i would think it would be 700lbs of force as opposed to the 1400+, but even that seems like a LOT of force required to lift 32.5 lbs. is something else wrong with my math, or do i really need an actuator rated for over 1000lbs of force to power this lift?

  4. #4
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    I have not checked your Math as I trust Zeke more than I trust myself.

    But, 5-degrees as a start angle is very low and will be prone to excessive loads on the actual lifting device components, lead screw and motor. With not too much wear it may not get started at all.

    See if you cannot at least get that to about 10-degrees otherwise starting loads will accelerate wear and tear alarmingly.

  5. #5
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    Thanks for the input PinkertonD. It needs to be compact at start, height needs to be no more than 30". that being said,
    (h/4) = Lsin(theta)
    h=4Lsin(theta)
    h=(4)(34)sin(10)= 24"
    which is definitely feasible.

    so using Zeke's formula at 10 degrees, the force at start up would be
    F=2 n((W +(Wa)/2)/tan(theta))
    = 4((25 + 7.5)/tan(10))= 4(32.5/0.1763) = ~737.38 lbs of force. this still seems high though.

    The reason i ask is because im working on a building projector screen lift, not a drop down like most systems, hence the height and length dimensions. but i see people powering TV lifts using actuators rated at 150lbs. TVs are a lot heavier than a projector screen. so using an actuator requiring ~700lbs of to lift a load of 32.5 lb seems like a lot. granted, those systems use an actuator pushing a rail as opposed to a scissor lift, but i didnt think it would require more force to use a scissor lift over rail lift. should i just look for alternative designs for my system or am i missing something in my calculations?

  6. #6
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    The scissor lift is basically a toggle. The force-to-lift -v- the load-lifted is inversely relational. As the arms approach the vertical, the load can theoretically be almost infinite for quite a tiny lifting force at that point.

    That's why the start lift angle is very important. Try doing the calculations for a start angle of say 30-degrees. Then work back and find an optimal angle that will work on force required for a compact start height.

    Life is all about compromise and Engineering is no different.
    Last edited by PinkertonD; 01-17-2013 at 04:30 PM. Reason: Rambling physics!! Dyslexia, doncha love it?

  7. #7
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    yeah, youre right. thanks for the help guys! i think i got it from here.

  8. #8
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    Dear members,
    I need a solution for designing scissor lift i want to lift the material upto 725 mm and in rest off position height to be 120mm and width should be 500mm

  9. #9
    Administrator Kelly Bramble's Avatar
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    Quote Originally Posted by ksenthil@uatpl.in View Post
    Dear members,
    I need a solution for designing scissor lift i want to lift the material upto 725 mm and in rest off position height to be 120mm and width should be 500mm
    See:

    Scissor Lift Design Equations

  10. #10
    Senior Engineer
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    Nice solution for the designs off scissor. Thanks

  11. #11
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    I am designing hydraulic scissor lift of 6T capacity. Details are in the attached pic. I am taking cylinder of 130mm bore dia. Output rod dia. 65mm. what will be the maximum pressure required in bar. I have calculated with 25% overload and got 500 bar pressure or 600KN of force to be required to lift the material.

    Please confirm if there is any changes or advise me what to do.
    Attached Images Attached Images

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