Spring too strong

In "A better drive mechanism", I looked for springs that could exert a force of 0.25 lb when compressed 5/16", and whose uncompressed length is 1.316". I had determined that McMaster-Carr did not have any such springs, but that Associated Spring Raymond did. It turns out that Associated also does not have a spring that fit the requirement.

Let me first review the calculations for McMaster-Carr's springs. We need a spring that will fit around a 1/8" x 1/4" steel rod. The diagonal of such a rod, that is its largest dimension, is 0.280". For a spring whose outer diameter is 3/8" (0.375"), we need to subtract twice the spring's wire diameter, and that must be greater than 0.280. Thus, our first criterion is that the wire diameter must be less than 0.0475", and preferably significantly less than that, since we don't want the inner side of the spring to scrape against the rod.

There are four continuous-length springs that meet this criterion. They are listed in the table below.

Item Length Wire size Constant
(lb coils/in)
n
(coils/in)
Force =
0.3125C/1.316n
Fully compressed length =
1.316n x wire size
9662K19 36" 0.035" 54.7 7.7 3.37 lb 0.355"
96565K39 12" 0.041" 109 6 8.64 lb 0.324"
9663K56 20" 0.032" 32.3 7.7 1.99 lb 0.324"
9663K23 20" 0.041" 95.4 9 5.03 lb 0.486"

So clearly, none of the springs exert a gentle-enough force. The weakest spring is at least eight times as strong as we need.

Taking a look now at the springs from Associated Spring Raymond (downloading the catalog PDF gives more information about the springs than the online tables do), the spring rate given is in lb/in, and the formula for computing the force is a bit easier, being 0.3125RL/1.316, where R is the rate and L is the original length of the spring (10" or 18"). This makes it much easier to solve for R. For 10" springs, R must be less than 0.10, and for 18" springs, less than 0.058.

Looking only at springs whose outer diameter is 3/8" or higher, we quickly find that not a single spring meets the force criterion.

Quickly checking a few other manufacturers of springs shows that nobody makes a weak compression spring.

What about expansion springs? I could reverse the drive and have the spring pull the rod in, at the expense of attaching the spring to the rod and the wall (instead of just sticking it on the rod and letting it push against the wall).

McMaster-Carr again, this time using the unexpanded length in the denominator for the force equation. The fully compressed length is always 0.691:

Item Length Wire size Constant
(lb coils/in)
n
(coils/in)
Force =
0.3125C/0.691n
9664K51 36" 0.028" 21.1 35.71 0.27 lb
9664K19 36" 0.035" 54.3 28.57 0.86 lb
9665K24 20" 0.032" 31.3 31.3 0.45 lb
9665K25 20" 0.041" 92.7 24.4 1.7 lb


We have gotten much closer!

Now, if we remove the constraint that the spring must fit over the rod, we can go with a smaller diameter spring. The reason we wanted the compression spring to fit over the rod is that such a spring needed an axial piece to remain straight. Expansion springs do not need that, so let us try 1/4" instead of 3/8":

Item Length Wire size Constant
(lb coils/in)
n
(coils/in)
Force =
0.3125C/0.691n
9664K46 36" 0.018" 12.2 55.56 0.10 lb
9664K47 36" 0.023" 34.8 43.48 0.36 lb
9665K84 20" 0.015" 5.3 66.7 0.036 lb
9665K16 20" 0.017" 8.2 58.8 0.063 lb
9665K57 20" 0.020" 20 50 0.18 lb


Here we have four candidates to choose from, ranging from 0.036 lb to 0.18 lb. Any of these springs should do.

Now, using an expansion spring will change the drive. Instead of the drive preventing the rod from being pulled in, now the drive must prevent the rod from being pushed out, unless we put the spring back on the drive side, and connect it between the follower bearing (the thing on the rod which follows the contour of the drive) and the wall. Thus again, the drive prevents the rod from being pulled in.

However, the dimensions of the rod on the drive side are different than those on the other side. The distance the rod moves is the same, 0.3125", but the spring at full compression is now 0.816" instead of 0.691". Looking at the force equation, we can see that this increase in the denominator only serves to decrease the force further, so we can remain with the same choices for springs.