-1
$\begingroup$

I'm trying to find the formula to plot the following spiral:

It grows at a rate of √2 for every 45 degrees of rotation, so for example starting at 0 degrees (at 45 degree increments), it expands thus:0, 0.7, 1, 1.41, 2, 2.82, 4 etc... But I want to be able to find the distance from the centre at any angle, not just at the 45 degree increments.

Can anyone help with this please?

Many thanks

Paul uk

$\endgroup$
0
$\begingroup$

Such a spiral has a polar representation of the form $$r(\phi)=C e^{\lambda\phi}\qquad(-\infty<\phi<\infty)$$ with a certain constant $\lambda\ne0$. Your condition amounts to $$e^{\lambda\pi/4}=\sqrt{2}\ ,$$which translates into $$\lambda={2\over\pi}\log 2\ .$$

$\endgroup$
0
$\begingroup$

The simple way to write what you are asking is $r=A (\sqrt 2)^{\frac \theta {45^\circ}}$ where $A$ is the radius at $\theta=0$. As $\theta$ increases by $45^\circ$ the exponent of $\sqrt 2$ increases by $1$.

You could be more sophisticated and use $e$ as the base, saying $r=Ae^{\ln \sqrt 2\cdot \frac {\theta}{45^\circ}}$. If you have an $e^x$ button and no $x^y$ button that will be preferable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.