attempt at finding center of arithmetic spiral to be tangent to two lines

Given the two orange lines in the picture, how can I calculate the center for an arithmetic spiral that will join the two lines whilst simultaneously being tangent to the ends of those lines. There is a similar question that has already been solved that deals with finding the center of logarithmic spirals in How to find the center of a log spiral? but the properties of logarithmic spirals being a constant angle means that it is inapplicable in this case though it does seem close. I believed that it could be possible to constrain the angles to find the center but an arithmetic spiral has different angles at every point of the curve so I would need to somehow use the fact that the radius changes linearly with the change in angle. I don't quite know how to do that though.

Any Ideas?

Update: I now have something that almost works, thanks to the image provided in the comments and with that I have this spiral that is tangent to one original line and needs a separate line shown in green that is parallel to the unused orange original line. I had just renamed aθ to be either r1 or r2 as that is more relevant in my use case than knowing what the θ is but it should not create the error that is shown in the image.


Here is what the tool for creating the spiral is doing. It requires a circle to be drawn first for the starting diameter of the spiral and takes in the variables of start angle, swept angle and pitch in order to fully define the spiral. All those values are pulled directly from the earlier sketch or calculated for the pitch by multiplying dimension a times 2pi.

  • $\begingroup$ This diagram might spark some ideas. $\endgroup$ – amd Sep 20 '18 at 20:43
  • $\begingroup$ @amd Thanks for the link. However, Maybe I'm doing something wrong so it didn't work perfectly. the spiral ended up creating an end parallel to what I believed it would actually land on. $\endgroup$ – betasniper Sep 21 '18 at 4:31
  • $\begingroup$ It’s a starting point for a solution to your problem. The spiral you’re trying to construct is likely going to need to be rotated ($a\ne0$ in $r=a+b\theta$) $\endgroup$ – amd Sep 21 '18 at 4:43
  • $\begingroup$ @amd What would the r = aθ + b do with the b value or how would it work with the tool shown? I believe that I had rotated the spiral correctly and if that is the case, then the software is actually not doing it correctly, but that shouldn't be so since it is a very well known program. $\endgroup$ – betasniper Sep 21 '18 at 7:47
  • $\begingroup$ actually, maybe it is terrible for constructing spirals. link $\endgroup$ – betasniper Sep 21 '18 at 8:00

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