Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Actuator to move 400 N 90 degrees in 0.2 sec

  1. May 24, 2009 #1
    I seek to construct an apparatus capable of moving a steel plate against gravity in 200 milliseconds.

    The plate is roughly the same size and shape as the hatch on a submarine:

    mass = 40 kg;
    dimension: 0.3 meter diameter--> = radius in calculations below;
    angle of rotation: 90 degrees--horizontal to vertical;
    time available for actuator to work against gravity: 200 milliseconds;
    distance traveled: 1/4 * Pi * Diameter = 1/4 * pi * 0.6m = approximately 0.5m;
    actuator return time constraint: within 0.6 seconds;
    life cycle: ideally: suitable for 100 million repetitions without fatigue.
    maximum DC power source available: 48 volts (four automotive batteries in series).


    So, to restate the problem in general: How to rotate 90 degrees, in 200 milliseconds, an object of mass 40 kg, a distance of 0.5m, against the force of gravity, relying upon not more than 4 auto batteries in series?

    Torque = r * F * sin theta, where F is approximately 400 Newtons (40 * 9.8), and r, radius is 0.3 meter, (not 0.15m), and theta is 90 degrees, hence sin(90) = 1.

    Thus the torque is 120 Nm.

    Torque is also defined as rate of change of angular mommentum, i.e. dL/dt, where L = r * p* sin theta, where p is momentum, i.e. mass * velocity

    So, p = 40kg * 0.5m/0.2 sec, which is about 100kg m/sec
    Therefore, L should be approximately equal to 0.3m * 100kgm/sec : 30 kg m^2/sec

    Using this value, then, one can estimate omega from the equation:
    L = I omega, where I is the moment of inertia, and omega the angular velocity,
    first simplifying by assuming, perhaps incorrectly, that I = mr^2, i.e. 40 * 0.3 ^2, or approximately 3.6 kg m^2

    This would yield a value for omega of 30/3.6, or about 8.3 radians/sec

    How to find, or construct, an actuator capable of meeting these requirements?
    (If the calculations are correct!!)

    Any ideas, suggestions, or improvements in the calculations would be welcomed.

    With that amount of weight to move, i.e. 400 N, there will also be non-negligible frictional forces opposing rotation, and also bearing constraints, which have thus far been neglected....

    Thanks for your advice, opinions, or comments
    CAI ENG
     
  2. jcsd
  3. May 24, 2009 #2
    The steel plate has to be accelerated to maximum angular velocity in about 100 milliseconds, and decelerated for the remaining 100 milliseconds. The torque is so high that the batteries cannot reliably provide it. I suggest using a spring-loaded hydraulic accumulator that is pumped up by the batteries, and provides the instantaneous power via a hydraulic mechanism to move the plate. The deceleration phase could be used to recharge the spring accumulator. See
    http://en.wikipedia.org/wiki/Hydraulic_accumulator
     
  4. May 24, 2009 #3
    Thank you Bob. I will investigate hydraulic mechanisms, further, along with electromagnetic actuators.
     
  5. May 25, 2009 #4
    The moment of inertia is comprised of two parts: The plate spun about a diameter is MR2/4, where M is the mass and R is the radius of the plate. The parallel axis theorem for moving the axis to the rim of the plate adds MR2, so the moment of inertia is I = 5MR2/4.
    The average rotational speed is 5 pi/4 radians per sec. If the plate is accelerated for half the time, and decelerated the remaining time, the peak rotational speed is 5 pi/2 radians per sec.
    The peak kinetic energy after 0.1 sec is then
    E = (1/2) I w2 = (1/2) (5/4) M R2 (25 pi2/4) = (125/32) M R2 pi2
    Putting in M=40 Kg, R=0.15 m, we get E= 34.7 joules.
    Because this energy has to be delivered in about 0.1 sec, the instantaneous power has to be about 347 watts (not including inefficiencies). So let's say 700 watts (15 amps at 48 volts). Not too bad. Another issue is whether this actuator will be exposed to outside weather conditions.
     
  6. May 25, 2009 #5
    Yes, it will.
    The anticipated temperature range is -20 C to + 40 C.
    Thank you Bob, for your very helpful calculations.

    The moving object will most certainly be constructed from non-magnetic materials, however, it will still weigh about 40kg.

    I also have not yet figured out how to recharge the batteries, nor, have I yet estimated how many repetitions the actuator can accomplish, before the batteries have become depleted.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Actuator to move 400 N 90 degrees in 0.2 sec
  1. Rotary Actuator (Replies: 9)

Loading...