## Graphing in point perspective:

This
was a quick experiment in point projection done before I drew
the leaning tower of Pisa, 2003. It was based off an old
2D version I graphed (2001) when I first learnt about mathematical functions (below,
Equations for 2D graph)

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))

More projects →