Hi,

this is physics, but it's the math that is a problem to me. Problem is added in pdf-file. Can anyone quide me from EQ1 to EQ2 and further to EQ3?

If I use a computer program it gives me arctan...

Thanks!

- Trond

Results 1 to 5 of 5

- Feb 25th 2010, 12:40 AM #1

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- Feb 25th 2010, 08:09 AM #2
To get from equation 1 to equation 2, you must differentiate the mean pressure function with respect to z:

$\displaystyle \frac{\partial}{\partial{z}} p(z) = \frac{0\left(1 + \left(\frac{z}{h}\right)^2\right) - p_0\left(0 + \left(\frac{2zh^2 - 0z^2}{h^4}\right)\right)}{\left(1 + \left(\frac{z}{h}\right)^2\right)^2}$

$\displaystyle \frac{\partial}{\partial{z}} p(z) = \frac{-p_0\frac{2z}{h^2}}{\left(1 + \left(\frac{z}{h}\right)^2\right)^2}$

$\displaystyle \frac{\partial}{\partial{z}} p(z) = -2\frac{p_0z}{\left(1 + \left(\frac{z}{h}\right)^2\right)^2h^2}$

That's to EQ2.

All they do for EQ3 is just substitute p_0 for P, and since I'm not too familiar with this topic, I'd just go with it.

- Feb 25th 2010, 10:17 AM #3

- Feb 25th 2010, 10:32 AM #4

- Feb 25th 2010, 12:09 PM #5

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EQ 2 and 3 are only the left side of EQ 1. Further I set the solution of the diff like right side of EQ 1. So g will be back in EQ 4.

Thanks, this helped me further on. (By the way, P is pressure. And the next steps will end up with a function of temperature with respect to z (after substitude with a gas law). z is height from sea level).