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The following is an error correcting equation for a sidereal astrophotography tracking mount I'm building and $t$ represents the amount of time before the tracking is off by a quarter of a stepper motor step. I need to solve for $t$, but I've having trouble solving this.

I can plot it with a value of $n$ and find an ok approximation, but the problem is that I need it for several values of $n$. Any good ways? $n$ is the integer number of corrections applied thus far. If it can't be solved in a general sense, what's a good way to generate approximate solutions for a few hundred values of $n$, starting at 0?

$-0.25=\frac{\sqrt{2\cdot 150^2-2\cdot 150^2\cdot \cos(t\cdot 0.000072733)} - (t\cdot 0.0109170306+(n\cdot -0.25\cdot 0.005))}{0.005}$

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    @DayLateDon - It would be if exact construction were possible. In an exact construction, 150 would actually be 150.0973588306.2011-03-28

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Here is a table of values of $t$ for integers $0\le n\le 500$, generated using Mathematica's FindRoot.

Here are the first few values:

n=  0:  t= 174.7 n=  1:  t= 339.8 n=  2:  t= 489.7 n=  3:  t= 623.7 n=  4:  t= 743.3 n=  5:  t= 850.4 n=  6:  t= 947.2 n=  7:  t=1035.4 n=  8:  t=1116.4 n=  9:  t=1191.3 n= 10:  t=1261.0 

Here's the rough source code:

Table[{n,    t /. FindRoot[-0.25 == (Sqrt[          2*150^2 -            2*150^2*Cos[             t*0.000072733]] - (t*0.0109170306 + (n*-0.25*.005)))/       0.005, {t, 200 + 150 n}]}, {n, 0, 500}] 
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    Just an FYI for anybody looking at this, what's interesting is that from an engineering perspective this means that starting out it will need to be error corrected every few minutes but after an hour its diverging fast enough at it needs error correction every few seconds. This is an isosceles Haig mount for those interested.2011-03-28