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Calculation of torque for motion control camera arm

  1. Jan 30, 2012 #1
    Hi,

    First post here, but I've been reading the forums for a while getting answers for various projects.

    I am currently working on a lightweight motion control camera arm (kind of a small jib arm, but with servo actuators). I've researched a lot to figure out what force I would need in order to move things smoothly... I would love if someone could have a look at my calculation and see if I forgot something.

    Here's a rough CAD of the arm:
    http://www.newreel.ca/arm.jpg [Broken]

    At the pivot point, the load is almost perfectly balanced (it will hold any position without trying to move, and requires very little effort to move around). The whole arm (including camera/lens and counter-weight) is about 6KG. Dual ball-bearings on all axis, so friction is minimal.

    In order to calculate the torque required to move (or stop the movement) of the arm, I've used the expression used to get the moment of inertia of a cylinder about its central diameter axis;

    I = (MR2)/4 + (ML2)/12

    The arm is 1.5m long (110cm on the long end, and 40cm back for the counterweigth), and about 20cm in diameter (a bit less, but since it's not quite a cylinder I figured it would be close). As I said, it's 6kg, so:

    I = (6*0.12)/4 + (6*1.52)/12

    Which gives me 1.14 kg m2

    I want the arm to cover 90° in 0.75 sec, so I'm asking for a 2.0944 rad/s acceleration, right?
    If this is right, then the required torque would be:

    T = I*A2;

    T = 1.14*2.09442;

    T = 5Nm = 708.14oz/in.


    The servo I plan on using can deliver 580oz/in at max power, and is driving a timing belt/pulley with a 2.16:1 ratio... In theory, that would give me over 1,200oz/in of torque which would cover (hopefully) for the friction induced at the timing belt/pulley and bearings... Does all that seem to make sense? Am I too close? I'm not sure about the cylinder calculation, since the pivot point is offset, but since the counterweight is moving the center of gravity, I thought it would be equivalent... I'm also not sure about the acceleration calculation, as 2.0944 rad/s is the average, but it has to accelerate and then stop at 90°, so it will be moving quite faster midway - do I need to take that into account, or the acceleration is effectively calculated by doing the average?

    Thanks for looking into it, any pointers and/or suggestion are much appreciated.

    -A
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
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