ANALYTICAL SOLUTION OF A SINGLE RESECTION

(for a mechanical calculator)

B. F. Krutoi

In developing new formulas for solving a single resection (Pothenot’s problem), the aim was to give formulas such that all computations could be carried out exclusively on a mechanical calculator, without the use of tables of trigonometric functions in the intermediate operations. This is possible if operations with angular quantities and their functions are not introduced explicitly into the intermediate formulas. Under this condition, all layouts become the simplest and most mechanical, which provides a guarantee against errors of logical order.

Guided by these considerations, after a number of trials we arrived at the following solution of a single resection (at the beginning of 1933).

Solution

Statement of the problem: At point \(1(x_1, y_1)\), which is to be determined, the angles \(\alpha\) and \(\beta\) between the directions to fixed (control) points \(2(x_2, y_2)\), \(3(x_3, y_3)\), \(4(x_4, y_4)\), whose coordinates \(x_2, y_2;\ x_3, y_3;\ x_4, y_4\) are known, have been measured.

Fig. 1

It is required, from these data, to find the coordinates \(x_1, y_1\) of the sought point 1 (Fig. 1).

Let us proceed to the solution of the problem. First of all, to simplify the subsequent computations, we transfer the origin of coordinates to point \(\overline{2}\), denoting the coordinates of the points relative to the new origin by a prime. We shall have (Table 1):

Table 1

We then express the tangents of the directional angles \(t_{1\cdot2}\), \(t_{1\cdot3}\), \(t_{1\cdot4}\) of the lines \(1.\overline{2}\), \(1.\overline{3}\), \(1.\overline{4}\) through the coordinates of the points 1, 2, 3, 4 in the new system:

\[ \operatorname{tg} t_{1\cdot2} =\frac{0-y'_1}{0-x'_1} =\frac{y'_1}{x'_1}; \qquad \operatorname{tg} t_{1\cdot3} =\frac{\overline{y}'_3-y'_1}{\overline{x}'_3-x'_1}; \qquad \operatorname{tg} t_{1\cdot4} =\frac{\overline{y}'_4-y'_1}{\overline{x}'_4-x'_1} \tag{1} \]

Then, observing that

\[ \alpha=t_{1\cdot3}-t_{1\cdot2}, \qquad \beta=t_{1\cdot4}-t_{1\cdot2}, \tag{2} \]

we may represent the tangents of the angles \(\alpha\) and \(\beta\) in the following form:

\[ \operatorname{tg}\alpha = \frac{\operatorname{tg}t_{1\cdot3}-\operatorname{tg}t_{1\cdot2}} {1+\operatorname{tg}t_{1\cdot2}\operatorname{tg}t_{1\cdot3}} = \frac{ \dfrac{\overline{y}'_3-y'_1}{\overline{x}'_3-x'_1} - \dfrac{y'_1}{x'_1} }{ 1+ \dfrac{\overline{y}'_3-y'_1}{\overline{x}'_3-x'_1} \cdot \dfrac{y'_1}{x'_1} } =a \]

\[ \operatorname{tg}\beta = \frac{\operatorname{tg}t_{1\cdot4}-\operatorname{tg}t_{1\cdot2}} {1+\operatorname{tg}t_{1\cdot2}\operatorname{tg}t_{1\cdot4}} = \frac{ \dfrac{\overline{y}'_4-y'_1}{\overline{x}'_4-x'_1} - \dfrac{y'_1}{x'_1} }{ 1+ \dfrac{\overline{y}'_4-y'_1}{\overline{x}'_4-x'_1} \cdot \dfrac{y'_1}{x'_1} } =b \tag{3} \]

Let us transform this system of two equations with the two unknowns \(x'_1, y'_1\):

\[ a\left(\overline{x}'_3x'_1-x_1'^2+\overline{y}'_3y'_1-y_1'^2\right) = \overline{y}'_3x'_1-y'_1x'_1-\overline{x}'_3y'_1 + x'_1y'_1 = \overline{y}'_3x'_1-\overline{x}'_3y'_1 \]

\[ b\left(\overline{x}'_4x'_1-x_1'^2+\overline{y}'_4y'_1-y_1'^2\right) = \overline{y}'_4x'_1-y'_1x'_1-\overline{x}'_4y'_1 + x'_1y'_1 = \overline{y}'_4x'_1-\overline{x}'_4y'_1 \]

\[ ax_1'^2-(a\overline{x}'_3-\overline{y}'_3)x'_1 -(a\overline{y}'_3+\overline{x}'_3)y'_1 +ay_1'^2=0 \]

\[ bx_1'^2-(b\overline{x}'_4-\overline{y}'_4)x'_1 -(b\overline{y}'_4+\overline{x}'_4)y'_1 +by_1'^2=0 \tag{4} \]

Introducing the notation:

\[ a\overline{x}'_3-\overline{y}'_3=k \qquad b\overline{x}'_4-\overline{y}'_4=m \]

\[ a\overline{y}'_3+\overline{x}'_3=l \qquad b\overline{y}'_4+\overline{x}'_4=n, \tag{5} \]

we write the preceding system as follows:

\[ \begin{aligned} a{x'_1}^{\,2}-kx'_1-ly'_1+a{y'_1}^{\,2}&=0\\ b{x'_1}^{\,2}-mx'_1-ny'_1+b{y'_1}^{\,2}&=0 \end{aligned} \tag{4a} \]

Multiplying the first equation by \(-b\), and the second by \(+a\), and adding, we express \(y'_1\) in terms of \(x'_1\):

\[ y'_1=-\frac{bk-am}{bl-an}\,x'_1=Mx'_1, \tag{6} \]

where

\[ M=+\frac{bk-am}{an-bl}. \tag{7} \]

Now substitute the value of \(y'_1\) from (6) into (4a):

\[ \begin{aligned} a{x'_1}^{\,2}-kx'_1-lMx'_1+aM^2{x'_1}^{\,2}&=0\\ b{x'_1}^{\,2}-mx'_1-nMx'_1+bM^2{x'_1}^{\,2}&=0 \end{aligned} \]

or, after cancellation by \(x'_1\):

\[ \begin{aligned} a(1+M^2)x'_1&=k+lM\\ b(1+M^2)x'_1&=m+nM \end{aligned} \]

Hence we obtain:

\[ x'_1=\frac{k+lM}{a(1+M^2)}=\frac{m+nM}{b(1+M^2)}. \tag{8} \]

Noting that \(x_1=\bar{x}_2+x'_1\) and \(y_1=\bar{y}_2+y'_1\), we finally find:

\[ x_1=\bar{x}_2+\frac{k+lM}{a(1+M^2)}=\bar{x}_2+\frac{m+nM}{b(1+M^2)} \tag{9} \]

\[ y_1=\bar{y}_2+Mx'_1 \]

The following formula may serve as a final check on the computations:

\[ a=\operatorname{tg}\alpha=\frac{d-c}{1+cd}, \tag{10} \]

where

\[ c=\frac{y'_1}{x'_1}\qquad d=\frac{\bar{y}_3-y_1}{x_3-x_1}. \tag{11} \]

An intermediate check—the double computation of \(x'_1\)—may be omitted, since it checks only the function \(M=M(k,l,m,n)\). Therefore, if an error is made in computing \(k,l,m,n\), the two values of \(x'_1\) will agree, but the problem will nevertheless have been solved incorrectly.

Let us collect all the formulas necessary for the computation in the order of successive determination of the corresponding quantities.

\[ \begin{array}{lll} 1)\ \operatorname{tg}\alpha=a & 4)\ a\bar{y}'_3+\bar{x}'_3=l & 7)\ \dfrac{bk-am}{an-bl}=M \\[1.1em] 2)\ \operatorname{tg}\beta=b & 5)\ b\bar{x}'_4-\bar{y}'_4=m & \\[1.1em] 3)\ a\bar{x}'_3-\bar{y}'_3=k & 6)\ b\bar{y}'_4+\bar{x}'_4=n & 8)\ x'_1=\dfrac{k+lM}{a(1+M^2)}=\dfrac{m+nM}{b(1+M^2)} \end{array} \]

Single Resection

For a Mechanical Calculator

According to B. F. Krutoy

\[ \alpha=38^\circ19'00'' \]

\[ \beta=150^\circ09'44'' \]

\[ \operatorname{tg}\alpha=-6.84082=a \]

\[ \operatorname{tg}\beta=+2.77187=b \]

\[ \begin{aligned} 1)\quad &\operatorname{tg}\alpha=a\\ 2)\quad &\operatorname{tg}\beta=b \end{aligned} \]

\[ \begin{aligned} 3)\quad &a\overline{x}'_3-\overline{y}'_3=k\\ 4)\quad &a\overline{y}'_3+\overline{x}'_3=l\\ 5)\quad &b\overline{x}'_4-\overline{y}'_4=m\\ 6)\quad &b\overline{y}'_4+\overline{x}'_4=n \end{aligned} \]

\[ 7)\quad \frac{bk-am}{an-bl}=M \]

\[ 8)\quad x'_1=\frac{k+lM}{a(1+M^2)}=\frac{m+nM}{b(1+M^2)} \]

\[ 9)\quad y'_1=Mx'_1 \]

\[ 10)\quad x_1=\overline{x}_2+x'_1 \]

\[ 11)\quad y_1=\overline{y}_2+y'_1 \]

\[ 12)\quad \frac{y'_1}{x'_1}=c \]

\[ 13)\quad \frac{\overline{y}_3-y_1}{\overline{x}_3-x_1}=d \]

Control.

\[ 14)\quad \operatorname{tg}\alpha=\frac{d-c}{1+cd}=a \]

Solution

\[ 9)\; y'_1 = Mx'_1 \]

\[ 10)\; x_1 = \overline{x}_2 + x'_1 \]

\[ 11)\; y_1 = \overline{y}_2 + y'_1 \]

\[ 12)\; \frac{y'_1}{x'_1} = c \]

\[ 13)\; \frac{y_3 - y_1}{x_3 - x_1} = d \]

Check:

\[ 14)\; \operatorname{tg}\alpha = \frac{d - c}{1 + cd} \]

In conclusion, let us note that formulas close to ours, but obtained on the basis of other considerations, were published in 1936 by Associate Professor of the Tomsk Polytechnic Institute V. S. Nuvar’ev.\(^{1}\) However, V. S. Nuvar’ev indicates the possibility of deriving his formulas also starting from expressions (3).

A computation form is appended to the present work.

\(^{1}\) Nuvar’ev V. S. Solution of the Pothenot and Hansen Problems on the Plane and in Gauss–Krüger Coordinates, GONTI, 1936.