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I am working a problem concerning a 3 axis gryoscope, the spinning mass is a magnet (dipole). This is part of a optical sensing device. The inner gimbal is for pitch rotation, and the outer gimbal is for yaw precession. The outer gimbal is fixed in the sensor casing. The whole assembly is well balanced and not affected by gravity. The casing contains 4 electrical coils located at 90 degree intervals. These coils are pulsed to make the magnetic rotor spin as in a motor. There is another coil that is wrapped around the casing and it is pulsed to precess the gyro in pitch and yaw. The spinning mass is on a bearing through which extends a non spinning axis (forward). On the end of that axis is an optical sensor (which does not spin). The outer casing clearly will limit the range of pitch and yaw motion. And technically, it is desirable to not come into contact with it. The part that would come into contact is the optical sensor. There are other sensors that keep up with spin speed, pitch and yaw gimbal angle, and dipole location. This is needed to be able to send proper spin and precession commands.

I have built a very good simulink model using Euler's torque equations, and all of the primary suff works. I even have a good idea of how to model the impact of the rotor precessing INTO the casing wall (reflect the pitch and yaw rotor velocities). But I cannot seem to figure out how to do the other case: Casing rotating into the rotor, where rotor precessions are essentially zero. How do I trasfer the casing momentum to the rotor and still keep the solution physically bounded by the casing itself?

Since the casing is mounted in a moving platform with 6 DOF, it can clearly rotate into the rotor. But remember, only the non-spinning part of the sensor will touch the casing, so the angular momentum of the rotor is not DIRECTLY involved with contact.

Thank you, and sorry for the long description, MikeO

Here is a graphic, certainly not scale. And the design is such that the optical sensor would contact the encasement before the rotor could. Two spin coils are not depicted. This drawing is purely conceptual. In the actual device, the gimbals are quite small and hidden inside a hollow spot within the center of the rotating gyro mass. You cannot see the gimbals.

Precession Commanded Gyro

The inertia tensors for the case and the rotor are known. The gimbals are taken as inconsequential. The intertia tensor of the case is much larger than that of the gyro and can be seen to be unaffected by the gyro. Whether the case rotates about the rotor, or the rotor rotates does not matter (by rotate, I mean pitch and yaw, not spin). The only parts of the entire assembly that can come into contact are the non-rotating optical sensor assembly and the case wall. Therefore the rotor will not touch the case. I am looking to write a conservation of angular momentum equation that would describe the initial rotations (pitch and yaw) of the rotor assuming the rotor was not pitching or yawing at the time of contact, in other words, the case rotated into the optical sensor which in turn transmits torques about the gimbal axes.

In the equations v1=u1*(m1-m2)/(m1+m2) + u2*(2*m2/(m1+m2)) and v2=u2*(m2-m1)/(m1+m2) + u1*(2*m1/(m1+m2))

May I subsitute the inertia tensors for mass and the body rates for velocities?

enter image description here enter image description here enter image description here

The problem I am trying to solve is how to compute the reactions at the the boundry (the cylinder that blocks motion). I think I have solved the problem of what to do when the gyro is driven into the wall... I am reflecting the precession values (pitch and yaw). But the other problem is harder... what to do when the cylinder is driven into the gyro. Remember, the rotating mass never comes into contact with the wall, just the non-rotating mass at the end of the axle (spin axis). At first glance, it might seem that the answer is the same, but it does not work. I need to be able to calculate the "imparted torque" from the collision, and confine computed location within the cylindrical boundary.

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Draw/sketch it please ;) –  Kvothe Feb 2 at 20:49
    
I will try to do that tomorrow. I understand it is difficult to visualise. –  user38780 Feb 3 at 3:01
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