Курсовой по машинной графике на тему Фрактальные рельефы

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> 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: [email protected] (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 256×256 grid of points, each having an 8-bit 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 co-ords: (u,v) co-ordinates that describe a position on the map. The
map can be thought of as a height function h=w(u,v) sampled discretely.

Screen co-ords: (x,y) co-ordinates 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 old-hat to many people, but the map is made more realistic
by blurring:

w(u,v)=k1*w(u,v)+k2*w(u+3,v-2)+k3*w(u-2,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
Now, down to business. The screen is as follows:

| |
| |
| Sky |
| |
| |
|a————————| Horizon
| |
| | Point (a)=screen co-ords (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 co-ordinates 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 co-ordinate of map co-ord (p,v,0) (perspective transform)

you:->———————— Horizontal view
r :
—————————–P Ground
……………………. (q-v)
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:

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 co-ords 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 co-ord corresponding to (0,y). v is constant along
each line.

4. Starting at the calculated (u,v), traverse the screen, incrementing the
x co-ordinate and adding on a constant, c, to u such that (u+c,v) are the map
co-ords corresponding to the screen co-ords (1,y). You then have 256 map
co-ords along a line of constant v. Get the height, w, at each map co-ord and
draw a spot at (x,y-w*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

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:

| |
| |
| |
| **** |
| ********* | 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.

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