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this is my first post on the mathematics node of stack exchange, so please forgive me if I'm not posting an appropriate question, but I'm not sure where else to address this. I'm trying to figure out what equation might generate the following pattern. (keep in mind the numbers might be slightly off, not exact)

 0  :: 0, 0  10 :: 19, -16 (0.17453)  20 :: 41, -28 (0.34906)  30 :: 64, -37 (0.52359)  40 :: 89, -41 (0.69813)  50 :: 113,-41 (0.87266)  60 :: 138,-37 (1.04719)  70 :: 162,-28 (1.22)  80 :: 182,-16 (1.392)  90 :: 201, 0  (1.570)  100:: 217,19 (1.745)  110:: 229,41 (1.919)  120:: 238,64 (2.094)  130:: 241,89 (2.268)  140:: 241,113 (2.443)  150:: 238,138 (2.617)  160:: 229,162 (2.792)  170:: 217,182 (2.792)  180:: 201,201 (2.967)  190:: 182,217 (3.316)  200:: 162,229 (3.490)  210:: 138,238 (3.665)  220:: 113,241 (3.839)  230:: 89,241 (4.0142)  240:: 64,238 (4.188)  250:: 41,229  260:: 19,217  270:: 0,201  280:: -16,182  290:: -28,160  300:: -37,137  310:: -41,113  320:: -41,89  330:: -37,64  340:: -28,41  350:: -16,19  360:: 0,0 

as you math wizez may have guessed it has to do with adjusting coordinates for rotation around an axis. Any help would be great!

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    @Zev: I think there's a qualitative difference from the examples you link to. In the typical questions that your criticism applies to there is only a few (up to 7 in your examples) data points, which isn't enough to decide the general shape of a fit with any degree of accuracy. But with 37 data points, as here, there is a good chance that someone with mathematical experience can look at a plot and see a solution type that can fit the data better than other solutions _of comparable simplicity_, whereas an OP with less experience perhaps can't. That makes it a _good_ question in my opinion.2012-06-23

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Just plot your data and make a guess.

Let $f(x) = 142 \sin \frac{\pi (t -45)}{180}+100$. Then your data is reasonably well approximated by $\{t, f(t), f(360-t)\}_{t\in \{0,\cdots,360\}}$.

Here is a plot of $f$ and the first two columns of the data above:

enter image description here

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    brilliant! Thank you so much!2012-06-23