Double resection, its generalization, and various methods of solution
Abstract Generated abstract
This article surveys and systematizes methods for solving double resection in geodesy, including the classical Hansen problem and its generalized form. It distinguishes logarithmic procedures from natural formulas intended for arithmometer computation, reviews established approaches such as auxiliary points, an auxiliary angle, conditional direction and base, and Nuvaryev’s analytic method, and comments on their practical advantages, checks, and computational labor. The paper then develops the author’s own solution for generalized double resection, including the case with three control points, and supplies formula schemes and numerical examples to guide practical calculation.
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DOUBLE RESECTION, ITS GENERALIZATION AND VARIOUS SOLUTION METHODS
B. F. Krutoi
Introduction
Double resection is known in old courses of elementary geodesy under the name of Hansen’s problem. In this problem, the coordinates of two points to be determined, 1 and 2, are sought from the known coordinates of two fixed (control) points 3 and 4 and from the four angles \(\alpha, \beta, \gamma, \delta\), measured at points 1 and 2, between the directions to the fixed and the points to be determined (Fig. 1).
Double resection, in this formulation of the initial conditions, has wide practical application, for example in ties. A rather large number of methods have been developed for its solution; of these we shall present only the principal ones, indicating their characteristic features.
According to the character of the formulas, these solutions may be divided into 2 groups:
1) formulas intended for the use of logarithms,
2) formulas for computation on an arithmometer (natural formulas).
Formulas of the first type have long been known and are included in some textbooks of elementary geodesy; formulas of the second type appeared comparatively recently, which is explained by the late introduction of arithmometers into computational practice.
In § 1 of the present work we shall first give a brief review of solutions with logarithmic formulas, then set forth the substance of the solutions by natural formulas available in the periodical literature. But double resection in the above-indicated formulation of the initial conditions is one of the special cases of the so-called generalized double resection. Therefore, after examining the various methods for solving the special double resection, in § 2 we shall generalize this problem. Having presented two existing methods for solving the generalized double resection, we shall not confine ourselves to them, and in § 3 we shall develop our own solution of the indicated problem, which will constitute the main content of the present article.
Fig. 1
In conclusion we shall point out that the present work may be regarded as a monograph on the question of solving double resection—both special and general. The necessity for such a work is caused by the circumstance that, from the large number of different methods for solving double resection—partial
of resection of the particular type only one method is usually included in existing textbooks on elementary geodesy—namely, the solution with an auxiliary angle. The solution of the generalized double resection, however, is entirely absent from existing textbooks. Meanwhile, at the present time a number of methods are known for solving double resection, each of which, under certain conditions, has one or another advantage in comparison with other methods of solving the indicated problem. These methods have for the most part been published in various periodicals, not always accessible to the broad mass of practitioners. It therefore seems useful to make a compilation of the methods known at present and having the greatest practical value for solving double resection, supplying them with the corresponding critical remarks. This will make it possible for the practitioner, depending on the computational skill he possesses and the available computational means, to choose that method for solving double resection which will lead him to the goal with the least expenditure of labor. An experienced computer will choose a method that contains no intermediate checks, but requires the smallest number of computational operations. A less experienced computer, on the other hand, will prefer a method which, although it contains a larger number of operations, nevertheless these operations are reliably checked at intermediate stages of the computations. Therefore, in the final analysis, the chosen longer method of solution will lead the less experienced computer to the goal with less expenditure of labor than a more abbreviated, but unchecked at the intermediate stages, method of solving the same problem.
§ 1. Various methods for solving a particular double resection
Solutions of Type I (by logarithmic formulas)
1) With auxiliary points (formulas 1). Assuming the problem solved, let us draw through points 3, 4, and 1 (Fig. 2) an auxiliary circle, which, in its intersection with line 1.2, will give the auxiliary
Fig. 2
point \(A\). Drawing a similar circle through points 3, 4, and 2, we obtain at its intersection with line 1.2 the second auxiliary point \(B\). The idea of the solution consists in finding the directional angle of the line \(AB\), which coincides with line 1.2, after which it is not difficult to find the coordinates of points 1 and 2.
The course of the solution is as follows. From the coordinates of points 3 and 4 we find the directional angle and the length of the straight line joining them. From the solution of triangle \(3A4\), in which the angles at points 4 and 3 are respectively equal to \(\alpha\) and \(\beta-\alpha\), as is clear from Fig. 2, we compute the directional angles and the lengths of the lines \(3A\) and \(4A\). From the elements thus found we then determine the coordinates of point \(A\). In an entirely analogous way the coordinates of point \(B\) are computed, after which we easily find the directional angle of the line \(AB\). Using the latter, we very simply find the angles \(m_1\) and \(n_1\) at points 4 and 3. When the angles \(m_1\) and \(n_1\) have been found, from the solution of triangle \(4.1.3\) we compute the coordinates of point 1. In the same way we lay off the angles \(m_2\) and \(n_2\) at point \(B\), the coordinates of point \(B\), and, finally, the coordinates of point 2.
The indicated solution is somewhat lengthy, but it is advantageous in the sense that it has many checks and therefore can be carried out even by an inexperienced computer.
Fig. 3
2) With an auxiliary angle (second formulas). The solution set forth below is somewhat shorter, but it requires greater experience on the part of the computer. We conduct the solution in the following way.
From triangles \(3.1.2\) and \(4.1.2\) (Fig. 3) we find the angles \(\gamma_1\) and \(\gamma_2\), and from triangle \(3.4.2\)—the half-sums of the angles \(\varphi\) and \(\psi\). Then, writing the sine condition at point 2 and passing to the sines of the opposite angles, we isolate from it the ratio
\[ \frac{\operatorname{tg}\psi}{\operatorname{tg}\varphi}=\operatorname{tg}Q, \tag{1.1} \]
where \(Q\) is the auxiliary angle.
Transforming the last formula by means of a derived proportion, we find the half-difference of the angles \(\varphi\) and \(\psi\); comparing it with the previously computed half-sum of the same angles, we find \(\varphi\) and \(\psi\). After this, from triangle \(3.4.1\) it is already not difficult to find the coordinates of point 1, and from triangle \(3.4.2\)—the coordinates of point 2.
3) With a conditional directional angle and a base (third formulas). The essence of the method is as follows. Taking for line 1.2 the conditional directional angle \(\tilde t_{1.2}=0\) and the conditional length \(\tilde D_{1.2}=1\), from triangles \(1.3.2\) and \(1.4.2\) (Fig. 4) we find, by these elements, the conditional coordinates \(\tilde x_1,\tilde y_1,\tilde x_2,\tilde y_2\) of points 1 and 2, as well as the conditional directional angle \(\tilde t_{3.4}\) and the conditional length \(\tilde D_{3.4}\) of line \(3.4\). Comparing the latter with their actual
...the true values; we compute the rotation angle \(\Theta=t_{1\cdot2}\) and the relative correction \(m\) of the side lengths, taking these into account we finally pass from the conditional quantities to the true quantities. After this, the true coordinates of points 1 and 2 are found in the usual way.
Formulas I
Formulas
for the solution of a particular double resection with the aid of auxiliary points
\[ \begin{aligned} 1)\quad \operatorname{tg}\,\overline{t}_{3\cdot4} &=\frac{\overline{y}_{4}-\overline{y}_{3}}{\overline{x}_{4}-\overline{x}_{3}} =\frac{\overline{\eta}_{3\cdot4}}{\overline{\xi}_{3\cdot4}} & 2)\quad \overline{D}_{3\cdot4} &=\frac{\overline{\eta}_{3\cdot4}}{\sin \overline{t}_{3\cdot4}} =\frac{\overline{\xi}_{3\cdot4}}{\cos \overline{t}_{3\cdot4}} \end{aligned} \]
\[ \begin{aligned} 3)\quad r_1&=\frac{\overline{D}_{3\cdot4}}{\sin\beta} & 4)\quad r_2&=\frac{\overline{D}_{3\cdot4}}{\sin\delta} \end{aligned} \]
\[ \begin{aligned} 5)\quad D_{3\cdot A}&=r_1\sin\alpha & 9)\quad D_{3\cdot B}&=r_2\sin\gamma \end{aligned} \]
\[ \begin{aligned} 6)\quad D_{4\cdot A}&=r_1\sin(\beta-\alpha) & 10)\quad D_{4\cdot B}&=r_2\sin(\delta-\gamma) \end{aligned} \]
\[ \begin{aligned} 7)\quad t_{3\cdot A}&=\overline{t}_{3\cdot4}-(\beta-\alpha) & 11)\quad t_{3\cdot B}&=\overline{t}_{3\cdot4}+\gamma \end{aligned} \]
\[ \begin{aligned} 8)\quad t_{4\cdot A}&=\overline{t}_{4\cdot3}+\alpha & 12)\quad t_{4\cdot B}&=\overline{t}_{4\cdot3}-(\delta-\gamma) \end{aligned} \]
\[ 13)\quad x_A=\overline{x}_3+D_{3\cdot A}\cos t_{3\cdot A} =\overline{x}_4+D_{4\cdot A}\cos t_{4\cdot A} \]
\[ 14)\quad y_A=\overline{y}_3+D_{3\cdot A}\sin t_{3\cdot A} =\overline{y}_4+D_{4\cdot A}\sin t_{4\cdot A} \]
\[ 15)\quad x_B=\overline{x}_3+D_{3\cdot B}\cos t_{3\cdot B} =\overline{x}_4+D_{4\cdot B}\cos t_{4\cdot B} \]
\[ 16)\quad y_B=\overline{y}_3+D_{3\cdot B}\sin t_{3\cdot B} =\overline{y}_4+D_{4\cdot B}\sin t_{4\cdot B} \]
\[ 17)\quad \operatorname{tg} t_{B\cdot A} =\frac{y_A-y_B}{x_A-x_B} =\frac{\eta_{B\cdot A}}{\xi_{B\cdot A}} \]
\[ \begin{aligned} 18)\quad m_1&=t_{A\cdot3}-t_{A\cdot B} & 24)\quad m_2&=t_{B\cdot3}-t_{B\cdot A} \end{aligned} \]
\[ \begin{aligned} 19)\quad n_1&=t_{A\cdot B}-t_{A\cdot4} & 25)\quad n_2&=t_{B\cdot A}-t_{B\cdot4} \end{aligned} \]
\[ \begin{aligned} 20)\quad t_{3\cdot1}&=\overline{t}_{3\cdot4}+n_1 & 26)\quad t_{3\cdot2}&=\overline{t}_{3\cdot4}-n_2 \end{aligned} \]
\[ \begin{aligned} 21)\quad t_{4\cdot1}&=\overline{t}_{4\cdot3}-m_1 & 27)\quad t_{4\cdot B}&=\overline{t}_{4\cdot3}+m_2 \end{aligned} \]
\[ \begin{aligned} 22)\quad D_{3\cdot1}&=r_1\sin(\delta-\gamma) & 28)\quad D_{3\cdot2}&=r_2\sin\alpha \end{aligned} \]
\[ \begin{aligned} 23)\quad D_{4\cdot1}&=r_1\sin\gamma & 29)\quad D_{4\cdot2}&=r_2\sin(\beta-\alpha) \end{aligned} \]
\[ 30)\quad x_1=\overline{x}_3+D_{3\cdot1}\cos t_{3\cdot1} =\overline{x}_4+D_{4\cdot1}\cos t_{4\cdot1} \]
\[ 31)\quad y_1=\overline{y}_3+D_{3\cdot1}\sin t_{3\cdot1} =\overline{y}_4+D_{4\cdot1}\sin t_{4\cdot1} \]
\[ 32)\quad x_2=\overline{x}_3+D_{3\cdot2}\cos t_{3\cdot2} =\overline{x}_4+D_{4\cdot2}\cos t_{4\cdot2} \]
\[ 33)\quad y_2=\overline{y}_3+D_{3\cdot2}\sin t_{3\cdot2} =\overline{y}_4+D_{4\cdot2}\sin t_{4\cdot2} \]
Formulas
for the solution of the particular double resection with the aid of an auxiliary angle
Formulas 2-a
\[ \begin{aligned} 1)\quad & \gamma_1 = 180^\circ - (\alpha_1 + \beta_1) \\ \\ 2)\quad & \gamma_2 = 180^\circ - (\alpha_2 + \beta_2) \\ \\ 3)\quad & \frac{1}{2}(\varphi+\psi)=90^\circ-\frac{1}{2}(\beta_2-\beta_1) \\ \\ 4)\quad & \frac{\sin \psi}{\sin \varphi}=\operatorname{tg} Q= \frac{\sin \alpha_1}{\sin \alpha_2}\cdot \frac{\sin \gamma_2}{\sin \gamma_1} \\ \\ 5)\quad & \operatorname{tg}\frac{1}{2}(\varphi-\psi) =\operatorname{tg}\frac{1}{2}(\varphi+\psi)\operatorname{ctg}(45^\circ+Q) \\ \\ 6)\quad & \operatorname{tg}\,\overline{t}_{3\cdot4} =\frac{\overline{y}_4-\overline{y}_3}{x_4-x_3} \\ \\ 7)\quad & t_{3\cdot1}=\overline{t}_{3\cdot4}+\varphi+\gamma_1 \\ \\ 8)\quad & t_{3\cdot2}=\overline{t}_{3\cdot4}+\varphi \\ \\ 9)\quad & t_{4\cdot1}=\overline{t}_{4\cdot3}-(\psi-\gamma_2) \\ \\ 10)\quad & t_{4\cdot2}=\overline{t}_{4\cdot3}-\psi \\ \\ 11)\quad & D_{3\cdot1}= \frac{\overline{D}_{3\cdot4}\sin(\psi-\gamma_2)} {\sin(\alpha_1-\alpha_2)} \\ \\ 12)\quad & D_{4\cdot1}= \frac{\overline{D}_{3\cdot4}\sin(\varphi+\gamma_1)} {\sin(\alpha_1-\alpha_2)} \\ \\ 13)\quad & D_{3\cdot2}= \frac{\overline{D}_{3\cdot4}\sin\psi} {\sin(\beta_1-\beta_2)} \\ \\ 14)\quad & D_{3\cdot4}= \frac{\overline{D}_{4\cdot2}\sin\varphi} {\sin(\beta_1-\beta_2)} \\ \\ 15)\quad & x_1=\overline{x}_3+D_{3\cdot1}\cos t_{3\cdot1} =\overline{x}_4+D_{4\cdot1}\cos t_{4\cdot1} \\ \\ 16)\quad & y_1=\overline{y}_3+D_{3\cdot1}\sin t_{3\cdot1} =\overline{y}_4+D_{4\cdot1}\sin t_{4\cdot1} \\ \\ 17)\quad & x_2=\overline{x}_3+D_{3\cdot2}\cos t_{3\cdot2} =\overline{x}_4+D_{4\cdot2}\cos t_{4\cdot2} \\ \\ 18)\quad & y_2=\overline{y}_3+D_{3\cdot2}\sin t_{3\cdot2} =\overline{y}_4+D_{4\cdot2}\sin t_{4\cdot2} \end{aligned} \]
Formulas
for the solution of the particular double resection with the aid of conditional direction angle and base
Formulas 3-i
We provisionally take \(\widetilde{D}_{1\cdot2}=1\) and \(\widetilde{t}_{1\cdot2}=0\). Then:
\[ \begin{aligned} 1)\quad & \widetilde{D}_{1\cdot3}=\frac{\sin\delta}{\sin(\alpha+\delta)} \\ \\ 2)\quad & \widetilde{D}_{2\cdot3}=\frac{\sin\alpha}{\sin(\alpha+\delta)} \\ \\ 3)\quad & \widetilde{\xi}_{1\cdot3}=+\widetilde{D}_{1\cdot3}\cos\alpha \\ \\ 4)\quad & \widetilde{\eta}_{1\cdot3}=-\widetilde{D}_{1\cdot3}\sin\alpha \\ \\ 5)\quad & \widetilde{\xi}_{2\cdot3}=-\widetilde{D}_{2\cdot3}\cos\delta \\ \\ 6)\quad & \widetilde{\eta}_{2\cdot3}=-\widetilde{D}_{2\cdot3}\sin\delta \\ \\ 7)\quad & \widetilde{D}_{1\cdot4}=\frac{\sin\gamma}{\sin(\beta+\gamma)} \\ \\ 8)\quad & \widetilde{D}_{2\cdot4}=\frac{\sin\beta}{\sin(\beta+\gamma)} \end{aligned} \]
Fig. 4
\[ 9)\quad \tilde{\xi}_{1\cdot4}=\tilde{D}_{1\cdot4}\cos\beta \qquad 10)\quad \tilde{\eta}_{1\cdot4}=-\tilde{D}_{1\cdot4}\sin\beta \]
\[ 11)\quad \tilde{\xi}_{2\cdot4}=-\tilde{D}_{2\cdot4}\cos\gamma \qquad 12)\quad \tilde{\eta}_{2\cdot4}=\tilde{D}_{2\cdot4}\sin\gamma \]
Check:
\[ 13)\quad \tilde{\xi}_{1\cdot3}-\tilde{\xi}_{2\cdot3} =\tilde{\xi}_{1\cdot4}-\tilde{\xi}_{2\cdot4}=1 \]
\[ 14)\quad \tilde{\eta}_{1\cdot3}-\tilde{\eta}_{2\cdot3} =\tilde{\eta}_{1\cdot4}-\tilde{\eta}_{2\cdot4}=0 \]
\[ 15)\quad \operatorname{tg}\tilde{t}_{3\cdot4} = \frac{\tilde{\eta}_{1\cdot4}-\tilde{\eta}_{3\cdot4}} {\tilde{\xi}_{1\cdot4}-\tilde{\xi}_{1\cdot3}} = \frac{\tilde{\eta}_{3\cdot4}}{\tilde{\xi}_{3\cdot4}} = \frac{\tilde{\eta}_{2\cdot4}-\tilde{\eta}_{2\cdot3}} {\tilde{\xi}_{2\cdot4}-\tilde{\xi}_{2\cdot3}} \]
\[ 16)\quad \tilde{D}_{3\cdot4} = \frac{\tilde{\xi}_{3\cdot4}}{\cos\tilde{t}_{3\cdot4}} = \frac{\tilde{\eta}_{3\cdot4}}{\sin\tilde{t}_{3\cdot4}} \]
\[ 17)\quad \operatorname{tg}\bar{t}_{3\cdot4} = \frac{\bar{y}_{4}-\bar{y}_{3}}{\bar{x}_{4}-\bar{x}_{3}} \]
\[ 18)\quad \bar{D}_{3\cdot4} = \frac{\bar{x}_{4}-\bar{x}_{3}}{\cos\bar{t}_{3\cdot4}} = \frac{\bar{y}_{4}-\bar{y}_{3}}{\sin\bar{t}_{3\cdot4}} \]
\[ 19)\quad \bar{t}_{3\cdot4}-\tilde{t}_{3\cdot4}=\Theta=t_{1\cdot2} \]
\[ 20)\quad m=\frac{\bar{D}_{3\cdot4}}{\tilde{D}_{3\cdot4}} \]
Solutions of Type II (by natural formulas)
Nuvaryev’s solution (formulas 4). In 1936, V. S. Nuvaryev, assistant professor of the Irkutsk Mining Institute, published an original solution of the double resection, designed for use with an arithmometer. The essence of the method is as follows.
First of all, for \(\operatorname{tg}\alpha\) (Fig. 5) its expression is composed through the tangents of the directional angles of the sides forming this angle. The initial formula, as is known from analytic geometry, will have the following form:
Fig. 5
\[ \operatorname{tg}\alpha= \frac{\operatorname{tg}t_{1\cdot4}-\operatorname{tg}t_{1\cdot3}} {1+\operatorname{tg}t_{1\cdot4}\operatorname{tg}t_{1\cdot3}}. \]
Next, the tangents of the direction angles on the right-hand side of equality (1.2) are replaced by their expressions through the coordinates of points 1, 3, 4 in a particular coordinate system with origin at point 1 and axes parallel to the axes of the original system. Taking as unknowns in this particular system the coordinates \(x'_3, y'_3\) of fixed point 3, and as known the coordinates \(x'_1=0\), \(y'_1=0\) of the point 1 being determined, after some simplifications we arrive at a quadratic equation with two unknowns, which will be the coordinates \(x'_3, y'_3\) of fixed point 3 in the particular system.
To obtain the second missing equation, one first finds, from the coordinates of the endpoints in the particular system, the length of side \(1.3\), and from the solution of triangle \(1.3.2\), the length of side \(1.2\). Next, from the same coordinates, \(\sin t_{1.4}\) and \(\cos t_{1.4}\) are computed; knowing these, \(\sin t_{1.2}\) and \(\cos t_{1.2}\) are then found from the formulas:
\[ \sin t_{1.2}=\sin(t_{1.4}+\gamma), \]
\[ \cos t_{1.2}=\cos(t_{1.4}+\gamma). \tag{1.3} \]
Finally, from the length of side \(1.2\) and its direction angle \(t_{1.2}\), the coordinates \(x'_2,y'_2\) of the second point 2 being determined are computed in the local system. Having then expressed the coordinates of point 4 as functions of the unknown coordinates \(x'_3,y'_3\) of point 3 in the particular system and of the given increments \(x_4-x_3=\xi_{3.4}\), \(y_4-y_3=\eta_{3.4}\) of side \(3.4\), one computes from these quantities \(\operatorname{tg} t_{2.3}\) and \(\operatorname{tg} t_{2.4}\).
In conclusion, an expression for \(\operatorname{tg}\beta\) is formed by the formula:
\[ \operatorname{tg}\beta= \frac{\operatorname{tg} t_{2.4}-\operatorname{tg} t_{2.3}} {1+\operatorname{tg} t_{2.4}\cdot \operatorname{tg} t_{2.3}}, \tag{1.4} \]
which, after simplification, gives one more quadratic equation with the same two unknowns \(x'_3,y'_3\). Solving both equations (1.2) and (1.4) simultaneously, we find the required coordinates \(x'_3,y'_3\) of point 3 in the particular system. After this it is no longer difficult to compute the coordinates of the first point 1 being determined in the general system.
To obtain the coordinates of the second point 2 being determined, we move the origin of coordinates to point 2, still preserving the direction of the axes parallel to the axes of the general system. Interchanging, in the preceding reasoning, the roles of points 3 and 4 and of the angles \(\alpha\) and \(\beta\), \(\gamma\) and \(\delta\), i.e., taking as unknown the coordinates \(x'_4, y'_4\) of point 4 in the particular system, we obtain for the coordinates of the second point 2 being determined in the general system final formulas entirely analogous to the formulas for the coordinates of the first point 1 being determined.
§ 2. Generalized double resection
Below we shall give two solutions of the so-called generalized double resection. The point is that the double resection of two points being determined from two fixed points is only a special case of a more general problem, since the initial conditions of the original problem can be extended.
The first generalization consists in the following: we regard point 4 as split into 2 points, 4 and 5, in consequence of which, in angle \(\gamma\), side \(2.4\) becomes \(2.5\) (Fig. 6). In this case the number of fixed points is 3, and the number of measured angles remains the same, equal to 4.
Assuming further that point \(\overline{3}\) splits into \(2:\overline{3}\) and \(\overline{6}\), while side \(2,\overline{3}\) passes into \(2,\overline{6}\), we obtain a second generalization, in which the number of fixed points is equal to 4 and the number of measured angles is also equal to 4 (Fig. 7). Further generalization of the initial conditions of the double resection will be impossible, since by the second generalization we have attained a one-to-one correspondence between the control points and the measured angles. Therefore, with further splitting of the control points we shall obtain points in general not connected with the determined points 1 and 2, provided only that the number of measured angles remains equal to 4.
Fig. 6
Fig. 7
The achieved generalization of the double resection with respect to 4 control points evidently includes, as special cases, the cases of 3 and 2 points. If
formulas for solving the generalized double resection have been found, then by the same token we shall also solve the last 2 cases. Geodesists have taken up the solution of the generalized double inverse resection comparatively recently; therefore in the periodical press there are only a few methods, and not always original ones. This happened, for example, with the solution of N. Nazarov, published in No. 9 of Geodezist for 1935, which, through unfamiliarity with the geodetic literature, repeated Sosna’s method, published as early as 1896. Below we give 2 original solutions of the generalized double inverse resection, given by Soviet geodesists.
- The solution of A. Kabenin (formulas 5). First of all, the author writes expressions for the tangents of the direction angles of the sides connecting the points to be determined with one another and with the fixed points (Fig. 8). In these equations, the unknowns are the coordinates of the points to be determined \(x_1, y_1, x_2, y_2\) and the direction angle \(t_{1\cdot 2}\) of the side connecting the points to be determined with one another. The constants are the values of the measured angles. Further, using the expressions for the tangents of the direction angles of the directions from the points to be determined to the fixed points, the author eliminates from the expression for \(\operatorname{tg} t_{1\cdot 2}\) the unknown coordinates of the points to be determined 1 and 2 and obtains
Fig. 8
Formulas 4
Formulas
for the solution of the particular double inverse resection according to Nuvariev
\[ \begin{aligned} 1)&\quad \xi_{3\cdot 4}=\bar{x}_4-\bar{x}_3,\\ 2)&\quad \eta_{3\cdot 4}=\bar{y}_4-\bar{y}_3,\\ 3)&\quad a=\xi_{3\cdot 4}-\eta_{3\cdot 4}\operatorname{ctg}\alpha,\\ 4)&\quad b=\eta_{3\cdot 4}+\xi_{3\cdot 4}\operatorname{ctg}\alpha,\\ 5)&\quad c=\xi_{3\cdot 4}-\eta_{3\cdot 4}\operatorname{ctg}\beta,\\ 6)&\quad d=\eta_{3\cdot 4}+\xi_{3\cdot 4}\operatorname{ctg}\beta,\\ 7)&\quad \varphi=180^\circ-(\gamma+\delta), \end{aligned} \]
\[ \begin{aligned} 11)&\quad t_{2\cdot 3}=t_{1\cdot 3}-\varphi,\\ 12)&\quad k_2=\operatorname{tg} t_{2\cdot 3},\\ 13)&\quad x_2-\bar{x}_3=\frac{c+k_2 d}{k_2^2+1},\\ 14)&\quad y_2-\bar{y}_3=k_2(x_2-\bar{x}_3). \end{aligned} \]
Check:
\[ 15)\quad \frac{\bar{y}_4-y_1}{x_4-x_1}=\operatorname{tg} t_{1\cdot 4} \qquad 16)\quad \frac{y_2-y_1}{x_2-x_1}=\operatorname{tg} t_{1\cdot 2} \]
\[ 17)\quad \frac{\bar{y}_4-y_2}{x_4-x_2}=\operatorname{tg} t_{2\cdot 4} \]
\[ 8)\quad k_1= \frac{(a-c)\operatorname{ctg}\varphi+a\operatorname{ctg}\delta+d} {(d-b)\operatorname{ctg}\varphi-b\operatorname{ctg}\delta+c} = \frac{M}{N} = \operatorname{tg} t_{1\cdot 3} \]
\[ \begin{aligned} 9)&\quad x_1-\bar{x}_3=\frac{a+k_1 b}{k_1^2+1},\\ 10)&\quad y_1-\bar{y}_3=k_1(x_1-\bar{x}_3), \end{aligned} \]
\[ 18)\quad t_{1\cdot 4}-t_{1\cdot 3}=\alpha \qquad 20)\quad t_{1\cdot 4}-t_{2\cdot 3}=\beta, \]
\[ 19)\quad t_{1\cdot 2}-t_{1\cdot 4}=\gamma \qquad 21)\quad t_{2\cdot 3}-t_{2\cdot 4}=\delta. \]
Formulas 5
Formulas
for solving the generalized double resection according to Kabeinin
\[ \begin{aligned} &1)\quad (\bar{x}_4-\bar{x}_3)\sin\beta+(\bar{y}_4-\bar{y}_3)\cos\beta=a \end{aligned} \]
\[ \begin{aligned} &2)\quad (\bar{x}_4-\bar{x}_3)\cos\beta-(\bar{y}_4-\bar{y}_3)\sin\beta=b \end{aligned} \]
\[ \begin{aligned} &3)\quad -(\bar{x}_6-\bar{x}_5)\sin(\gamma+\delta)+(\bar{y}_6-\bar{y}_5)\cos(\gamma+\delta)=c \end{aligned} \]
\[ \begin{aligned} &4)\quad (\bar{x}_6-\bar{x}_5)\cos(\gamma+\delta)+(\bar{y}_6-\bar{y}_5)\sin(\gamma+\delta)=d \end{aligned} \]
\[ \begin{aligned} &5)\quad \frac{y_2-y_1}{x_2-x_1} =\operatorname{tg}t_{1\cdot 2} =\frac{a\sin(\alpha+\beta)\sin\delta+c\sin\alpha\sin\gamma+(\bar{y}_6-\bar{y}_4)\sin\alpha\sin\delta} {b\sin(\alpha+\beta)\sin\delta+d\sin\alpha\sin\gamma+(\bar{x}_6-\bar{x}_4)\sin\alpha\sin\delta} \end{aligned} \]
\[ \begin{aligned} &6)\quad x_1-\bar{x}_4= \frac{(\bar{y}_4-\bar{y}_3)-(\bar{x}_4-\bar{x}_3)\operatorname{tg}(t_{1\cdot 2}-\beta)} {\operatorname{tg}(t_{1\cdot 2}-\beta)-\operatorname{tg}[t_{1\cdot 2}-(\alpha+\beta)]} \end{aligned} \]
\[ \begin{aligned} &7)\quad x_2-\bar{x}_6= \frac{(\bar{y}_6-\bar{y}_5)-(\bar{x}_6-\bar{x}_5)\operatorname{tg}[t_{1\cdot 2}+(\gamma+\delta)]} {\operatorname{tg}[t_{1\cdot 2}+(\gamma+\delta)]-\operatorname{tg}(t_{1\cdot 2}+\gamma)} \end{aligned} \]
\[ \begin{aligned} &8)\quad y_1-\bar{y}_4=(x_1-\bar{x}_4)\operatorname{tg}[t_{1\cdot 2}-(\alpha+\beta)] \end{aligned} \]
\[ \begin{aligned} &9)\quad y_2-\bar{y}_6=(x_2-\bar{x}_6)\operatorname{tg}(t_{1\cdot 2}+\gamma) \end{aligned} \]
\[ \begin{aligned} &10)\quad x_1=(x_1-\bar{x}_4)+\bar{x}_4 \end{aligned} \]
\[ \begin{aligned} &11)\quad y_1=(y_1-\bar{y}_4)+\bar{y}_4 \end{aligned} \]
\[ \begin{aligned} &12)\quad x_2=(x_2-\bar{x}_6)+\bar{x}_6 \end{aligned} \]
\[ \begin{aligned} &13)\quad y_2=(y_2-\bar{y}_6)+\bar{y}_6 \end{aligned} \]
to a first-degree equation with one unknown \(t_{1\cdot 2}\). After finding the direction angle \(t_{1\cdot 2}\), the value of the latter is substituted into the principal system of equations, from which we find the coordinates of points 1 and 2. Thus, we see that Kabeinin’s solution is, in idea, very simple. A certain disadvantage of the method is that the formulas contain a rather large number of trigonometric functions of various angles.
Fig. 9
2. Gan’shin’s solution (formulas 6) is based on introducing the auxiliary angle
\[ \omega=\alpha+\varphi=\delta+\psi, \]
which, in the author’s opinion, considerably simplifies the solution (Fig. 9).
Proceeding from the equality of the vertical angles at the point \(O\) of intersection of the sides \(\overline{1.2}\) and \(\overline{3.6}\), we find the relation between the angles of the triangles \(1.3.O\) and \(2.6.O\). Next we determine the angle \(1.4.3\) as a function of the angles \(\omega, \mu\) and \(\beta\), after which, from triangle \(\overline{3.1.4}\), we find the length \(D_{1.3}\) of the side \(1.3\), and from triangle \(1.3.O\) the length \(D_{3.O}\) of the segment \(3.O\). In the same way we compute the length \(D_{6.O}\) of the segment \(6.O\). Since \(D_{3.6} = D_{3.O} + D_{6.O}\), and the length \(D_{3.6}\) is known, then, substituting in the last equality the values of \(D_{3.O}\) and \(D_{6.O}\) as functions of the known angles \(\alpha, \beta, \gamma, \delta\) and of the auxiliary angle \(\omega\), we finally arrive at an equation from which the quantity \(\omega\) is determined. When the angle \(\omega\) has been found, from the triangles \(\overline{3.1.O}\) and \(\overline{6.2.O}\) we find the coordinates of points 1 and 2. It is very regrettable that the author did not give a check formula for computing the angle \(\omega\), which is absolutely necessary for practical computations.
**Formulae** for the solution of the generalized double resection according to Ganyshin **6 formulae**\[ \text{1) }\quad \alpha+\varphi=\delta+\psi=\omega \]
\[ \text{2) }\quad P=D_{3.4}\frac{\sin\alpha}{\sin(\alpha+\beta)} \]
\[ \text{3) }\quad Q=D_{5.6}\frac{\sin\delta}{\sin(\gamma+\delta)} \]
\[ \text{4) }\quad \mu=\bar t_{3.4}-\bar t_{3.6} \]
\[ \text{5) }\quad \nu=\bar t_{6.5}-\bar t_{6.3} \]
\[ \text{6) }\quad \operatorname{ctg}\omega= \frac{D_{3.6}-P\cos(\beta-\mu)-Q\cos(\gamma-\nu)} {P\sin(\beta-\mu)+Q\sin(\gamma-\nu)} \]
The further solution of the problem (after finding the angle \(\omega\)) reduces to the usual case of solving triangles by two angles and one side. The directional angles \(t_{3.4}, t_{3.6}, t_{6.5}\) and the lengths \(D_{3.4}, D_{3.6}, D_{5.6}\) of the sides \(\overline{3.4}, \overline{3.6}, \overline{5.6}\) are found from the coordinates of the endpoints of the fixed points \(\overline{3}, \overline{4}, \overline{5}, \overline{6}\).
After this brief survey of the existing solutions of double resection, below a new solution of the same problem is presented, based on analytic considerations.
§ 3. New solution of the generalized double resection
(by natural formulae)
To facilitate the derivations and simplify the formulae, one of the fixed points, for example \(\overline{3}\) (Fig. 7), is taken as a particular origin of coordinates. Denoting the coordinates of the points relative to the new origin with a prime, we obtain the following relation between the old and new coordinates, which we express in the form of Table 1.
**Table 1**| \(\overline{3}\) | \(\overline{4}\) | \(\overline{5}\) | \(\overline{6}\) | \(1\) | \(2\) |
|---|---|---|---|---|---|
| \(x'_3=x_3-x_3=0\) \(y'_3=y_3-y_3=0\) |
\(x'_4=x_4-x_3\) \(y'_4=y_4-y_3\) |
\(x'_5=x_5-x_3\) \(y'_5=y_5-y_3\) |
\(x'_6=x_6-x_3\) \(y'_6=y_6-y_3\) |
\(x'_1=x_1-x_3\) \(y'_1=y_1-y_3\) |
\(x'_2=x_2-x_3\) \(y'_2=y_2-y_3\) |
Having transformed the coordinates, we shall now find the dependence between the measured angles and the coordinates of the fixed and determined points. To do this we shall use formula (1.2), given earlier. The tangents of the direction angles of lines 1.3, 1.4, 1.5, 2.6, needed for compiling the indicated formula, will be expressed through the coordinates of the ends of these lines (Table 2).
Table 2.
| $\operatorname{tg} t_{1.3}=\dfrac{o-y'_1}{o-x'_1}=\dfrac{y'_1}{x'_1}$ | $\operatorname{tg} t_{1.2}=\dfrac{y'_2-y'_1}{x'_2-x'_1}$ | $\operatorname{tg} t_{1.4}=\dfrac{y'_4-y'_1}{x'_4-x'_1}$ | $\operatorname{tg} t_{2.5}=\dfrac{y'_5-y'_2}{x'_5-x'_2}$ | $\operatorname{tg} t_{2.6}=\dfrac{y'_6-y'_2}{x'_6-x'_2}$ |
Having found expressions for the direction angles of the lines, we proceed to compose equations between the measured angles and the coordinates of the points, using formula (1.2). The number of these equations will be equal to 4—corresponding to the 4 measured angles. But, as our investigations have shown, it is far from immaterial which combination of these angles we take for substitution into formula (1.2), since the greater or lesser complexity of the final formulas for the coordinates of the determined points 1 and 2 depends on this. The most advantageous combination in the indicated sense is the following (Fig. 7):
\[ 1)\ \alpha;\qquad 2)\ \alpha+\beta;\qquad 3)\ \pm[180^\circ-(\alpha+\delta)];\qquad 4)\ \gamma+\delta \tag{3.1} \]
In the third group, before the square brackets there is a double sign $\pm$. The sign must be chosen so that the inequality
\[ \pm[180^\circ-(\alpha+\delta)]>0, \tag{3.2} \]
is always satisfied, since geometrically the quantity $\pm[180^\circ-(\alpha+\delta)]$ is the numerical value of the angle $\mu$, formed at the point $S$ of intersection of the prolonged directions 1.3 and 2.5. The double sign before the brackets arises because two positions of the point of intersection are possible: $S_1$ and $S_2$ (Figs. 10 and 11). From consideration of the triangles $S_1.1.2$ and $S_2.1.2$ and of the condition $\mu>0$, we obtain precisely the general formula (3.2).
Fig. 10
Fig. 11
Let us now express, by the formula of the form (1.2), the tangents of the angles entering into combination (3.1), substituting for this into the right-hand side of formula (1.2) the values of the tangents of the corresponding direction angles according to Table 2.
Carrying out the indicated substitution, we obtain the following system of 4 equations with 4 unknowns \(x'_1, y'_1, x'_2, y'_2\):
\[ \begin{aligned} 1)\quad \tg\alpha &=a= \frac{\dfrac{y'_2-y'_1}{x'_2-x'_1}-\dfrac{y'_1}{x'_1}} {1+\dfrac{y'_2-y'_1}{x'_2-x'_1}\cdot\dfrac{y'_1}{x'_1}}; \\[1ex] 2)\quad \tg(\alpha+\beta) &=b= \frac{\dfrac{\bar y'_4-y'_1}{\bar x'_4-x'_1}-\dfrac{y'_1}{x'_1}} {1+\dfrac{\bar y'_4-y'_1}{\bar x'_4-x'_1}\cdot\dfrac{y'_1}{x'_1}}; \\[1ex] 3)\quad \tg\{\pm[180^\circ-(\alpha+\delta)]\} &=c= \frac{\dfrac{y'_2-\bar y'_6}{x'_2-\bar x'_6}-\dfrac{y'_1}{x'_1}} {1+\dfrac{y'_2-\bar y'_6}{x'_2-\bar x'_6}\cdot\dfrac{y'_1}{x'_1}}; \\[1ex] 4)\quad \tg(\gamma+\delta) &=d= \frac{\dfrac{\bar y'_6-y'_2}{\bar x'_6-x'_2}-\dfrac{\bar y'_5-y'_2}{\bar x'_5-x'_2}} {1+\dfrac{\bar y'_6-y'_2}{\bar x'_6-x'_2}\cdot\dfrac{\bar y'_5-y'_2}{\bar x'_5-x'_2}} . \end{aligned} \]
Freeing ourselves from denominators and multiplying the factors, we obtain:
\[ \begin{aligned} 1)\quad& a(x'_2x'_1-x_1'^2+y'_2y'_1-y_1'^2) -x'_1y'_2-x'_1y'_1-y'_1x'_2+x'_1y'_1 =x'_1y'_2-y'_1x'_2, \\ 2)\quad& b(\bar x'_4x'_1-x_1'^2+\bar y'_4y'_1-y_1'^2) =\bar y'_4x'_1-x'_1y'_1-\bar x'_4y'_1+x'_1y'_1 =\bar y'_4x'_1-\bar x'_4y'_1, \\ 3)\quad& c(x'_1x'_2-\bar x'_6x'_1+y'_1y'_2-\bar y'_6y'_1) =y'_1x'_2-\bar x'_6y'_1-x'_1y'_2+\bar y'_6x'_1, \\ 4)\quad& d(\bar x'_5\bar x'_6-\bar x'_5x'_2-\bar x'_6x'_2+x_2'^2 +\bar y'_5\bar y'_6-\bar y'_5y'_2-\bar y'_6y'_2+y_2'^2) =\bar x'_5\bar y'_6 \\ &\qquad -\bar x'_5y'_2+\bar y'_6x'_2+y'_2x'_2-\bar y'_5\bar x'_6 +\bar y'_5x'_2+\bar x'_6y'_2-y'_2x'_2 . \end{aligned} \]
We perform the following regrouping of terms:
\[ \begin{aligned} 1)\quad& ax_1'^2-(ax'_2-y'_2)x'_1-(ay'_2+x'_2)y'_1+ay_1'^2=0, \\ 2)\quad& bx_1'^2-(b\bar x'_4-\bar y'_4)x'_1-(b\bar y'_4+\bar x'_4)y'_1+by_1'^2=0, \\ 3)\quad& \bigl[(cx'_2+y'_2)-(c\bar x'_6+\bar y'_6)\bigr]x'_1 +\bigl[(cy'_2-x'_2)-(\bar c\bar y'_6-\bar x'_6)\bigr]y'_1=0, \\ 4)\quad& dx_2'^2-\bigl[d(\bar x'_5+\bar x'_6)-(\bar y'_6-\bar y'_5)\bigr]x'_2 -\bigl[d(\bar y'_5+\bar y'_6)+(\bar x'_6-\bar x'_5)\bigr]y'_2 \\ &\qquad +\bigl[d(\bar x'_5\bar x'_6+\bar y'_5\bar y'_6) +(\bar x'_6\bar y'_5-\bar y'_6\bar x'_5)\bigr]=0 . \end{aligned} \tag{3.4} \]
Introduce the notation:
\[ \begin{gathered} b\bar x'_4-\bar y'_4=e, \qquad c\bar x'_6+\bar y'_6=g, \qquad d(\bar x'_5+\bar x'_6)-(\bar y'_6-\bar y'_5)=i, \\ b\bar y'_4+\bar x'_4=f, \qquad c\bar y'_6-\bar x'_6=h, \qquad d(\bar y'_5+\bar y'_6)+(\bar x'_6-\bar x'_5)=j, \\ d(\bar x'_5\bar x'_6+\bar y'_5\bar y'_6) +(\bar y'_5\bar x'_6-\bar x'_5\bar y'_6)=k . \end{gathered} \tag{3.5} \]
In the new notation, system (3.4) will be written as follows:
\[ \begin{aligned} 1)\quad& ax_1'^2-(ax'_2-y'_2)x'_1-(ay'_2+x'_2)y'_1+ay_1'^2=0, \\ 2)\quad& bx_1'^2-ex'_1-fy'_1+by_1'^2=0, \\ 3)\quad& (cx'_2+y'_2-g)x'_1+(cy'_2-x'_2-h)y'_1=0, \\ 4)\quad& dx_2'^2-ix'_2-jy'_2+dy_2'^2+k=0 . \end{aligned} \tag{3.4.1} \]
Multiplying the 1st equation by \((-b)\), and the 2nd by \((+a)\), and adding, we find:
\[ (abx'_2-by'_2-ae)x'_1+(aby'_2+bx'_2-af)y'_1=0. \tag{3.6} \]
From equation (3.6) and the 3rd equation of system (3.4.1) we obtain:
\[ \frac{y'_1}{x'_1} = -\frac{abx'_2-by'_2-ae}{aby'_2+bx'_2-af} = -\frac{cx'_2+y'_2-g}{cy'_2-x'_2-h}. \tag{3.7} \]
In equation (3.7) we eliminate the denominators and multiply the trinomials:
\[ (abx'_2-by'_2-ae)(cy'_2-x'_2-h)=(aby'_2+bx'_2-af)(cx'_2+y'_2-g) \]
\[ abcx'_2y'_2+bcy_2'^2-acey'_2+abx_2'^2+bx'_2y'_2+aex'_2-abhx'_2+bhy'_2+aeh= \]
\[ =abcx'_2y'_2+bcx_2'^2-afcx'_2+aby_2'^2+bx'_2y'_2-afy'_2-abgy'_2-bgx'_2+afg. \]
After collecting like terms, we move all terms to the left-hand side of the equation and group by powers of \(x'_2, y'_2\):
\[ b(a+c)x_1'^2-\bigl[a(cf-bh)+(ae+bg)\bigr]x'_2-\bigl[a(bg-ce)+(af+bh)\bigr]y'_2+ \]
\[ +b(a+c)y_2'^2+a(fg-eh)=0. \tag{3.7.1} \]
Multiply equation (3.7.1) by \((-d)\), and the 4th equation of system (3.4.1) by \(b(a+c)\) and add them:
\[ \left\{d\bigl[a(cf-bh)+(ae+bg)\bigr]-b(a+c)i\right\}x'_2+ \left\{d\bigl[a(bg-ce)+(af+bh)\bigr]-\right. \]
\[ \left.-b(a+c)j\right\}y'_2+\bigl[b(a+c)k-ad(fg-eh)\bigr]=0. \tag{3.8} \]
Introduce the notation:
\[ d\bigl[a(cf-bh)+(ae+bg)\bigr]-b(a+c)i=P \]
\[ d\bigl[a(bd-ce)+(af+bh)\bigr]-b(a+c)j=Q \tag{3.9} \]
\[ b(c+c)k-ad(fg-eh)=R \]
In the new notation, equation (3.8) will have the following form:
\[ Px'_2+Qy'_2+R=0, \tag{3.8.1} \]
whence
\[ y'_2=-\frac{R+Px'_2}{Q} = -\left(\frac{R}{Q}+\frac{P}{Q}x'_2\right) = M+Nx'_2, \tag{3.10} \]
where
\[ M=-\frac{R}{Q} \qquad N=-\frac{P}{Q}. \tag{3.11} \]
Substituting in the 4th equation of system (3.4) \(y'_2\) by \(M+Nx'_2\), we arrive at the quadratic equation with unknown \(x'_2\):
\[ dx_2'^2-ix'_2-j(M+Nx'_2)+d(M+Nx'_2)^2+k=0 \]
\[ dx_2'^2-ix'_2-jM-jNx'_2+dM^2+2dMNx'_2+dN^2x_2'^2+k=0 \]
\[ d(1+N^2)x_2'^2-(i+jN+2dMN)x'_2+(dM^2-jM+k)=0 \tag{3.12} \]
or
\[ Sx_2'^2-2Tx'_2+U=0, \tag{3.12.1} \]
where
\[ \begin{gathered} d(1+N^{2})=S,\\ i+jN+2dMN=2T,\\ dM^{2}-jM+k=U \end{gathered} \tag{3.13} \]
Finally, solving equation (3.12.1), we find \(x'_2\):
\[ x'_2=\frac{1}{S}\left(T\pm\sqrt{T^{2}-SU}\right). \tag{3.14} \]
Let us analyze the last formula. In it, before the square root there stands the double sign \(\pm\), which in general indicates a duality in the solution of the problem. But this duality is only apparent, since the appearance of the double sign in formula (3.14) is connected with the presence of the double sign in formula (3.2). With regard to the latter, however, we have already indicated that in a concrete problem only one particular sign is possible: either \(+\), or \(-\). Therefore in formula (3.14) one particular sign must be taken, and we need only establish the correspondence between the signs of formulas (3.2) and (3.14). We shall find this connection by applying formula (3.14) to a special case: for example, to the case of two control points. Then, first, from formula (3.14) we obtain:
\[ x'_2=\frac{1}{S}(T\pm T), \tag{3.14.1} \]
since in this case \(U=0\), which follows from formulas (3.5), (3.9), (3.11) and (3.13) when
\[ \bar{x}'_3=\bar{x}'_6=0 \qquad \bar{x}'_4=\bar{x}'_5 \]
\[ \bar{y}'_3=\bar{y}'_6=0 \qquad \bar{y}'_4=\bar{y}'_5 \]
On the other hand, solving directly, under the changed conditions, the system (3.4), we find:
\[ x'_2=\frac{2T}{S}. \tag{3.15} \]
(the derivation is not given here, in view of its complete analogy with the preceding one, for the general case). The two solutions must be identically equal to each other, i.e.
\[ \frac{1}{S}(T\pm T)=\frac{2T}{S}, \tag{3.16} \]
which is fulfilled if the sign \(+\) is taken before the root. Since with 2 control points in (3.2) we also take the sign \(+\), we conclude from this that the correspondence between the signs of formulas (3.14) and (3.2) will be direct, i.e.
\[ \begin{gathered} \text{before the root we take } +,\ \text{if } 180^\circ-(\alpha+\delta)>0,\\ \text{before the root we take } -,\ \text{if } 180^\circ-(\alpha+\delta)<0. \end{gathered} \tag{3.17} \]
Having found the values \(x'_2\) and \(y'_2\) from formulas (3.14) and (3.10), and having passed from \(x'_2,y'_2\), on the basis of Table 1, to \(x_2,y_2\), we check the obtained solution as follows:
\[ \operatorname{tg}(\gamma+\delta)=d=\frac{m-l}{1+lm}, \tag{3.18} \]
where
\[ l=\frac{\bar y_5-y_2}{x_5-x_2}, \qquad m=\frac{\bar y_6-y_2}{x_6-x_2}. \tag{3.19} \]
This completes the first stage of the solution—the determination of the coordinates \(x_2,y_2\) of point 2.
We now proceed to determine the values \(x'_1,y'_1\). Substituting the values \(x'_2,y'_2\) into formula (3.7), we express \(y'_1\) through \(x'_1\):
\[ y'_1=-\frac{(M-g)+x'_2(c+N)}{(cM-h)+x'_2(cN-1)}x'_1=Wx'_1, \tag{3.20} \]
where
\[ W=-\frac{(M-g)+x'_2(c+N)}{(cM-h)+x'_2(cN-1)}. \tag{3.21} \]
Then substituting the found value of \(y'_1\) into the second equation of system (3.4), we arrive at an equation with the unknown \(x'_1\):
\[ bx_1'^2-ex'_1-fWx'_1+bW^2x_1'^2=b(1+W^2)x_1'^2-(e+fW)x'_1=0, \]
whence:
\[ x'_1=\frac{e+fW}{b(1+W^2)}. \tag{3.22} \]
Having found \(x'_1\), we find the unknown \(y'_1\) from formula (3.7):
\[ y'_1=-\frac{cx'_2+y'_2-g}{cy'_2-x'_2-h} \tag{3.23} \]
Then, having computed according to Table 1 the coordinates \(x_1,y_1\) of point 1, we check the found values \(x_1,y_1\) by the formula:
\[ \operatorname{tg}(\alpha+\beta)=b=\frac{s-r}{1+rs}, \tag{3.24} \]
where
\[ r=\frac{y'_1}{x'_1}, \qquad s=\frac{\bar y_4-y_1}{x_4-x_1}. \tag{3.25} \]
This completes the solution of the generalized double resection. It remains only to collect together the necessary apparatus of formulas (Tables 3, 4, and 5).
Table 3
I. Case of 4 control points
| Initial quantities | ||
|---|---|---|
| 1) \(\operatorname{tg}\alpha=a\) 2) \(\operatorname{tg}(\alpha+\beta)=b\) 3) \(\operatorname{tg}\{\pm[180^\circ-(\alpha+\delta)]\}=0\) The sign before \(\sqrt{\ }\) is chosen so that always \(\{\pm[180^\circ-(\alpha+\delta)]\}>0\) 4) \(\operatorname{tg}(\gamma+\delta)=d\) |
We transfer the origin of coordinates to point \(\overline{3}\), denoting the coordinates of the points relative to the new origin by a prime. With the new origin \(\overline{x}'_3=\overline{x}_3-\overline{x}_3=0\) \(\overline{y}'_3=\overline{y}_3-\overline{y}_3=0\) |
(Fig. 7) |
| 5) \(b\overline{x}'_4-\overline{y}'_4=e\) 6) \(b y'_4+\overline{x}'_4=f\) 7) \(c x'_6+y'_6=g\) 8) \(c\overline{y}'_6-\overline{x}'_6=h\) 9) \(d(\overline{x}'_5+\overline{x}'_6)-(y'_6-y'_5)=i\) 10) \(d(\overline{y}'_5+\overline{y}'_6)+(\overline{x}'_6-\overline{x}'_5)=j\) 11) \(d(\overline{x}'_5 x'_6+y'_5\overline{y}'_6)+(y'_5 x'_6-x'_5 y'_6)=k\) 12) \(d[a(cf-bh)+(ae+bg)-b(a+c)i]=P\) 13) \(d[a(bg-ce)+(af+bh)-b(a+c)j]=Q\) 14) \(b(a+c)k-ad(fg-eh)=R\) 15) \(-\dfrac{R}{Q}=M\) 16) \(-\dfrac{P}{Q}=N\) |
17) \(d(1+N^2)=S\) 18) \(i+jN+2dMN=2T\) 19) \(dM^2-jM-k=U\) 20) \(x_2=\overline{x}_3+\dfrac{1}{S}\left(T\pm\sqrt{T^2-SU}\right)=\overline{x}_3+x'_2\) where before \(\sqrt{\ }\) we take \(+\), if \(180^\circ-(\alpha+\delta)>0\), and before \(\sqrt{\ }\) we take \(-\), if \(180^\circ-(\alpha+\delta)<0\) 21) \(y_2=\overline{y}_3+(M+Nx'_2)=\overline{y}_3+y'_2\) 22) \(l=\dfrac{y_5-y_2}{x_5-x_2}\) 23) \(m=\dfrac{\overline{y}_6-y_2}{\overline{x}_6-x_2}\) Check 1 24) \(\operatorname{tg}(\gamma+\delta)=d=\dfrac{m-l}{1+lm}\) |
25) \(\dfrac{(M-g)+x'_2(c+N)}{(cM-h)+x'_2(cN-1)}=W\) 26) \(x_1=\overline{x}_3+\dfrac{e+fW}{b(1+W^2)}=\overline{x}_3+x'_1\) 27) \(y_1=\overline{y}_3-\dfrac{cx'_2+y'_2-g}{cy'_2-x'_2-h}\,x'_1=\overline{y}_3+y'_1\) 28) \(\dfrac{y'_1}{x'_1}=r\) 29) \(\dfrac{\overline{y}_4-y_1}{x_4-x_1}=s\) Check 2 30) \(\operatorname{tg}(\alpha+\beta)=b=\dfrac{s-r}{1+rs}\) |
Such is the apparatus of formulas for solving the double resection under the most general initial conditions. From this we easily obtain, for the case of 3 control points, i.e., putting
\[ \overline{x}_3=\overline{x}_6 \]
\[ \overline{y}_3=\overline{y}_6,\qquad c=\operatorname{tg}[180^\circ-(\alpha+\delta)] \]
the following set of formulas:
Table 4
II. Case of 3 control points
| Initial data | ||
|---|---|---|
| \(1)\ \tg\alpha=a\) \(2)\ \tg(\alpha+\beta)=b\) \(3)\ \tg[180^\circ-(\alpha+\delta)]=c\) \(4)\ \tg(\gamma+\delta)=d\) |
Source data | We transfer the origin of coordinates to point \(\overline{3}\), denoting the coordinates of points relative to the new origin by a prime. With the new origin \(\overline{x}'_3=\overline{x}_3-\overline{x}_3=0\) \(\overline{y}'_3=\overline{y}_3-\overline{y}_3=0\) (Fig. 6) |
| \(5)\ b\overline{x}'_4-\overline{y}'_4=e\) \(6)\ b\overline{y}'_4+\overline{x}'_4=f\) \(7)\ d\overline{x}'_5+\overline{y}'_5=g\) \(8)\ d\overline{y}'_5-\overline{x}'_5=h\) \(9)\ ad(cf-e)=P\) \(10)\ bi(a+c)=Q\) \(11)\ ad(f-ce)=R\) \(12)\ bi(a+c)=S\) \(13)\ \dfrac{P-Q}{S-R}=N\) \(14)\ x_2=\overline{x}_3+\dfrac{g+hN}{d(1+N^2)}=\overline{x}_3+x'_2\) |
\(15)\ y_2=\overline{y}_3+Nx'_2=\overline{y}_3+y'_2\) \(16)\ l=\dfrac{\overline{y}_5-y_2}{x_5-x_2}\) \(17)\ m=\dfrac{y'_2}{x'_2}=N\) Check 1 \(18)\ \tg(\gamma+\delta)=d=\dfrac{m-l}{1+lm}\) |
\(19)\ cx'_2+y'_2=T\) \(20)\ x'_2-cy'_2=U\) \(21)\ \dfrac{T}{U}=M\) \(22)\ x_1=\overline{x}_3+\dfrac{e+fM}{b(1+M^2)}=\overline{x}_3+x'_1\) \(23)\ y_1=\overline{y}_3+Mx'_1=\overline{y}_3+y'_1\) |
\(24)\ \dfrac{y'_1}{x'_1}=r=M\) \(25)\ \dfrac{\overline{y}_4-y_1}{x_4-x_1}=s\) Check 2 \(26)\ \tg(\alpha+\beta)=b=\dfrac{s-r}{1+rs}\) |
We note that here we have introduced somewhat different notation as compared with the preceding case of 4 control points, in order to reduce the number of formulas.
Finally, with 2 control points, when
\[ \overline{x}_3=\overline{x}_6 \qquad \overline{x}_4=\overline{x}_5 \]
\[ \overline{y}_3=\overline{y}_6 \qquad \overline{y}_4=\overline{y}_5 \]
we shall have a solution scheme almost analogous to that of case 2.
Table 5
III. Case of 2 control points
| Initial data | ||
|---|---|---|
| \(1)\ \tg\alpha=a\) \(2)\ \tg(\alpha+\beta)=b\) \(3)\ \tg[(180^\circ-(\alpha+\delta)]=c\) \(4)\ \tg(\gamma+\delta)=d\) |
Source data | We transfer the origin of coordinates to point \(\overline{3}\), denoting the coordinates of points relative to the new origin by a prime. With the new origin \(\overline{x}'_3=\overline{x}_3-\overline{x}_3=0\) \(\overline{y}'_3=\overline{y}_3-\overline{y}_3=0\) (Fig. 1) |
| \(5)\ b\overline{x}'_4-\overline{y}'_4=e\) \(6)\ b\overline{y}'_4+\overline{x}'_4=f\) \(7)\ d\overline{x}'_4+\overline{y}'_4=g\) \(8)\ d\overline{y}'_4-\overline{x}'_4=h\) \(9)\ ad(cf+e)=P\) \(10)\ bg(a+c)=Q\) |
\(11)\ ad(f-ce)=R\) \(12)\ bh(a+c)=S\) \(13)\ \dfrac{P-Q}{S-R}=N\) \(14)\ x_2=\overline{x}_3+\dfrac{g+hN}{d(1+N^2)}=\overline{x}_3+x'_2\) \(15)\ y_2=\overline{y}_3+Nx'_2=\overline{y}_3-y'_2\) |
\(16)\ l=\dfrac{\overline{y}_4-y_2}{x_4-x_2}\) \(17)\ m=\dfrac{y'_2}{x'_2}=N\) Check 1 \(18)\ \tg(\gamma+\delta)=d=\dfrac{m-l}{1+lm}\) |
\(19)\ cx'_2+y'_2=T\) \(20)\ x'_2-cy'_2=U\) \(21)\ \dfrac{T}{U}=M\) \(22)\ x_1=\overline{x}_3+\dfrac{e+fM}{b(1+M^2)}=\overline{x}_3+x'_1\) \(23)\ y_1=\overline{y}_3+Mx'_1=\overline{y}_3+y'_1\) |
\(24)\ \dfrac{y'_1}{x'_1}=r=M\) \(25)\ \dfrac{\overline{y}_4-y_1}{x_4-x_1}=s\) Check 2 \(26)\ \tg(\alpha+\beta)=b=\dfrac{s-r}{1+sr}\) |
Appendix 1
Double resection from 2 points
(for an arithmometer)
According to B. F. Krutoy
| Drawing | Formulas |
|---|---|
| (schematic drawing with points \(1(x_1,y_1)\), \(2(x_2,y_2)\), \(\overline{3}(\bar x_3,\bar y_3)\), \(\overline{4}(\bar x_4,\bar y_4)\), and angles \(\alpha,\beta,\gamma,\delta\)) | We transfer the origin of coordinates to point \(\overline{3}\), denoting the coordinates of the remaining points relative to the new origin with a prime. At the new origin \[x'_3=\bar x_3-\bar x_3=0\] \[y'_3=\bar y_3-\bar y_3=0\] |
| 1) | \(\operatorname{tg}\alpha=a\) | 13) | \(\dfrac{P-Q}{S-R}=N\) |
| 2) | \(\operatorname{tg}(\alpha+\beta)=b\) | 14) | \(x_2=\bar x_3+\dfrac{g+hN}{d(1+N^2)}=\bar x_3+x'_2\) |
| 3) | \(\operatorname{tg}[180^\circ-(\alpha+\delta)]=c\) | 15) | \(y_2=\bar y_3+Nx'_2=\bar y_3+y'_2\) |
| 4) | \(\operatorname{tg}(\gamma+\delta)=d\) | 16) | \(l=\dfrac{\bar y_4-y_2}{\bar x_4-x_2}\) |
| 5) | \(bx'_4-y'_4=e\) | 17) | \(m=\dfrac{y'_2}{x'_2}=N\) |
| 6) | \(by'_4+x'_4=f\) | Check 1 | |
| 7) | \(dx'_4+y'_4=g\) | 18) | \(\operatorname{tg}(\gamma+\delta)=d=\dfrac{m-l}{1+lm}\) |
| 8) | \(dy'_4-x'_4=h\) | 19) | \(cx'_2+y'_2=T\) |
| 9) | \(ad(cf+e)=P\) | 20) | \(x'_2-cy'_2=U\) |
| 10) | \(bg(a+c)=Q\) | 21) | \(\dfrac{T}{U}=M\) |
| 11) | \(ad(f-ce)=R\) | 22) | \(x'_1=\bar x_3+\dfrac{e+fM}{b(1+M^2)}=\bar x_3+x'_1\) |
| 12) | \(bh(a+c)=S\) | 23) | \(y'_1=\bar y_3+Mx'_1=\bar y_3+y'_1\) |
| Initial quantities | 24) | \(\dfrac{y'_1}{x'_1}=r\) | |
| 25) | \(\dfrac{\bar y_4-y_1}{\bar x_4-x_1}=s\) | ||
| Check 2 | |||
| 26) | \(\operatorname{tg}(\alpha+\beta)=b=\dfrac{s-r}{1+rs}\) |
| Angle | Value | Sum | Value |
|---|---|---|---|
| \(\alpha\) | \(54^\circ40'40''.3\) | \(\alpha+\beta\) | \(102^\circ56'50''.3\) |
| \(\beta\) | \(48^\circ16'10''.0\) | \(\alpha+\delta\) | \(94^\circ26'39''.4\) |
| \(\gamma\) | \(33^\circ41'15''.8\) | \(180^\circ-(\alpha+\delta)\) | \(85^\circ33'20''.6\) |
| \(\delta\) | \(39^\circ45'59''.1\) | \(\gamma+\delta\) | \(73^\circ27'14''.9\) |
\[ \operatorname{tg}\alpha=1.141194=a \]
\[ \operatorname{tg}(\alpha+\beta)=-4.349723=b \]
\[ \operatorname{tg}[180^\circ-(\alpha+\delta)]=+12.86617=c \]
\[ \operatorname{tg}(\gamma+\delta)=+3.366047=d \]
Solution
| No. | Quantity | Value | No. | Quantity | Value |
|---|---|---|---|---|---|
| 59 | \(x_1\) | \(+6221940.335\) | 60 | \(y_1\) | \(-63408.317\) |
| 36 | \(x_2\) | \(+6223975.646\) | 37 | \(y_2\) | \(-69365.641\) |
| 1 | \(\bar x_4\) | \(+6222263.350\) | 2 | \(\bar y_4\) | \(-62168.674\) |
| 3 | \(\bar x_3\) | \(+6223241.151\) | 4 | \(\bar y_3\) | \(-64086.985\) |
| 57 | \(x'_1\) | \(-1300.816\) | 58 | \(y'_1\) | \(+673.668\) |
| 34 | \(x'_2\) | \(+734.495\) | 35 | \(y'_2\) | \(+1721.341\) |
| 5 | \(x'_4\) | \(-977.801\) | 6 | \(y'_4\) | \(+1918.311\) |
| 7 | \(b\,y'_4\) | \(-8344.1215\) | 9 | \(b x'_4\) | \(+4253.1635\) |
| 8 | \(d\,y'_4\) | \(+6457.1250\) | 10 | \(d x'_4\) | \(-3291.3241\) |
| 11 | \(b y'_4+x'_4=f\) | \(-9321.9225\) | 13 | \(b x'_4-y'_4=e\) | \(+2334.8525\) |
| 12 | \(d y'_4-x'_4=h\) | \(+7434.9260\) | 14 | \(d x'_4+y'_4=g\) | \(-1373.0131\) |
| 15 | \(bg\) | \(+5972.227\) | 18 | \(bh\) | \(-32339.869\) |
| 16 | \(ad\) | \(+4.7501453\) | 19 | \(a+c\) | \(+14277.364\) |
| 17 | \(cf\) | \(-119937.440\) | 20 | \(ce\) | \(+30040.609\) |
| 21 | \(cf+e\) | \(-117602.587\) | 24 | \(f-ce\) | \(-39362.532\) |
| 22 | \(ad(cf+e)=P\) | \(-558629.376\) | 25 | \(ad(f-ce)=R\) | \(-186977.745\) |
| 23 | \(bg(a+c)=Q\) | \(+85267.655\) | 26 | \(bh(a+c)=S\) | \(-461728.076\) |
| 27 | \(P-Q\) | \(-643897.03\) | 28 | \(S-R\) | \(-274750.33\) |
| No. | Quantity | Value |
|---|---|---|
| 29 | \(N=\dfrac{P-Q}{S-R}\) | \(+2.343571\) |
| 30 | \(hN\) | \(+17424.277\) |
| 31 | \(N^2\) | \(+5.492325\) |
| 32 | \(g+hN\) | \(+16051.264\) |
| 33 | \(d(1+N^2)\) | \(+21.853471\) |
| 34 | \(x'_2\) | \(+734.495\) |
| 35 | \(Nx'_2=y'_2\) | \(+1721.341\) |
| 38 | \(\bar y_4-y_2\) | \(+196.970\) |
| 39 | \(\bar x_4-x_2\) | \(+1712.296\) |
| 40 | \(l\) | \(-0.115033\) |
| 41 | \(m\) | \(+2.343571\) |
| 42 | \(m-l\) | \(+2.458604\) |
| 43 | \(lm\) | \(-0.269588\) |
| 44 | \(1+lm\) | \(+0.730412\) |
| 45 | \(\dfrac{m-l}{1+lm}=d\) | \(+3.366051\) |
| 46 | \((\gamma+\delta)'\) | \(73^\circ27'15''.0\) |
| 47 | \(\omega_2\) | \(+0''.1\) |
| No. | Quantity | Value |
|---|---|---|
| 48 | \(cx'_2\) | \(+9450.1375\) |
| 49 | \(cy'_2\) | \(+22147.066\) |
| 50 | \(cx'_2+y'_2=T\) | \(+11171.479\) |
| 51 | \(x'_2-cy'_2=U\) | \(-21412.571\) |
| 52 | \(M=\dfrac{T}{U}\) | \(-0.521725\) |
| 53 | \(fM\) | \(+4863.480\) |
| 54 | \(M^2\) | \(+0.272197\) |
| 55 | \(e+fM\) | \(+7198.332\) |
| 56 | \(b(1+M^2)\) | \(-5.533705\) |
| 57 | \(x'_1\) | \(-1300.816\) |
| 58 | \(Mx'_1=y'_1\) | \(+678.668\) |
| 61 | \(\bar y_4-y_1\) | \(-1239.643\) |
| 62 | \(\bar x_4-x_1\) | \(-323.015\) |
| 63 | \(s\) | \(+3.837725\) |
| 64 | \(r\) | \(-0.521725\) |
| 65 | \(s-r\) | \(+4.359450\) |
| 66 | \(rs\) | \(-2.002237\) |
| 67 | \(1+rs\) | \(-1.002237\) |
| 68 | \(\dfrac{s-r}{1+rs}=b\) | \(-4.349720\) |
| 69 | \((\alpha+\beta)'\) | \(102^\circ56'50''.3\) |
| 70 | \(\omega_1\) | \(0''0\) |
Appendix 2
Double resection from 3 points
(for an arithmometer)
According to B. F. Krutoy
| Drawing | Formulas |
|---|---|
| Diagram with points \(1(x_1,y_1)\), \(2(x_2,y_2)\), \(\overline{3}(\overline{x}_3,\overline{y}_3)\), \(\overline{4}(\overline{x}_4,\overline{y}_4)\), \(\overline{5}(\overline{x}_5,\overline{y}_5)\), and angles \(\alpha,\beta,\gamma,\delta\). | We transfer the origin of coordinates to point \(\overline{3}\), denoting the coordinates of the remaining points relative to the new origin with a prime. With the new origin: \[\overline{x}'_3=\overline{x}_3-\overline{x}_3=0\] \[\overline{y}'_3=\overline{y}_3-\overline{y}_3=0\] |
| Given angles | Value | Combined angles | Value |
|---|---|---|---|
| \(\alpha\) | \(54^\circ40'40''.3\) | \(\alpha+\beta\) | \(102^\circ56'50''.3\) |
| \(\beta\) | \(48^\circ16'10''.0\) | \(\alpha+\delta\) | \(94^\circ26'39''.4\) |
| \(\gamma\) | \(92^\circ14'35''.9\) | \(180^\circ-(\alpha+\delta)\) | \(85^\circ33'20''.6\) |
| \(\delta\) | \(39^\circ45'59''.1\) | \(\gamma+\delta\) | \(132^\circ00'35''.0\) |
\[ \tg\alpha=+1.411194=a \]
\[ \tg(\alpha+\beta)=-4.349723=b \]
\[ \tg[180^\circ-(\alpha+\delta)]=+12.86717=c \]
\[ \tg(\gamma+\delta)=-1.110233=d \]
\[ \begin{aligned} 1)&\quad \tg\alpha=a\\ 2)&\quad \tg(\alpha+\beta)=b\\ 3)&\quad \tg[180^\circ-(\alpha+\delta)]=c\\ 4)&\quad \tg(\gamma+\delta)=d \end{aligned} \qquad \left\}\ \begin{array}{c} \text{initial}\\ \text{quantities} \end{array}\right. \]
\[ 5)\quad b\overline{x}'_4-\overline{y}'_4=e \]
\[ 6)\quad b\overline{y}'_4+\overline{x}'_4=f \]
\[ 7)\quad b\overline{x}'_5+\overline{y}'_5=g \]
\[ 8)\quad d\overline{y}'_5-\overline{x}'_5=h \]
\[ 9)\quad ad(cf+e)=P \]
\[ 10)\quad bg(a+c)=Q \]
\[ 11)\quad ad(f-ce)=R \]
\[ 12)\quad bh(a+c)=S \]
\[ 13)\quad \frac{P-Q}{S-R}=N \]
\[ 14)\quad x_2=\overline{x}_3+\frac{g+hN}{d(1+N^2)}=\overline{x}_3+x'_2 \]
\[ 15)\quad y_2=\overline{y}_3+Nx'_2=\overline{y}_3+y'_2 \]
\[ 16)\quad l=\frac{\overline{y}_5-y_2}{\overline{x}_5-x_2} \]
\[ 17)\quad m=\frac{y'_2}{x'_2}=N \]
Check 1
\[ 18)\quad \tg(\gamma+\delta)=d=\frac{m-l}{1+lm} \]
\[ 19)\quad cx_2'^2+y_2'^2=T \]
\[ 20)\quad x_2'^2-cy_2'^2=U \]
\[ 21)\quad \frac{T}{U}=M \]
\[ 22)\quad x_1=\overline{x}_3+\frac{e+fM}{b(1+M^2)}=\overline{x}_3+x'_1 \]
\[ 23)\quad y_1=\overline{y}_3+Mx'_1=\overline{y}_3+y'_1 \]
\[ 24)\quad \frac{y'_1}{x'_1}=r \]
\[ 25)\quad \frac{\overline{y}_4-y_1}{\overline{x}_4-x_1}=s \]
Check 2
\[ 26)\quad \tg(\alpha+\beta)=b=\frac{s-r}{1+rs} \]
Solution
| No. | Quantity | Value | No. | Quantity | Value | No. | Quantity | Value | No. | Quantity | Value |
|---|---|---|---|---|---|---|---|---|---|---|---|
| 63 | \(x_1\) | \(+6221989.779\) | 64 | \(y_1\) | \(-63519.425\) | 33 | \(N=\dfrac{P-Q}{S-R}\) | \(+2.754596\) | 52 | \(cx_2'^2\) | \(+7695.0504\) |
| 40 | \(x_2\) | \(+6223839.235\) | 41 | \(y_2\) | \(-62439.505\) | 34 | \(hN\) | \(-9031.684\) | 53 | \(cy_2'^2\) | \(+2196.7578\) |
| 1 | \(\overline{x}_5\) | \(+6223054.149\) | 2 | \(\overline{y}_5\) | \(-60965.324\) | 35 | \(N^2\) | \(+7.587799\) | 54 | \(cx_2'^2+y_2'^2=T\) | \(+9312.5304\) |
| 3 | \(\overline{x}_4\) | \(+6222263.350\) | 4 | \(\overline{y}_4\) | \(-62168.674\) | 36 | \(g+hN\) | \(-5702.407\) | 55 | \(x_2'^2-cy_2'^2=U\) | \(-20598.6738\) |
| 5 | \(\overline{x}_3\) | \(+6223241.151\) | 6 | \(\overline{y}_3\) | \(-64086.985\) | 37 | \(d(1+N^2)\) | \(-9.534458\) | 56 | \(\dfrac{T}{U}=M\) | \(-0.4535500\) |
| \(61'\) | \(x'_1\) | \(-1251.3715\) | \(62'\) | \(y'_1\) | \(+567.5595\) | 38 | \(x'_2\) | \(+598.084\) | 57 | \(fM\) | \(+4227.9579\) |
| \(38'\) | \(x'_2\) | \(+598.084\) | \(39'\) | \(y'_2\) | \(+1647.480\) | 39 | \(Nx'_2=y'_2\) | \(+1647.480\) | 58 | \(M^2\) | \(+0.2057076\) |
| 7 | \(x'_5\) | \(+187.002\) | 9 | \(y'_5\) | \(+3121.661\) | 42 | \(\overline{y}_5-y_2\) | \(-1474.181\) | 59 | \(e+fM\) | \(+6562.8105\) |
| 8 | \(x'_4\) | \(-977.801\) | 10 | \(y'_4\) | \(+1918.311\) | 43 | \(\overline{x}_5-x_2\) | \(+785.086\) | 60 | \(b(1+M^2)\) | \(-5.2444941\) |
| 11 | \(by'_4\) | \(-8344.1215\) | 13 | \(bx'_4\) | \(+4253.1635\) | 44 | \(l\) | \(-1.877732\) | 61 | \(x'_1\) | \(-1251.3715\) |
| 12 | \(dy'_5\) | \(-3465.7711\) | 14 | \(dx'_5\) | \(+207.6158\) | 45 | \(m\) | \(+2.754596\) | 62 | \(y'_1\) | \(+567.5595\) |
| 15 | \(by'_4+x'_4=f\) | \(-9321.9225\) | 17 | \(bx'_4-y'_4=e\) | \(+2334.8525\) | 46 | \(m-l\) | \(+4.632328\) | 65 | \(\overline{y}_4-y_1\) | \(-1350.7515\) |
| 16 | \(dy'_5-x'_5=h\) | \(-3278.7691\) | 18 | \(dx'_5+y'_5=g\) | \(+3329.2768\) | 47 | \(lm\) | \(-5.172393\) | 66 | \(\overline{x}_4-x_1\) | \(-273.5705\) |
| 19 | \(bg\) | \(-14481.432\) | 22 | \(bh\) | \(+4261.737\) | 48 | \(1+lm\) | \(-4.172393\) | 67 | \(s\) | \(+4.937489\) |
| 20 | \(ad\) | \(-1.5668541\) | 23 | \(a+c\) | \(+14.277364\) | 49 | \(\dfrac{m-l}{1+lm}=d\) | \(-1.110233\) | 68 | \(r\) | \(-0.453550\) |
| 21 | \(cf\) | \(-119937.440\) | 24 | \(ce\) | \(+30040.609\) | 50 | \((\gamma+\delta)'\) | \(132^\circ00'35''.0\) | 69 | \(s-r\) | \(+5.391039\) |
| 25 | \(cf+e\) | \(-117602.587\) | 28 | \(f-ce\) | \(-39362.532\) | 51 | \(w_2\) | \(0''.0\) | 70 | \(rs\) | \(-2.239398\) |
| 26 | \(ad(cf+e)=P\) | \(+184254.335\) | 29 | \(ad(f-ce)=S\) | \(+61671.408\) | 71 | \(1+rs\) | \(-1.239398\) | |||
| 27 | \(bg(a+c)=Q\) | \(-206256.676\) | 30 | \(bh(a+c)=R\) | \(+203620.010\) | 72 | \(\dfrac{s-r}{1+rs}=b\) | \(-4.349724\) | |||
| 31 | \(P-Q\) | \(+391011.01\) | 32 | \(S-R\) | \(+141948.60\) | 73 | \((\alpha+\beta)'\) | \(102^\circ56'50''.3\) | |||
| 74 | \(w_1\) | \(0''0\) |
As for the actual computation of the coordinates of points 1 and 2, it is carried out with the aid of an arithmometer; the initial quantities \(a, b, c, d\) are taken with 5, 6, or 7 decimal places, as determined by the accuracy of the observations. The same remark also applies to the required number of digits in the intermediate computations; it may be advised to take everywhere one decimal place more than is required in the final result.
Below are given 2 forms with solved examples for double resection: the 1st—for the case of 2 control points, the 2nd—for the case of 3 control points (appendices 1 and 2).
Literature
- Olkhovskii E. M.—Vertical surveying, Tomsk, 1929.
- Course of Geodesy, edited by Prof. Krasovskii, issue 3, 1934.
- Chebotarev A.—Traversing, lecture notes, 1927.
- Nuvariev V. S.—Solution of the Pothenot and Hansen problems on the plane and in Gauss–Krüger coordinates, GONTI, 1935.
- “GEODEZIST,” Nos. 3–4, 1934.
- “GEODEZIST,” No. 5, 1935.
- “GEODEZIST,” No. 9, 1935.