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Re: Volume Size [message #11464 is a reply to message #11383]

Sun, 12 April 1998 00:00

a2652099
Messages: 10
Registered: August 1996

Junior Member

davidf@dfanning.com (David Fanning) wrote:

>> Presumably you can distinguish the pores by some "value"
>> in the 3D volume. Suppose the pores had values less than
>> 10 and the rock (rest of the 3D volume) had values greater
>> than 10. Then calculating the percentage volume of the pores
>> is as simple as this:
>>
>> pores = Where(volume LE 10)
>> percent_pores = N_Elements(volume[pores])/N_Elements(volume)
>
> And what I *should* have written is this:
>
> percent_pores = FLOAT(N_Elements(volume[pores]))/N_Elements(volume)
>
> Or you will be looking at a 0 percent, always!

To avoid the slow WHERE function I'd suggest

percent_pores = 1.0 * Total( volume LE 10 ) / N_Elements( volume )

Alex
--
Alex Schuster Wonko@weird.cologne.de PGP Key available
alex@pet.mpin-koeln.mpg.de

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		Volume Size By: Les Butler on Thu, 09 April 1998 00:00
		Re: Volume Size By: a2652099 on Sun, 12 April 1998 00:00

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