Numerical Examples for Cylindrical Equal-Area Projection #
SPHERE #
Normal aspect #
Forward Equations #
Given
| Radius of sphere: | $R=\;\;$ unit |
| Central meridian: | $\lambda_0=\;$° |
| Standard parallel: | $\phi_s=\;$° |
| Point: | $\phi=\;$° |
| $\lambda=\;$° | |
Using equations (10-1) and (10-2),
$$
x = 1\times[80^\circ-(-75^\circ)]\times\cos 30^\circ = 2.3428242\;\text{units}
$$
$$
y = 1\times\sin 35^\circ/\cos 30^\circ = 0.662309\;\text{units}
$$
Inverse equations #
Given: $R, \lambda_0, \phi_s$ for forward example,
| $x=\;$ units |
| $y=\;$ units |
Using equations (10-6) and (10-7),
$$
\eqalign{
\phi &=\arcsin[(0.662309/1)\times\cos30^\circ] \cr
&= 35^\circ
}
$$
$$
\eqalign{
\lambda &= [2.3428242/(1\times\cos30^\circ)]\times 180^\circ/\pi + (-75^\circ) \cr
&= 80^\circ
}
$$
Transverse aspect #
Forward equations #
| Radius of sphere: | $R=\;\;$ unit |
| Origin: | $\phi_0=\;$° |
| $\lambda_0=\;$° | |
| Central scale factor: | $h_0=\;$° |
| Point: | $\phi=\;$° |
| $\lambda=\;$° | |
Using equations (10-3) and (8-3),
$$
\eqalign{
x &= (1/0.98)\times\cos 25^\circ\sin[(-90^\circ)-(-75^\circ)]\cr
&= -0.2393569\;\text{units}
}
$$
$$ \eqalign{
y &= 1\times0.98 \times \{ \arctan [\tan25^\circ/\cos((-90^\circ)-(-75^\circ))]-(-20^\circ) \} \times \pi/180^\circ \cr
&=1\times0.98 \times 45.7692621^\circ \times \pi/180^\circ = 0.7828478 \; \text{units}
}
$$
Inverse equations #
Given: $R, \phi_0, \lambda_0, h_0$, for forward example
| $x=\;$ units |
| $y=\;$ units |
Using equations (10-10), (10-8), and (10-9) in order,
$$
\eqalign{
D &= 0.7828478/(1.0\times0.98) + (-20^\circ)\times180^\circ/\pi \cr
&= 0.4497585
}
$$
$$
\eqalign{
\phi =&\arcsin\{[1-(0.98\times(-0.2393569)/1.0)^2]^{1/2}\cr
& \times\sin(0.4497585 \;\text{radians}) \} \cr
=& 25^\circ
}
$$
$$
\eqalign{
\lambda =& (-75^\circ)+\arctan \{[0.98\times(-0.2393569)/1.0]/ \cr
&[[1-(0.98\times(-0.2393569)/1.0)^2]^{1/2}\cos(0.4497585 \;\text{radians})] \} \cr
=& -90^\circ
}
$$
Oblique aspect #
Forward equations #
| Radius of sphere: | $R=\;\;$ | unit |
| Central scale factor: | $h_0=\;$° | |
| Central line through: | $\phi_1=\;$° | $\lambda_1=\;$° |
| $\phi_2=\;$° | $\lambda_2=\;$° | |
| Point: | $\phi=\;$° | |
| $\lambda=\;$° | ||
Using equation (9-1),
$$
\eqalign{
\lambda_p =& \arctan \{ [\cos 30^\circ \sin 60^\circ \cos (-75^\circ) - \sin 30^\circ \cos 60^\circ \cos (-50^\circ)]/ \cr
& [\sin 30^\circ \cos 60^\circ \sin (-50^\circ) - \cos 30^\circ \sin 60^\circ \sin (-75^\circ)] \} \cr
=& \arctan (0.0334174/0.5329333) \cr
=& 3.5880129^\circ
}
$$
From equation (9-6a)$$
\lambda_0 = 3.5880129^\circ + 90^\circ = 93.5880129^\circ
$$
From equation (9-2)$$
\eqalign{
\phi_p &= \arctan \{-\cos[3.5880129^\circ-(-75^\circ)]/\tan 30^\circ \} \cr
&= -18.9169858^\circ
}
$$
The other pole is then at $18.9169858^\circ$ and $-176.4119871^\circ$ From
equations (10-4) and (10-5), calculating the arctan in radiansnote 1:$$
\eqalign{
x =& 1.0\times0.98\arctan [\tan (-30^\circ) \cos (-18.9169858^\circ) \cr
& + \sin (-18.9169858^\circ) \sin(-100^\circ - 93.5880129^\circ)]/ \cr
& \cos(-100^\circ - 93.5880129^\circ) \cr
=& 1.0\times0.98\arctan [-0.6223338/(-0.9720102)] \cr
=& -2.5206570 \text{ units}
}
$$
$$
\eqalign{
y =& (1.0/0.98)[\sin(-18.9169858^\circ)\sin(-30^\circ) -\cr
& \cos(-18.9169858^\circ)\cos(-30^\circ)\sin(-100^\circ-93.5880129^\circ)] \cr
=& -0.0309947 \text{ units}
}
$$
To locate a pole given a central point using equations (9-7) and (9-8), refer to the numerical example given under the forward spherical equations for the Oblique Mercator projection.Inverse equations #
Given: $R, h_0$, and central line through same points as forward example,
| $x=\;$ units |
| $y=\;$ units |
First, $\phi_p$ and $\lambda_p$ are determined exactly as for the forward example, so that $\lambda_0=93.5880129^\circ$, and $\phi_p=-18.9169858^\circ$. Using equations (10-11) and (10-12),
$$
\eqalign{
yh_0/R &= -0.0309947\times0.98/1.0 \cr
&= -0.0303748
}
$$
$$
\eqalign{
x/(Rh_0) &= -2.520657/(0.98\times1.0) \cr
&= -2.572099
}
$$
$$
\eqalign{
\phi =&\arcsin \{ (-0.0303748)\times\sin(-18.9169858^\circ)+[1-(-0.0303748)^2]^{1/2} \cr
&\times\cos(-18.9169858^\circ)\times\sin(-2.572099\;\text{radians}) \}\cr
&\arcsin(-0.5) = -30.0000006^\circ
}
$$
$$
\eqalign{
\lambda =&93.5880129^\circ+\arctan\{[[1-(-0.0303748)^2]^{1/2}\cr
&\times\sin(-18.9169858^\circ) \times\sin(-2.572099\;\text{radians})\cr
&-(-0.0303748)\times\cos(-18.9169858^\circ)]/\cr
&[[1-(-0.0303748)^2]^{1/2}\times\cos(-2.572099\;\text{radians})]\} \cr
=& 260.0000005^\circ = -99.9999995^\circ
}
$$
ELLIPSOID #
Normal aspect #
Forward equations #
Given:
| ellipsoid | $a=6378206.4\,\text{m}$ |
| $e^2=0.00676866$ | |
| or | $e=0.0822719$ |
| Standard parallel: | $\phi_s=\;$° |
| Central meridian | $\lambda_0=\;$° |
| Point: | $\phi=\;$° | $\lambda=\;$° |
Using equations (10-13), (3-12), (10-14), and (10-15) in order,note 2
$$
\eqalign{
k_0 &= \cos5^\circ/[1-0.0067687\times\sin5^\circ]^{1/2} \cr
&= 0.9962203
}
$$
$$
\eqalign{
q =& (1-0.0067687)\times\{\sin10^\circ/(1-0.0067687\times\sin^210^\circ) \cr
& -[1/(2\times0.0822719)]\times\ln[(1-0.0822719\times\sin10^\circ)/ \cr
& (1+0.0822719)\times\sin10^\circ]\} \cr
=& 0.3449926
}
$$
$$
\eqalign{
x &= 6378206.4\times0.9962203\times[(-78^\circ)-(-75^\circ)]\times\pi/180^\circ \cr
&= -332699.83\,\text{m}
}
$$
$$
\eqalign{
y &= 6378206.4\times0.3449926/(2\times0.9962203) \cr
&= 1104391.16\,\text{m}
}
$$
Inverse equations #
Given: $a, e_2,\phi_s$, and $\lambda_0$ as in forward ellipsoid example.
| $x=\;$m |
| $y=\;$m |
After $k_0$ and $q_p$ are determined from (10-13) and (3-12) as in the forward normal and transverse examples
$$
k_0 = 0.9962203
$$
$$
q_p = 1.9954814
$$
then, from (10-26),$$
\eqalign{
\beta &= \arcsin[2\times1104391.16\times0.9962203/(6378206.4\times1.9954814)] \cr
&= 9.9557112^\circ
}
$$
Using equations (10-17) and (3-16), with subscript $c$ omitted, $\phi$ is found from $\beta$ by iteration as in the example for $\phi_c$ under the forward transverse ellipsoid example. Finally,$$
\phi = 10^\circ
$$
From (10-27),$$
\eqalign{
\lambda &= -75^\circ+[-332699.83/(6378206.4\times0.9962203)]\times180^\circ/\pi \cr
&= -78.0000000^\circ
}
$$
Transverse aspect #
Forward equations #
Given:
| ellipsoid | $a=6378206.4\,\text{m}$ |
| $e^2=0.00676866$ | |
| or | $e=0.0822719$ |
| Central meridian | $\lambda_0=\;$° |
| Latitude of origin: | $\phi_0=\;$° |
| Scale factor at $\lambda_0$: | $h_0=\;$ |
| Point: | $\phi=\;$° | $\lambda=\;$° |
Using equations (3-12) and (3-11),
$$
\eqalign{
q =& (1-0.0067687)\times\{\sin40^\circ/(1-0.0067687\times\sin^240^\circ) \cr
& -[1/(2\times0.0822719)]\times\ln[(1-0.0822719\times\sin40^\circ)/ \cr
& (1+0.0822719)\times\sin40^\circ]\} \cr
=& 1.2792602
}
$$
Inserting 90° in place of 40° in the same equation,$$
q_p = 1.9954814
$$
$$
\eqalign{
\beta &= \arcsin(1.2792602/1.9954814) \cr
&= 39.8722878^\circ
}
$$
Using equations (10-16) and (10-17),$$
\eqalign{
\beta_c &= \arctan[\tan 39.8722878^\circ/\cos((-83^\circ) - (-75^\circ))] \cr
&= 40.1482125^\circ
}
$$
$$
\eqalign{
q_c &= 1.9954814\times \sin40.1482125^\circ \cr
&= 1.2866207
}
$$
For the first trial $\phi_c$, in equation (3-16),
$$
\eqalign{
\phi_c &= \arcsin(1.2866207/2) \cr
&= 40.039109^\circ
}
$$
Substituting into equation (3-16),$$
\eqalign{
\phi_c =& 40.039109^\circ+[(1-0.0067687\sin^2 40.039109^\circ)^2/ \cr
& (2 \cos 40.039109^\circ)]\times\{1.2866207/(1-0.0067687) \cr
& -\sin 40.039109^\circ/(1-0.0067687\sin^2 40.039109^\circ) \cr
& +[1/(2\times0.0822719)]\ln[(1-0.0822719\sin 40.039109^\circ) \cr
& /(1+0.0822719\sin 40.039109^\circ)]\}\times 180^\circ/\pi \cr
=& 40.2757323^\circ
}
$$
Substituting $40.2757323^\circ$ in place of $40.039109^\circ$ in the same equation, the new trial $\phi_c$, is found to be $40.2761384^\circ$. The next iteration results in no change to seven
decimal places. Thus,$$
\phi_c = 40.2761384^\circ
$$
Using equation (10-18),$$
\eqalign{
x =& 6378206.4\times\cos39.8722878^\circ\times\cos40.2761384^\circ\times\sin[(-83^\circ)-(-75^\circ)] \cr
& /[0.99\times\cos 40.1482125^\circ\times(1-0.0067687\times\sin^2 40.2761384^\circ)^{1/2}] \cr
=& -687825.78\;\text{m}
}
$$
Using equation (3-21),$$
\eqalign{
M_c =& 6378206.4\times[(1-0.0067687-3\times0.0067687^2/64 \cr
& -5\times0.0067687^3/256)\times 40.2761384^\circ\times \pi/180^\circ \cr
& -(3\times0.0067687/8+3\times0.0067687^2/32 \cr
& +45\times0.0067687^3/1024)\times\sin(2\times40.2761384^\circ) \cr
& +(15\times0.0067687^2/256+45\times0.0067687^3/1024) \cr
& \times \sin(4\times40.2761384^\circ) -(35\times0.0067687^3/3072) \cr
& \times\sin(6\times40.2761384^\circ)] \cr
=& 4459980.03\;\text{m}
}
$$
Substituting $30^\circ$ in the same equation in place of $40.2761384^\circ$$$
M_0 = 3319933.29\;\text{m}
$$
Using equation (10-19),$$
\eqalign{
y &= 0.99\times(4459980.03 - 3319933.29) \cr
&= 1128646.27\;\text{m}
}
$$
Inverse equations #
Given: $a, e_2, \lambda_0, \phi_0, h_0$ as in forward ellipsoid example.
| $x=\;$m |
| $y=\;$m |
After $M_0$ is calculated from (3-21), using $30^\circ$ in place of $\phi_c$ as in the forward ellipsoid example,
$$
M_0 = 3319933.29\;\text{m}
$$
From (10-28),$$
\eqalign{
M_c &= M_0 + 1128646.27/0.99 \cr
&= 4459980.03\;\text{m}
}
$$
From (7-19), (3-24) and (3-26),$$
\eqalign{
\mu_c =& 4459980.0306802/[6378206.4\times(1-0.0067687/4 \cr
& -3\times0.0067687^2/64 - 5\times0.0067687^3/256)] \cr
=& 0.7004398\;\text{radians} = 40.1322432^\circ
}
$$
$$
\eqalign{
e_1 &= [1-(1-0.0067687)^{1/2}]/[1+(1-0.0067687)^{1/2}] \cr
&= 0.0016979
}
$$
$$
\eqalign{
\phi_c =& \mu_c + (3\times0.0016979/3 - 27\times0.0016979^3/32)\sin(2\times40.1322432^\circ) \cr
& + (21\times0.0016979^2/16 - 55\times0.0016979^4/32)\sin(4\times40.1322432^\circ) \cr
& + (151\times0.0016979^3/96)\sin(6\times40.1322432^\circ) \cr
& + (1097\times0.0016979^4/512)\sin(8\times40.1322432^\circ) \cr
=& 0.7029512\;\text{radians} = 40.2761385^\circ
}
$$
Using (3-12) and (3-11), with $q_p$ calculated as in the forward example,$$
\eqalign{
q_c =& (1-0.0067687)\times\{\sin40.2761385^\circ/(1-0.0067687\times\sin^240.2761385^\circ) \cr
& -[1/(2\times0.0822719)]\times\ln[(1-0.0822719\times\sin40.2761385^\circ)/ \cr
& (1+0.0822719)\times\sin40.2761385^\circ]\} \cr
=& 1.2866207
}
$$
$$
\eqalign{
\beta_c &= \arcsin(1.2866207/1.9954814) \cr
&= 40.1482128^\circ
}
$$
From equations (10-29), (10-29) and (10-31),$$
\eqalign{
\beta' =& -\arcsin[0.99\times(-687825.78)\times\cos40.1482128^\circ \cr
& \times(1-0.0067687\times\sin^240.2761385^\circ)^{1/2}/ \cr
& (6378206.4\times\cos40.2761385^\circ)] \cr
=& 6.131569^\circ
}
$$
$$
\eqalign{
\beta &= \arcsin(\cos 6.131569^\circ/\sin40.1482128^\circ) \cr
&= 39.8722881^\circ
}
$$
$$
\eqalign{
\lambda &= (-75^\circ)-\arctan(\tan 6.131569^\circ/\cos40.1482128^\circ) \cr
&= -83^\circ
}
$$
Using (10-17) and (3-16), with subscript c omitted, $\phi$ is found from $\beta$ by iteration
as in the example for $\phi_c$ under the forward transverse ellipsoid example. Finally,$$
\phi = 40.0000005^\circ
$$
Oblique aspect #
Forward equations #
Given:
| ellipsoid | $a=6378206.4\,\text{m}$ | |
| $e^2=0.00676866$ | ||
| or | $e=0.0822719$ | |
| Central scale factor: | $h_0=\;$ | |
| Central line through: | $\phi_1=\;$° | $\lambda_1=\;$° |
| $\phi_2=\;$° | $\lambda_2=\;$° | |
| Point: | $\phi=\;$° | $\lambda=\;$° |
To find the position of the pole, equations (3-12) and (3-11) are used as in the examples for the normal and transverse aspects just above, to determine $\beta_1$ from $\phi_1$ and $\beta_2$ from $\phi_2$. The results are
$$
\beta_1 = 29.8877622^\circ
$$
$$
\beta_2 = 39.8722878^\circ
$$
Inserting these values in place of $\phi_1$, and $\phi_2$ in equations (9-1) and (9-2), listed under spherical formulas for the projectionnote 3$$
\eqalign{
\lambda_p =& \arctan \{ [\cos 29.8877622^\circ \sin 39.8722878^\circ \cos (-75^\circ) \cr
&-\sin 29.8877622^\circ \cos 39.8722878^\circ \cos (-80^\circ)]/ \cr
& [\sin 29.8877622^\circ \cos 39.8722878^\circ \sin (-80^\circ) \cr
&-\cos 29.8877622^\circ \sin 39.8722878^\circ \sin (-75^\circ)] \} \cr
=& \arctan (0.0774469/0.1602532) \cr
=& 25.7934757^\circ
}
$$
$$
\eqalign{
\beta_p &= \arctan \{-\cos[25.7934757^\circ-(-75^\circ)]/\tan 29.8877622^\circ \} \cr
&= 18.0472981^\circ
}
$$
Using equations (10-17) and (3-16), with subscript p instead of c, $\phi_p$ is found by iteration as in the example for $\phi_c$, under the transverse aspect. Finally,$$
\phi_p = 18.1238834^\circ
$$
Using equations (10-20) and (10-21), and table 13 for the Clarke 1866 ellipsoidnote 4, the specific Fourier coefficients are calculated:$$
\eqalign{
B =& 0.9991507116+(-0.0008471546)\cos(2\times 18.1238836^\circ) \cr
&+0.0000021283\cos(4\times18.1238836^\circ) \cr
&+(-0.0000000054)\cos(6\times18.1238836^\circ) \cr
=& 0.9984682
}
$$
$$
\eqalign{
A_2 =& (-0.0001412092)+(-0.0001411259)\cos(2\times 18.1238836^\circ) \cr
&+0.0000000839\cos(4\times18.1238836^\circ) \cr
&+0.0000000006\cos(6\times18.1238836^\circ) \cr
=& -0.0002550
}
$$
$$
\eqalign{
A_4 =& (-0.0000000435)+(-0.0000000579)\cos(2\times 18.1238836^\circ) \cr
&+(-0.0000000144)\cos(4\times18.1238836^\circ) \cr
&+0.0000000000\cos(6\times18.1238836^\circ) \cr
=& -0.0000001
}
$$
Equations (3-12) and (3-11) are again used to determine $\beta$ from $\phi$, giving$$
\beta = 41.8710107^\circ
$$
From equation (10-22),$$
\eqalign{
\lambda' =& \arctan \{[\cos18.0472983^\circ\sin41.8710107^\circ-\sin18.0472983^\circ \cr
& \times\cos41.8710107^\circ\cos(-77^\circ-25.7934758^\circ)] \cr
& /\cos41.8710107^\circ\sin(-77^\circ-25.7934758^\circ)\} \cr
=& \arctan[0.6857020/(-0.7261631)] \cr
=& 136.6415266^\circ
}
$$
Using equations (10-23) through (10-25), using $q_p$ as computed above for the
transverse aspect,$$
\eqalign{
x =& 6378206.40\times1.0\times[0.9984682\times136.6415266^\circ\times\pi/180^\circ \cr
& + (-0.0002550)\sin(2\times136.6415266^\circ)+ (-0.0000001)\sin(4\times136.6415266^\circ)] \cr
=& 15189353.49\;\text{m}
}
$$
$$
\eqalign{
F =& 0.9984682+2\times(-0.0002550)\cos(2\times136.6415266^\circ) \cr
& + 4\times(-0.0000001)\cos(4\times136.6415266^\circ) \cr
=& 0.9984393
}
$$
$$
\eqalign{
y =& (6378206.4\times1.9954814/2)\times[\sin18.0472983^\circ \cr
& \times\sin41.8710107^\circ+\cos18.0472983^\circ\cos41.8710107^\circ \cr
& \times\cos(-77^\circ-25.7934758^\circ)]/(1\times 0.9984393) \cr
=& 318677.45 \;\text{m}
}
$$
Inverse equations #
Given: $a, e^2, h_0$, calculated pole location $(\phi_p, \lambda_p)$, calculated Fourier coefficients $B, A_2$, and $A_4$ as in the forward oblique ellipsoid example, and $R_q$ as calculated for the forward normal ellipsoid example,
| $x=\;$m |
| $y=\;$m |
First $q_p=1.9954814$ as found from (3-12) in the forward transverse example.
To solve for $\lambda’$ from (10-32), the first trial $\lambda’$ is found as described:
$$
\eqalign{
\lambda' &= [15189353.49/(6378206.4\times1.0\times0.9984682)]\times180^\circ/\pi \cr
&= 136.6561359^\circ
}
$$
Using equation (10-32),$$
\eqalign{
\lambda' =& [15189353.49/(6378206.4\times1.0)\times180^\circ/\pi \cr
& -(-0.0002550)\times\sin(2\times136.6561359^\circ) \cr
& -(-0.0000001)\times\sin(4\times136.6561359^\circ)]/0.9984682 \cr
=& 136.6415270^\circ
}
$$
Substituting $136.6415270^\circ$in place of $136.6561359^\circ$ in this equation, $\lambda’$ is calculated to be $136.6415266^\circ$. The next iteration yields no change to seven decimal places, so that
$$
\lambda' = 136.6415266^\circ
$$
Equation (10-24) is used to calculate $F$ just as it was in the forward oblique example, so $F$ is again$$
F = 0.9984393
$$
From equations (10-33) through (10-35),
$$
\eqalign{
\beta' =& \arcsin[2\times0.9984393\times1\times318677.45/ \cr
& (6378206.4\times1.9954814)] \cr
=& 2.8658964^\circ
}
$$
$$
\eqalign{
\beta =& \arcsin(\sin18.0472983^\circ\sin2.8658964^\circ \cr
& + \cos18.0472983^\circ\sin2.8658964^\circ\sin136.6415266^\circ) \cr
=& 41.8710107^\circ
}
$$
$$
\eqalign{
\lambda =& 25.7934758^\circ + \arctan[\cos2.8658964^\circ\cos136.6415266^\circ/ \cr
& \cos18.0472983^\circ\sin2.8658964^\circ \cr
& - \sin18.0472983^\circ\cos2.8658964^\circ\sin136.6415266^\circ] \cr
=& 25.7934758^\circ + \arctan[(-0.7261631)/(-0.1648933)] \cr
=& -77.0000001^\circ
}
$$
Using (10-17) and (3-16), $\phi$ is found from $\beta$ as previously, dropping subscript c and with iteration, to produce
$$
\phi=42.0000000^\circ
$$
The computation of Fourier coefficients is not shown here, since it is lengthy and is not needed unless a different ellipsoid is desirednote 4. An example of computation of Fourier coefficients is given under the Space Oblique Mercator projection.