Releasing winch system without shifters

[Ether]

Originally Posted by androb4

He's referring to the cam they used to power their kicker.

I've heard it called a da vici cam.. Like this.

http://img.rlt.com/x/12131.jpg

One nice feature of a cam is that, with a little math, you can derive the equation for the spiral so that the torque required to drive it remains constant as it is pulling back the arm. Then you can use that equation to cut out the cam.

[androb4]

Originally Posted by Ether

One nice feature of a cam is that, with a little math, you can derive the equation for the spiral so that the torque required to drive it remains constant as it is pulling back the arm. Then you can use that equation to cut out the cam.

That's sweet!

I've made this (https://www.dropbox.com/s/vmkm7jq1km...%2012%20AM.png) the other day using Inventor's Create Spiral tool. What do you think?

[Ether]

Originally Posted by androb4

That's sweet!

I've made this (https://www.dropbox.com/s/vmkm7jq1km...%2012%20AM.png) the other day using Inventor's Create Spiral tool. What do you think?

I can't tell for sure but it doesn't look quite right. Can you post the generating equation?

[magnets]

Originally Posted by Ether

One nice feature of a cam is that, with a little math, you can derive the equation for the spiral so that the torque required to drive it remains constant as it is pulling back the arm. Then you can use that equation to cut out the cam.

Or, you could derive the equation so you gained mechanical advantage (and torque) as you wound back. It takes more force to stretch the surgical tubing at the end than it does at the beginning.

For another type of release mechanism, use a doorknob or gate latch.

[Ether]

Originally Posted by magnets

Or, you could derive the equation so you gained mechanical advantage (and torque) as you wound back.

What you want is for the torque required to drive the spiral cam to remain constant (think about it):

One nice feature of a cam is that, with a little math, you can derive the equation for the spiral so that the torque required to drive it remains constant as it is pulling back the arm. Then you can use that equation to cut out the cam.

It takes more force to stretch the surgical tubing at the end than it does at the beginning

True for many (most) designs but not necessarily true for all. It depends on how the mechanical advantage of the linkage changes as the kicker is being pulled back. In any event, what you want is for the torque required to drive the spiral cam to remain constant. To achieve that you would need to take into account the F=k*x spring constant of the tubing as well as any linkage geometry changes

[Lil' Lavery]

Originally Posted by magnets

It takes more force to stretch the surgical tubing at the end than it does at the beginning.

That depends on your wording and if the surgical tubing (or other spring mechanism) is strained beyond the linear elastic region. While, yes, it will take more total force to stretch the material from 0 to Y than 0 to X (assuming Y>X), the force to stretch from X to X+1 may not be more than the force required to stretch from X-1 to X.

Point 1 on that diagram represents the yield stress, where the linear elastic region stops and any further strain on the object will result in some permanent deformation.

Ideally, teams would keep their surgical turbing (or other spring) entirely within the linear elastic region so the characteristics of their launch remain consistent over time. However, the reality of the situation is often that, in order to achieve enough initial force given a limited travel distance, the tubing is stretched beyond the yield stress, and some permanent deformation occurs to the tubing each time it is stretched.