Arc length of a Spiral

[Dr. Zimski]

I might have already posted this in this forum quite some time ago, but the problem just came back to me, and now I want to solve it again, or at least know the solution.

 

Here it is, more in depth: Arc length of a spiral

 

Say you are givin a spiral with an initial radius r' (read r-prime), where r' is any real number greater than 0. As the radius "spins" it grows at the same rate as it "spins". So for every degree it rotates, the radius gains 1/360th of it's initial length (r'); exapmle- if r' = 1 cm, after 90 degrees the new radius is 1.25 cm, then after a full revolution it is 2 cm. And this goes on for however many revolutions.

 

What I need it a solution for any number of revolutions, and partial revolutions. Or a nudge in the right direction at the very least.

[EvoN1020v]

I remember this thread pretty well. It was a rather lengthy argument. I suggest this link: http://www.ctre.iastate.edu/educweb/ce353/lec05/lecture.htm. It looks really complicated, so I don't know how I can help.

[CPL.Luke]

well in polar coordinates the equation is r= r'/360 (theta) +r'

 

 

because arc length is the integral of r with respect to theta you just integrate that equation to get

 

(arc length)= r'/520 (theta)^2+r' (theta)

 

 

so then you just plug in the angle theta eg. 360 degrees, 520 degrees etc. and you'll get the arc length up until that angle.

 

^it should e noted that the integral above is technically a definate integra from 0-theta hence no constant. and theta should be entered in in degrees as thats what the original post called for.

 

 

EDIT: it appears that the link evon gave is for a method of calculating curvature without calculus, that or a method for projecting circles when surveying a landscape etc.

[Dr. Zimski]

^^^I already thought of that, but it seemed too easy an answer to be true. I'll look more into that though... if it turns out to be right, thumbs up.

[CPL.Luke]

it is, most answers are really simple and easy if their right, when they get complicated, thats when you know you messed up.

[Dave]

Take great care when using degrees in integrals and other such calculations. It is a much better approach to use radians when dealing with this, and then convert the angle over to degrees afterwards. Trig functions tend to break easily when not handled under radians in terms of calculus.

[Dr. Zimski]

well in polar coordinates the equation is r= r'/360 (theta) +r'

 

 

because arc length is the integral of r with respect to theta you just integrate that equation to get

 

(arc length)= r'/520 (theta)^2+r' (theta)

 

 

so then you just plug in the angle theta eg. 360 degrees, 520 degrees etc. and you'll get the arc length up until that angle.

 

^it should e noted that the integral above is technically a definate integra from 0-theta hence no constant. and theta should be entered in in degrees as thats what the original post called for.

 

 

EDIT: it appears that the link evon gave is for a method of calculating curvature without calculus, that or a method for projecting circles when surveying a landscape etc.

If you integrated that, wouln't it be r'/720(theta)+r'(theta), since 1/2 of 1/360 is 1/720?

[CPL.Luke]

you'd be corect sir. I made a istake on te arithmtic.

[CPL.Luke]

also Dave what do you mean by break?

 

and sorry for the very poor spelling in my last post, I was running a fever of about 101 when I wrote it.

[Dave]

Well, when we differentiate say, sin(x), we assume that x is given in terms of radians. If, on the other hand, x was given in terms of degrees, then the derivative we get would be:

 

[math]\frac{d}{dx} \sin \left( \frac{\pi x}{180} \right) = \frac{\pi}{180} \cos \left( \frac{\pi x}{180} \right)[/math]

 

This isn't a very nice function to be dealing with, so we only consider x as being done in terms of radians. Really, you should use radians exclusively over degrees, as they're a far more useful and natural way of describing the notion of an angle.

[Garvielloken]

I might have already posted this in this forum quite some time ago, but the problem just came back to me, and now I want to solve it again, or at least know the solution.

 

Here it is, more in depth: Arc length of a spiral

 

Say you are givin a spiral with an initial radius r' (read r-prime), where r' is any real number greater than 0. As the radius "spins" it grows at the same rate as it "spins". So for every degree it rotates, the radius gains 1/360th of it's initial length (r'); exapmle- if r' = 1 cm, after 90 degrees the new radius is 1.25 cm, then after a full revolution it is 2 cm. And this goes on for however many revolutions.

 

What I need it a solution for any number of revolutions, and partial revolutions. Or a nudge in the right direction at the very least.

 

the polar equation would simply be r=C(the initial radius) + theta/2pie

[kavlas]

the polar equation would simply be r=C(the initial radius) + theta/2pie

the arc length in polar coordinates is :

 

S = [math]\int_{0\to\theta}(\sqrt{(\frac{dr}{d\theta})^2+ r^2})d\theta[/math]

 

So if you plug in the function r = Cr'+ [math]\frac{\theta}{2\pi}[/math] and do the appropriate calculation you will get the desired result