Hi All..

I wanted to know how do you make the Weibull equation below into the form y=mx+c. The solution to this has the logs taken twice..

$\displaystyle P_f= 1 - exp\left[-\left(\frac{\sigma}{\sigma{_0}}\right)^m\right] $

Thank you

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- Oct 30th 2008, 11:21 AMdadonweibull parameter
Hi All..

I wanted to know how do you make the Weibull equation below into the form y=mx+c. The solution to this has the logs taken twice..

$\displaystyle P_f= 1 - exp\left[-\left(\frac{\sigma}{\sigma{_0}}\right)^m\right] $

Thank you - Oct 30th 2008, 12:38 PMmr fantastic
- Oct 30th 2008, 12:54 PMdadon
Hi.. thanks for your reply.. there is no question just from notes where the weibull is in the form y=mx+c.. just wanted to know the process on how it is derived?

I will put the solution..

$\displaystyle \ln\ln\left(\frac{1}{1-P_f}\right) = m\ln\sigma - m\ln\sigma_0 $

Thanks - Oct 30th 2008, 01:09 PMSean12345
Hi dadon,

We have

$\displaystyle P_f=1-exp\left[-\left(\frac{\sigma}{\sigma{_0}}\right)^m\right]$

Note that we can re-write this as

$\displaystyle exp\left[-\left(\frac{\sigma}{\sigma{_0}}\right)^m\right]=1-P_f$

Taking logs to the base $\displaystyle e$ of both sides gives

$\displaystyle -\left(\frac{\sigma}{\sigma{_0}}\right)^m=\ln(1-P_f)$

Take the minus over to make it easier to deal with....

$\displaystyle \left(\frac{\sigma}{\sigma{_0}}\right)^m=-\ln(1-P_f)$

Now we know that $\displaystyle a\ln(b)\equiv\ln(b^a)$ thus we are left with

$\displaystyle \left(\frac{\sigma}{\sigma{_0}}\right)^m=\ln\left( \frac{1}{1-P_f}\right)$

Taking logs of base $\displaystyle e$ again and using that same principle leads to

$\displaystyle

m\ln\left[\left(\frac{\sigma}{\sigma{_0}}\right)\right]=\ln\left[\ln\left(\frac{1}{1-P_f}\right)\right]

$

Using the fact that $\displaystyle \ln\left(\frac{a}{b}\right)\equiv \ln(a)-\ln(b)$ gives the required result.

Hope this helps. - Oct 30th 2008, 01:19 PMdadonthanks
Thank you.. that's exactly what I needed..You made it very clear line by line..:)