User:Hgeron/sandbox

A while ago I sent a detailed info about my TRUE solution. I talked about the Funnel computation about how I know LOG works. I want to add this to show I know how to solve the Colebrook equation. For how this works, let me show you with a real easy way to find the solution to an equation. This is just a explanation of how it works.

Lets say... What is the X value in the equation X=10+Log(X). This is kind of like the Colebrook equation, and I will show how the Log works. Find X, well lets guess first the X=5. Solve it like this ... we enter in Excel  =10+Log(5) and that gives 10.6989... So let's solve for =10+Log(10.6989) and you will get 11.29...  Keep doing that... you will find (in about 10 loops) the when X= 11.043090636672800 the solution will be solved. So we know that

11.043090636672800=10+Log(11.043090636672800)

This is the same method you can use for the Colebook equation that is 1/sqrt(f)=-2*Log(Rr/3.7+2.51/Re*1/sqrt(f)), but we first make it simple by letting X be "1/sqrt(f)" like this. X =-2*Log(Rr/3.7+2.51/Re*1/ X ) Once X is found then  f = 1/X^2

Please forward this to the Wikipedia guy (Ac44ck) I replied to a few hours ago to help him understand the "math" method.

Harrell Geron