GENERAL METHODS FOR SOLVING THE MAIN COMPUTATIONAL PROBLEMS ON THE SURFACE OF THE EARTH’S SPHEROID

B. F. Krutoi

(Presented by the scientific seminar of the Department of Mine Surveying and Geodesy)

1. Let us introduce the notation:

\(B_i, L_i\) — the geodetic latitude and longitude of point \(i\) of the Earth’s spheroid;

\(A_{ik}\) — the geodetic azimuth at point \(i\) of the adjusted (geodesic) curve \(\Gamma_{ik}\), drawn on the spheroid through point \(i\) and the neighboring point \(k\);

\(s_{ik}\) — the length of the adjusted curve \(\Gamma_{ik}\) between the indicated points \(i, k\).

Then the main computational problems on the surface of the Earth’s spheroid may be called the following three problems:

a) Direct problem \((i = 1,\ k = 2)\). Given \(B_1, L_1, A_{1.2}, s_{1.2}\); find \(B_2, L_2, A_{2.1}\).

b) Inverse problem \((i = 1,\ k = 2)\). Given \(B_1, L_1\) and \(B_2, L_2\); find \(A_{1.2}, A_{2.1}, s_{1.2}\).

c) Direct intersection \((i = 1,2;\ k = 3)\). Given \(B_1, L_1, A_{1.3}\) and \(B_2, L_2, A_{2.3}\). Find \(B_3, L_3\), as well as \(A_{3.1}, s_{1.3}\) and \(A_{3.2}, s_{2.3}\).

1. For solving the first two problems, more than ten special methods have been proposed [1, 2, 3], suitable for distances \(s_{1.2}\) not exceeding 1000–3000 km, and one general method, due to Bessel, for any distances \(s_{1.2}\). As for the method of A. M. Virovets [4], on the basis of the investigations of N. A. Urmaev [5] and A. V. Butkevich [6], it should be regarded merely as a certain modification of the same Bessel method.

Below, one more general method is proposed for solving the first two problems, in which the question of possible relations between the initial and the determined quantities is considered with due completeness. The expressions derived here are then applied to the solution of the third problem.

1. The new method is based on a system of three first-order differential equations with variable coefficients, determining the adjusted curve \(\Gamma_{ik}\) of the spheroid:

\[ \begin{aligned} 1)\quad & M\,dB + 0.dL - \cos A\,ds + 0.dA = 0.\\ 2)\quad & 0.dB + r\,dL - \sin A\,ds + 0.dA = 0, \tag{1}\\ 3)\quad & 0.dB + \sin B\,dL + 0.ds - 1.dA = 0 \end{aligned} \]

and the well-known Clairaut equation following from this system

\[ \frac{r}{a}\sin A=\nu=\text{const.} \tag{2} \]

Here \(M, r\) are the radii of curvature of the meridian and the parallel at the variable point \(C\) of the given rectified curve \(\Gamma\), issuing from \(C\) with azimuth \(A\).

1. Using the indicated set of equalities (1), (2), after certain transformations we obtain the following final expressions, which solve the first two main computational problems on the terrestrial spheroid:

\[ \begin{aligned} 1)\quad & \eta\cdot s_{1.2}=a\mu\int_{\varphi_1}^{\varphi_2} \frac{d\varphi}{(1-k^2\sin^2\varphi)^{3/2}};\\[6pt] 2)\quad & s_{1.2}=a\nu\int_{A_{1.2}}^{A'_{2.1}} \sqrt{\frac{\sin^2 A-e^2\nu^2}{\sin^2 A-\nu^2}}\, \csc^2 A\,dA;\\[6pt] 3)\quad & \eta\cdot \Delta L_{1.2}=\nu\mu\int_{\varphi_1}^{\varphi_2} \frac{d\varphi}{(1-m^2\sin^2\varphi)\sqrt{1-k^2\sin^2\varphi}};\\[6pt] 4)\quad & \Delta L_{1.2}=\int_{A_{1.2}}^{A'_{2.1}} \sqrt{\frac{\sin^2 A-e^2\nu^2}{\sin^2 A-\nu^2}}\,dA;\\[6pt] 5)\quad & 2\Delta A_{1.2}= \left.\arc\sin\left[2\nu^2(1-e^2)\sec^2 B-(1-2e^2\nu^2)\right]\right|_{B_1}^{B_2};\\[6pt] 5a)\quad & 2\cdot\Delta A_{1.2}= \arc\sin(1-2\sin^2 A_{1.2}) -\arc\sin(1-2g^2\sin^2 A_{1.2});\\[6pt] 6)\quad & \operatorname{ctg} A_{1.2}= \frac{V_2\cos B_1-V_1\cos B_1\cos\Delta A_{1.2}} {V_1\cos B_2\sin\Delta A_{1.2}} =\frac{g-\cos\Delta A_{1.2}}{\sin\Delta A_{1.2}};\\[6pt] 7)\quad & -\operatorname{ctg} A'_{2.1}= \frac{V_1\cos B_2-V_2\cos B_1\cos\Delta A_{1.2}} {V_2\cos B_1\sin\Delta A_{1.2}} =\frac{(1:g)-\cos\Delta A_{1.2}}{\sin\Delta A_{1.2}};\\[6pt] 8)\quad & \sin A_{1.2}=\frac{a\nu}{r_1} =\sqrt{1-e^2}\,\frac{\nu V_1}{\cos B_1};\\[6pt] 9)\quad & \sin A'_{2.1}=\frac{a\nu}{r_2} =\sqrt{1-e^2}\,\frac{\nu V_2}{\cos B_2};\\[6pt] 10)\quad & L_2=L_1+\Delta L_{1.2};\qquad 11)\quad A_{2.1}=A'_{2.1}\pm 180^\circ =(A_{1.2}+\Delta A_{1.2})\pm 180^\circ. \end{aligned} \tag{3} \]

In equalities (3) the following notations have been introduced:

\[ \begin{aligned} 1)\quad & V=\sqrt{1+e'^2\cos^2 B};\\[6pt] 2)\quad & \nu=\frac{r_1}{a}\sin A_{1.2} =\sqrt{1+e'^2}\,\frac{\cos B_1\sin A_{1.2}}{V_1} =\nu_{1.2};\\[6pt] 3)\quad & \mu=\frac{1-e^2}{\sqrt{1-e^2\nu^2}}=\mu_{1.2};\qquad 4)\quad \tau^2=\frac{1-e^2\nu^2}{1-\nu^2}=\tau^2_{1.2}; \end{aligned} \]

\[ \left\{ \begin{aligned} &5)\quad k^2=e^2\tau^2=k^2_{1.2}; \qquad &&6)\quad m^2=1:\tau^2=m^2_{1.2};\\[4pt] &7)\quad \sin\varphi=\tau\sin B; \qquad &&8)\quad g=\frac{V_2\cos B_1}{V_1\cos B_2};\\[6pt] &9)\quad e^2=\frac{a^2-b^2}{a^2}; \qquad &&10)\quad e'^2=\frac{a^2-b^2}{b^2}.\\[6pt] &\eta= \begin{cases} +1, & \text{if } \varphi_2>\varphi_1 \quad (A'_{2.1}>A_{1.2}),\\ -1, & \text{if } \varphi_2<\varphi_1 \quad (A'_{2.1}<A_{1.2}). \end{cases} \end{aligned} \right. \tag{4} \]

\(a\) and \(b\) are the major and minor semiaxes of the terrestrial spheroid.

The set (3) has two noteworthy features:

a) equality (3.1) for \(s_{1.2}\) has the same form as the well-known expression for the length \(x_{1.2}\) of a meridian arc;

b) the quantities \(s_{1.2}\) and \(\Delta L_{1.2}\) are expressed not only in terms of the latitude \(B\) at the current point \(C\) of the adjusted curve \(\Gamma_{1.2}\), but also as a function of the azimuth \(A\) of the curve \(\Gamma_{1.2}\) at the same current point \(C\).

1. Let us consider the application of the set (3) for computational purposes.

Integral (3.1) is evaluated by the usual expansion in a series in powers of \(k^2\) \((k^2\le e^2=0.0066934)\).

Integral (3.3), after expansion in a series in powers of \(k^2\), is evaluated by the method proposed in 1935 by Prof. V. P. Vetchinkin in [7].

Integrals (3.2) and (3.4) are elliptic integrals of general type, and their solution is most simply obtained by numerical integration by the Gauss method, or by expanding the numerator \((\sin^2 A-e^2\nu^2)^{1/2}\) in a series by Newton’s binomial.

The working expressions that follow from the set (3), and their application to the solution of the direct and inverse problems, as well as other relations used here, are given in Appendices 1 and 2.

1. To find \(B_3\) and \(\Delta L_{1.3}, \Delta L_{2.3}\) in the third problem—the direct intersection—we solve, by successive approximation, the equation

\[ \left\{ \begin{aligned} \eta_1\Delta L_{1.2} &=\eta_1\Delta L_{1.3}-\eta_1\Delta L_{2.3}\\ &=\gamma_{1.3}\mu_{1.3} \int_{\varphi_1}^{\varphi_2} \frac{d\varphi} {(1-m^2_{1.3}\sin^2\varphi)\sqrt{1-k^2_{1.3}\sin^2\varphi}}\\ &\quad-\gamma_{2.3}\mu_{2.3} \int_{\psi_1}^{\psi_2} \frac{d\psi} {(1-m^2_{2.3}\sin^2\psi)\sqrt{1-k^2_{2.3}\sin^2\psi}}, \end{aligned} \right. \tag{5} \]

where \(\gamma_{ij}, \mu_{ij}, k^2_{ij}, m^2_{ij}\) are found according to (4.2)—(4.6), while \(\varphi\) and \(\psi\) are connected by the relation

\[ \sin\psi=-\frac{\tau_{2.3}}{\tau_{1.3}}\sin\varphi, \tag{6} \]

in which \(\tau_{ij}\) is computed from (4.4).

The actual solution of the direct intersection on the spheroid, for different formulations of the conditions of this problem, is given in Appendix 3.

1. Note that the reduced elliptic integrals of the third kind \(P(\varphi,k,m^2)\), entering into (3.1), (3.3), and (5), contain, by virtue of (4.5) and (4.6), only two independent variables. Therefore the indicated integrals can be represented in the form of tables with two arguments, similar to

of the reduced elliptic integral of the first kind \(F(\varphi, k)\). The availability of such tables will substantially facilitate the solution, by the proposed method, of all three of the above-mentioned principal computational problems on the spheroid.

REFERENCES

1. F. R. Helmert. Die mathematischen und physikalischen Theorien der höheren Geodäsie. I Theil, Leipzig, 1880.

2. Jordan–Eggert. Handbuch der Vermessungskunde, dritter Band. Stuttgart, 1923.

3. F. N. Krasovskii, A Manual of Higher Geodesy, Part II, Moscow, 1942.

4. N. A. Urmaev, Spheroidal Geodesy, Moscow, 1955.

5. A. M. Virovets, Solution of the direct geodetic problem for considerable distances between geodetic points, Geodesist, No. 4, 1935.

6. A. V. Butkevich and V. N. Gan’shin, On the solution of the direct geodetic problem for long distances by the methods of Bessel and Virovets, Transactions of NIIGAiK, Vol. XI, Novosibirsk, 1958.

7. V. P. Vetchinkin, New Formulas and Tables of Elliptic Integrals and Functions, Publishing House of the Air Force Academy of the RKKA, Moscow, 1935.

Appendix 1

Solution

of the direct problem for the rectified curve \(\Gamma_{1.2}\)

Statement of the problem: given \(B_1, L_1, A_{1.2}, s_{1.2}\); find \(B_2, L_2, A_{2.1}\).

I. Determination of \(B_2\)

\[ \text{1) }\quad V=\sqrt{1+e'^2\cos^2 B_1}\quad \text{— from geodetic tables.} \]

\[ \text{2) }\quad \nu=\sqrt{1+e'^2}\,\frac{\cos B_1\sin A_{1.2}}{V_1}; \qquad \text{5) }\quad k^2=\frac{e^2}{\tau^2}<e^2; \]

\[ \text{3) }\quad \mu=\frac{1-e^2}{\sqrt{1-e^2\nu^2}}; \qquad \text{6) }\quad \sin\varphi_1=\tau\sin B; \]

\[ \text{4) }\quad \tau^2=\frac{1-e^2\nu^2}{1-\nu^2}>1; \qquad \text{7) }\quad m^2=\frac{1}{\tau^2}<1; \]

\[ \text{8) }\quad C_0=1+\sum_{\lambda=1}^{n}(-1)^\lambda \frac{\left({-3/2 \atop \lambda}\right)\left({2\lambda \atop \lambda}\right)}{2^{2\lambda}}\,k^{2\lambda} = 1+\sum_{\lambda=1}^{n}c_{0.2\lambda}k^{2\lambda}; \]

\[ \text{9) }\quad C_{2u}=\sum_{\lambda=u}^{n}(-1)^{\lambda-u} \frac{\left({-3/2 \atop \lambda}\right)\left({2\lambda \atop \lambda-u}\right)}{2^{2\lambda}\cdot u}\, k^{2\lambda} = \sum_{\lambda=u}^{n}c_{2u.2\lambda}k^{2\lambda} \qquad (u=1,2,\ldots,n); \]

\[ \text{10) }\quad \varphi_2-\varphi_1=\Delta\varphi_{1.2} = \frac{s_{1.2}}{C_0a\mu} - \sum_{u=1}^{n}\frac{C_{2u}}{C_0} \left(\sin 2u\varphi_2-\sin 2u\varphi_1\right); \]

\[ \text{11) }\quad \varphi_2=\varphi_1+\Delta\varphi_{1.2}; \qquad \text{12) }\quad \sin B_2=\frac{1}{\tau}\sin\varphi_2. \]

In the case when \(\Delta\varphi_{1.2}\) is small, \(B_2\) is computed as follows:

\[ \text{11a) }\quad \Delta B''_{1.2} = (\operatorname{tg}B_1\operatorname{ctg}\varphi_1)\Delta\varphi''_{1.2} - \frac{\operatorname{tg}B_1}{2\rho} \left(\Delta\varphi_{1.2}^{\prime 2}-\Delta B_{1.2}^{2}\right)'' \]

\[ - \frac{1}{6\rho^2} \left[ (\operatorname{tg}B_1\operatorname{ctg}\varphi_1)\Delta\varphi_{1.2}^{3} - \Delta B_{1.3}^{3} \right]''; \]

\[ \text{12a) }\quad B_2=B_1+\Delta B_{1.2} \qquad \rho=206264.8. \]

II. Determination of \(L_{1.2}\)

\[ \text{1) }\quad \eta\cdot\Delta L_{1.2} = \psi\mu\sum_{\lambda=0}^{n}F(\lambda) = \psi\mu\sum_{\lambda=0}^{n}g(\lambda)\Phi(\lambda); \]

\[ \text{2) }\quad L_2=L_1+\Delta L_{1.2}; \]

Here:

\[ \text{a) }\quad g(\lambda)=(-1)^\lambda \binom{-1/2}{\lambda} k^{2\lambda}; \qquad \text{b) }\quad \Phi(\lambda)=\int_{\varphi_1}^{\varphi_2} \frac{\sin^{2\lambda}\varphi\,d\varphi}{1-m^2\sin^2\varphi}; \]

\[ \text{c) }\quad F(\lambda)=g(\lambda)\Phi(\lambda). \]

The factor \(\eta=+1\), if \(\varphi_2>\varphi_1\), \(\eta=-1\), if \(\varphi_2<\varphi_1\).

From (b) the value \(\Phi(0)=F(0)\) is found by direct integration, and for \(F(\lambda)\) with \(\lambda>0\) we shall have the following general expression:

\[ F(\lambda)=E(\lambda)[S(\lambda-1)-\Phi(\lambda-1)] =E(\omega+1)[S(\omega)-\Phi(\omega)], \]

\[ (\omega=\lambda-1=0,\ 1,\ 2,\ldots,n), \]

where

\[ \text{a) }\quad E(\lambda)=-\frac{g(\lambda)}{m^2} =(-1)^{\lambda+1}\binom{-1/2}{\lambda}\frac{k^{2\lambda}}{m^2} =(-1)^{\lambda+1}\binom{-1/2}{\lambda}e^2k^{2(\lambda-1)}; \]

\[ \text{b) }\quad S(0)=\varphi_2-\varphi_1; \]

\[ \text{c) }\quad S(\lambda-1)=S(\omega)=\int_{\varphi_1}^{\varphi_2}\sin^{2\omega}\varphi\,d\varphi = \frac{\binom{2\omega}{\omega}}{2^{2\omega}}(\varphi_2-\varphi_1) + \]

\[ +\frac{(-1)^\omega}{2^{2\omega}} \sum_{x=0}^{\omega-1} (-1)^x \frac{\binom{2\omega}{x}}{\omega-x} \left[\sin 2(\omega-x)\varphi_2-\sin 2(\omega-x)\varphi_1\right] \]

\[ \left(\omega=\lambda-1=1,\ 2,\ldots,n\right). \]

Hence for \(\psi F(\lambda)\) we obtain successively:

\[ \text{1) }\quad \psi F(0)= \frac{\nu}{|\nu|}\sqrt{1-e^2}\,[\arctan pt_2-\arctan pt_1] =\psi\Phi(0)=R(0) \]

\[ \text{2) }\quad \psi F(1)= -\psi\frac{e^2}{2}(\varphi_2-\varphi_1) +\frac{1}{2}e^2R(0) = \frac{e^2}{2} \left[ \psi F(0) -\frac{\nu}{|\nu|}p\sqrt{1-e^2}(\varphi_2-\varphi_1) \right] =R(1) \]

\[ \text{3) }\quad \psi F(2)= -\psi\frac{3}{16}e^2k^2 \left[ (\varphi_2-\varphi_1)-\frac{1}{2}(\sin2\varphi_2-\sin2\varphi_1) \right] \]

\[ +\frac{3}{4}e^2R(1)=R(2) \]

\[ \text{4) }\quad \psi F(3)= -\psi\frac{5}{16}e^2k^4 \left[ \frac{3}{8}(\varphi_2-\varphi_1) -\frac{1}{4}(\sin2\varphi_2-\sin2\varphi_1) + \right. \]

\[ \left. +\frac{1}{32}(\sin4\varphi_2-\sin4\varphi_1) \right] +\frac{5}{6}e^2R(2)=R(3) \]

and so on, where:

\[ \text{a) }\quad p=\sqrt{1-m^2}; \qquad \text{b) }\quad t=\tan\varphi \]

If \(\varphi_2-\varphi_1=\Delta\varphi_{1.2}\) is small, then \(\psi F(0)\) should be computed as follows:

\[ \text{1a) }\quad \psi F(0)= \frac{\nu}{|\nu|}\sqrt{1-e^2}\, \arctan \frac{p\sin\Delta\varphi_{1.2}} {\cos\varphi_1\cos\varphi_2-p^2\sin\varphi_1\sin\varphi_2} = \]

\[ = \frac{\nu}{|\nu|}\sqrt{1-e^2}\, \arctan \frac{p\sin\Delta\varphi_{1.2}} {\cos\Delta\varphi_{1.2}-m^2\sin\varphi_1\sin\varphi_2}. \]

III. Determination of \(A_{2.1}\)

\[ 1)\quad \sin A'_{2.1}=\sqrt{1-e^2}\,\frac{\nu V_2}{\cos B_2}; \]

\[ 2)\quad A_{2.1}=A'_{2.1}\pm 180^\circ=(A_{1.2}+\Delta A_{1.2})\pm 180^\circ. \]

If \(A'_{2.1}\) is close to \(90^\circ\) or \(270^\circ\), but \((B_2-B_1)\) is large, then instead of \(A'_{2.1}\) one computes \(\Delta A_{1.2}\):

\[ 3)\quad g=\frac{V_2\cos B_1}{V_1\cos B_2}; \]

\[ 4)\quad 2.\Delta A_{1.2}=\operatorname{arc}\sin(1-2\sin^2 A_{1.2})-\operatorname{arc}\sin(1-2g^2\sin^2 A_{1.2}). \]

If \((B_2-B_1)=\Delta B_{1.2}\) is small, then instead of \(A'_{2.1}\) one may also compute \(\Delta A_{1.2}\):

\[ 2.\quad \Delta A''_{1.2}= \frac{\rho''}{\sin 2A_{1.2}}\, \varepsilon \left[ 1+\frac{\cos 2A_{1.2}}{2\sin^2 2A_{1.2}}\,\varepsilon + \frac{1+2\cos^2 2A_{1.2}}{6\sin^4 2A_{1.2}}\,\varepsilon^2 \right], \]

where

\[ 5)\quad \varepsilon=2(g^2-1)\sin^2 A_{1.2} \]

Appendix 2

Solution of the inverse problem for the rectified curve \(\Gamma_{1.2}\)

Statement of the problem: given \(B_1, L_1\) and \(B_2, L_2\); find \(A_{1.2}, A_{2.1}\) and \(s_{1.2}\).

First method for determining \(A_{1.2}\) and \(A_{2.1}\) (for \(s_{1.2}>1000\) km).

First of all, we solve, by 2 approximations, the equation (see Appendix 1):

\[ \eta\cdot \Delta L_{1.2} = \psi \sum_{\lambda=0}^{n} F(\lambda) = \psi F(0)+\psi \sum_{\lambda=1}^{n} F(\lambda) \]

\[ = \frac{\nu}{|\nu|}\sqrt{1-e^2}\,[\operatorname{arc\,tg}pt_2-\operatorname{arc\,tg}pt_1] + \psi\sum_{\lambda=1}^{n}F(\lambda) = \tag{*} \]

\[ = \frac{\nu}{|\nu|}\sqrt{1-e^2} \left[ \operatorname{arc\,tg}\frac{p\sin B_2}{\sqrt{\cos^2 B_2-p^2}} - \operatorname{arc\,tg}\frac{p\sin B_1}{\sqrt{\cos^2 B_1-p^2}} \right] + \]

\[ +\psi\sum_{\lambda=1}^{n}F(\lambda), \]

with respect to the unknown \(p\), where the sum \(\psi\sum_{\lambda=1}^{n}F(\lambda)\) is a small quantity of order \(e^2\).

The initial sufficiently accurate value \(A^{(0)}_{12}\) of the azimuth \(A_{1.2}\), entering into the computation of the quantity \(\nu\), is found from the relations:

$$ \begin{aligned} 1)\quad &\operatorname{tg}\frac{1}{2}\left(\alpha'_{2.1}+\alpha_{1.2}\right) = \frac{\cos \frac{1}{2}(B_2+B_1)} {\sin \frac{1}{2}(B_2-B_1)} \operatorname{tg}\frac{\Delta L_{1.2}}{2}, \\[1.5em] 2)\quad &\operatorname{tg}\frac{1}{2}\left(\alpha'_{2.1}-\alpha_{1.2}\right) = \frac{\sin \frac{1}{2}(B_2+B_1)} {\cos \frac{1}{2}(B_2-B_1)} \operatorname{tg}\frac{\Delta L_{1.2}}{2}, \end{aligned} $$

which we solve with respect to $\alpha_{1.2}$ and then set $a_{1.2}=A^{(0)}_{1.2}$. The corresponding approximate values of the unknown $p$ and of the quantities $\psi$, $k^2$, and $R(0)$ are then determined from the expressions:

$$ 1)\quad p=\frac{\nu\sqrt{1-\gamma^2}}{\sqrt{1-e^2\gamma^2}} =\cos B_0 \simeq \sin A^{(0)}_{1.2}\cos B_1=p^{(0)}; $$

$$ 2)\quad \psi=\frac{\gamma}{|\gamma|}\,p\sqrt{1-e^2}; \qquad 3)\quad k^2=e^2\sin B_0; $$

$$ 4)\quad R(0)\simeq \frac{\nu}{|\gamma|}\sqrt{1-e^2}\,\Delta L_{1.2} $$

for $p=p^{(0)}$.

The improved value $p^{(s+1)}$ of the unknown $p$ as a result of the $s$-th approximation is found as follows:

$$ \text{a)}\quad w^{(s)}= \left[\psi\sum_{\lambda=1}^{n}F(\lambda)\right]^{(s)} -\eta\cdot \Delta L_{1.2} $$

$$ \text{b)}\quad \frac{dw^{(s)}}{dp} = \left[ \frac{\sqrt{1-e^2}}{1-p^2}\,(t_2-t_1) \right]^{(s)} $$

$$ \text{c)}\quad p^{(s+1)}=p^{(s)}-\frac{dw^{(s)}}{dp}\cdot w^{(s)}. $$

Having solved equation (*), we then compute $\gamma$, $A_{1.2}$, $A_{2.1}$, and for the quantity $\gamma$ from (1) we shall have, conversely:

$$ 4)\quad \gamma=\frac{p}{\sqrt{(1-e^2)+p^2}}. $$

Remark. If between the end points 1, 2 of the rectifying curve $\Gamma_{1.2}$ lies its vertex $O$, then equation (*) must be represented in the form

$$ \eta\cdot\Delta L_{1.2} = \eta(\Delta L_{1.0}+\Delta L_{0.2}) = \psi\sum_{\lambda=0}^{n}F_{1.0}(\lambda) + \psi\sum_{\lambda=0}^{n}F_{0.2}(\lambda) +\eta\pi, \tag{**} $$

where the latitude $B_0$ of the point $O$ is determined by the expressions:

$$ 5)\quad \frac{\gamma}{|\psi|}\sin B_0 = \sqrt{\frac{1-\gamma^2}{1-e^2\gamma^2}} = \frac{1}{\tau} = m = \sqrt{1-p^2} $$

$$ 6)\quad \cos B_0=\sqrt{1-m^2}=p. $$

Second method for determining \(A_{1.2}\) and \(A_{2.1}\) (for \(s_{1.2} \leqslant 1000\) km)

\[ \left. \begin{aligned} 1)\quad \tg \frac{1}{2}(\alpha'_{2.1}+\alpha_{1.2}) &= \frac{\cos \frac{1}{2}(B_2+B_1)} {\sin \frac{1}{2}(B_2-B_1)} \,\tg \frac{\Delta L_{1.2}}{2} = \tg \frac{\overline{\alpha}_{1.2}}{2}, \\[1.2em] 2)\quad \tg \frac{1}{2}(\alpha'_{2.1}-\alpha_{1.2}) &= \frac{\sin \frac{1}{2}(B_2+B_1)} {\cos \frac{1}{2}(B_2-B_1)} \,\tg \frac{\Delta L_{1.2}}{2} = \tg \frac{\Delta \alpha_{1.2}}{2}. \end{aligned} \right\} \quad \begin{gathered} \text{Hence find}\\ \alpha_{1.2}\ \text{and}\ \alpha'_{1.2}\\ \text{separately.} \end{gathered} \]

\[ 3)\quad Q=\frac{e^2}{\sin \Delta L_{1.2}}\,(V_1\sin B_2 - V_2\sin B_1); \]

\[ 4)\quad \alpha_{2.1}=\alpha'_{2.1}\pm 180^\circ=(\alpha_{1.2}+\Delta \alpha_{1.2})\pm 180^\circ \]

\[ 5)\quad \sin \sigma_{1.2} = \frac{\cos B_2}{\sin \alpha_{1.2}}\sin \Delta L_{1.2} = -\frac{\cos B_1}{\sin \alpha_{2.1}}\sin \Delta L_{1.2} \]

\[ 6)\quad \ctg A^\circ_{1.2} = \ctg \alpha_{1.2} - Q\,\frac{\cos B_1}{V_1\cos B_2}; \]

\[ 7)\quad \ctg A^\circ_{2.1} = \ctg \alpha_{2.1} - Q\,\frac{\cos B_2}{V_2\cos B_1} \]

\[ 8)\quad \eta''_{1.2} = \frac{e^2}{6\rho''}(\sigma''_{1.2})^2 \sin A_{1.2}\cos^2 B_1 \left( \cos A_{1.2} - \frac{\sigma''_{1.2}}{4\rho''}\tg B_1 \right) \]

\[ 9)\quad \eta''_{2.1} = \frac{e^2}{6\rho''}(\sigma''_{1.2})^2 \sin A_{2.1}\cos^2 B_2 \left( \cos A_{2.1} - \frac{\sigma''_{1.2}}{4\rho''}\tg B_2 \right) \]

\[ 10)\quad A_{1.2}=A^\circ_{1.2}-\eta''_{1.2} \]

\[ 11)\quad A_{2.1}=A^\circ_{2.1}-\eta''_{2.1} \]

Determination of \(s_{1.2}\)

We compute \(\gamma, \tau, \psi, k^2, \varphi_1, \varphi_2, C_0\) and \(C_{2u}\) \((u=1,2,\ldots,n)\)—as in the direct problem.

\[ 10)\quad s_{1.2} = C_0 a\mu(\varphi_2-\varphi_1) + a\mu \sum_{u=1}^{n} C_{2u}\,(\sin 2u\varphi_2-\sin 2u\varphi_1) \]

Appendix 3

Solution of the direct intersection on the spheroid

Two formulations of this problem are possible: general and particular.

General formulation. Given the geodetic coordinates \(B_1, L_1\) and \(B_2, L_2\) of the rays \(\delta L_{1.3}, \delta L_{2.3}\) from points 1, 2 to the point 3 to be determined. Find the geodetic coordinates \(B_3, L_3\) of point 3.

Particular formulation—it is assumed that the intersecting angle \(\gamma_3\) at the point 3 being determined is also given, and that the distances \(s_{1.3}, s_{2.3}\) do not exceed 300–400 km.

A) Solution for the general case (for arbitrary distances \(s_{1.3}, s_{2.3}\))

Writing, for \(L_1<L_2\), the basic relation

\[ \eta \Delta L_{1.2}=\eta \Delta L_{1.3}-\eta \Delta L_{2.3} = \nu_{1.3}\psi_{1.3}\sum_{\lambda=0}^{n}F_{1.3}(\lambda) - \nu_{2.3}\psi_{2.3}\sum_{\lambda=0}^{n}F_{2.3}(\lambda) \]

and transforming it according to Appendix 1, we arrive at the equation

\[ \sqrt{1-e^2}\left[ \frac{\nu_{1.3}}{|\nu_{1.3}|}\operatorname{arc\,tg} \frac{p_{1.3}\sin B_3}{\sqrt{(1-p_{1.3}^2)-\sin^2 B_3}} - \operatorname{arc\,tg} \frac{p_{2.3}\sin B_3}{\sqrt{(1-p_{2.3}^2)-\sin^2 B_3}} +\right. \]

\[ \left. +\nu_{1.3}\psi_{1.3}\sum_{\lambda=1}^{n}F_{1.3}(\lambda) - \nu_{2.3}\psi_{2.3}\sum_{\lambda=1}^{n}F_{2.3}(\lambda) \right] -\mathfrak{w} = \Phi(B_3)+\varepsilon(B_3)-\mathfrak{w}=0 \tag{1} \]

with unknown \(B_3\), where the correction term \(\varepsilon(B_3)\) is a small quantity of order \(e^2\), and the free term \(\mathfrak{w}\) is determined by the expression:

\[ \mathfrak{w} = \eta \Delta L_{1.2} - \sqrt{1-e^2} \left[ \frac{\nu_{2.3}}{|\nu_{2.3}|}\operatorname{arc\,tg} \frac{p_{2.3}\sin B_2}{\sqrt{(1-p_{2.3}^{2})-\sin^2 B_2}} -\right. \]

\[ \left. - \frac{\nu_{1.3}}{|\nu_{1.3}|}\operatorname{arc\,tg} \frac{p_{1.3}\sin B_1}{\sqrt{(1-p_{1.3}^{2})-\sin^2 B_1}} \right]. \tag{2} \]

We solve equation (1) by the method of successive approximations, and for this purpose in the separate approximations we represent the principal term \(\Phi(B_3)\) in the form

\[ \Phi(B_3)=\Phi(B_3^{(s+1)})=\Phi(B_3^{(s)})+\Phi'(B_3^{(s)})\cdot\delta B_3^{(s)},\quad (s=1,2,\ldots n), \tag{3} \]

where \(\Phi(B_3^{(s)})\) has been found from the preceding approximation, \(\delta B_3^{(s)}\) is the new unknown, and

\[ \text{1) }\quad \Phi'(B_3)=(1-e^2)\cos B_3 \left[ \frac{\left(\dfrac{\nu}{|\nu|}p^{\tau}\right)_{1.3}\sec^3\varphi_3}{1+p_{1.3}^{2}\operatorname{tg}^{2}\varphi_3} - \frac{\left(\dfrac{\nu}{|\nu|}p^{\tau}\right)_{2.3}\sec^3\psi_3}{1+p_{2.3}^{2}\operatorname{tg}^{2}\psi_3} \right], \]

\[ \text{2) }\quad \sin\varphi_3=\tau_{1.3}\sin B_3, \qquad \text{3) }\quad \sin\psi_3=\tau_{2.3}\sin B_3. \]

We take the initial value \(B_3^{(0)}\) of the latitude \(B_3\) from the map.

B) Solution for the particular case (for a given angle \(\gamma_3\))

\[ \text{0) }\quad i=1,2=4,n; \qquad \text{1) }\quad \nu_{i3}=\frac{\cos B_i\sin A_{i3}}{V_i\sqrt{1-e^2}} \]

\[ \text{2) }\quad \operatorname{ctg} A_{3.1} = \left(\cos\gamma_3-\frac{\nu_{2.3}}{\nu_{1.3}}\right)\csc\gamma_3 \]

\[ \text{3) }\quad \operatorname{ctg} A_{3.2} = \left(\frac{\nu_{1.3}}{\nu_{2.3}}-\cos\gamma_3\right)\csc\gamma_3 \]

\[ \left. \begin{array}{l} \\[-2.2em] \\ \end{array} \right\} \quad \text{Check: } A_{3.1}=A_{3.2}+\gamma_3 \]

\[ \text{4a)}\quad \sin^{2} B_{3} = \frac{\sin^{2} A_{3i} - \nu_{i3}^{2}} {\sin^{2} A_{3i} - e^{2}\nu_{i3}^{2}}; \]

\[ \text{4b)}\quad \cos^{2} B_{3} = \frac{(1-e^{2})\nu_{i3}^{2}} {\sin^{2} A_{3i} - e^{2}\nu_{i3}^{2}}; \]

\[ \text{4c)}\quad \operatorname{tg}^{2} B_{3} = \frac{\sin^{2} A_{3i} - \nu_{i3}^{2}} {(1-e^{2})\nu_{i3}^{2}}; \]

\[ \text{5)}\quad \operatorname{tg}\frac{1}{2}\Delta L_{i3} = \frac{\cos \frac{1}{2}(B_{3}-B_{i})} {\sin \frac{1}{2}(B_{3}+B_{i})} \operatorname{tg}\frac{1}{2}\Delta A_{i3}, \]

where

\[ \Delta A_{i3}=A_{3i}-A_{i3}\pm180^\circ . \]

Check:

\[ \Delta L_{1.3}-\Delta L_{2.3}=\Delta L_{1.2}. \]

Note. If the lengths \(s_{1.3}, s_{2.3}\) exceed 250–400 km, then in especially precise computations the corresponding small corrections must be introduced into the differences \(\Delta A_{i3}\) entering into (5).