# Rearrangement of Zhang's Equation

• Jul 31st 2008, 01:57 AM
Rearrangement of Zhang's Equation
Hello. I am having trouble rearranging this equation:

$\displaystyle Q=\frac{\pi t f_p f_m^2}{2(f_m^2 - f_p^2)}$

to get an equation for $\displaystyle f_p$ in terms of $\displaystyle t$. Only $\displaystyle f_p$ and $\displaystyle t$ are variables, the other symbols are constants.

I think I have got it as far as:

$\displaystyle \frac{f_p}{f_m^2}-\frac{1}{f_p}=\frac{-\pi t}{2Q}$

reasonably safely, but am not sure where to go from there! This looks distinctly like it should be quite do-able, but has been frustrating me all morning..

Any help would be appreciated.
• Jul 31st 2008, 02:23 AM
mr fantastic
Quote:

Hello. I am having trouble rearranging this equation:

$\displaystyle Q=\frac{\pi t f_p f_m^2}{2(f_m^2 - f_p^2)}$

to get an equation for $\displaystyle f_p$ in terms of $\displaystyle t$. Only $\displaystyle f_p$ and $\displaystyle t$ are variables, the other symbols are constants.

I think I have got it as far as:

$\displaystyle \frac{f_p}{f_m^2}-\frac{1}{f_p}=\frac{-\pi t}{2Q}$

reasonably safely, but am not sure where to go from there! This looks distinctly like it should be quite do-able, but has been frustrating me all morning..

Any help would be appreciated.

Streamline things a bit:

$\displaystyle Q = \frac{A f_p}{B - f_p^2}$

and it should be obvious what the constants A and B represent.

Then:

$\displaystyle QB - Qf_p^2 = A f_p \Rightarrow Q f_p^2 + A f_p - QB = 0$.

Solve the quadratic for $\displaystyle f_p$ and then substitute the appropriate expressions for A and B.
• Jul 31st 2008, 02:42 AM
I did this initially, but $\displaystyle t$ is not a constant, so I didn't think that it was valid?
I did this initially, but $\displaystyle t$ is not a constant, so I didn't think that it was valid?