Path: chuka.playstation.co.uk!news
From: "Tom" <codemonkey@cwcom.net>
Newsgroups: scea.yaroze.programming.3d_graphics,scee.yaroze.programming.3d_graphics
Subject: Re: 3D Splines
Date: Thu, 4 Nov 1999 23:54:11 -0800
Organization: PlayStation Net Yaroze (SCEE)
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Hello again,

I've been having a quick poke through some of my old textbooks and I found a
partial answer to you question in a book by AK Dewdney called "The New
Turing Omnibus". It's packed full of pointless mathematical things like
logic, fractals and things. Anyway here's the important stuff...

A spline is a cubic equation which can be thought of as a point moving
through space for a time t (this is taken over the range 0..1 for
simplicity). The equations are:

    x(t) = ax + bx*t + cx*(t^2) + dx*(t^3)
    y(t) = ay + by*t + cy*(t^2) + dy*(t^3)
    z(t) = az + bz*t + cz*(t^2) + dz*(t^3)

I'll just cover how to find a pratial solution for the x(t) equation as it's
more or less the same for y(t) and z(t). The curve is defined by the
position of the two end points (at t=0 and t=1) and their respective
gradients. The equation of the gradients can easily be found:

    let x(t) now be:
        x(t) = a + b*t + c*(t^2) + d*(t^3)

    then dx(t) is:
        dx(t) = b + 2*c*t + 3*d*(t^2)

To get a solution, we can find the values of a, b, c and d by taking the
values of x(t) and dx(t) and t=0 and t=1:

    x(0) = a
    dx(0) = b
    x(1) = a + b + c + d
    dx(1) = b + 2*c + 3*d

Simple or what? Now according to Mr Dewdney the values of a, b, c and d are
as follows:

    a = x(0)
    b = dx(0)
    c = 3*(x(1) - x(0)) - 2*dx(0) - dx(1)
    d = 2*(x(0) - x(1)) + dx(0) + dx(1)

I don't know how he got here, probably with some insanely complex maths but
I trust his maths more than mine. Now then, you have the values of x(0) and
x(1) (they're the end point values) and so the shape of the curve is now
defined by the gradients at these endpoints. Here's where the trouble lies.
All I know about finding the gradients is that the gradient of xi(1) should
be the same as the gradient of xi+1(0). In other words, the 2 gradients that
each point on your track must be the same to give a smooth edge. I'll try
writing a PC program tomorrow to see how the gradient values affect the
shape of the curve. I've a sneaking suspicion that the gradient values may
be the coordinates of the other two control points, but I could be wrong.

Hope this helps (and that someone else knows more than I do on this subject)

Cheers,

    Tom