Numerical Examples for Bonne Projection #
SPHERE #
Forward Equations #
Given
| Radius of sphere: | $R=\;\;$ units |
| Standard parallel: | $\phi_1=\;$ ° |
| Central meridian: | $\lambda_0=\;$ ° |
| Point: | $\phi=\;$ ° |
| $\lambda=\;$ ° | |
Find $x, y$
Using equations (19-1) through (19-4) in order,
$$
\eqalign{
\rho &= 1.0\times[\cot 40^\circ +(40^\circ - 30^\circ)\times \pi/180^\circ] \cr
&= 1.3662865\;\text{units}
}
$$
$$
\eqalign{
E &= 1.0\times[-85^\circ-(-75^\circ)]\cos30^\circ/1.3662865 \cr
&= -6.3385344^\circ
}
$$
$$
\eqalign{
x &= 1.3662865\sin (-6.3385344^\circ) \cr
&= -0.1508418\;\text{units}
}
$$
$$
\eqalign{
y &= 1.0\cot 40^\circ - 1.3662865\cos(-6.3385344^\circ) \cr
&= -0.1661807\;\text{units}
}
$$
Inverse Equations #
Inversing forward example: Given $R, \phi_1, \lambda_0$ for forward example
| Point: | $x=\;$ units |
| $y=\;$ units | |
Find $\phi, \lambda$
Using equations (19-5) through (19-7) in order
$$
\eqalign{
\rho &= [(-0.1508418)^2 + (1\cot40^\circ-(-0.1661807))^2]^{1/2} \cr
&= 1.3662865\;\text{units}
}
$$
$$
\eqalign{
\phi &= (\cot40^\circ)\times 180^\circ/\pi + 40^\circ -(1.3662865/1.0)\times 180^\circ/\pi \cr
&= 30.0000012^\circ
}
$$
$$
\eqalign{
\lambda =& -75^\circ + 1.3662865\times\{\arctan[-0.1508418/(1.0\cot40^\circ) \cr
& - (-0.1661807)]\}/(1.0\cos30.0000012^\circ) \cr
=& -84.9999985^\circ
}
$$
ELLIPSOID #
Forward Equations #
Given:
| ellipsoid | $a=$ | 6378206.4 m |
| $e^2=$ | 0.00676866 | |
| Standard parallel: | $\phi_1=$ | ° |
| Central meridian: | $\lambda_0=$ | ° |
| Point: | $\phi=$ | ° | $\lambda=$ | ° |
Using equations (14-15) and (3-21),
$$
\eqalign{
m &= \cos 30^\circ/(1-0.0067687\sin^230^\circ) \cr
&= 0.8667591
}
$$
$$
\eqalign{
M =&6378206.4[(1-0.0067687/4-3\times 0.0067687^2/64 - 5\times 0.0067687^3/256)\times 30^\circ\times\pi/180^\circ \cr
&-(3\times 0.0067687/8+3\times 0.0067687^2/32 +45\times0.0067687^3/1024)\sin(2\times30^\circ) \cr
&+(15\times 0.0067687^2/256 +45\times 0.0067687^3/1024)\sin(4\times 30^\circ) \cr
&-(35\times 0.0067687^3/3072)\sin(6\times30^\circ)] \cr
=& 3319933.29\;\text{m}
}
$$
Using the same equations, but with
$ \phi_1=40^\circ $
in place of
$ 30^\circ $
,$$
m_1 = 0.7671179
$$
$$
M_1 = 4429318.90\;\text{m}
$$
Using equations (19-8) through (19-11) in order,$$
\eqalign{
\rho &= 6378206.4\times0.7671179/\sin40^\circ+4429318.90-3319933.29 \cr
&= 8721287.35\;\text{m}
}
$$
$$
\eqalign{
E &= 6378206.40\times0.8667591\times[-75^\circ - (-85^\circ)]/8721287.35 \cr
&= -6.3389360^\circ
}
$$
$$
\eqalign{
x &= 8721287.35\sin(-6.3389360^\circ) \cr
&= -962915.09\;\text{m}
}
$$
$$
\eqalign{
y &= 6378206.40\times0.7671179\sin40^\circ-8721287.35\cos(-6.3389360^\circ) \cr
&= -1056065.01\;\text{m}
}
$$
Inverse Equations #
Inversing forward example:
Given
| $x=\;$m |
| $y=\;$m |
Using equations (14-15) and (3-21), $m_1$, and $M_1$ are calculated as in the forward example:
$$
m_1 = 0.7671179
$$
$$
M_1 = 4429318.90\;\text{m}
$$
Using equations (19-12), (19-13), (7-19), (3-24), and (3-26) in order,$$
\eqalign{
\rho &= [(-962915.09)^2+(6378206.4\times0.7671179/\sin40^\circ - (-1056065.01))^2]^{1/2} \cr
&= 8721287.36\;\text{m}
}
$$
$$
\eqalign{
M &= 6378206.4\times0.7671179/\sin40^\circ + 4429318.90 - 8721287.36 \cr
&= 3319933.29\;\text{m}
}
$$
$$
\eqalign{
\mu =& \{3319933.29/[6378206.4\times(1-0.0067687/4 \cr
& -3\times0.0067687^2/64 - 5\times0.0067687^3/256)]\} \times 180^\circ/\pi \cr
=& 29.8737593^\circ
}
$$
$$
\eqalign{
e_1 &= [1-(1-0.0067687)^{1/2}]/[1+(1-0.0067687)^{1/2}] \cr
&= 0.001697916
}
$$
$$
\eqalign{
\phi =& 29.8737593^\circ +[(3\times0.0016979/2 - 27\times 0.0016979^3/32)\sin(2\times29.8737593^\circ) \cr
& + (21\times 0.0016979^2/16-55\times0.0016979^4/32)\sin(4\times29.8737593^\circ) \cr
& + (151\times0.0016979^3/96)\sin(6\times29.8737593^\circ) \cr
& + (1097\times0.0016979^4/512)\sin(8\times29.8737593^\circ)]\times180^\circ/\pi \cr
=& 30.0000000^\circ
}
$$
Using equation (14-15),$$
\eqalign{
m &= \cos30^\circ/(1-0.0067687\sin^2 30^\circ)^{1/2} \cr
&= 0.8667591
}
$$
Using equation (19-14),$$
\eqalign{
\lambda =& -75^\circ + \{ 8721287.356278036\times \arctan[-962915.09/ \cr
& (6378206.4\times0.7671179/\sin40^\circ-(-1056065.01))]/ \cr
& (6378206.4\times 0.8667591)\}\times 180^\circ/\pi \cr
=& -85.0000000^\circ
}
$$