Jonathan Lansey
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
Matlab has been very unfriendly with me so I haven't been able to draw the
graphs efficiently. My older brother Eli has been gracious enough to
graph it for me in a program clearly superior to Matlab:
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.
If you wish to defend Matlab, write me an easy script, and I will formally apologize.
y=x^2/sin(x) Click for larger picture.
