- #1
erobz
Gold Member
- 3,937
- 1,678
I was trying to figure out how much landscape fabric I have left on a roll, given a known thickness ##T## of the fabric and the radius ##r## of that roll. I obviously don't want to do an integral to calculate it while I'm working, but was looking for some approximation that can be readily determined by comparing it to the integral. I'm proposing a simple Archimedean Spiral ##r = k \theta##
So I start with the arclength formula in Cartesian coordinates:
$$ ds = \sqrt{1 + \left( \frac{dy}{dx} \right)^2 } dx $$
With ##x= r \cos \theta, y = r \sin \theta ##
I start by making a change of variables to ##r##:
$$ \left( \frac{dy}{dx} \right)^2 = \left( \tan \theta \right)^2 $$
for the differential I get:
$$ dx = \left[ k \cos \theta - r \sin \theta \right] d\theta $$
this goes to:
$$ ds = \sqrt{1 + \left( \tan \theta \right)^2 } \left[ k \cos \theta - r \sin \theta \right] d\theta $$
If Iv'e done this correctly up to this point I see that it is integrable by researching it and I can get ##s## as a funtion of ##\theta##. But I realize its not really what I want... I want to measure the radius in the field.
I guess its true by definition of the Archimedean spiral that ##\theta## ( the upper limit of the integration ) is just the ##\frac{r}{T }2 \pi ##?
So if that is true, Is there a reasonable measure of some radius I can take in the field that would approximate the integral?
So I start with the arclength formula in Cartesian coordinates:
$$ ds = \sqrt{1 + \left( \frac{dy}{dx} \right)^2 } dx $$
With ##x= r \cos \theta, y = r \sin \theta ##
I start by making a change of variables to ##r##:
$$ \left( \frac{dy}{dx} \right)^2 = \left( \tan \theta \right)^2 $$
for the differential I get:
$$ dx = \left[ k \cos \theta - r \sin \theta \right] d\theta $$
this goes to:
$$ ds = \sqrt{1 + \left( \tan \theta \right)^2 } \left[ k \cos \theta - r \sin \theta \right] d\theta $$
If Iv'e done this correctly up to this point I see that it is integrable by researching it and I can get ##s## as a funtion of ##\theta##. But I realize its not really what I want... I want to measure the radius in the field.
I guess its true by definition of the Archimedean spiral that ##\theta## ( the upper limit of the integration ) is just the ##\frac{r}{T }2 \pi ##?
So if that is true, Is there a reasonable measure of some radius I can take in the field that would approximate the integral?