有关一个亮度的问题
通过一个WINDOWS的标准调色板，RGB=0,0,255可以得到一个标准的蓝色，显示的亮度是120；而一个RGB=100,100,255的颜色同样是蓝色，不过亮度是167。此时，如果在调色板里把167的亮度改回120，又可以得到RGB=0,0,255的标准蓝色了，说明这两个颜色只是亮度不同，色调是相同的。如果任意给一个颜色RGB＝X,Y,Z，可以求出它的亮度L,如果此时指定一个亮度L'，问当亮度为L'时相同色调的颜色的X,Y,Z分别是多少？ 问题点数：41、回复次数：4Top
1 楼kwxx（东京大屠杀万岁!）回复于 2005-08-03 23:59:22 得分 0
或者有哪位高手提供RGB转到色调和饱和度的代码也可以。上面所需要的只改变亮度色调与饱和度是不变的。Top
2 楼kwxx（东京大屠杀万岁!）回复于 2005-08-07 00:55:52 得分 0
upTop
3 楼fishmans（金脚指）回复于 2005-08-07 01:02:37 得分 41
[转贴]
代码～
Private Sub RGBToHSL( _
r As Long, g As Long, b As Long, _
h As Single, s As Single, l As Single _
)
Dim Max As Single
Dim Min As Single
Dim delta As Single
Dim rR As Single, rG As Single, rB As Single
rR = r / 255: rG = g / 255: rB = b / 255
'{Given: rgb each in [0,1].
' Desired: h in [0,360] and s in [0,1], except if s=0, then h=UNDEFINED.}
Max = Maximum(rR, rG, rB)
Min = Minimum(rR, rG, rB)
l = (Max + Min) / 2 '{This is the lightness}
'{Next calculate saturation}
If Max = Min Then
'begin {Acrhomatic case}
s = 0
h = 0
'end {Acrhomatic case}
Else
'begin {Chromatic case}
'{First calculate the saturation.}
If l <= 0.5 Then
s = (Max - Min) / (Max + Min)
Else
s = (Max - Min) / (2 - Max - Min)
End If
'{Next calculate the hue.}
delta = Max - Min
If rR = Max Then
h = (rG - rB) / delta '{Resulting color is between yellow and magenta}
ElseIf rG = Max Then
h = 2 + (rB - rR) / delta '{Resulting color is between cyan and yellow}
ElseIf rB = Max Then
h = 4 + (rR - rG) / delta '{Resulting color is between magenta and cyan}
End If
'Debug.Print h
'h = h * 60
'If h < 0# Then
' h = h + 360 '{Make degrees be nonnegative}
'End If
'end {Chromatic Case}
End If
'end {RGB_to_HLS}
End Sub
Public Sub HLSToRGB( _
h As Single, s As Single, l As Single, _
r As Long, g As Long, b As Long _
)
Dim rR As Single, rG As Single, rB As Single
Dim Min As Single, Max As Single
If s = 0 Then
' Achromatic case:
rR = l: rG = l: rB = l
Else
' Chromatic case:
' delta = Max-Min
If l <= 0.5 Then
's = (Max - Min) / (Max + Min)
' Get Min value:
Min = l * (1 - s)
Else
's = (Max - Min) / (2 - Max - Min)
' Get Min value:
Min = l - s * (1 - l)
End If
' Get the Max value:
Max = 2 * l - Min
' Now depending on sector we can evaluate the h,l,s:
If (h < 1) Then
rR = Max
If (h < 0) Then
rG = Min
rB = rG - h * (Max - Min)
Else
rB = Min
rG = h * (Max - Min) + rB
End If
ElseIf (h < 3) Then
rG = Max
If (h < 2) Then
rB = Min
rR = rB - (h - 2) * (Max - Min)
Else
rR = Min
rB = (h - 2) * (Max - Min) + rR
End If
Else
rB = Max
If (h < 4) Then
rR = Min
rG = rR - (h - 4) * (Max - Min)
Else
rG = Min
rR = (h - 4) * (Max - Min) + rG
End If
End If
End If
r = rR * 255: g = rG * 255: b = rB * 255
End Sub
Private Function Maximum(rR As Single, rG As Single, rB As Single) As Single
If (rR > rG) Then
If (rR > rB) Then
Maximum = rR
Else
Maximum = rB
End If
Else
If (rB > rG) Then
Maximum = rB
Else
Maximum = rG
End If
End If
End Function
Private Function Minimum(rR As Single, rG As Single, rB As Single) As Single
If (rR < rG) Then
If (rR < rB) Then
Minimum = rR
Else
Minimum = rB
End If
Else
If (rB < rG) Then
Minimum = rB
Else
Minimum = rG
End If
End If
End Function
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4 楼kwxx（东京大屠杀万岁!）回复于 2005-08-09 09:35:46 得分 0
谢了楼上的，我已经找到了这个算法，并且已经实现了。还是感谢！Top
