Length of spiral

[opinsky]

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

[Bignose]

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.

[opinsky]

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...

[Amr Morsi]

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.

[Bignose]

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.

[Amr Morsi]

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!

[ewmon]

Newton's method?

[Amr Morsi]

I really don't know what does "Newton's method" imply?

[the tree]

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.

[Amr Morsi]

We are not in need for any method. It is obvious that at s=0, theta=0. But, the derivatives aren't zero.

[Bignose]

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.

[psychlone]

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.

[Amr Morsi]

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]