Is this a real tortion spring?

[Matt H.]

Torsion springs produce torque so pounds per rotation is not a unit which can be applied to them. What the unit on McMaster car is referring to is the maximum torque when the spring is fully wound. It should have a linear decay from that point.

So for a 180 degree spring with 40in lbs of torque it would have 20in-lbs at 90 degrees and 0 in-lbs at 0 degrees.

I'm not sure if you know this so I also give a brief explanation of torque.
Essentially 1 ft-lb can support 1 lb on a one ft radius. so a 40 in-lb spring would be able to just barely support a 40lb load at 1 in from the point of rotation.

To figure out how much torque you need to produce a given speed you will have to used rotational physics equations such as torque=rotational acceleration*moment of inertia.

[DonRotolo]

[quote=Matt H.;686342] so a 40 in-lb spring would be able to just barely support a 40lb load at 4in from the point of rotation.
QUOTE]You mean a 40 lb load at 1 inch (or 10 lb at 4 inches). 40 lb at 4 inches is 160 in-lbs.

Torque in ft-lbs is exactly what it sounds like - a force of one pound at 1 foot from the center of rotation.

If you pushed with the same force at 2 feet from the CoR, you'd have 2 ft-lb, and you know that makes sense from when you use a wrench to tighten a bolt - a longer wrench puts more force on the bolt with the same push.

If you decided that 20 pounds would push a ball to a certain height, and your torsion spring was 40 ft-lb at 180 degrees of wind-up, then an arm two feet long would push with 20 pounds (for the first instant). As the spring unwinds, the force decreases linearly to zero at no rotation.

Does that make sense??

[Leav]

thanks guys...

so my calculations show that I need 173 inch-lbs.

which is about 4.5 more than what mcmaster carries.... I hope the local spring shop has what we need... thanks!

-Leav

[Matt H.]

You could also try and solve that by using several springs together.