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November 19, 2005, 20:28 |
Custom Field Functions
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#1 |
Guest
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Does anyone know about any peculiarities of the LOG function under the CFF menu?
I try to define the function cf = 0.08/log(0.06*Rex*F) where F is a constant and Rex is a linear non-zero function of x. When I try to plot the function I get a division by zero error. But I am able to export it. Any suggestions? |
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November 20, 2005, 11:52 |
Re: Custom Field Functions
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#2 |
Guest
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have a try! cf = 0.08/(log(0.06*F)+logRex)
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November 21, 2005, 02:36 |
Re: Custom Field Functions
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#3 |
Guest
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For (0.06*Rex*F)=1 you will get a division by zero.
RoM |
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November 21, 2005, 09:13 |
Re: Custom Field Functions
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#4 |
Guest
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Icw, thank you for your reply. I appreciate your help. I tried entring the formula in the form you specified, but I am getting the same error.
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November 21, 2005, 09:15 |
Re: Custom Field Functions
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#5 |
Guest
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RoM, thank you for your reply. Could you please explain why that makes sense that this function produces a division by zero error? How do I fix it? The formulation 0.08/(log(ReX)+log(0.06*Fre)) does not resolve the division by zero error. Thanks again.
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November 21, 2005, 09:31 |
Re: Custom Field Functions
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#6 |
Guest
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I just wanted to point out that log 1 = 0. I dont know what physical meaning stands behind this singularity in your function. You could use Icw's formulation and define 2 cffs, one for log(ReX) and the other log(0.06*Fre). Plot them and look in which regions of your domain log(ReX)+log(0.06*Fre)=1. Maybe this will help to judge whats going on and if there is any physical meaning behind.
Good Luck, RoM |
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