# Torque of a structure in zero G (space shuttle)

1. May 6, 2009

### mongolianbbq

I am just brainstorming here, but I am wondering about the following:

Look at this picture of the payload bay doors on the space shuttle.

Each door has a torque shaft driven by an actuator. Lets say the required torque to drive the shaft is around 500 in-lb while the shuttle is in orbit.
If I know the angle that the door rotates from closed to open and the dimensions, can I solve for the force required for that torque? Is the correct equation T=r*F*sin(deg)? Do I even need to know any thing about the door itself since this is zero G? Or would I need the dimensions of the torque shaft?

If I am just making a rough estimate (very rough) would it be possible to solve this?

This is not homework, this is just for my personal gain in engineering knowledge. Any ideas, insults, etc, are welcome.

2. May 6, 2009

### belliott4488

I'm not sure what you're asking. If you are just asking what force is required at a given distance from an axis in order to produce a desired torque, then yes, torque = FRsin(theta) is all you need, where theta is the angle between the direction of the force and the line from the axis to the point where the force is applied (it's really F X R, where F and R are vectors). The door is irrelevant since that is what the torque acts upon, not what produces the torque.

If you want to know how fast the door will open up when acted upon by that torque, then yes, you need to know about the door, specifically its moment of inertia.

Last edited: May 6, 2009
3. May 6, 2009

### mongolianbbq

I guess what I'm asking is what is the R? Is that the radius of the torque shaft? Since that is the item being directly driven by the actuator.

4. May 6, 2009

### belliott4488

That depends on where the force is being applied. If it's being applied to the surface of the torque shaft, say by a cable that has been wrapped around it, then yes, R is the radius of the shaft, and theta = 90 degrees.

5. May 6, 2009

### mongolianbbq

I think the force is being applied via a ring gear assembly inside the actuator. So with the teeth of the shaft on the surface i guess the force is being applied to the surface.
Thanks!