Cylindrical Equal-Area Projection

# 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: $\phi=\;$° $\lambda=\;$°
Find $x, y$

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
Find $\phi, \lambda$.

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: $\phi=\;$° $\lambda=\;$°
Find $x, y$.

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
Find $\phi, \lambda$

Using equations (10-10), (10-8), and (10-9)1 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: $\phi=\;$° $\lambda=\;$°
Find $x, y$.

Using equation (9-1)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)1 and (10-5), calculating the arctan in radiansnote 1:
\eqalign{ x =& 1\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\times0.98\arctan [(-0.6223338)/(-0.9720102)] \cr =& -2.520657 \;\text{units} }
\eqalign{ y =& (1/0.98)[\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
Find $\phi, \lambda$

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)1,

\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 #

Given:
 Clarke 1866 WGS-84 ellipsoid $a=6378206.4\,\text{m}$ $e^2=0.00676866$ or $e=0.0822719$ Standard parallel: $\phi_s=\;$° Central meridian $\lambda_0=\;$°

#### Forward equations #

 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
Find $\phi, \lambda$. 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,
$$\phi = 10^\circ$$

### Transverse aspect #

Given:
 Clarke 1866 WGS-84 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 $\lambda_0$: $h_0=\;$

#### Forward equations #

 Point: $\phi=$° $\lambda=$°
Find $x, y$.

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)1 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} }

Given: $a, e_2, \lambda_0, \phi_0$, \h_0$as in forward ellipsoid example. $x=\;$m$y=\;$m Find$\phi, \lambda$. After$M_0$is calculated from (3-21), using$30^\circ$in place of$\phi_cas 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), withq_pcalculated 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)1 and (10-31)1, \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 # Given:  Clarke 1866 WGS-84 ellipsoid$a=6378206.4\,\text{m}e^2=0.00676866$or$e=0.0822719$Central scale factor:$h_0=\;$Central line through:$\phi_1=\;$°$\lambda_1=\;$°$\phi_2=\;$°$\lambda_2=\;$° #### Forward equations #  Point:$\phi=$°$\lambda=$° Find$x, y$. 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$\phi_2in equations (9-1)1 and (9-2), listed under spherical formulas for the projectionnote 3 \eqalign{ \lambda_p =& \arctan \{ [\cos 29.8877622^\circ \sin 39.8722878^\circ \cos (-75^\circ) \cr &-\sin 29.8877622^\circ \cos 39.8722878^\circ \cos (-80^\circ)]/ \cr & [\sin 29.8877622^\circ \cos 39.8722878^\circ \sin (-80^\circ) \cr &-\cos 29.8877622^\circ \sin 39.8722878^\circ \sin (-75^\circ)] \} \cr =& \arctan (0.0774469/0.1602532) \cr =& 25.7934757^\circ } \eqalign{ \beta_p &= \arctan \{-\cos[25.7934757^\circ-(-75^\circ)]/\tan 29.8877622^\circ \} \cr &= 18.0472981^\circ } Using equations (10-17) and (3-16), with subscript p instead of c,\phi_p$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)1, \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), usingq_pas 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 Find$\phi, \lambda$. 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$\lambda’is found as described: \eqalign{ \lambda' &= [15189353.49/(6378206.4\times1.0\times0.9984682)]\times180^\circ/\pi \cr &= 136.6561359^\circ } Using equation (10-32), \eqalign{ \lambda' =& [15189353.49/(6378206.4\times1.0)\times180^\circ/\pi \cr & -(-0.0002550)\times\sin(2\times136.6561359^\circ) \cr & -(-0.0000001)\times\sin(4\times136.6561359^\circ)]/0.9984682 \cr =& 136.6415270^\circ } Substituting136.6415270^\circ$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$Fis again $$F = 0.9984393$$ From equations (10-33) through (10-35)1, \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.

1. This computation uses the ATAN2 function. ↩︎ ↩︎ ↩︎ ↩︎ ↩︎ ↩︎ ↩︎ ↩︎ ↩︎ ↩︎