Lansey's Alternate Resume :: Equations for Graphing in Point Perspective

Jonathan Lansey's Alternate Resume

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Graphing things in Point Perspective:

This was a quick experiment in point projection done before I drew Pisa, 2003. It was based off an old 2D version I graphed (2001) when I first learnt about functions (at right, Equations for 2D graph). Thank you Eli Lansey for Graphing some of these for me.

Imagine that the bridge is like a long door on hinges. The door post is “H” units to the left of the viewer and “L” units in front of him/her. “A” is how far (rad or degrees) the door has spun around. “t” is the length along the door.
The scene is projected onto a plane that is 1 unit away from the viewer (to make things easy and scale is only effected anyway).

I used 6 for H and L. a good value for “A” is anywhere between 90 and 20 degrees.

The general equation can be found quickly using similar triangles. it is:
x=(right-left distance)/(forward distance)=(-H+t*cos(A))/(L+t*sin(A))
y=(vertical height)/(forward distance)=f(t)/(L+t*sin(A))
in this case, f(t)={a line, or a circle}

A line in the above graph:
X1=(-H+t*cos(A))/(L+t*sin(A))
Y1=3/(L+t*sin(A))

An arch in the above graph:
X3=(-H+t*cos(A))/(L+t*sin(A))
Y3=(sqrt(3^2-(t-3)^2)-1)/(L+t*sin(A))

All the 3D Equations

My older brother Eli has been gracious enough to graph it for me in Mathematica (from the generous host of wolfram.integrator.com). If you are lucky enough to have it then try out Eli's script. Play with the variables.

Note: this was before I knew how to use Matlab.