TWO METHODS FOR SOLVING THE DIRECT RAY INTERSECTION ON A SPHERE

B. F. Krutoi

(Presented by the Department of Engineering and Computational Mathematics)

By a direct ray intersection on a sphere we shall mean the following problem: given the latitudes \(\varphi_1, \varphi_2\) and longitudes \(\lambda_1, \lambda_2\) of the initial points 1, 2 on the sphere, and also given the azimuths \(\alpha_{1.3}, \alpha_{2.3}\) of the intersecting rays \(\overset{\uparrow}{\odot}L_{1.3}, \overset{\uparrow}{\odot}L_{2.3}\) from points 1, 2 to the point 3 being determined; find the latitude \(\varphi_3\) and longitude \(\lambda_3\) of point 3, and also find the intersecting sides \(\sigma_{1.3}, \sigma_{2.3}\) (Fig. 1).

For the solution of the indicated problem two methods may be proposed.

a) From the initial data \(\varphi_1, \lambda_1, \alpha_{1.3}\) and \(\varphi_2, \lambda_2, \alpha_{2.3}\), only the coordinates \(\varphi_3, \lambda_3\) of the point 3 being determined are computed directly; the intersecting sides \(\sigma_{1.3}, \sigma_{2.3}\) are found from the initial data and the obtained values \(\varphi_3, \lambda_3\).

b) From the indicated initial data, the intersecting sides \(\sigma_{1.3}, \sigma_{2.3}\) are first computed, and only then the coordinates \(\varphi_3, \lambda_3\) of point 3.

Below, the solution of the direct ray intersection on a sphere by both methods is given. Let us note that the final expressions derived here were used in [1] in obtaining approximate values for the geodetic latitude \(B_3\) of the point 3 being determined and the distances \(s_{1.3}, s_{2.3}\) in a direct ray intersection specified on the terrestrial spheroid.

Fig. 1.

1. First method

In Fig. 1 the points 1, 2 are the initial points, and point 3 is the point being determined. The point \(P\) is the pole for the spherical coordinates \(\varphi, \lambda\); \(PQ\) is the initial meridian in the reckoning of longitudes \(\lambda\). In addition, the following notations have been introduced:

\[ 1)\quad \Theta_i = 90^\circ - \varphi_i; \qquad 2)\quad \Delta\lambda_{ij} = \lambda_j - \lambda_i. \tag{1} \]

Our problem will be solved with a check if, for computing the two unknown quantities \(\varphi_3, \lambda_3\), we take the following 3 initial expressions arising from consideration of Fig. 1:

\[ \begin{aligned} 1)\quad & \operatorname{ctg}\Theta_3 \sin\Theta_1 - \operatorname{ctg}\alpha_{1.3}\sin\Delta\lambda_{1.3} = \cos\Theta_1\cos\Delta\lambda_{1.3},\\ 2)\quad & \operatorname{ctg}\Theta_3 \sin\Theta_2 - \operatorname{ctg}\alpha_{2.3}\sin\Delta\lambda_{2.3} = \cos\Theta_2\cos\Delta\lambda_{2.3},\\ 3)\quad & \Delta\lambda_{1.2}=\Delta\lambda_{1.3}-\Delta\lambda_{2.3}. \end{aligned} \tag{2} \]

Hence we find a twofold representation for \(\operatorname{ctg}\Theta_3\):

\[ \operatorname{ctg}\Theta_3 = \frac{\operatorname{ctg}\alpha_{1.3}\sin\Delta\lambda_{1.3} +\cos\Theta_1\cos\Delta\lambda_{1.3}}{\sin\Theta_1} = \frac{\operatorname{ctg}\alpha_{2.3}\sin\Delta\lambda_{2.3} +\cos\Theta_2\cos\Delta\lambda_{2.3}}{\sin\Theta_2}. \tag{3} \]

Relation (3), however, cannot be used at once to determine \(\Theta_3\), since it expresses \(\Theta_3\) either through the unknown \(\Delta\lambda_{1.3}\), or through the unknown \(\Delta\lambda_{2.3}\). Therefore, from the last link in the chain of equalities (3), we first eliminate \(\Delta\lambda_{2.3}\), using relation (2.3). We shall have successively:

\[ \frac{\operatorname{ctg}\alpha_{1.3}\sin\Delta\lambda_{1.3}}{\sin\Theta_1} +\operatorname{ctg}\Theta_1\cos\Delta\lambda_{1.3} = \frac{\operatorname{ctg}\alpha_{2.3}}{\sin\Theta_2} \left(\sin\Delta\lambda_{1.3}\cos\Delta\lambda_{1.2} -\right. \]

\[ \left. -\sin\Delta\lambda_{1.2}\cos\Delta\lambda_{1.3} \right) +\operatorname{ctg}\Theta_2 \left(\cos\Delta\lambda_{1.3}\cos\Delta\lambda_{1.2} +\sin\Delta\lambda_{1.3}\sin\Delta\lambda_{1.2}\right) = \]

\[ = \left( \frac{\operatorname{ctg}\alpha_{1.3}}{\sin\Theta_1} - \frac{\operatorname{ctg}\alpha_{2.3}}{\sin\Theta_2}\cos\Delta\lambda_{1.2} - \operatorname{ctg}\Theta_2\sin\Delta\lambda_{1.2} \right)\sin\Delta\lambda_{1.3} = \]

\[ = \left( -\frac{\operatorname{ctg}\alpha_{2.3}}{\sin\Theta_2}\sin\Delta\lambda_{1.2} +\operatorname{ctg}\Theta_2\cos\Delta\lambda_{1.2} -\operatorname{ctg}\Theta_1 \right)\cos\Delta\lambda_{1.3}. \]

Hence we obtain the following expression for the direct computation of the unknown \(\Delta\lambda_{1.3}\) through the initial data:

\[ \operatorname{tg}\Delta\lambda_{1.3} = \frac{ \cos\Theta_2\cos\Delta\lambda_{1.2} -\operatorname{ctg}\alpha_{2.3}\sin\Delta\lambda_{1.2} -\operatorname{ctg}\Theta_1\sin\Theta_2 }{ \operatorname{ctg}\alpha_{1.3}\dfrac{\sin\Theta_2}{\sin\Theta_1} -\operatorname{ctg}\alpha_{2.3}\cos\Delta\lambda_{1.2} -\cos\Theta_2\sin\Delta\lambda_{1.2} }. \tag{4} \]

Having determined the unknown \(\Delta\lambda_{1.3}\) according to (4), we then find from (2.3) the second unknown \(\Delta\lambda_{2.3}\):

\[ \Delta\lambda_{2.3}=\Delta\lambda_{1.3}-\Delta\lambda_{1.2}, \tag{2.3;5} \]

after which we twice compute the last unknown \(\Theta_3\) from relation (3). The desired \(\varphi_3,\lambda_3\) are obtained from relations (1)—also twice:

\[ 1)\quad \varphi_3=90^\circ-\Theta_3^{1}=90^\circ-\Theta_3^{2}; \qquad 2)\quad \lambda_3=\lambda_1+\Delta\lambda_{1.3} =\lambda_2+\Delta\lambda_{2.3}, \tag{6} \]

where \(\Theta_3^{1}, \Theta_3^{2}\) are the values of \(\Theta_3\) found according to (3) from the solution of triangles \(1P3\) and \(2P3\).

In conclusion, from the same triangles \(1P3\) and \(2P3\) we compute the intersecting sides \(\sigma_{1.3}\) and \(\sigma_{2.3}\):

\[ 1)\quad \sin\sigma_{1.3} = \frac{\sin\Theta_3}{\sin\Delta\lambda_{1.3}}\sin\alpha_{1.3}; \qquad 2)\quad \sin\sigma_{2.3} = \frac{\sin\Theta_3}{\sin\Delta\lambda_{2.3}}\sin\alpha_{2.3}. \tag{7} \]

2. Second method

Knowing the latitudes \(\varphi_1,\varphi_2\) and longitudes \(\lambda_1,\lambda_2\) of the control points 1, 2, we first solve triangle \(1P2\), in which we find the angles \(\beta_1,\beta_2\) at points 1, 2 and the side \(\sigma_{1.2}\). In order to shorten the notation, in addition to the previously used designations (1), we introduce the following additional designations:

\[ 1)\quad \frac{1}{2}(\Theta_1+\Theta_2)=\Theta_{1.2}; \qquad 2)\quad \frac{1}{2}(\Theta_1-\Theta_2)=\delta\Theta_{1.2}; \tag{8} \]

\[ \begin{gathered} 3)\ \frac{1}{2}\Delta\lambda_{1.2}=\delta\lambda_{1.2};\qquad 4)\ \frac{1}{2}(\beta_1+\beta_2)=\beta_{1.2};\qquad 5)\ \frac{1}{2}(\beta_1-\beta_2)=\delta\beta_{1.2}. \end{gathered} \tag{8} \]

Then the sought angles \(\beta_1,\beta_2\) and the side \(\sigma_{1.2}\) in triangle \(1P2\) can be obtained from the chain of relations:

\[ \begin{gathered} 1)\ \frac{\sin\delta\Theta_{1.2}\operatorname{ctg}\delta\lambda_{1.2}}{\sin\Theta_{1.2}} =\operatorname{tg}\delta\beta_{1.2};\qquad 2)\ \frac{\cos\Theta_{1.2}\operatorname{ctg}\delta\lambda_{1.2}}{\cos\Theta_{1.2}} =\operatorname{tg}\beta_{1.2};\\[4pt] 3)\ \beta_{1.2}+\delta\beta_{1.2}=\beta_1;\qquad 4)\ \beta_{1.2}-\delta\beta_{1.2}=\beta_2;\\[4pt] 5)\ \frac{\sin\Theta_1}{\sin\beta_2}\sin\Delta\lambda_{1.2} =\frac{\sin\Theta_2}{\sin\beta_1}\sin\Delta\lambda_{1.2} =\sin\sigma_{1.2}. \end{gathered} \tag{9} \]

Now let us compute the remaining two sides \(\sigma_{1.3}, \sigma_{2.3}\) in triangle 123. For this purpose we first find the angles \(\gamma_1, \gamma_2\) at vertices 1 and 2 of the indicated triangle. From Fig. 1 it is seen that

\[ 1)\ \gamma_1=\beta_1-\alpha_{1.3};\qquad 2)\ \gamma_2=\beta_2+\alpha_{2.3}. \tag{10} \]

Introducing further the notation:

\[ 1)\ \frac{1}{2}(\gamma_1+\gamma_2)=\gamma_{1.2};\qquad 2)\ \frac{1}{2}(\gamma_1-\gamma_2)=\delta\gamma_{1.2};\qquad 3)\ \frac{1}{2}\sigma_{1.2}=\delta\sigma_{1.2}, \tag{11} \]

we determine the last two sides \(\sigma_{1.3}, \sigma_{2.3}\) and the third angle \(\gamma_3\) at vertex 3 in triangle 123 from the following sequence of equalities:

\[ \begin{gathered} 1)\ \frac{\sin\delta\gamma_{1.2}\operatorname{tg}\delta\sigma_{1.2}}{\sin\gamma_{1.2}} =\operatorname{tg}\delta\sigma_0;\qquad 2)\ \frac{\cos\gamma_{1.2}\operatorname{tg}\delta\sigma_{1.2}}{\cos\gamma_{1.2}} =\operatorname{tg}\sigma_0;\\[4pt] 3)\ \sigma_0+\delta\sigma_{1.2}=\sigma_{1.3};\qquad 4)\ \sigma_0-\delta\sigma_{1.2}=\sigma_{2.3};\\[4pt] 5)\ \frac{\sin\gamma_1}{\sin\sigma_{2.3}}\sin\sigma_{1.2} =\frac{\sin\gamma_2}{\sin\sigma_{1.3}}\sin\sigma_{1.2} =\sin\gamma_3. \end{gathered} \tag{12} \]

In addition, we compute the spherical excess \(\varepsilon\) of triangle 123:

\[ \varepsilon=\gamma_1+\gamma_2+\gamma_3. \tag{13} \]

Having carried out all the preceding calculations, we can finally find, twice, the latitude \(\varphi_3\) and longitude \(\lambda_3\) of the point 3 being determined:

\[ \begin{gathered} 1)\ \operatorname{ctg}\Delta\lambda_{1.3} =\frac{\operatorname{ctg}\sigma_{1.3}\sin\Theta_1-\cos\alpha_{1.3}\cos\Theta_1}{\sin\alpha_{1.3}};\\[6pt] 2)\ \operatorname{ctg}\Delta\lambda_{2.3} =\frac{\operatorname{ctg}\sigma_{2.3}\sin\Theta_2-\cos\alpha_{2.3}\cos\Theta_2}{\sin\alpha_{2.3}};\\[6pt] 3)\ \cos\varphi_3=\sin\Theta_3^1 =\frac{\sin\Delta\lambda_{1.3}}{\sin\sigma_{1.3}}\sin\alpha_{1.3} =\frac{\sin\Delta\lambda_{2.3}}{\sin\sigma_{2.3}}\sin\alpha_{2.3} =\sin\Theta_3^2;\\[6pt] 4)\ \lambda_3=\lambda_3^1=\lambda_1+\Delta\lambda_{1.3} =\lambda_2+\Delta\lambda_{2.3}=\lambda_3^2. \end{gathered} \tag{14} \]

The convergence of the two values \(\varphi_3^1,\varphi_3^2\) and \(\lambda_3^1,\lambda_3^2\) of the latitude \(\varphi_3\) and longitude \(\lambda_3\) for the point 3 being determined serves as a good overall check on the correctness of all the computations carried out above.

Literature

1. B. F. Krutoi. General methods for solving the basic computational problems on the terrestrial spheroid. Izvestiya TPI, vol. 131, 1965.
2. N. N. Stepanov. Spherical Trigonometry. Gostekhizdat, 1948.