Numerical Examples for Orthographic Projection #
SPHERE #
Forward Equations #
Given
| Radius of sphere: | $R=\;\;$ units |
| Center: | $\phi_1=\;$ ° |
| $\lambda_0=\;$ ° | |
| Point: | $\phi=\;$ ° |
| $\lambda=\;$ ° | |
Find $x, y$
In general calculations, to determine whether this point is beyond viewing, using equation (5-3),
$$
\eqalign{
\cos c &= \sin 40^\circ\sin30^\circ+\cos 40^\circ\cos 30^\circ\cos(-110^\circ-(-100^\circ)) \cr
&= 0.9747290
}
$$
Since this is positive, the point is within view.
Using equations (20-3) and (20-4),
$$
\eqalign{
x &= 1.0\cos30^\circ\sin(-110^\circ-(-100^\circ)) \cr
&= -0.1503837
}
$$
$$
\eqalign{
y &= 1.0[\cos40^\circ\sin30^\circ - \sin40^\circ\cos30^\circ\cos(-110^\circ-(-100^\circ))] \cr
&= -0.1651911
}
$$
Examples of other forward equations are omitted, since the formulas for the oblique aspect apply generally.Inverse Equations #
Inversing forward example: Given $R, \phi_1, \lambda_0$ for forward example
| Point: | $x=\;$ units |
| $y=\;$ units | |
Using equations (20-18) and (20-19),
$$
\eqalign{
\rho &= [(-0.1503837)^2 + (-0.1651911)]^{1/2} \cr
&= 0.2233906
}
$$
$$
\eqalign{
c &= \arcsin(0.2233906/1.0) \cr
&= 12.9082572^\circ
}
$$
Using equations (20-14) and (20-15),$$
\eqalign{
\phi =& \arcsin[\cos12.9082572^\circ\sin40^\circ + (-0.1651911)\sin12.9082572^\circ \cr
& \cos40^\circ/0.2233906] \cr
=& 30.0000004^\circ
}
$$
$$
\eqalign{
\lambda =& -100^\circ + \arctan[-0.1503837\sin12.9082572^\circ/(0.2233906 \cr
&\cos40^\circ\cos12.9082572^\circ - (-0.1651911)\sin40^\circ \sin12.9082572^\circ)] \cr
=& -109.9999978^\circ
}
$$