Denthor/Asphyxia’s VGA trainers: 3D Solids

On to today’s tutorial: 3D solids. That is what the people wanted, that is what the people get! This tutorial is mainly on how to draw the polygon on screen. For details on how the 3D stuff works, check out part 8.

DENTHOR, coder for ...
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smith9@batis.bis.und.ac.za
The great South African Demo Team! Contact us for info/code exchange!

Grant Smith, alias Denthor of Asphyxia, wrote up several articles on the creation of demo effects in the 90s. I reproduce them here, as they offer so much insight into the demo scene of the time.

These articles apply some formatting to Denthor's original ASCII files, plus a few typo fixes.

How to draw a polygon

Sounds easy enough, right? WRONG! There are many, many different ways to go about this, and today I’ll only be showing you one. Please don’t take what is written here as anything approaching the best method, it is just here to get you on your way…

The procedure I will be using here is based on something most of us learned in standard eight … I think. I seem to recall doing something like this in Mrs. Reid’s maths class all those years ago ;)

Take two points, x1,y1 and x2,y2. Draw them:

                  + (x1,y1)
                   \
                     \  <-- Point a somewhere along the line
                       \
                         + (x2,y2)

Right, so what we have to do is this: if we know the y-coord of a, what is its x-coord? To prove the method we will give the points random values.

                 + (2,10)
                  \
                    \  <-- a.y = 12
                      \
                        +  (15,30)

Right. Simple enough problem. This is how we do it:

   (a.y-y1) = (12 - 10)  {to get a.y as though y1 was zero}
   *(x2-x1) = *(15 - 2)  {the total x-length of the line}
   /(y2-y1) = /(30 - 10) {the total y-length of the line}
        +x1 = +2         { to get the equation back to real coords}

So our equation is:

(a.y-y1)*(x2-x1)/(y2-y1)+x4 or
(12-10)*(15-2)/(30-10)+2

which gives you:

2*13/20+2 = 26/20+2
          = 3.3

That means that along the line with y=12, x is equal to 3.3. Since we are not concerned with the decimal place, we replace the / with a div, which in Pascal gives us an integer result, and is faster too. “All well and good,” I hear you cry, “but what does this have to do with life and how it relates to polygons in general?” The answer is simple. For each of the four sides of the polygon we do the above test for each y line. We store the smallest and the largest x values into separate variables for each line, and draw a horizontal line between them. Ta-Dah! We have a cool polygon!

For example - two lines going down:

    
                +             +
               / <-x1     x2->|   <--For this y line
             /                |
           +                  +

Find x1 and x2 for that y, then draw a line between them. Repeat for all y values.

Of course, it’s not as simple as that. We have to make sure we only check those y lines that contain the polygon (a simple min y, max y test for all the points). We also have to check that the line we are calculating actually extends as far as where our current y is (check that the point is between both y’s). We have to compare each x to see whether it is smaller then the minimum x value so far, or bigger then the maximum (the original x min is set as a high number, and the x max is set as a small number). We must also check that we only draw to the place that we can see (0-319 on the x; 0-199 on the y (the size of the MCGA screen)).

To see how this looks in practice, have a look at the sample code provided. (Mrs. Reid would probably kill me for the above explanation, so when you learn it in school, split it up into thousands of smaller equations to get the same answer ;))

Okay, that’s it! What’s that? How do you draw a vertical line? Thats simple…

Drawing a vertical line

Right, this is a lot easier than drawing a normal line (part 5 .. I think), because you stay on the same y value. So, what you do is you set ES to the screen you want to write to, and get DI to the start of the y-line (see earlier trainers for a description of how SEGMENT:OFFSET works).

    
IN   : x1 , x2, y, color, where

           asm
             mov    ax,where
             mov    es,ax
             mov    di,y
             mov    ax,y
             shl    di,8   { di:=di*256 }
             shl    ax,6   { ax:=ax*64 }
             add    di,ax  { di := (y*256)+(y*64) := y*320 Faster then a
                             straight multiplication }

Right, now you add the first x value to get your startoff.

    
             add    di,x1

Move the color to store into ah and al:

    
             mov    al,color
             mov    ah,al       { ah:=al:=color }

then get CX equal to how many pixels across you want to go:

    
             mov    cx,x2
             sub    cx,x1   { cx:=x2-x1 }

Okay, as we all know, moving a word is a lot faster thán moving a byte, so we halve CX:

    
             shr    cx,1    { cx:=cx/2 }

but what happens if CX was an odd number? After a shift, the value of the last number is placed in the carry flag, so what we do is jump over a single byte move if the carry flag is zero, or execute it if it is one.

   
            jnc     @Start  { If there is no carry, jump to label Start }
            stosb           { ES:[DI]:=al ; increment DI }
        @Start :            { Label Start }
            rep     stosw   { ES:[DI]:=ax ; DI:=DI+2; repeat CX times }

Right, the finished product looks like this:

   
Procedure Hline (x1,x2,y:word;col:byte;where:word); assembler;
  { This draws a horizontal line from x1 to x2 on line y in color col }
asm
  mov   ax,where
  mov   es,ax
  mov   ax,y
  mov   di,ax
  shl   ax,8
  shl   di,6
  add   di,ax
  add   di,x1

  mov   al,col
  mov   ah,al
  mov   cx,x2
  sub   cx,x1
  shr   cx,1
  jnc   @start
  stosb
@Start :
  rep   stosw
end;

Done!

In closing

This 3D system is still not perfect. It needs to be faster, and now I have also dumped the problem of face-sorting on you! Nyahahahaha!