Counting 1/3 stops

I'm preparing an Excel table that counts reciprocity failure compensation for long exposures for Kodak E100 stuff, thanks to the calculation formula and exp. factor I was hinted in the corresponding thread here.

It is well known that each full stop is twice exposure time (at given aperture), half stop is exp. time*sqrt(2). But what is 1/3 stop ? How it is calculated ?

Thanks in advance, Alex

If you think of the equation as T1=T0*(2^1/x), it's obvious that for 1/3 stop use an exponent of 1/3. For half stop it's 1/2, quarter is 1/4, and so on.

Dead easy, when you set up the equation the correct way.

Like Ole says - switch to fractional exponents and you can plug it into a spreadsheet which does not allow N-th root equations. On a calculator for example you would find the 3rd root of 2 for 1/3 stop or 6th root of 2 for 1/6 stop etc directly - you can't in Excel without re-expressing the equation in terms Excel can handle.

Have fun, Bob.

Thank you guys.
Assuming Ole is correct in the generalization of stops calculation representation, this is indeed quite straight-forward. Frankly, my question was intended to deal with basic principle rather then its formal representation. If Ole is right claiming that each partial stop is correspondingly partial root multiplier, then its acceptable for Excel representation is only technical issue...
Thanks guys once again.
Now I'll be able to dilute my Excel table to show only necessary incerements (spaced by third stop).

Well, running 1/3 stop flag over the spreadsheet of long exposures ranging from 1 sec to 300 sec in 1 sec steps it turned that given exp. factor of 0.95 (as was advised in Kodak reciprocity thread), I only get the exposures of longer then 76 seconds being equal or greater then 1/3 steps thus requiring reciprocity factoring in (and only by third stop).
Frankly, I would expect much more compensation (and at earlier stages) will be required....(or perhaps the exp. factor of 0.95 wasn't really correct for the emulsion..)