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By Lithium | Published Jan 11 2001 12:05 PM in Math and Physics

value plane shading light dot distance word product falloff
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Hopefully you have read the companion document 3D Rotations, as this one will build upon the concepts presented in my attempt to teach some of the math need to make 3D graphics a reality. This file will cover such important topics as the Dot Product and how routines are best constructed for real-time 3D rotations and planar shading.

Our Friend, The Dot Product

The Dot Product is a neat relation that will allow you to quickly find the angle between any two vectors. It's easiest to explain graphically, so I will exercise my extended-ASCII keys.

Two Vectors A & B:

A (Xa, Ya, Za) |A| = sqrt( (Xa)^2 + (Ya)^2 + (Za)^2 )
B (Xb, Yb, Zb) |B| = sqrt( (Xb)^2 + (Yb)^2 + (Zb)^2 )

Where Xa, and the others coorispond to some value on their respective Axis's

|A
/
/
/
/
\ angle   <-- Angle Theta between vector A and B
\
\
\
|B

Cos(angle) =  Xa * Xb + Ya * Yb + Za * Zb
----------------------------
|A|*|B|

Example:

A (1,2,3)         |A| = sqrt( 1^2 + 2^2 + 3^2) = sqrt(14) = 3.7417
B (4,5,6)         |b| = sqrt( 4^2 + 5^2 + 6^2) = sqrt(77) = 8.7750

Cos(angle) =  1 * 4 + 2 * 5 + 3 * 6   =  4 + 10 + 18    =    32    =  0.9746
---------------------     ---------------    -----
(3.7417)*(8.7750)           32.8334       32.8334

ArcCos (.9746) = 12.9ø

So, your wondering how this revolutionizes you code, huh? Well, remember our other friend, the Normal vector? You use Normal vectors that define the directions of everything in our 3D world. Let's say that vector A was the Normal vector from my plane, and B is a vector that shows the direction that the light in my scene is pointing. If I do the Dot Product of them, you will get the angle between them, if that angle is >= 90o and <= 270o then no light falls on the visible surface and it doesn't need to be displayed.

Also notice, the way the values of the Cosine orient themselves

90                  Cos 000 =  1
Cos 090 =  0
|                   Cos 180 = -1
Negative  |  Positive         Cos 270 =  0
|
|
180 -------|-------  0         An angle between a light and a plane that
|                   is less than 90 or greater than 270 will
|                   be visible, so you can check if the Cos(theta)
Negative  |  Positive         is greater than 0 to see if it is visible.
|
|

270

How Do You Implement The Code? Easy As Pie.

We will define our points like this:

STRUC XYZs
Xpos	dd  ?
Ypos	dd  ?
Zpos	dd  ?
Dist	dd  ?
ENDS  XYZs          	;size is 16 bytes

The X,Y,Zpos define a point in 3D space, Dist is the distance from the origin:

Dist = sqrt( X^2 + Y^2 + Z^2 )

Precalculate these values and have them handy in your data area. Our planes should look something like this:

STRUC PlaneSt
NumPts  	db  	?         	;3 or 4
NormIndex   dw  	?
PtsIndex	dw  	?
dw  	?
dw  	?
dw  	?
ENDS  PlaneSt

The number of points that in the plane depends on the number your fill routines can handle you must have at least 3 and more than 6 is not suggested. Then we set up our data like this:

MaxPoints   =       	100
MaxPlanes   =       	100

PointList   XYZs    	MaxPoints DUP()
PlaneList   PlaneSt 	MaxPlanes DUP()
NormalList  XYZs    	<0,0,0, 10000h> , MaxPlanes DUP()

Non-ASM User Note:I set up points in a structure that had an X,Y,Z and Distance value. I set up a plane structure that had the number of points the index number of the normal vector for that plane and the index numbers for the points in the plane. The next lines set up arrays of these points in PointList, and the number of points was defined as MaxPoints. An array of planes was created as PlaneList with MaxPlanes as the total number of plane structures in the array. NormalList is an array of the vectors that are normal to the planes, one is set up initially (I'll explain that next) and then one for each possible plane is allocated.

You'll notice that I defined the first Normal and then created space for the rest of the possible normals. I'll call this first normal, the Zero Normal. It will have special properties for planes that don't shade and are never hidden.

Well, before I start telling all the tricks to the writing code, let me make sure a couple of points are clear.

In 3D Rotations I said that you could set your view point on the Z-Axis and then figure out if planes were visible by the post-rotation Normal vectors, if their Z was greater than 0 then display, if not, don't.

That is an easy way to set up the data, and I didn't feel like going into the Dot Product at the time, so I generalized. So, what if you don't view your plane from the Z-Axis, the answer is you use the...Dot Product! That's right. The angle will be used now to figure whether or not to display the plane.

I have been mentioning lights and view points as vectors that I can use with the Normal vector from my plane. To work correctly, these vectors for the lights and view should point in the direction that you are looking or the direction that the light is pointing, *NOT* a vector drawn from the origin to the viewer position or light position.

True Normal vectors only state a direction, and should therefore have a unit distance of 1. This will have the advantage of simplifying the math involved to figure you values. Also, for God's sake, pre-compute your normal, don't do this every time. Just rotate them when you do your points and that will update their direction.

If the Normal's have a length of 1 then |A|*|B| = 1 * 1 = 1. So:

Cos(angle) = Xa * Xb + Ya * Yb + Za * Zb
---------------------------
|A|*|B|

Is reduced to:

Cos(angle) = Xa * Xb + Ya * Yb + Za * Zb

We eliminated a multiply and a divide! Pat yourself on the back.

You ASM users might be wondering why I defined my Zero Normal as: < 0,0,0,10000h > How does 10000h equal a length of 1 ?

Well, this is a trick you can do in ASM, instead of using floating point values that will be slow on computers without math co-processors, we can use a double word to hold our value. The high word holds the integer value, and the low word is our decimal. You do all of your computations with the whole register, but only pull the high word when you go to display the point. So, with that under consideration, 10000h = 1.00000. Not bad for integers.

How does the Zero Normal work? Since the X,Y,and Z are all 0, the Cos(angle) = 0, so if you always display when Cos(angle) = 0, then that plane will always be seen.

So, Beyond The Babble...

How To Set Up Your Code
• Define Data Points, Normals, and Planes
• Pre-Calculate as many values as possible
• Rotate Points and Normals
• Determine Visible Planes With Dot Product (Save this value if you want to shade)
• Sort Visible Planes Back to Front (Determine Shade From Dot Product)
• Clip Plane to fit scene
• Draw to the screen
• Change Angles
• Goto Rotation
A quick way to figure out which color to shade your plane if you are using the double word values like I described before is to take the Dot Product result, it will lie between 10000h - 0h if you would like say 16 shades over the angles, then take that value and shr ,12 that will give you a value from 0h - 10h (0-16, or 17 colors) if you make 10h into 0fh, add that offset to a gradient in your palette, then you will have the color to fill your polygon with.

Note also that the Cosine function is weighted toward the extremes. If you want a smooth palette change as the angles change, your palette should weight the gradient accordingly.

A useful little relation for depth sorting is to be able to find the center of a triangle.

E         The center C = (D + E + F)/3
^
/ \        Divide each coordinate by (Xd + Xe + Xf)/3 = Xc
/ C \         and do the same for the Y's and Z's if you
/     \        choose to sort with this method.  Then rotate
D---------F      that point and use it to depth sort the planes

Recently, someone asked me about the practicality of real-time phong and gouraud shading. The technique is common to ray-tracers and requires a great deal of calculation when working with individual rays cast from each pixel, but when only using this for each plane, it is possible. This type of shading involves taking into account the reduced luminosity of light as distance increases. For each light, you define a falloff value. This value should be the distance a which the light will be at full intensity. Then at 2*FallOff you will have 1/2 intensity, 3*FallOff will yield 1/3 and so on. To implement this type of shading, you will need to determine the distance from the light to the center of the plane. If distance < FallOff, then use the normal intensity. If it is greater, divide the FallOff value by the distance. This will give you a scalar value that you can multiple by the shading color that the plane should have. Use that offset and it will be darker since it is further away from the light source.

However, to determine the distance form the light to each plane, you must use a Square Root function, these are inherently slow unless you don't care about accuracy. Also, it would be difficult to notice the use of this technique unless you have a relatively small FallOff value and your objects move about in the low intensity boundaries.