Numerical Examples for Modified Stereographic Conformal Projection #
SPHERE #
Forward Equations #
Given1
| Radius of sphere: | $R=\;\;$ unit |
| Order of equation: | $m=\;\;6$ |
| Center: | $\phi_1=\;64.0^\circ$ |
| $\lambda_0=\;-152.0^\circ$ | |
| Constants $A_1-A_m$ $B_1-B_m$ | See Table 33, using constants for sphere. |
| Point: | $\phi=\;$° |
| $\lambda=\;$° | |
Find $x, y, k$
Using equations (26-1) through (26-3) in order,
$$
\eqalign{
k' &= 2/\{1+\sin64^\circ\sin60^\circ+\cos64^\circ\cos60^\circ\cos[-150^\circ - (-152^\circ)]\}\cr
&= 1.0012864
}
$$
$$
\eqalign{
x' &= 1.0012864\cos60^\circ\sin[-150^\circ-(-152^\circ)] \cr
&= 0.0174722
}
$$
$$
\eqalign{
y' &= 1.0012864\{ \cos64^\circ\sin60^\circ - \sin64^\circ\cos60^\circ\cos[-150^\circ-(-152^\circ)] \} \cr
&= -0.0695721
}
$$
Using equations in (26-6), with $j = 2$, in order,$$
\eqalign{
r &= 2\times0.0174722 \cr
&= 0.0349444
}
$$
$$
\eqalign{
s' &= 0.0174722^2 + (-0.0695721)^2 \cr
&= 0.0051456
}
$$
$$
g_0 = 0
$$
$$
\eqalign{
a_1 &= A_6 + B_6\,i \cr
&= 0.3660976-0.2937382\,i
}
$$
$$
\eqalign{
b_1 &= A_5 + B_5\,i \cr
&= 0.0636871-0.1408027\,i
}
$$
$$
\eqalign{
c_1 &= 6.0\times(0.3660976-0.2937382\,i) \cr
&= 2.1965856-1.7624292\,i
}
$$
$$
\eqalign{
d_1 &= 5.0\times(0.0636871-0.1408027\,i) \cr
&= 0.3184355-0.7040135\,i
}
$$
$$
\eqalign{
a_2 =& b_1 + ra_1 \cr
=& (0.0636871-0.1408027\,i)+0.0349444\times(0.3660976-0.2937382\,i) \cr
=& 0.0764802-0.1510672\,i
}
$$
$$
\eqalign{
b_2 =& A_4 +iB_4 - s'a_1\cr
=& (-0.0153783-0.1968253\,i)-0.0051456\times(0.3660976-0.2937382\,i) \cr
=& -0.0172621-0.1953139\,i
}
$$
$$
\eqalign{
c_2 =& d_1 + rc_1 \cr
=& (0.3184355-0.7040135\,i)+0.0349444\times(2.1965856-1.7624292\,i) \cr
=& 0.3951938-0.7656005\,i
}
$$
$$
\eqalign{
d_2 =& 4\times(A_4+iB_4) - s'c_1 \cr
=& 4.0\times(-0.0153783-0.1968253\,i) - 0.0051456\times(2.1965856-1.7624292\,i) \cr
=& -0.0728158-0.7782325\,i
}
$$
Incrementing $j$ to 3,4, and 5 for the four variables $a_j,b_j,c_j, \text{ and } d_j$ in the same set of equations,$$
a_3 = b_2+ra_2 = -0.0145895-0.2005928\,i
$$
$$
b_3 = A_3 +iB_3 - s'a_2 = 0.0070671+0.0055898\,i
$$
$$
c_3 = d_2+rc_2 = -0.0590060-0.8049860\,i
$$
$$
d_3 = 3\times(A_3+iB_3) - s'c_2 = 0.0203483+0.0183769\,i
$$
$$
a_4 = b_3+ra_3 = 0.0065572-0.0014198\,i
$$
$$
b_4 = A_2 +iB_2 - s'a_3 = 0.0053264-0.0030853\,i
$$
$$
c_4 = d_3+rc_3 = 0.0182864-0.0097528\,i
$$
$$
d_4 = 2\times(A_2+iB_2) - s'c_3 = 0.0108062-0.0040929\,i
$$
$$
a_5 = b_4+ra_4 = 0.0055555-0.0031350\,i
$$
$$
b_5 = A_1 +iB_1 - s'a_4 = 0.9972186+0.0000073\,i
$$
$$
c_5 = d_4+rc_4 = 0.0114452-0.0044337\,i
$$
$$
d_5 = 1\times(A_1+iB_1) - s'c_4 = 0.9971582+0.0000502\,i
$$
Incrementing $j$ to 6 for $a_i \text{ and } b_j$ only,$$
a_6 = b_5+ra_5 = 0.9974127-0.0001022\,i
$$
$$
b_6 = g_0 - s'a_5 = -0.0000286+0.0000161\,i
$$
Using equations (26-7) through (26-9) in order, and with the relationship $i^2 = -1$,$$
\eqalign{
x+i\,y =& 1.0\times[(0.0174722-0.0695721\,i)(0.9974127-0.0001022\,i)+ \cr
& (-0.0000286+0.0000161\,i)] \cr
=& 0.0173913-0.0693777\,i
}
$$
$$
x = 0.01739129\text{ units}
$$
$$
y = -0.06937775 \text{ units}
$$
$$
\eqalign{
F_2+i\,F_1 =& (0.0174722-0.0695721\,i)(0.0114452-0.0044337\,i) + \cr
& (0.9971582+0.0000502\,i) \cr
=& 0.9970497-0.0008236\,i
}
$$
$$
\eqalign{
k &= [0.9970497^2 + (-0.0008236)^2]^{1/2}\times1.0012864 \cr
&= 0.9983327
}
$$
Inverse Equations #
Inversing forward example:
Given: $R,m, \phi_1, \lambda_0, A_1-A_6,\text{ and }B_1-B_6,$ for forward example
| Point: | $x=\;$ units |
| $y=\;$ units | |
Find $\phi, \lambda$
Using the Knuth algorithm equations (26-6) with (26 -10), (26-13), and (26-8), but not in that order, the first trial $ x' = 0.0173913/1.0 $ and $ y' = -0.0693777/1.0 $ . Except for the values of $x’$ and $y’$,equations (26-6) are used in the same manner as they were in the forward example, resulting in
$$
a_6 = 0.9974119-0.0001021\,i
$$
$$
b_6 = -0.0000284+0.0000161\,i
$$
$$
c_5 = 0.0114414-0.0044528\,i
$$
$$
d_5 = 0.9971586+0.0000493\,i
$$
Using equations (26-13), (26-8), and (26-10) in order,
$$
\eqalign {
f(x'+i\,y') =& (0.0173913-0.0693777\,i)(0.9974119-0.0001021\,i) +\cr
& (-0.0000284+0.0000161\,i) - (0.0173913-0.0693777\,i)/1.0 \cr
=& -0.0000805+0.0001938\,i
}
$$
$$
\eqalign{
F_2+i\,F_1 =& (0.0173913-0.0693777\,i)(0.0114414-0.0044528\,i) + \cr
& (0.9971586+0.0000493\,i) \cr
=& 0.9970487-0.0008219\,i
}
$$
$$
\eqalign{
\Delta(x'+i\,y') &= -(-0.0000805+0.0001938\,i)/(0.9970487-0.0008219\,i) \cr
&= 0.0000809-0.0001943\,i
}
$$
The division in equation (26-10) uses the relationship that
$$
(a + b\,i)/(c + d\,i) = (ac + bd)/(c^2 + d^2) + [(bc-ad)/(c^2 + d^2)]\,i
$$Adding $\Delta (x’ + iy’) \text{ to } (x’ + i y’)$,
$$
\eqalign{
(x'+i\,y') &= (0.0173913-0.0693777\,i)+(0.0000809-0.0001943\,i) \cr
&= 0.0174722-0.0695720\,i
}
$$
Repeating the above steps with the new values of $(x’, y’)$ until the value of $|\Delta(x’+i\,y’)|$ is 0 to eight decimal places.Finally,
$$
x' = 0.017472205
$$
$$
y' = -0.069572049
$$
Equations (26-14) through (26-17) may be used in order,
$$
\eqalign{
\rho &= [0.0174722^2 + (-0.0695720)^2]^{1/2} \cr
&= 0.0717325
}
$$
$$
\eqalign{
c &= 2\arctan(0.0717325/2) \cr
&= 4.1082071^\circ
}
$$
$$
\eqalign{
\phi =& \arcsin[\cos4.1082071^\circ\sin64^\circ +(-0.0695720)\times \cr
& \sin4.1082071^\circ\cos64^\circ/0.0717325] \cr
=& 60.0000026^\circ
}
$$
$$
\eqalign{
\lambda =& -152^\circ+\arctan[0.0174722\sin4.1082071^\circ/(0.0717325 \cr
& \cos64^\circ\cos4.1082071^\circ - (-0.0695720)\sin64^\circ\sin4.1082071^\circ)]\cr
=& -149.9999987^\circ
}
$$
Some of parameters for this projection are fixed ↩︎