MacHack 1996/MacHack 1996.toast/Hacks/Hacks ’95/Venus/mars

Internet Message Format | 1994-05-17 | 8.1 KB | [TEXT/ttxt]

From oauld@ponder Sun Apr 10 14:48:00 1994
Resent-From: "John Booth    " <[e-mail]>
Resent-To: oleg@ponder
Resent-Date:   Sun, 10 Apr 1994 14:47:31 CST6CDT
Date: Sat, 26 Mar 94 12:19:31 -0600
From: oauld@ponder (Orion Auld)
To: [e-mail], elliott@ponder
Subject: Mars demo guts
Status: OR

>Newsgroups: rec.games.programmer
>Path: news.unt.edu!cs.utexas.edu!swrinde!emory!news-feed-2.peachnet.edu!concert!hearst.acc.Virginia.EDU!murdoch!usenet
>From: [e-mail] (Colin  Prepscius)
>Subject: Re: Recursive Map Generation
>Date: Fri, 25 Mar 1994 00:44:59 GMT

In article <[e-mail]>
[e-mail] (Brian Hacking) writes:
> In article <2mq013$[e-mail]>, [e-mail]
> (Marwan Taher) wrote:
>
> > Also, in his paper Tim suggested 'blurring' the map with:
> >     w(u,v)=k1*w(u,v)+k2*(u+3,v-2)+k3*w(u-2,v+4)
> > where k1+k2+k3=1. ( w(u,v) contains the height of point at coords
(u,v). )
>
> I and several other readers missed this post on Mars guts.  Can someone
> report it or email it to me????  Puhlease....
>
> *** These are my words, not Novell's ***

Here ya go.  I asked the same thing about 2 weeks ago.  Note - corrections
to this (below) were also posted - basically just replace r w/ q in the
following sentence...
line of map from v=r+100 (say) down to v=r, do this:

--
!@#$%^&*()_++_)(*&^%$#@!

Colin Prepscius
[e-mail]
homephone:804-295-2108
labphone:804-924-0356


*****************************************


        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 and 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 w=f(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 horizon).
 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=r+100 (say) down to v=r, do this:

  1. Calculate the y co-ordinate of map co-ord (p,v,0) (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.

  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.

 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 half-drawn.
     |-----**************------|
     |*************************|
     |*********       *********|
     |******             ******|
     |.........................| <- The row of dots is at screen co-ord y
     |                         |   corresponding to an altitude of 0 for
that
     ---------------------------   particular distance v.

 Now the screen-scanning 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 p-something 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.

 Thank you for the response I have received since uploading this program.

 Tim Clarke, [e-mail]