# help re-arranging a complex formula

• Oct 20th 2009, 02:55 PM
markanswrth
help re-arranging a complex formula
Hi,
I'm having some trouble rearranging the formula below in terms of t, i'm sure its much easier than I think but if would be great if i could be pointed in the right direction. Its been a while since I've had to do something like this so im quite rusty!

http://i176.photobucket.com/albums/w...h/formula1.gif

I've got several equations like this which i wil need to arrange and im wondering if it can be done in Mathcad, which I have but im not familer with?

any help much appreciated

mark
• Oct 20th 2009, 08:07 PM
tonio
Quote:

Originally Posted by markanswrth
Hi,
I'm having some trouble rearranging the formula below in terms of t, i'm sure its much easier than I think but if would be great if i could be pointed in the right direction. Its been a while since I've had to do something like this so im quite rusty!

http://i176.photobucket.com/albums/w...h/formula1.gif

I've got several equations like this which i wil need to arrange and im wondering if it can be done in Mathcad, which I have but im not familer with?

any help much appreciated

mark

First of all you can cancellate $\displaystyle t^4$ with the $\displaystyle t$ under $\displaystyle s$ as there's a multiplication, and then we get:

$\displaystyle F=\frac{4E}{1-\mu^2}\cdot \frac{t^3}{K_1\cdot D_e^2}\cdot s\left[\left(\frac{h_0-s}{t}\right)\left(\frac{2h_0-s}{2t}\right)+1\right]$ $\displaystyle =\frac{4E}{1-\mu^2}\cdot \frac{t^3}{K_1\cdot D_e^2}\cdot s\left[\frac{2h_0^2-3h_0s+s^2}{2t^2}+1\right]=$

$\displaystyle \frac{4E}{1-\mu^2}\cdot \frac{t^3}{K_1\cdot D_e^2}\cdot s\left(\frac{2h_0^2-3h_0s+s^2+2t^2}{2t^2}\right)$

Now cancellate $\displaystyle t^3$ with the $\displaystyle t^2$ inside the parentheses, getting

$\displaystyle \frac{4E}{1-\mu^2}\cdot \frac{t}{K_1\cdot D_e^2}\cdot s\left(\frac{2h_0^2-3h_0s+s^2+2t^2}{2}\right)=$ $\displaystyle \frac{4E}{1-\mu^2}\cdot \frac{t}{K_1\cdot D_e^2}\cdot \frac{s}{2}\left(2h_0^2-3h_0s+s^2+2t^2\right)$[/tex]

leaving now all what's attached to $\displaystyle t$ in the right side and passing to the left one all the rest we get a cubic in $\displaystyle t$ which, still, is pretty nasty, but at least closer to what you, perhaps, want.

Tonio