Solving for a constant in a power

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SUMMARY

The discussion focuses on solving for the constant z0 in the context of wind turbine height calculations using the formula vh = vref . (h/href)^(1/(ln(h/z0))). The user, AA, seeks assistance in manipulating logarithmic expressions to isolate z0. The solution provided by another user, tiny tim, outlines the steps to derive z0 as z0 = 105/eln(13.125)/ln(1.4035), demonstrating the application of logarithmic properties and exponential functions in solving the equation.

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Anti-alias
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Homework Statement



Hi all, this isn't exactly a homework question (I'm a law student long out of high school maths!) but has arisen in the context of some of my legal work analysing the difference in height of windspeeds re: wind turbines.

The "problem" is solve for the constant z0.

Homework Equations



The main formula is as follows:

vh = vref . (h/href)^(1/(ln(h/z0)), where vh = velocity at desired height, vref = velocity at measured height, h = desired height, href = measured height, z0 = aerodynamic roughness.

After inserting a known answer from a previous calculation, the formula reduces to:

(6.4/4.56) = 13.125^(1/(ln(105/z0))

The Attempt at a Solution



As I stated earlier, I'm a couple of years out of high school maths and cannot remember for the life of me how to solve for powers involving logs themselves. I can barely remember the simplest of log rules.

Any help would be much appreciated.

Many thanks all,
AA
 
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Welcome to PF!

Anti-alias said:
(6.4/4.56) = 13.125(1/(ln(105/z0))

Hi Anti-alias ! Welcome to PF! :smile:

Let's see … 6.4/4.56 = 1.4035.

So 1.4035 = 13.125(1/(ln(105) - ln(z0)))

= e(ln(13.125))(1/(ln(105) - ln(z0)))

so ln(1.4035) = (ln(13.125))/(ln(105) - ln(z0))

so (ln(105) - ln(z0) = ln(13.125)/ln(1.4035)

so ln(z0) = ln(105) - ln(13.125)/ln(1.4035)

so z0 = 105/eln(13.125)/ln(1.4035) :smile:
 
You're a legend tiny tim, huge thanks :approve:
 

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