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| Back to square one on histograms |
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Sat, 19 Jan 2008 22:24:38 -050 |
Hi Everybody,
Tell me if I have this straight.
1. I have a full gradient with tone values from 0 to 255.
2, I compress this - one way or another - to a spread from a tone value of
32 to a tone value of 64.
3. I now have only 64 tone values because granularity can only be in
integers.
4. I expand - one way or another - to tones from 0 to 255.
5. The expansion does NOT fill in missing tone values.
6. The resulting expansion has only 64 tone values.
7. Thus 3 out of 4 of the original tone values are not present.
Is this correct.
Phil
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| Re: Back to square one on histograms |
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Sun, 20 Jan 2008 10:05:55 -050 |
YES!
--
Alfred
http://www.alfredky.com
"Philip K" <PhilipK@aol.com> wrote in message
news:4792bec3_2@cnews...
>
> Hi Everybody,
> Tell me if I have this straight.
> 1. I have a full gradient with tone values from 0 to 255.
> 2, I compress this - one way or another - to a spread from a tone value
> of 32 to a tone value of 64.
> 3. I now have only 64 tone values because granularity can only be in
> integers.
> 4. I expand - one way or another - to tones from 0 to 255.
> 5. The expansion does NOT fill in missing tone values.
> 6. The resulting expansion has only 64 tone values.
> 7. Thus 3 out of 4 of the original tone values are not present.
> Is this correct.
> Phil
>
>
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| Post Reply
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| Re: Back to square one on histograms |
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Sun, 20 Jan 2008 10:08:36 -050 |
If you spread the tone values from 32 to 64, you will have a total of 64
values, but the will have the observed spaces..
--
Alfred
http://www.alfredky.com
"Gareth Hawker"
<gareth@_removethis.bit_garethhawker.freeserve.co.uk> wrote
in message news:47933672$1_1@cnews...
>
> Hi Phil,
>
> That all makes sense to me (except that I think a spread from 32 to 64
> would
> leave a total of 32, not 64).
>
> Regards
>
> Gareth> Hi Everybody,
>> Tell me if I have this straight.
>> 1. I have a full gradient with tone values from 0 to 255.
>> 2, I compress this - one way or another - to a spread from a tone
value
>> of 32 to a tone value of 64.
>> 3. I now have only 64 tone values because granularity can only be in
>> integers.
>> 4. I expand - one way or another - to tones from 0 to 255.
>> 5. The expansion does NOT fill in missing tone values.
>> 6. The resulting expansion has only 64 tone values.
>> 7. Thus 3 out of 4 of the original tone values are not present.
>> Is this correct.
>> Phil
>>
>>
>
>
>
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| Re: Back to square one on histograms |
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Sun, 20 Jan 2008 11:50:33 -000 |
Hi Phil,
That all makes sense to me (except that I think a spread from 32 to 64 would
leave a total of 32, not 64).
Regards
Gareth
> Hi Everybody,
> Tell me if I have this straight.
> 1. I have a full gradient with tone values from 0 to 255.
> 2, I compress this - one way or another - to a spread from a tone value
> of 32 to a tone value of 64.
> 3. I now have only 64 tone values because granularity can only be in
> integers.
> 4. I expand - one way or another - to tones from 0 to 255.
> 5. The expansion does NOT fill in missing tone values.
> 6. The resulting expansion has only 64 tone values.
> 7. Thus 3 out of 4 of the original tone values are not present.
> Is this correct.
> Phil
>
>
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| Post Reply
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| Re: Back to square one on histograms |
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Sun, 20 Jan 2008 11:55:11 -000 |
Hi Phil,
That all makes sense to me (except that I think a spread from 32 to 64 would
leave a total of 32, not 64).
Regards
Gareth> Hi Everybody,
> Tell me if I have this straight.
> 1. I have a full gradient with tone values from 0 to 255.
> 2, I compress this - one way or another - to a spread from a tone value
> of 32 to a tone value of 64.
> 3. I now have only 64 tone values because granularity can only be in
> integers.
> 4. I expand - one way or another - to tones from 0 to 255.
> 5. The expansion does NOT fill in missing tone values.
> 6. The resulting expansion has only 64 tone values.
> 7. Thus 3 out of 4 of the original tone values are not present.
> Is this correct.
> Phil
>
>
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| Post Reply
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