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View Full Version : How does one calculate tsub0


Joe
September 1st, 2000, 03:48 PM
[Originally Posted: 1/20/00-- Transferred by ReliaSoft Moderator]

Hi,

My boss is fascinated with the 6 page web article, Weibull's Distribution Application (Step by Step) found at the hyperlink: http://www.geocities.com/CapeCanaveral/Lab/1418/info2.htm

This article starts by providing a data table consisting of equal length times to failure intervals of 100 hours each with the corresponding observed frequency. It then proceeds onto page 2 where it provides the basic graphical formula for t0 as

t0 = t2 - ( ((t3-t2) * (t2-t1)) / ((t3-t2) - (t2-t1)) ) which always becomes t0 = 1,200 - (10,000 / 0 ) when t1=1,100 and t2=1,200 and t3=1,300.

Of course, this is not the case!!! Obviously, we never divide by zero! Then in the Computer Program case, the article states that t0 should be 900 hours.

I would really appreciate some expert advice and insight on the correct application on the formulas given in the article and on the computer program so that I too can crank out the numbers provided in this excellent article to arrive at the same answers! Does Weibull++ provide t0 calculations ability and would it also provide 900 hours as its answer too?

Thank you very much for your kind help, Joe

ReliaSoft Support
September 1st, 2000, 03:49 PM
[Originally Posted: 1/21/00-- Transferred by ReliaSoft Moderator]

The Gamma Variable in a three parameter Weibull is the t-sub-0. The html document at -- http://www.reliasoft.com/support/supq012100.htm -- provides a detailed answer including reworking the referenced example using Weibull++.