How to solve a pow equation?

Question

I have an formula that generates a number from 20 to 180 (degree).

var a = 20 * Math.pow(str/dist, 1.5));
    a = Math.max(180, w);

Using this formulae, i can generates some arcs and draw them just fine and depending on dist (which is cursor distance from target), the arc is either very wide, or narrow (pow basicly makes it so the arc expands non-linear).

This approach works in one direction: Having a location, and a distant point, to create arc. I need a counter-equation that gives me the required DIST for a given angle.

i.e. assuming str is 800, what is the value of dist so a would be 114 (or 68 or anything).

114 = Math.pow(900/dist, 1.5)) // ????

How can i solve this equation for dist ?

thanks,

Answer 1

I'm going to disregard your max line for a moment (did you mean min? Written with a max, any values in the 20-180 range you described will come out as 180, since max returns the greatest argument)

We can solve this step by step by reversing each operation. Starting with the multiplication by 20:

$$ \begin{align} a &= 20 \times{(\frac{str}{dist})}^{1.5} \\ \frac a {20} &= {(\frac{str}{dist})}^{1.5} \end{align} $$

The opposite operation of a power is a corresponding root. So to undo a square \$ x^2 \$ we'd take a square root \$ \sqrt {x^2} = x \$, often available as its own function, sqrt(). For less common roots, we can use pow as its own inverse, using the fact that the nth root of x, \$ \sqrt[n]{x} \$, is just x to the power of 1/n:

$$ \begin{align} \frac a {20} &= {(\frac{str}{dist})}^{1.5} \\ {(\frac a {20})}^{\frac 1 {1.5}} &= \frac {str} {dist} \end{align} $$

And taking the reciprocal to get \$dist\$ on top (which is the same as negating the power), then multiplying both sides by \$str\$...

$$ \begin{align} {(\frac a {20})}^{-\frac 1 {1.5}} &= \frac {dist} {str} \\ str \times {(\frac a {20})}^{-\frac 1 {1.5} } &= dist \end{align} $$

Or in code:

dist = str * pow(a / 20.0, -1/1.5)