I have 2 colors #DCE7FA and #CADBF7. Want the intermediate color(a kind of Arithmetic mean).
Hexadecimal arithmetic median does not work.
How to proceed?
+2
A:
Yes the normal hex median wont work.. !!
try spliting to to R, G , B and find individual medians..
r1 = DC ; r2 = CA
g1 = E7 ; g2 = DB
b1 = FA ; b2 = F7
now find individual medians..
now,
r3 = (r1+r2)/2 = D3 ;
g3 = E1
b3 = F5
now ur intermediate color = #D3E1F5..
echo
2010-03-31 09:38:23
not good: seehttp://www.colortools.net/color_compare_colors.html visually `(#DCE7FA + #CADBF7) / 2 != #ABCDEF`
serhio
2010-03-31 09:47:26
@seriho.. i gave an EXAMPLE on how to split to RGB and find individial medians,, now i took my time and calculated the actual values and re-posted.. check it now..
echo
2010-03-31 09:50:04
and again : the median of colores SHOULD NOT be calculated as `(HexColor1 + HexColor2)/2` . Its calculated as the median of individual channels like i said in the answer.
echo
2010-03-31 09:54:14
F8, not F5 :) thanks.
serhio
2010-03-31 09:55:20
F8...! yeah!! typo!
echo
2010-03-31 09:56:53
A:
Found an online tool: http://www.colortools.net/color_combination.html
serhio
2010-03-31 09:57:47
That combines (finds the actual `(HexColor1 + HexColor2)/2`), guess thats not you want.. This type of finding mean is not at all reccomended.. even in one channel (say grayscale) coloring, they make `(color1+color2)%255` [note they take the remainder with the maximum color.]
echo
2010-03-31 10:02:40
I say the combination seems to work on this site. It's not `(hex1+hex2) / 2`...
serhio
2010-03-31 12:32:43