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Topic : "Values & planes question" |
RoyaleWithCheese
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Joined: 19 Mar 2006
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 Posted: Thu Apr 06, 2006 1:21 am |
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How do I assign values to planes based on their respective angle to the light source? I'm looking at these images off gnomon from Scott Robertson's matte object rendering lessons and he seems to have some kind of system for applying values to planes based on their angle to the source. Seems to be something they teach well at Art center as lots of well known people from there have done these exercises with rendering odd shaped geometry. More difficult than cubes/spheres (which themselves require a lot of study).
Has anyone bought these dvds and are they worth the price?
http://www.thegnomonworkshop.com/img/screengrabs/sro07/sro07k.jpg
http://www.thegnomonworkshop.com/img/screengrabs/sro07/sro07h.jpg |
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jfrancis
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| Posted: Thu May 04, 2006 6:16 pm |
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There are very precise formulas for assigning vlues to planes based on their angle to the light. They are not particularly complex formulas, either. They aren't used by Painters, though. They are used by 3D rendering apps like Maya.
For painting, you use observation, intuition, and imagination more than math.
If you want the math -- here it is -- at least the Lambert and Phong illumination models
http://developer.apple.com/documentation/QuickTime/QD3D/qd3dshading.4.htm |
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Jin
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| Posted: Thu May 04, 2006 8:22 pm |
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Hi,
Maybe this tutorial will help. The author hasn't completed all of the planned pages, but there are several full of good information and images.
Light - A Detailed Tutorial
Jinny Brown
Corel Painter Instructor, TutorAlley Forums
Corel Painter Focused Sites:
http://www.pixelalley.com
http://www.tutoralley.com
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Last edited by Jin on Thu May 04, 2006 8:27 pm; edited 1 time in total |
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Jin
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| Posted: Thu May 04, 2006 8:22 pm |
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| Duplicate Post. Sorry. |
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Ranath
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| Posted: Fri May 05, 2006 5:12 am |
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stuff in that jfrancis' link was crystal clear to me. No worries..  |
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jfrancis
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| Posted: Wed May 10, 2006 11:11 am |
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They make it look complicated, but for diffuse lighting it's like this -
You take the ANGLE to the light and turn that into a lighting PERCENTAGE.
When the normal of a surface points straight at the light, the illumination from the light is at 100%. When the normal points at 90 degrees (or more) from the light, you are by definition talking about the edge of the object or the back side of the object (from the light's point of view) and so no light from that source directly reaches the surface. (it could bounce there indirectly, or another source could light it directly)
The actual calculation that turns ANGLE into PERCENTAGE is called the dot product.
http://en.wikipedia.org/wiki/Dot_product
Here is a Java applet that lets you visually turn angles into percentages.
http://www.falstad.com/dotproduct/
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Specular lighting obeys slightly more complicated rules.
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The 100% straight-at-light percentage assumes no inverse-square distance falloff. That's a little bit extra calculation. |
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JeffZugale
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| Posted: Sat Jun 10, 2006 1:16 pm |
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| Quote: |
| The 100% straight-at-light percentage assumes no inverse-square distance falloff. That's a little bit extra calculation. |
Actually that's quite applicable to lighting in a drawn or painted illustration, in the sense that when you're lighting an object you first figure out how much light is going to reach the object surface in the first place, and THEN you figure out how much of that light will hit the surfaces as they turn away from the light.
So you treat the light that hits the part of your object that directly faces the light source as 100% of the light that is going to hit the object, and do your figuring about how much light hits all the other surfaces from that, with surfaces turned 90 degrees or more away from the source getting zero light, surfaces turned 45 degrees getting half the light, etc.
The math relationship is that rotation away from source of 90 degrees = 100% change in light intensity.
The equation is thus:
Intensity at surface (in %) = 100 - 90/degrees rotated from source
That's a useful approximation for an artist, although it would utterly fail in a computer program. 
That's because, as jfrancis says, it's not really accurate in terms of how light actually works, because you need to take into account that the surface which is rotated away is also further from the source, so inverse square comes into effect; and even more complex than that is the fact that a surface that turns away from the light has more light falling on the part of it closest to the source than the part farthest away, even though the light is hitting at the same angle. Draw an octagon, you'll see what I mean.
And finally, this only applies if the light rays are parallel when they reach the object, which only applies to something like sunlight.
That's why the equations on jfrancis's link are so much more complex... not the kind of easy rule of thumb you can do in your head!
If you start with that easy linear relationship and then sort of "fudge" an approximation of the inverse square falloff in places where it would be obvious to the eye, you can get nice results.
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Drunken Monkey
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| Posted: Mon Jun 26, 2006 12:00 am |
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I don't know about the java applet math stuff... but robertsons method was as simple as you think. 90 deg angle is your lightest value, 45 deg angle is your middle value and anything thats not hit by the light is your darkest.
your darkest value is determined by 'half way to black' formula which is for example if you have 0-10 scale 0 being white 10 being black - if your lightest value is 2 half way to black(half way to 10) is 6, if the lightest is 4 hwtb is 7.
So nothing in cast shadow goes below half way to black value.
the lightest value is also not necessarity white, its whatever your choose based on environmental conditions. So nothing goes above lightest value unless its emiting light.
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