2
$\begingroup$

Some trampoline springs extend only 40-50% before reaching their elastic limit. (e.g. a 10 inch spring can extend to 14 or 15 inches. Good springs can extend to 100% (E.g. a 10 inch spring can extend to 20 inches.

What is the difference?

Is this solely a matter of the alloy of spring wire used; Is it the shape of the spring?

What goes into the design of a long extension spring?

$\endgroup$
2
  • $\begingroup$ I imagine you would have more coils per length in a spring with longer extension since the material itself can only handle limited deformation. $\endgroup$
    – DKNguyen
    Jul 25, 2023 at 23:52
  • $\begingroup$ Obvious when you think about it. But thinner wires mean less force. Larger diameter should extend more too. $\endgroup$ Jul 27, 2023 at 1:56

3 Answers 3

4
$\begingroup$

For a given free length and spring rate there are an infinite number of designs of coil spring. However, there are many practical limitations that limit the options. A spring with a few coils, of thin wire, will reach its yield point before one made of thicker wire, with many coils, even if it is the same material.

In answer to OP's comment, here's the math. Smath v4

$\endgroup$
2
  • $\begingroup$ So if you have a given spring 10" long with 1" diameter with a max 40% extension, how do you design a 10" long spring with 100% extension. $\endgroup$ Jul 27, 2023 at 2:05
  • $\begingroup$ Too complicated to explain in a comment, I'll edit my answer $\endgroup$ Jul 27, 2023 at 3:12
2
$\begingroup$

The ratio of the spring diameter to the wire diameter is the primary factor.

$\endgroup$
1
  • $\begingroup$ Ok. but go on. What are the downsides of a wider spring? $\endgroup$ Jul 27, 2023 at 1:58
1
$\begingroup$

Let's compare two springs of the same alloy but with different numbers of turns

$N_1 =2 N_2$

.

over the length of the coil, $L$

Now if we extend the coils to a length $$L=2L_{inital}$$ The coils with more turns, $N_1$ Get half deformed per turn, so they can be stretched twice deforming less.

This usually means a thinner spring wire!

$\endgroup$
2
  • $\begingroup$ So what is the tradeoff for the same spring constant? If you use a thinner wire, it has to be both stronger and springier. Since we're talking bending, I would expect the spring constant to increase as the square of diameter of the wire and decrease with the length of the spring. (Linear with the extension per turn) $\endgroup$ Jul 27, 2023 at 2:03
  • $\begingroup$ If you had a really clever spring forming machine you should be able to overlap coils, one outside, one inside to get twice as many coils in the same space. $\endgroup$ Jul 27, 2023 at 2:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.