Actuator to move 400 N 90 degrees in 0.2 sec

In summary, the problem is to construct an apparatus capable of rotating a 40kg steel plate 90 degrees against the force of gravity in 200 milliseconds. This would require a torque of 120 Nm and an angular velocity of 8.3 radians/sec. To provide this power, a hydraulic accumulator or electromagnetic actuator could be used, with an estimated power requirement of 700 watts. The actuator will also need to be able to withstand weather conditions and have a way to recharge the batteries.
  • #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
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  • #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
  • #3
Thank you Bob. I will investigate hydraulic mechanisms, further, along with electromagnetic actuators.
  • #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.
  • #5
Bob S said:
Another issue is whether this actuator will be exposed to outside weather conditions.
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.

What is an actuator?

An actuator is a mechanical or electronic device that is responsible for moving or controlling a mechanism or system. It converts energy into motion to perform a specific task.

How much weight can this actuator move?

The actuator is designed to move up to 400 N (newtons) of weight. This is equivalent to approximately 90 pounds.

How far can this actuator move?

The actuator can move up to 90 degrees in rotation. This is a commonly used unit of measurement for rotational movement.

How fast can this actuator move?

This actuator is capable of completing its movement of 90 degrees in just 0.2 seconds. This is a very fast speed and is ideal for applications that require quick and precise movements.

What type of energy does this actuator use?

This actuator uses mechanical energy to convert into motion. This can be achieved through various mechanisms such as electric motors, hydraulic systems, or pneumatic systems.

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