Mark Allanson
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| posted on 29/1/03 at 08:53 PM |
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Excell fromulae help
Just found a copy of "Designing and bulding special cars" by Andre Jute as recommended by Rorty, there are some fantastic formlae and equations to
work out most things in a self buld car.
I am transfering them into excel, but cannot find out how to do a cube root, I have tried doing them a a logarithm and dividing by three, but then
there is no antlog to get the result.
There must be an excel pro out there who knows the answer?
Mark
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Alan B
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| posted on 29/1/03 at 09:18 PM |
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its ^0.333333333
i.e to the power 1/3
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CairB
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| posted on 29/1/03 at 09:26 PM |
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I'm not an Ecel pro, but try a VB macro function
Function CubeRoot(Value)
CubeRoot = Value ^ (1 / 3)
End Function
Then in your spreadsheet cell put:
CuberRoot(X)
Where X is the numeber you want the root to.
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Mark Allanson
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| posted on 29/1/03 at 09:52 PM |
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Ideal,
I'll give it a try
Thanks
Mark
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Rorty
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| posted on 30/1/03 at 02:21 AM |
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Well done Mark. Is it still in print, and can you post the ISBN please?
While you have it in your grubby likes, can you confirm the figures he mentions in regard to scrub radius?
Cheers, Rorty.
"Faster than a speeding Pullet".
PLEASE DON'T U2U ME IF YOU WANT A QUICK RESPONSE. TRY EMAILING ME INSTEAD!
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wicket
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| posted on 30/1/03 at 08:31 AM |
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Its available through Amazon details below
Designing and Building Special Cars
Andre Jute
Availability: Seller usually dispatches in 1-2 working days
Hardcover - 192 pages (31 December, 1985)
B.T. Batsford; ISBN: 0713407786
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StuartA
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| posted on 30/1/03 at 03:15 PM |
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I'm no car building professional, but have spent a number of years plugging stuff into Excel....
What you want is a cell that has the following formula in it:
=POWER(A1,1/3)
... where cell A1 contains the number that you want to do the operation on. Of course, you can use this formula to raise any number to any power (the
second parameter in the brackets).
Alternatively you can do what Alan suggested with either:
=A1/0.3333
or
=A1/(1/3)
The latter is exactly the same as the power function.... depends how accurate you want your answer.
[Edited on 30/1/03 by StuartA]
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