# Arc subtended by stretching of a conical spiral spring

I'm trying to work out the angle subtended by the stretching of a conical spring and I have a solution but I'm not 100% sure about it so would appreciate some feedback.

An Archimedes spiral spring is shown above with radii at various points. What I would like to come up with is an equation for the angle that each part of the spring gets stretched by as a function of the radius and the distance from rest $$x$$. I know that for a normal spiral spring of constant radius $$R_{ss}$$ if it is stretched by a small amount $$dx$$ then it will subtend an angle $$d\theta$$ therefore $$dx=R_{ss}d\theta \\ \int_0^{x}dx=R_{ss}\int_0^\theta d\theta \\ \therefore \theta=\frac{x}{R_{ss}}$$

For a spiral conical spring with radius $$R_{cs}$$ that starts at $$R_{cs0}$$ and grows to $$R_{cs1}$$ what I did was integrate the radius: $$dx=dR_{cs}d\theta \\ \int_0^{x}dx=\int_{R_{cs0}}^{R_{cs1}} \int_0^\theta d\theta dR_{cs} \\ \therefore \theta=\frac{x}{R_{cs1}-R_{cs0}}$$ It seems too simple though hence why I would like to check here what the thought is on the solution.

## 1 Answer

No, Its not right. If the cone becomes more, steep at the limit when $$R_{cs1} \Longrightarrow R_{cs0} \quad \theta \Longrightarrow \infty$$

• My apologies but I'm not sure I understand your answer. Since it's an Archimedean spiral the cone doesn't become more steep because the slope of the cone is constant. Oct 16 '19 at 7:42
• @enea19 ok fine, but lets assume we started with a more steep cone, it is going to have smaller theta, but by your calcs it will have a bigger one. Oct 16 '19 at 19:53
• Aah, I see what you mean. But isn't that ok? I mean, a conical spiral spring that has less than a single turn is not very useful. Mathematically I understand the problem, but in practice it would never happen. Oct 22 '19 at 16:41