Re: [Bodington-developers] CSS Processing

[Antony C. <an...@sm...>]

We will be holding a 3 day seminar in HSV processing here in late June  
- to register, click on the link below:
http://www.whatthefeckishetalkingabout.com/

On 6 Apr 2006, at 19:37, Jon Maber wrote:

> I must bow to your superior mathologicity.
>
> Many thanks for taking on this work package, good luck.....
>
> Jon
>
> Sean Mehan wrote:
>
>> Yes, all fine Jon, BUT what about compliance with the critical and   
>> well known Hausdorff's Metric Top, you know:
>>
>> Let (S,d) be a metric space, and let X be the collection of all   
>> nonempty bounded closed subsets of S. Let f:S * X -> R+ be defined by  
>>  f(s,B) = inf _(b in B) d(s,b), and let g: X * X -> R+ be given by g  
>> (A,B) = sup_(a in A) f(a,B), and let delta(A,B) = max {g(A,B),g  
>> (B,A)}. You must concern yourself with (X, delta)! Your method   
>> clearly fails here! Pah!
>>
>> And for anyone wanting to quibble with my counterexample, yes, I   
>> know, but I couldn't write the vector cross product adequately in   
>> this crap test format, so had to resort to *.
>>
>> -s
>>
>>
>> On 6 Apr 2006, at 18:59, Jon Maber wrote:
>>
>>> Nearly all user colour preference work will operate in Hue-  
>>> Saturation-Value space, not Red-Green-Blue space. However, the   
>>> standard Java HSV colour space (which Bodington currently uses) is   
>>> not very good so I've devised a better method. (Probably reinvented   
>>> but it only took a couple of days work.) The standard HSV space is   
>>> cylindrical where H is the perpendicular distance from the axis to   
>>> the colour point, H is the angle of that line from the red plane   
>>> and V is the distance up the cylinder.  All of the bottom face is   
>>> black. (which means that if V=0 then the colour is black regardless   
>>> of the values of H or S) However, the top face has white only at   
>>> the center and has bright rainbow colours around the edge.  This   
>>> means that if V is non-zero, V and H are kept constant and S is   
>>> varied then not only does the perceived saturation change but the   
>>> brightness does too. The method used to map the cube rgb space onto   
>>> the cylinder is also dodgy since for certain values of H changing S   
>>> distorts the perceived hue.
>>>
>>> My method also maps to a cylinder but the top face of the cylinder   
>>> is completely white.  This means that V==100% always produces white   
>>> no matter what the value of H or S. It also means that if H and V   
>>> are kept constant and S is varied, the perceived brightness and hue   
>>> keep constant. This is acheived by two rotations of the rgb cube so   
>>> that the white to black vector lines up with the z axis. Then the   
>>> shape is distorted so that vectors which run from the z axis to the   
>>> cube surface and which lie parallel to the x,y plane are stretched   
>>> uniformly to unit length.  The resulting cylinder is squashed down   
>>> the z axis to unit length.
>>
>>
>>
>
>
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