Hand Wheel Rim Pull Calculation

[Actuators1]

Does anyone know how to calculate hand wheel rim pull on a worm gear system?

The gear box turns as follow:
Hand Wheel -> Worm -> Worm Gear -> Acme Threaded Shaft ->

How can I determine the rim pull force required to attain a certain linear force at the end of the system? See a drawing of the system here:

http://www.freeimagehosting.net/mkp9q.

Questions are welcomed. Thanks!

 

[Cockroach]

Basically it is the thread torque or input load required at the pitch diameter of the mating gear system to the rim radius.

Regards,
Cockroach

 

[drawoh]

ActuatorsATI,

Unless I am missing something, this looks like very basic machine design stuff. Torque into the worm equals force times radius. Torque out of the worm gear is multiplied by the ratio, and you subtract significant amounts of friction. The acme thread is similar. You have not provided details.

This is a simple problem for an engineer or an engineering technologist. What is your background?

JHG

 

[chicopee]

He probably did not take machine design in college.

 

[Actuators1]

Solved it, thanks just needed some reassurance.
Yes I took machine design.

 

[rmw]

Caution about the hand wheel torque. Most actuaters I have seen with manual overdrives, because of the ratio-ing so that it could be hand operated, would overtorque and break the final drive train gears. I've seen very few actuator gears broken by the actuator drive mechanism itself. Most were broken by operators hand-driving the actuator into the stops and then 'bowing' up on it to try to open or shut it. Or worse, getting some help from a 'cheater' like a pipe wrench.

rmw

 

[drawoh]

rmw,

Lots of small gear-motors are capable of self destruction, without hand-wheels.

JHG