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Gimbal table; tilt calculation given acceleration

  1. Apr 16, 2013 #1
    I am working on an electronics project which is sort of an auto-correcting two-axis gimbal table. There are two degrees of freedom, being two square frames - one can rotate on a pitch or y-axis, the outer frame can rotate about a roll or x-axis. There is a third outer frame which is fixed to a base. These frames are connected together by ball bearing pillow blocks and attached to stepper motors, which in turn are connected to a microprocessor to control rotation of these frames, so that I can move them at a specific angle.
    The purpose of this device is to keep a mass which is placed on the center frame, on the frame without it sliding off - in a moving vehicle.
    I am monitoring the velocity of the vehicle with an accelerometer, so I can get the g-force in the x and y direction.

    So I know I need to tilt, x and y a certain degree to compensate for the acceleration and deceleration of the vehicle, but I have no idea how to get the relationship between the vehicle's velocity and the angle that the gimbal tables should be.

    With no object on the table, and with the velocity changing, I'm thinking the table should not move. But placing something on the table will raise the center of gravity and it will want to topple over, like an inverted pendulum - at least this is what I'm envisioning. I need to tilt the pendulum to keep it from falling over.

    Can someone point me in the right direction???

    Thanks in advance!
  2. jcsd
  3. Apr 17, 2013 #2


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    If the tilting table was fixed to the vehicle it would be quite easy. You just need to tilt the top it so that the total force on the object is zero. The maths is similar to the "vehicle on a banked track" problem.

    If the tilting table itself can move (eg tip over) then I think you need more degrees of freedom so that you can shift the center of gravity relative to the "feet" of the table...or somehow allow the object to slide on the table so the weight is more over some feet than others... but I think that would be very hard to do.
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