Talk:Laser flash analysis

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I'm not sure if the formula for alpha is right. The original paper says 1.388*L^2/ (\pi^2*t_{1/2}), which is quite the same, but just quite! --134.176.16.145 (talk) 08:56, 12 June 2014 (UTC)[reply]

The formula is right. The original paper says

1.38L^{2}/(\pi ^{2}t_{1/2})

, but it seems to be a mistake; the numerical constant is the ω solving equation 6 for V = 1/2 which is about 1.370. The correct formula would thus be

1.370L^{2}/(\pi ^{2}t_{1/2})

which is the same as

0.388L^{2}/t_{1/2}

Skåningiexil (talk) 22:33, 8 August 2015 (UTC)[reply]

Indeed. The solution is as follows.

0.5\Theta _{\infty }=\Theta _{\infty }\left[1+2\sum _{n=1}^{\infty }(-1)^{n}\exp \left(-n^{2}{\frac {t_{0.5}}{\tau }}\right)\right]

, where

\tau ={\frac {l^{2}}{\pi ^{2}\alpha }}

. The equation is a special case of the Theta function where

0.5=\vartheta _{4}(0,\exp(^{-t_{0.5}}/_{\tau }))

. Solving,

^{t_{0.5}}/_{\tau }=1.36975597849933\ldots

and,

{\frac {1.36975597849933\ldots }{\pi ^{2}}}=0.13878529704272\ldots \approx 0.1388

. rolandog (talk) 23:57, 1 March 2017 (UTC)[reply]

What are the units fo these equations? This is missing in this article and is wrong till they are given. Kakila (talk) 15:12, 19 March 2017 (UTC)[reply]

It seems that units are cm^2/s based on Eq.1 of the first reference. Kakila (talk) 15:26, 19 March 2017 (UTC)[reply]