Calculating colour between multiple lights?

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TheComet

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Joined: 18th Oct 2007

Location: I`m under ur bridge eating ur goatz.

Posted: 15th Aug 2011 21:06 Edited at: 15th Aug 2011 21:08

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Hey everyone

Light 1 is positioned at x1,y1,z1, has range1, color r1,g1,b1.
Light 2 is positioned at x2,y2,z2, has range2, color r2,g2,b2.
Light 3 is positioned at x3,y3,z3, has range3, color r3,g3,b3.

Point x is at px,py,pz.

How do I calculate the color you'd see at point x? I don't think it's linear, is it?

Thanks,

TheComet

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Max P

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Joined: 23rd Jan 2010

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Posted: 15th Aug 2011 21:13 Edited at: 16th Aug 2011 12:51

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It's untested code, but it should work:

DistanceLight1X# = px - x1
DistanceLight1Y# = py - y1
DistanceLight1Z# = pz - z1
DistanceLight1#  = sqrt(DistanceLight1X#*DistanceLight1X# + DistanceLight1Y#*DistanceLight1Y# + DistanceLight1Z#*DistanceLight1Z#)

DistanceLight2X# = px - x2
DistanceLight2Y# = py - y2
DistanceLight2Z# = pz - z2
DistanceLight2#  = sqrt(DistanceLight2X#*DistanceLight2X# + DistanceLight2Y#*DistanceLight2Y# + DistanceLight2Z#*DistanceLight2Z#)

DistanceLight3X# = px - x3
DistanceLight3Y# = py - y3
DistanceLight3Z# = pz - z3
DistanceLight3#  = sqrt(DistanceLight3X#*DistanceLight3X# + DistanceLight3Y#*DistanceLight3Y# + DistanceLight3Z#*DistanceLight3Z#)

Color1R# = (range1 - DistanceLight1#) / range1 * r1
Color1G# = (range1 - DistanceLight1#) / range1 * g1
Color1B# = (range1 - DistanceLight1#) / range1) * b1

Color2R# = (range2 - DistanceLight2#) / range2 * r2
Color2G# = (range2 - DistanceLight2#) / range2 * g2
Color2B# = (range2 - DistanceLight2#) / range2 * b2

Color3R# = (range3 - DistanceLight3#) / range3 * r3
Color3G# = (range3 - DistanceLight3#) / range3 * g3
Color3B# = (range3 - DistanceLight3#) / range3 * b3

FinalColorR# = (Color1R# + Color2R# + Color3R#)/3.0
FinalColorG# = (Color1G# + Color2G# + Color3G#)/3.0
FinalColorB# = (Color1B# + Color2B# + Color3B#)/3.0

FinalColor = rgb(FinalColorR#, FinalColorG#, FinalColorB#)

+ Code Snippet

DistanceLight1X# = px - x1
DistanceLight1Y# = py - y1
DistanceLight1Z# = pz - z1
DistanceLight1#  = sqrt(DistanceLight1X#*DistanceLight1X# + DistanceLight1Y#*DistanceLight1Y# + DistanceLight1Z#*DistanceLight1Z#)

DistanceLight2X# = px - x2
DistanceLight2Y# = py - y2
DistanceLight2Z# = pz - z2
DistanceLight2#  = sqrt(DistanceLight2X#*DistanceLight2X# + DistanceLight2Y#*DistanceLight2Y# + DistanceLight2Z#*DistanceLight2Z#)

DistanceLight3X# = px - x3
DistanceLight3Y# = py - y3
DistanceLight3Z# = pz - z3
DistanceLight3#  = sqrt(DistanceLight3X#*DistanceLight3X# + DistanceLight3Y#*DistanceLight3Y# + DistanceLight3Z#*DistanceLight3Z#)

Color1R# = (range1 - DistanceLight1#) / range1 * r1
Color1G# = (range1 - DistanceLight1#) / range1 * g1
Color1B# = (range1 - DistanceLight1#) / range1) * b1

Color2R# = (range2 - DistanceLight2#) / range2 * r2
Color2G# = (range2 - DistanceLight2#) / range2 * g2
Color2B# = (range2 - DistanceLight2#) / range2 * b2

Color3R# = (range3 - DistanceLight3#) / range3 * r3
Color3G# = (range3 - DistanceLight3#) / range3 * g3
Color3B# = (range3 - DistanceLight3#) / range3 * b3

FinalColorR# = (Color1R# + Color2R# + Color3R#)/3.0
FinalColorG# = (Color1G# + Color2G# + Color3G#)/3.0
FinalColorB# = (Color1B# + Color2B# + Color3B#)/3.0

FinalColor = rgb(FinalColorR#, FinalColorG#, FinalColorB#)

edit Made a small mistake, updated the code.

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SJPro

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Joined: 3rd Jul 2003

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Posted: 16th Aug 2011 14:00

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It's worth noting that the above assumes a linear relationship. Real light follows the inverse square law. I'm not sure how lighting is implemented in darkbasic/directx especially wrt to when it has a range.

www.freewarehub.net

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DarkBASIC Professional Discussion / Calculating colour between multiple lights?