Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do I solve this expression for $\frac{R}{r}$?

$\frac{1}{(r-R)^3}-\frac{y}{R^2(r-R)}=\frac{1}{r^3}$

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Rearrange the equation to get

$$1 - y \left ( 1 - \frac{r}{R} \right )^2 = \left ( 1 - \frac{r}{R} \right )^3 $$

Set $z = 1-r/R$ and get the equation:

$z^3+y z^2-1=0$

Get a root at $z_0(y)$ numerically or otherwise. Then $R/r$ is then

$$\frac{R}{r} = \frac{1}{1-z_0(y)} $$

NB There is an analytical expression for $z_0$:

$$ z_0(y) = \frac{1}{3} \left(\frac{\sqrt[3]{-2 y^3+3 \sqrt{3} \sqrt{27-4 y^3}+27}}{\sqrt[3]{2}}+\frac{\sqrt[3]{2} y^2}{\sqrt[3]{-2 y^3+3 \sqrt{3} \sqrt{27-4 y^3}+27}}-y\right) $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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