opinsky Posted April 3, 2010 Share Posted April 3, 2010 Hello, if the arc-length of an archimdes spiral (polar formula is r = a * theta) is s(theta) = 1/2 a (theta * sqrt (1 + thetaˆ2) + asinh theta) what is the formula calculate theta as a function of S? Best regards, Oren Link to comment Share on other sites More sharing options...
Bignose Posted April 3, 2010 Share Posted April 3, 2010 You'll have to try to find the inverse. Given the form of the function, I highly doubt that it will be a "nice" form. However, since the formula is 1 to 1 valued (a given [math]\theta[/math] results in 1 and only 1 value of s), an inverse should exist. Link to comment Share on other sites More sharing options...
opinsky Posted April 3, 2010 Author Share Posted April 3, 2010 Thanks - any tips on how to find the inverse. Mathematica doesn´t seem to help (i used Solve[L==function(theta),theta) ). I´ve also tried to taylor expand to no avail. Regards... Link to comment Share on other sites More sharing options...
Amr Morsi Posted April 3, 2010 Share Posted April 3, 2010 asinh(theta) contains a ln function which cannot be solved with "theta * sqrt (1 + thetaˆ2)". So, I think the inverse function will not be a closed form. Link to comment Share on other sites More sharing options...
Bignose Posted April 3, 2010 Share Posted April 3, 2010 Thanks - any tips on how to find the inverse. Mathematica doesn´t seem to help (i used Solve[L==function(theta),theta) ). I´ve also tried to taylor expand to no avail. Regards... Right, as I said and Amr said, it is incredibly unlikely that it will be in a "nice" form. I won't say impossible, but nothing pops out at me. However, if you want to use it for computations, a computer program can do the trick for you. A straightforward "brute force" method to find what the angle is for a given arc length would be to guess a theta, and then add or subtract from the guess based on whether the guessed theta resulted in an arc length above or below the value you are aiming for. Then iterate following that same plan until you achieve the accuracy you desire. Link to comment Share on other sites More sharing options...
Amr Morsi Posted April 4, 2010 Share Posted April 4, 2010 I think you can get the Taylor Series. Try to get d(theta)/ds, theta'', theta''', .......etc from the relation given with s=0. Tn=nth-derivative * s^n /n! Link to comment Share on other sites More sharing options...
ewmon Posted April 4, 2010 Share Posted April 4, 2010 Newton's method? Link to comment Share on other sites More sharing options...
Amr Morsi Posted April 5, 2010 Share Posted April 5, 2010 I really don't know what does "Newton's method" imply? Link to comment Share on other sites More sharing options...
the tree Posted April 5, 2010 Share Posted April 5, 2010 Well for a given value of S you could approximate theta via the Newton-Raphson method but you'd still need an initial guess and it's not exactly a speedy approach when looking for answers for different theta's. Link to comment Share on other sites More sharing options...
Amr Morsi Posted April 6, 2010 Share Posted April 6, 2010 We are not in need for any method. It is obvious that at s=0, theta=0. But, the derivatives aren't zero. Link to comment Share on other sites More sharing options...
Bignose Posted April 6, 2010 Share Posted April 6, 2010 Newton's method? The wikipedia article on it is pretty good, really: http://en.wikipedia.org/wiki/Newton%27s_method It is a computational method wherein you make an initial guess. Then, the derivative of the function at that guessed value provides you with the next guess. And then you repeat until you get the desired accuracy. It isn't a perfect method. For functions that oscillate a lot, the next guessed value may overshoot the target value by too much and the subsequent guesses never converge. There are ways of testing and minimizing these issues, however. Furthermore, the theta versus arc length function will be nicely behaved. It won't dip or oscillate or anything like that. It is monotonically increasing, so I think that Newton's method is guaranteed to be convergent for a monotonic function. Link to comment Share on other sites More sharing options...
psychlone Posted April 7, 2010 Share Posted April 7, 2010 I think you can get the Taylor Series. Try to get d(theta)/ds, theta'', theta''', .......etc from the relation given with s=0. Tn=nth-derivative * s^n /n! Amr Morsi, I'm just curious, If you were to rewrite your above equation, it would look like be the following, Or not; I'm just pondering your post. [math]\theta_{n}= S_{n} - \frac{f^{(n)}(0)}{n!}S^{n}[/math] where [math]\theta = theta [/math] [math]a = \theta[/math] [math]x = S[/math] Taking the formula notation from the following Wiki page: http://en.wikipedia.org/wiki/Taylor_series. Link to comment Share on other sites More sharing options...
Amr Morsi Posted April 7, 2010 Share Posted April 7, 2010 If [math]\theta_{n}[/math] is the nth term in Taylor Series of [math]\theta[/math], then it will be as follows: [math] \theta_{n}= \frac{\theta^{(n)}(0)}{n!}S^{n} [/math] Link to comment Share on other sites More sharing options...
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