Talk:Laser flash analysis: Difference between revisions
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:The formula is right. The original paper says <math>1.38 L^2/ (\pi^2 t_{1/2})</math>, 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 <math>1.370 L^2/ (\pi^2 t_{1/2})</math> which is the same as <math>0.388 L^2/ t_{1/2}</math> [[User:Skåningiexil|Skåningiexil]] ([[User talk:Skåningiexil|talk]]) 22:33, 8 August 2015 (UTC) |
:The formula is right. The original paper says <math>1.38 L^2/ (\pi^2 t_{1/2})</math>, 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 <math>1.370 L^2/ (\pi^2 t_{1/2})</math> which is the same as <math>0.388 L^2/ t_{1/2}</math> [[User:Skåningiexil|Skåningiexil]] ([[User talk:Skåningiexil|talk]]) 22:33, 8 August 2015 (UTC) |
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:Indeed. The solution is as follows. <math>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 ]</math>, where <math>\tau = \frac{l^2}{\pi^2 \alpha}</math>. The equation is a special case of the [[Theta function]] where <math>0.5 = \vartheta_4(0, \exp(^{-t_{0.5}}/_{\tau}))</math>. [http://www.wolframalpha.com/input/?i=0.5%3DEllipticTheta%5B4,0,e%5E(-c)%5D+solve+for+c Solving], <math>^{t_{0.5}}/_{\tau} = 1.36975597849933 \ldots</math> and, <math>\frac{1.36975597849933 \ldots}{\pi^2} = 0.13878529704272 \ldots \approx 0.1388</math>. [[User:Rolandog|rolandog]] ([[User talk:Rolandog|talk]]) 23:57, 1 March 2017 (UTC) |
:Indeed. The solution is as follows. <math>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 ]</math>, where <math>\tau = \frac{l^2}{\pi^2 \alpha}</math>. The equation is a special case of the [[Theta function]] where <math>0.5 = \vartheta_4(0, \exp(^{-t_{0.5}}/_{\tau}))</math>. [http://www.wolframalpha.com/input/?i=0.5%3DEllipticTheta%5B4,0,e%5E(-c)%5D+solve+for+c Solving], <math>^{t_{0.5}}/_{\tau} = 1.36975597849933 \ldots</math> and, <math>\frac{1.36975597849933 \ldots}{\pi^2} = 0.13878529704272 \ldots \approx 0.1388</math>. [[User:Rolandog|rolandog]] ([[User talk:Rolandog|talk]]) 23:57, 1 March 2017 (UTC) |
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What are the units fo these equations? This is missing in this article and is wrong till they are given. [[User:Kakila|Kakila]] ([[User talk:Kakila|talk]]) 15:12, 19 March 2017 (UTC) |
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Revision as of 15:12, 19 March 2017
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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)
- The formula is right. The original paper says , 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 which is the same as Skåningiexil (talk) 22:33, 8 August 2015 (UTC)
- Indeed. The solution is as follows. , where . The equation is a special case of the Theta function where . Solving, and, . rolandog (talk) 23:57, 1 March 2017 (UTC)
![{\displaystyle 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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c0eb95f50a4cc838673392371936bd22cd613be)
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)