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    Discussion in 'Calculations' started by Mastino, Mar 15, 2013.

    1. Mastino

      Mastino New Member

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      Hello all!
      Like you can see on photos and video I want to make construction like a little crane. I want to buy linear actuator one of these: http://www.robotshop.com/eu/standard-actuators.html
      Now the problem, how much pull power in lbs do I need? I know it must be very easy to calculate, but I am afraid to make mistake and to buy a bad one. They are not cheap.. A lot of thanks!
      [​IMG]
      [​IMG]
      [​IMG]
      [video=youtube_share;xSqYBIdnJLs]http://youtu.be/xSqYBIdnJLs[/video]
       
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    3. mhjones12

      mhjones12 Well-Known Member

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      Well, you need to give a few more details before your problem is fully constrained, but I’ll make a few assumptions to simplify things and get you going. I’ll assume that the actuator is set up to pull perpendicular to the short arm. Start by balancing the torques on the system:

      Torque_left_side = 1kg*g*2.1m + 1.2kg*g*Lcmg + 1.3kg*g*Lcmb

      where:
      g = gravity = 9.81m/s/s
      Lcmg = length to center mass of green bar, measured from pivot point
      Lcmb = length to center mass of blue bar, measured from pivot point

      Torque_right_side = Force_needed*0.079m

      where:
      Force_needed = force needed to maintain equilibrium

      Set the two torques equal to each other and solve for Force_needed. Note that the Torque_left_side equation is approximate (since the pivot point is not in line with the crane limb), and the Torque_right_side equation assumes that you are pulling perpendicular to the short lever. Also, the force solved for is just enough hold the system in equilibrium (i.e. not move up or down), so you need to make sure to pick an actuator that can supply more than this. How much more? Well, it depends on how fast you want to move your load (Extra_force*0.079m = Moment_of_Inertia*angular_acceleration if you want an idea of what the acceleration will be). Also realize that as soon as the arm moves a bit, all of the angles change and thus the torque balance will change also, so this calculation really only applies to the instant the mass begins to move (though I expect the force required will lessen as the arm is lifted, so it may not be a concern). Last, if you want to solve for the exact position dependent torques, then you have to use:
      [​IMG]
      or since it's planar:
      [​IMG]
      where:
      r = the radius from the center of the pivot to the applied mass (or center mass)
      F = the force (or mass*gravity)
      theta = angle between r and F

      Hope this helps.
       

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