Фрактальные рельефы
Фрактальные рельефы
{
> Can someone explain the details of
how to make game graphics like
> in commanche over kill ?
It's called Voxel. Don't ask me why. Below
is a text that explains it.

From: pleung@cs.buffalo.edu (Patrick Leung)
Newsgroups: rec.games.programmer
Subject: How To: Mars Landscape (Repost)
Well, since I posted something about the:
'Voxel landscapes and
How I did it' post by Tim Clarke, a coupla
people have asked about
it. So, here it is, Tim Clarke's : Voxel
landscapes and How I did it
with some corrections and modifications he
added in later postings.
Hope I dont infringe on any of his rights
in doing this...
The 'I' s in the following represent Tim
Clarke, not me :)
Voxel landscapes and How I did it

This document describes the method I used
in my demo of a Martian terrain,
which can be found at
garbo.uwasa.fi:/pc/demo/mars10.zip.
It's similar to a floating horizon hidden
line removal algorithm, so you'll
find discussion of the salient points in
many computer graphics books. The
difference is the vertical line
interpolation.
First, some general points:

The map is a 256x256 grid of points, each
having an 8bit integer height
and a colour. The map wraps round such
that, calling w(u,v) the height at
(u,v), then
w(0,0)=w(256,0)=w(0,256)=w(256,256). w(1,1)=w(257,257), etc.
Map coords: (u,v) coordinates that
describe a position on the map. The
map can be thought of as a height function
h=w(u,v) sampled discretely.
Screen coords: (x,y) coordinates for a
pixel on the screen.
To generate the map:

This is a recursive subdivision, or
plasma, fractal. You start of with
a random height at (0,0) and therefore also
at (256,0), (0,256), (256,256).
Call a routine that takes as input the size
and position of a square, in the
first case the entire map.
This routine get the heights from the
corners of the square it gets given.
Across each edge (if the map has not been
written to at the point halfway
along that edge), it takes the average of
the heights of the 2 corners on that
edge, applies some noise proportional to
the length of the edge, and writes
the result into the map at a position
halfway along the edge. The centre of
the square is the average of the four
corners+noise.
The routine then calls itself recursively,
splitting each square into four
quadrants, calling itself for each quadrant
until the length of the side is
2 pixels.
This is probably oldhat to many people,
but the map is made more realistic
by blurring:
w(u,v)=k1*w(u,v)+k2*w(u+3,v2)+k3*w(u2,v+4) or something.
Choose k1,k2,k3 such that k1+k2+k3=1. The
points at which the map is sampled
for the blurring filter do not really
matter  they give different effects,
and you don't need any theoretical reason
to choose one lot as long as it
looks good. Of course do everything in
fixed point integer arithmetic.
The colours are done so that the sun is on
the horizon to the East:
Colour=A*[ w(u+1,v)w(u,v) ]+B
with A and B chosen so that the full range
of the palette is used.
The sky is a similar fractal but without
the colour transformation.
How to draw each frame

First, draw the sky, and blank off about
50 or so scan lines below the
horizon since the routine may not write to
all of them (eg. if you are on top
of a high mountain looking onto a flat
plane, the plane will not go to the
horizon).
Now, down to business. The screen is as
follows:

 
 
 Sky 
 
 
a Horizon
 
  Point
(a)=screen coords (0,0)
 Ground  x
increases horizontally
  y
increases downwards
 

Imagine the viewpoint is at a position
(p,q,r) where (p,q) are the (u,v)
map coordinates and r is the altitude.
Now, for each horizontal (constant v)
line of map from v=q+100 (say) down to v=q,
do this:
1. Calculate the y coordinate of map
coord (p,v,0) (perspective transform)
you:>
Horizontal view
:
r :
:
:
P Ground
......................... (qv)
q v
You have to find where the line between P
and you intersects with the
screen (vertical, just in front of 'you').
This is the perspective transform:
y=r/(qv).
2. Calculate scale factor f which is how
many screen pixels high a mountain
of constant height would be if at distance
v from q. Therefore, f is small
for map coords far away (v>>q) and
gets bigger as v comes down towards q.
So, f is a number such that if you
multiply a height from the map by f, you
get the number of pixels on the screen high
that height would be. For
example, take a spot height of 250 on the
map. If this was very close, it
could occupy 500 pixels on the screen
(before clipping)>f=2.
3. Work out the map u coord
corresponding to (0,y). v is constant along
each line.
4. Starting at the calculated (u,v),
traverse the screen, incrementing the
x coordinate and adding on a constant, c,
to u such that (u+c,v) are the map
coords corresponding to the screen coords
(1,y). You then have 256 map
coords along a line of constant v. Get the
height, w, at each map coord and
draw a spot at (x,yw*f) for all x.
I.e. the further away the scan line is,
the more to the "left" u will start,
and the larger c will be (possibly skipping
some u columns if c > 1); the
closer the scan line, the lesser u will
start on the "left", and c will be
smaller.
Sorry, but that probably doesn't make much
sense. Here's an example:
Imagine sometime in the middle of drawing
the frame, everything behind a
point (say v=q+50) will have been drawn:

 
 
 
 **** 
 *********  < A
mountain halfdrawn.
**************
*************************
********* *********
****** ******
......................... < The
row of dots is at screen coord y
 
corresponding to an altitude of 0 for that

particular distance v.
Now the screenscanning routine will get called
for v=q+50. It draws in a
point for every x corresponding to heights
at map positions (u,v) where u
goes from psomething to p+something, v
constant. The routine would put points
at these positions: (ignoring what was
there before)

 
 
 
 
 

 ***** 
 *** *** 
******* *******
.........................
 

So, you can see that the screen gets drawn
from the back, one vertical
section after another. In fact, there's
more to it than drawing one pixel
at every x during the scan  you need to
draw a vertical line between
(x,y old) to (x,y new), so you have to have
a buffer containing the y values
for every x that were calculated in the
previous pass. You interpolate
along this line (Gouraud style) from the
old colour to the new colour also,
so you have to keep a buffer of the colours
done in the last pass.
Only draw the vertical lines if they are
visible (ie. going down,
y new>y old). The screen is drawn from
the back so that objects can be drawn
inbetween drawing each vertical section at
the appropriate time.
If you need further information or
details, mail me or post here... Posting
will allow others to benefit from your points
and my replies, though.
