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
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
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.