Why Diffuse Light use max(N · H, 0) instead of just letting it be negative?
In Cg tuts, Diffuse Section
Surfaces that face away from the light will produce negative dot-product values, so the max(N · L, 0) in the equation ensures that these surfaces show no diffuse lighting.
My question is why not just leave it be negative? IMO, both the negative and zero will make the color be black.
lighting cg
asked 1 hour ago
AlexWei
1554
add a comment |
In Cg tuts, Diffuse Section
Surfaces that face away from the light will produce negative dot-product values, so the max(N · L, 0) in the equation ensures that these surfaces show no diffuse lighting.
My question is why not just leave it be negative? IMO, both the negative and zero will make the color be black.
lighting cg
asked 1 hour ago
AlexWei
1554
add a comment |
In Cg tuts, Diffuse Section
Surfaces that face away from the light will produce negative dot-product values, so the max(N · L, 0) in the equation ensures that these surfaces show no diffuse lighting.
My question is why not just leave it be negative? IMO, both the negative and zero will make the color be black.
lighting cg
asked 1 hour ago
AlexWei
1554
In Cg tuts, Diffuse Section
Surfaces that face away from the light will produce negative dot-product values, so the max(N · L, 0) in the equation ensures that these surfaces show no diffuse lighting.
My question is why not just leave it be negative? IMO, both the negative and zero will make the color be black.
lighting cg
lighting cg
asked 1 hour ago
AlexWei
1554
asked 1 hour ago
AlexWei
1554
asked 1 hour ago
AlexWei
1554
asked 1 hour ago
AlexWei
1554
asked 1 hour ago
AlexWei
1554
1554
add a comment |
add a comment |
2 Answers
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If lighting with multiple lights, they add together to make the final lighting value. A negative light would darken other lights, which is incorrect. This is also true if you add in specular reflections, emissive lighting, or other sources of lighting. The negative lighting from one source of lighting would darken some other source of non negative lighting.
answered 40 mins ago
Alan Wolfe
4,74621247
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Because negative shading values are undefined and don't play well with tonemapping. It is preferable to use values in $[0,1]$ for low-dynamic-range or $[0,+infty)$ for high-dynamic-range images.
answered 1 hour ago
Hubble
1415
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If lighting with multiple lights, they add together to make the final lighting value. A negative light would darken other lights, which is incorrect. This is also true if you add in specular reflections, emissive lighting, or other sources of lighting. The negative lighting from one source of lighting would darken some other source of non negative lighting.
answered 40 mins ago
Alan Wolfe
4,74621247
add a comment |
If lighting with multiple lights, they add together to make the final lighting value. A negative light would darken other lights, which is incorrect. This is also true if you add in specular reflections, emissive lighting, or other sources of lighting. The negative lighting from one source of lighting would darken some other source of non negative lighting.
answered 40 mins ago
Alan Wolfe
4,74621247
add a comment |
If lighting with multiple lights, they add together to make the final lighting value. A negative light would darken other lights, which is incorrect. This is also true if you add in specular reflections, emissive lighting, or other sources of lighting. The negative lighting from one source of lighting would darken some other source of non negative lighting.
answered 40 mins ago
Alan Wolfe
4,74621247
If lighting with multiple lights, they add together to make the final lighting value. A negative light would darken other lights, which is incorrect. This is also true if you add in specular reflections, emissive lighting, or other sources of lighting. The negative lighting from one source of lighting would darken some other source of non negative lighting.
answered 40 mins ago
Alan Wolfe
4,74621247
answered 40 mins ago
Alan Wolfe
4,74621247
answered 40 mins ago
Alan Wolfe
4,74621247
answered 40 mins ago
Alan Wolfe
4,74621247
4,74621247
add a comment |
add a comment |
Because negative shading values are undefined and don't play well with tonemapping. It is preferable to use values in $[0,1]$ for low-dynamic-range or $[0,+infty)$ for high-dynamic-range images.
answered 1 hour ago
Hubble
1415
add a comment |
Because negative shading values are undefined and don't play well with tonemapping. It is preferable to use values in $[0,1]$ for low-dynamic-range or $[0,+infty)$ for high-dynamic-range images.
answered 1 hour ago
Hubble
1415
add a comment |
Because negative shading values are undefined and don't play well with tonemapping. It is preferable to use values in $[0,1]$ for low-dynamic-range or $[0,+infty)$ for high-dynamic-range images.
answered 1 hour ago
Hubble
1415
Because negative shading values are undefined and don't play well with tonemapping. It is preferable to use values in $[0,1]$ for low-dynamic-range or $[0,+infty)$ for high-dynamic-range images.
answered 1 hour ago
Hubble
1415
answered 1 hour ago
Hubble
1415
answered 1 hour ago
Hubble
1415
answered 1 hour ago
Hubble
1415
1415
add a comment |
add a comment |
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