Conversion from Geodetic Coordinates to Rectangular Coordinates and Back Using Special Tables

B. I. Bolshanin
Submitted 1958 | SovietRxiv: ru-195801.08787 | Translated from Russian

Abstract Generated abstract

This paper develops a tabular interpolation method for converting between geodetic coordinates and rectangular coordinates in the Gauss projection, intended to reduce the labor and expertise required by classical formula-based computations. It derives correction terms for interpolation along parallels and meridians, estimates their magnitudes for 1:10000 trapezia, and indicates which terms must be retained to reach millimeter-level accuracy. Worked direct and inverse examples show how coordinates can be obtained from precomputed trapezium corner tables with small curvature corrections. The method is presented as suitable for field use, coordinate checks, and transitions between coordinate zones, provided sufficiently accurate rectangular coordinate tables are available.

Full Text

Conversion from Geodetic Coordinates to Rectangular Coordinates and Back Using Special Tables

B. I. Bolshanin

(Presented to the scientific seminar of the departments of mine surveying and geodesy)

Computational work connected with the conversion of point coordinates from geodetic to rectangular coordinates and back takes much time and requires highly qualified computers, as well as the availability of special tables and the use of calculating machines. The solution of direct and inverse geodetic problems in the practice of topographic-geodetic work has to be carried out rather often and even under field conditions; therefore it must be performed simply and be accessible to specialists of average qualification. The solution of these problems can be significantly simplified and facilitated if, for this purpose, special tables of rectangular coordinates are used, compiled with the necessary accuracy and supplemented by a nomogram or a small table for obtaining corrections.

In the geodetic literature [1], [2], [3], [4], [5] a similar question is considered, but only for solving the direct problem with an accuracy of no more than \(0.1\) m and for very narrow purposes—for determining the rectangular coordinates of the corners of survey trapezia having quite definite geodetic coordinates. The question considered by us, that of accurately determining the coordinates of arbitrary points, is new and has not been raised in the Soviet and foreign literature known to us.

The present work considers the theory of the question, the practical possibility and expediency of this method of conversion from geodetic coordinates to rectangular coordinates and back with an accuracy up to \(0.001\) m.

Solution of the Direct Problem

It is known that a curvilinear trapezium of the terrestrial ellipsoid, bounded by meridians and parallels, is represented on the plane in the conformal Gauss projection also by a curvilinear figure close to a trapezium. If, in solving the problem posed, one uses a trapezium of small dimensions, then the sides of this figure will be close to straight lines. Therefore the coordinates of points lying inside the trapezium or on its boundary may be obtained by linear interpolation, with the introduction of small corrections for the curvature of the images of meridians and parallels on the plane. Let us find the values of these corrections for interpolation along meridians and parallels separately.

Suppose the rectangular coordinates are given for points in the Gauss projection \(A(x_1,y_1)\) and \(B(x_2,y_2)\) (Fig. 1), lying on the same parallel with latitude \(B\), and with longitudes respectively \(L_1\) and \(L_2\). The curve \(AMB\) is the representation on the plane of the parallel; the straight line \(AM'B\) is the chord. The coordinates of the point \(M\), whose longitude is \(L\), may be obtained as follows:

\[ x=x'-\Delta_x^l,\qquad y=y'+\Delta_y^l, \tag{1} \]

where \(x'\) and \(y'\) are coordinates obtained by means of linear interpolation

\[ x'=x_1+(x_2-x_1)\frac{l-l_1}{l_2-l_1}, \]

\[ y'=y_1+(y_2-y_1)\frac{l-l_1}{l_2-l_1}; \tag{2} \]

\(\Delta_x^l\) and \(\Delta_y^l\) are corrections to the interpolated coordinates for the curvature of the representation of the parallel.

Fig. 1

Fig. 1

In formulas (2), \(l_1\), \(l_2\), and \(l\) are distances from the axial meridian of the coordinate zone, equal to:

\[ l_1=L_1-L_0, \]

\[ l_2=L_2-L_0, \]

\[ l=L-L_0, \]

where \(L_0\) is the longitude of the axial meridian of the zone.

Rectangular coordinates are computed by the following formulas of higher geodesy [1].

\[ \begin{aligned} x = X_B &+ \frac{N}{2\rho^{\prime\prime 2}} \sin B \cos^2 B \cdot l^{\prime\prime 2} \\ &+ \frac{N}{24\rho^{\prime\prime 4}} \sin B \cdot \cos^3 B \left(5 - t^2 + 9\eta^2 + 4\eta^4\right) l^{\prime\prime 4} \\ &+ \frac{N}{720\rho^{\prime\prime 6}} \sin B \cos^5 B \left(61 - 58t^2 + t^4\right) l^{\prime\prime 6} + \ldots, \tag{3} \end{aligned} \]

\[ \begin{aligned} y &= \frac{N}{\rho^{\prime\prime}} \cos B \cdot l^{\prime\prime} + \frac{N}{6\rho^{\prime\prime 3}} \cos^3 B \left(1 - t^2 + \eta^2\right) l^{\prime\prime 3} \\ &\quad + \frac{N}{120\rho^{\prime\prime 5}} \cos^5 B \left(5 - 18t^2 + t^4 + 14\eta^2 - 58t^2\eta^2\right) l^{\prime\prime 5} + \ldots \end{aligned} \]

Introduce the notation:

\[ \alpha'=\frac{N}{2\rho^{\prime\prime 2}}\sin B\cos B; \]

\[ \alpha''=\frac{N}{24\rho^{\prime\prime 4}}\sin B\cos^3 B\left(5-t^2+9\eta^2+4\eta^4\right); \]

\[ \beta'=\frac{N}{\rho^{\prime\prime}}\cos B; \tag{4} \]

\[ \beta''=\frac{N}{6\rho^{\prime\prime 3}}\cos^3 B\left(1-t^2+\eta^2\right). \]

Then, without taking into account the terms with \(l^5\) and \(l^6\):

\[ x=X_B+\alpha' l^2+\alpha'' l^4,\qquad y=\beta' l+\beta'' l^3, \tag{5} \]

\[ x_1=X_B+\alpha' l_1^2+\alpha'' l_1^4,\qquad y_1=\beta' l_1+\beta'' l_1^3, \]

\[ x_2=X_B+\alpha' l_2^2+\alpha'' l_2^4,\qquad y_2=\beta' l_2+\beta'' l_2^3. \]

Using formulas (1) and (2), write the expression for the correction \(\Delta_x^l\):

\[ \Delta_x^l=(x_1-x)+(x_2-x_1)\frac{\delta l}{\Delta l}, \tag{6} \]

where

\[ \delta l=l-l_1,\quad \Delta l=l_2-l_1. \]

Substituting the values of \(x\) from (5), we obtain:

\[ \Delta l_x^{\,1}=\alpha^1\left[(l_1^2-l^2)+(l_2^2-l_1^2)\frac{\delta l}{\Delta l}\right]+ \]

\[ +\alpha^2\left[(l_1^4-l^4)+(l_2^4-l_1^4)\frac{\delta l}{\Delta l}\right], \tag{6a} \]

or, in abbreviated form,

\[ \Delta l_x^{\,1}=I_x^l+II_x^l . \]

After transforming the expressions in square brackets and taking (4) into account, we obtain:

\[ I_x^l=\frac{N}{2\rho^2}\sin B\cos B(\Delta l\,\delta l-\delta l^2), \tag{7} \]

\[ II_x^l=\frac{N}{4\rho^4}\sin B\cos^3 B(5-t^2+9\eta^2+4\eta^4)(\Delta l\,\delta l-\delta l^2)l_x^2 . \tag{8} \]

The factor \(\Delta l\,\delta l-\delta l^2\) is symmetric and does not depend on \(l\). Therefore the first correction term \(I_x^l\) for all points at latitude \(B\) in trapezia of equal size will be the same and may be presented in the form of a small table. The maximum value of this correction term will occur when \(\delta l=\frac{1}{2}\Delta l\), that is, at the middle of the trapezium sheet.

The second correction term \(II_x^l\) depends both on the dimensions of the trapezium and on its distance from the axial meridian of the coordinate zone. Its maximum values for the corresponding \(l\) will also occur when \(\delta l=\frac{1}{2}\Delta l\).

The numerical values of the correction terms \((I_x^l)_{\max}\) and \((II_x^l)_{\max}\) are given in Tables 1 and 2 for a trapezium at a scale of 1:10000, having \(\Delta l=3'45''=225''\).

Table 1

\(B\) \((I_x^l)_{\max}\) \(B\) \((I_x^l)_{\max}\)
m m
\(30^\circ\) \(+0,411\) \(55^\circ\) \(+0,447\)
35 0,446 60 0,412
40 0,468 65 1,364
45 0,475 70 0,306
50 0,468 75 0,238

Table 2

\[ (II_x^l)_{\max} \]

\(B\) \(l\) \(2^\circ 00'\) \(2^\circ 30'\) \(3^\circ 00'\) \(3^\circ 30'\)
mm mm mm mm
\(30^\circ\) \(+1\) \(+1\) \(+2\) \(+3\)
35 1 1 2 3
40 1 1 2 2
45 1 1 1 2
50 0 1 1 1
55 0 0 1 1

Similarly to (6), let us write the expression for the correction \(\Delta_y^l\):

\[ \Delta_y^l=(y-y_1)-(y_2-y_1)\frac{\delta l}{\Delta l} \tag{9} \]

and, taking (5) into account, we obtain:

\[ \Delta_y^l=\beta'\left[(l-l_1)-(l_2-l_1)\frac{\delta l}{\Delta l}\right]+ \]

\[ +\beta''\left[(l^3-l_1^3)-(l_2^3-l_1^3)\frac{\delta l}{\Delta l}\right] \tag{9a} \]

or

\[ \Delta_y^l=I_y^l+II_y^l. \]

\[ I_y^l=\beta'\left[(l-l_1)-(l_2-l_1)\frac{\delta l}{\Delta l}\right]=0. \tag{10} \]

\[ II_y^l=-\frac{N}{6\rho^3}\cos^3 B(1-t^2+\eta^2)(l_1+l+l_2)(\Delta l\,\delta l-\delta l^2). \tag{11} \]

The values of \((II_y^l)_{\max}\) are given in Table 3.

Let us now find the corrections to the coordinates when interpolating by latitude. Let the points \(A\) and \(B\), whose coordinates are \(x_1, y_1\) and \(x_2, y_2\), lie on one meridian with longitude \(l\) (Fig. 2). The curve \(AMB\) is the representation of the meridian on the plane; the straight line \(AM'B\) is the chord. According to the drawing:

\[ x=x'+\Delta_x^B,\qquad y=y'+\Delta_y^B, \tag{12} \]

where

\[ x'=x_1+(x_2-x_1)\frac{B-B_1}{B_2-B_1},\qquad y'=y_1+(y_2-y_1)\frac{B-B_1}{B_2-B_1}; \tag{13} \]

\(\Delta_x^B\) and \(\Delta_y^B\)—corrections for the curvature of the representation of the meridian on the plane.

Table 3

\(B\) \(l\) \(l\) \(l\) \(l\) \(l\) \(l\) \(l\)
\(0^\circ 30'\) \(1^\circ 00'\) \(1^\circ 30'\) \(2^\circ 00'\) \(2^\circ 30'\) \(3^\circ 00'\) \(3^\circ 30'\)
mm mm mm mm mm mm mm
\(30^\circ\) −4 −7 −11 −14 −18 −22 −25
35 2 5 7 9 12 14 16
40 1 2 3 4 6 7 8
45 0 0 0 0 0 0 0
50 +1 +2 +3 +4 +5 +6 +6
55 2 3 5 6 8 10 11
60 2 4 6 8 10 12 15
65 2 5 7 9 11 14 16
70 2 4 7 9 11 13 15
75 2 4 6 7 9 11 13

Introducing the notation

\[ B - B_1 = \delta B, \qquad B_2 - B_1 = \Delta B; \]

from formulas (12) and (13) we compose expressions for the corrections \(\Delta_x^B\) and \(\Delta_y^B\):

\[ \Delta_x^B = (x - x_1) - (x_2 - x_1)\frac{\delta B}{\Delta B}; \]

\[ \Delta_y^B = (y - y_1) - (y_2 - y_1)\frac{\delta B}{\Delta B}. \tag{14} \]

According to the notation introduced in (4), we write:

\[ \begin{aligned} x &= X_B + \alpha' l^2 + \alpha'' l^4,\\ x_1 &= X_{B_1} + \alpha'_1 l^2 + \alpha''_1 l^4,\\ x_2 &= X_{B_2} + \alpha'_2 l^2 + \alpha''_2 l^4,\\ y &= \beta' l + \beta'' l^3,\\ y_1 &= \beta'_1 l + \beta''_1 l^3,\\ y_2 &= \beta'_2 l + \beta''_2 l^3. \end{aligned} \tag{15} \]

Fig. 2

After substituting the values of \(x\) into (14), we obtain an expression for \(\Delta_x^B\)

\[ \Delta_x^B = \left[ (X_B - X_{B_1}) - (X_{B_2} - X_{B_1}) \frac{\delta B}{\Delta B} \right] + \]

\[ {}+ \left[ (\alpha' - \alpha'_1) - (\alpha'_2 - \alpha'_1)\frac{\delta B}{\Delta B} \right] l^2 + \left[ (\alpha'' - \alpha''_1) - (\alpha''_2 - \alpha''_1)\frac{\delta B}{\Delta B} \right] l^4 \tag{16} \]

or

\[ \Delta_x^B = I_x^B + II_x^B + III_x^B . \tag{16a} \]

In the first term, the differences \(X_B - X_{B_1}\) and \(X_{B_2} - X_{B_1}\) are determined as the lengths of meridian arcs:

\[ X_{B_2} - X_{B_1} = M_m \cdot \Delta B, \]

\[ X_B - X_{B_1} = M'_m \cdot \delta B, \]

where \(M_m\) and \(M'_m\) are the mean radii of curvature of the meridian ellipse, corresponding to the mean latitudes of these arcs:

\[ \frac{1}{2}(B_1 + B_2) \quad \text{and} \quad \frac{1}{2}(B_1 + B). \]

\[ I_x^B = (X_B - X_{B_1}) - (X_{B_2} - X_{B_1})\frac{\delta B}{\Delta B} = (M'_m - M_m)\,\delta B. \]

The difference \(M'_m - M_m\) represents the change in the radius of curvature of the meridian \(M\) when the latitude changes by

\[ \frac{1}{2}(B + B_1) - \frac{1}{2}(B_1 + B_2) = \frac{1}{2}(\delta B - \Delta B). \]

\[ M = \frac{a(1 - e^2)}{(1 - e^2 \sin^2 B)^{3/2}}, \qquad dM = \frac{3}{2}ae^2 \sin 2B \cdot dB. \]

Taking

\[ dM = M'_m - M_m \quad \text{and} \quad dB = \frac{1}{2}(\delta B - \Delta B), \]

we obtain

\[ I_x^B = -\frac{3}{4\rho^2}ae^2 \sin 2B\,(\Delta B\,\delta B - \delta B^2). \tag{17} \]

Table 4 gives the values of \(I_x^B\), the first term of formula (16). For trapezia at the scale \(1:10000\), \(\Delta B\) is equal to \(2'30''\), or \(150''\).

Let us consider the second term of formula (16).

\[ a'=\frac{N}{2\rho^2}\sin B\cos B=\frac{N}{4\rho^2}\sin 2B. \]

\[ II_x^B=-\frac{l^2}{4\rho^2} \left[ \left(N\sin 2B-N_1\sin 2B_1\right) - \left(N_2\sin 2B_2-N_1\sin 2B_1\right)\frac{\delta B}{\Delta B} \right]. \]

Table 4

\(B\) \(\delta B=75''\) \(\delta B=60''\) \(\delta B=45''\) \(\delta B=30''\) \(\delta B=15''\)
mm mm mm mm mm
35 −4 −4 −3 −3 −1
45 4 4 4 3 2
55 4 4 3 3 1
65 3 3 3 2 1
75 2 2 2 1 1

Replace \(N\) and \(N_2\) by \(N_1+\delta N\) and \(N_1+\Delta N\).

\[ II_x^B= \frac{l^2}{4\rho^2} \left\{ N_1 \left[ (\sin 2B-\sin 2B_1) - (\sin 2B_2-\sin 2B_1)\frac{\delta B}{\Delta B} \right] + \right. \]

\[ \left. + \left[ \delta N\sin 2B - \Delta N\sin 2B_2\frac{\delta B}{\Delta B} \right] \right\}. \tag{A} \]

\[ (\sin 2B-\sin 2B_1) - (\sin 2B_2-\sin 2B_1)\frac{\delta B}{\Delta B} = \]

\[ =2\sin 2B_m(\Delta B\,\delta B-\delta B^2), \]

where

\[ 2B_m \simeq \frac{1}{2}(B+2B_1+B_2). \]

The radius of curvature of the prime vertical is equal to

\[ N=\frac{a}{(1-e^2\sin^2 B)^{1/2}}, \qquad dN=\frac{1}{2}ae^2\sin 2B\cdot dB. \]

For our case we take

\[ \delta N=\frac{1}{2}ae^2\sin 2B\cdot \delta B, \]

\[ \Delta N=\frac{1}{2}ae^2\sin 2B\cdot \Delta B. \]

and substitute into the second term of the braces of expression (A).

\[ \frac{1}{2}ae^2 \sin 2B\left(\sin 2B\,\delta B-\sin 2B_2\,\Delta B\,\frac{\delta B}{\Delta B}\right)= \]

\[ =-\frac{1}{2}ae^2\sin 4B\,(\Delta B\,\delta B-\delta B^2). \]

\[ II_x^B=-\frac{N_1}{2\rho^4}\sin 2B(\Delta B\,\delta B-\delta B^2)\,l^2- \]

\[ -\frac{ae^2}{8\rho^4}\sin 4B(\Delta B\,\delta B-\delta B^2)\,l^2. \tag{18} \]

The maximum value of the first term in \(II_x^B\) will occur for

\[ B=45^\circ \quad \text{and} \quad \delta B=\frac{1}{2}\Delta B; \]

for \(l=3^\circ30'\) it is equal to \(+0.0016\) m,

for \(l=3^\circ00'\) it is equal to \(+0.0012\) m,

for \(l=2^\circ30'\) it is equal to \(+0.0008\) m.

The second term in (18), which arose as a consequence of replacing \(N\) and \(N_2\) by \(N_1\), will be smaller than the first by \(\dfrac{4}{e^2}\simeq 600\) times; it may be neglected. Then

\[ II_x^B=-\frac{N}{2\rho^4}\sin 2B(\Delta B\,\delta B-\delta B^2)\,l^2. \tag{18a} \]

The third correction term, \(III_x^B\), in (16) will be considerably smaller than the second; therefore we shall discard it without considering it.

Let us now find the correction \(\Delta_y^B\). For this purpose we substitute the values of \(y\) from (15) into (14):

\[ \Delta_y^B=\left[(\beta'-\beta'_1)-(\beta'_2-\beta'_1)\frac{\delta B}{\Delta B}\right]l+ \]

\[ +\left[(\beta''-\beta''_1)-(\beta''_2-\beta''_1)\frac{\delta B}{\Delta B}\right]l^3 \tag{19} \]

or

\[ \Delta_y^B = I_y^B + II_y^B. \]

Knowing that \(\beta'=\dfrac{N}{\rho}\cos B\), we obtain the expression for the first term in \(\Delta_y^B\)

\[ I_y^B=\frac{1}{\rho}\left[(N\cos B-N_1\cos B_1)-(N_2\cos B_2-N_1\cos B_1)\frac{\delta B}{\Delta B}\right]l. \]

On the basis of the derivation made above, let us replace \(N_1\) and \(N_2\) by \(N\) and, after some transformations, obtain

\[ I_y^B=\frac{N}{2\rho^3}\left[\cos B_m(\Delta B\,\delta B-\delta B^2)\right]l. \tag{20} \]

The maximum values of \(I_y^B\) for various \(B\) and \(l\) are given in Table 5.

Table 5

\(B \backslash l\) \(0^\circ 30'\) \(1^\circ 00'\) \(1^\circ 30'\) \(2^\circ 00'\) \(2^\circ 30'\) \(3^\circ 00'\) \(3^\circ 30'\)
mm mm mm mm mm mm mm
\(30^\circ\) +3 +6 +10 +13 +16 +19 +22
\(40\) 3 6 8 11 14 17 20
\(50\) 2 5 7 9 12 14 17
\(60\) 2 4 6 7 9 11 13
\(70\) 1 3 4 5 6 8 9

Let us consider the second correction term in (19).

\[ II_y^B=\left[(\beta''-\beta_1'')-(\beta_2''-\beta_1'')\frac{\delta B}{\Delta B}\right]l^3. \tag{Б} \]

\[ \beta''=\frac{N}{6\rho^3}\cos^3 B(1-t^2+\eta^2); \]

\[ t^2=\operatorname{tg}^2 B;\qquad \eta^2=e'^2\cos^2 B. \]

Since \(\eta^2\) in absolute value is no more than \(1:200\), it may be neglected:

\[ \cos^3 B(1-\operatorname{tg}^2 B)=\frac{1}{2}(\cos 3B+\cos B); \]

\[ \beta''=-\frac{N}{12\rho^3}(\cos 3B+\cos B). \]

Substituting the expression obtained for \(\beta''\) into (B), after some transformations we shall have:

\[ II_y^B=-\frac{N}{24\rho^3}(9\cos 3B-\cos B)(\Delta B\,\delta B-\delta B^2). \tag{21} \]

The maximum value of \(II_y^B\) is less than 0.0001; it may be neglected.

Thus, for interpolation along a parallel we have the formulas:

\[ x=x'-\Delta_x^l;\qquad y=y'+\Delta_y^l \]

and for interpolation along a meridian:

\[ x=x'+\Delta_x^B;\qquad y=y'+\Delta_y^B; \]

in which:

\[ \Delta_x^l=I_x^l+II_x^l,\qquad \Delta_y^l=III_y^l, \]

\[ \Delta_x^B=I_x^B+II_x^B,\qquad \Delta_y^B=I_y^B. \]

Below is given a summary table of the maximum values of the correction terms in the interpolated coordinates of points for a trapezium of scale 1:10000, having dimensions in latitude \(3'45''\) and in longitude \(2'30''\).

Table 6

Correction Correction term Maximum values of the correction terms Conditions:
a) for the middle of the trapezium,
b) at mean latitudes
\(\Delta_x^l\) \(I_x^l\) m
0,350–0,475
a) for any values of \(l\)
\(\Delta_x^l\) \(II_x^l\) 0,001–0,003 for \(l\) from \(2^\circ\) to \(3^\circ 30'\)
\(\Delta_x^B\) \(I_x^B\) 0,003–0,004 for any values of \(l\)
\(\Delta_x^B\) \(II_x^B\) 0,001–0,0016 for \(l\) from \(2^\circ\) to \(3^\circ 30'\)
\(\Delta_y^l\) \(III_y^l\) 0,002–0,015 for \(l\) from \(30'\) to \(3^\circ 30'\)
\(\Delta_y^B\) \(I_y^B\) 0,003–0,020 for \(l\) from \(30'\) to \(3^\circ 30'\)

If the computations are carried out using the coordinates of the corners of trapezia of larger or smaller dimensions, then the correction terms will respectively increase or decrease: \(I_x^l\) and \(I_x^B\)—proportionally to the square, \(II_y^l\) and \(II_y^B\)—proportionally to the cube, and \(III_x^l\) and \(II_x^B\)—proportionally to the fourth power of the increase or decrease of the trapezium.

Thus, if one has tables of rectangular coordinates with a one-minute interval in latitude and longitude, then even when computing coordinates to an accuracy of \(0.001\) m, only one correction term, \(I_x^l\), will have to be taken into account, and its value will be 14 times smaller than that given in Table 6. Bearing in mind that in many cases of practice it is sufficient to know the coordinates of points to an accuracy of \(1\)—\(2\) cm, they can be obtained by linear interpolation without introducing corrections.

In practice, the computation of the rectangular coordinates of points from geodetic coordinates, when a table of coordinates of the corners of trapezia is available, may be performed as follows.

Let \(x_1, x_2, x'_1\), and \(x'_2\) be the abscissae of the corners of the trapezium corresponding to latitudes \(B_1\) and \(B_2\) and longitudes \(l_1\) and \(l_2\); \(B\) and \(l\) are the geodetic coordinates of the point being determined. For clarity, let us arrange these data in a scheme.

\(l_1\) \(l\) \(l_2\)
\(B_2\) \(x_2\) \(x'_2\)
\(B\) \(x_0\) \(x'\) \(x'_0\)
\(B_1\) \(x_1\) \(x'_1\)

Formulas for computing \(x\):

\[ x_0 = x_1 + (x_2 - x_1)\frac{\delta B}{\Delta B}; \]

\[ x'_0 = x'_1 + (x'_2 - x'_1)\frac{\delta B}{\Delta B}; \]

\[ x' = x_0 + (x'_0 - x_0)\frac{\delta l}{\Delta l}; \]

\[ x = x' + \Delta x, \]

where

\[ \Delta x = -\Delta_x^l + \Delta_x^B. \]

To compute the ordinate, in the scheme one should substitute the values \(y\) for the values \(x\), and introduce into the interpolated value the correction \(\Delta_y\).

Formulas for computing \(y\):

\[ y_0 = y_1 + (y_2-y_1)\frac{\delta B}{\Delta B}, \]

\[ y'_0 = y'_1 + (y'_2-y'_1)\frac{\delta B}{\Delta B}, \]

\[ y' = y_0 + (y'_0-y_0)\frac{\delta l}{\Delta l}, \]

\[ y = y' + \Delta_y,\quad \Delta_y=\Delta_y^l+\Delta_y^B. \]

Solution of the inverse problem

The solution of the inverse problem, that is, the determination of geodetic coordinates from rectangular ones, will consist in finding the quantities \(\delta B\) and \(\delta l\) by the method of successive approximations. The quantity \(\delta B\) should be determined from the abscissas, and \(\delta l\)—from the ordinates. All computations are carried out in two schemes simultaneously, in the following order.

\(l_1\) \(l\) \(l_2\)
\(B_2\) \(x_2\) \(x_{0,2}\) \(x'_2\)
\(B\) \(x, x'\)
\(B_1\) \(x_1\) \(x_{0,1}\) \(x'_1\)
\(l_1\) \(l\) \(l_2\)
\(B_2\) \(y_2\) \(y'_2\)
\(B\) \(y_0\) \(y, y'\) \(y'_0\)
\(B_1\) \(y_1\) \(y'\)

First approximation:

\[ (\delta B)'=\frac{x-x_1}{x_2-x_1}\,\Delta B', \]

\[ (y_0)'=y_1+(y_2-y_1)\frac{(\delta B)'}{\Delta B}, \]

\[ (y'_0)'=y'_1+(y'_2-y'_1)\frac{(\delta B)'}{\Delta B}, \]

\[ (\delta l)'=\frac{y-(y_0)'}{(y'_0)'-(y_0)'}\,\Delta l. \]

From the \(\delta B\) and \(\delta l\) obtained in the first approximation, one can select from the tables all correction terms and, consequently, obtain the quantities \(x'\) and \(y'\).

Second approximation:

\[ (x_{01})' = x_1 + (x_1' - x_1)\frac{(\delta l)'}{\Delta l}, \]

\[ (x_{02})' = x_2 + (x_2' - x_2)\frac{(\delta l)'}{\Delta l}, \]

\[ (\delta B)'' = \frac{x' - (x_{01})'}{(x_{02})' - (x_{01})'} \Delta B, \]

\[ (y_0)'' = y_1 + (y_2 - y_1)\frac{(\delta B)''}{\Delta B}, \]

\[ (y_0')'' = y_1' + (y_2' - y_1')\frac{(\delta B)''}{\Delta B}, \]

\[ (\delta l)'' = \frac{y' - (y_0)''}{(y_0')'' - (y_0)''}\Delta l = \delta l. \]

And, finally,

\[ (x_{01})'' = x_1 + (x_1' - x_1)\frac{\delta l}{\Delta l}, \]

\[ (x_{02})'' = x_2 + (x_2' - x_2)\frac{\delta l}{\Delta l}, \]

\[ (\delta B)''' = \frac{x' - (x_{01})''}{(x_{02})'' - (x_{01})''}\Delta B = \delta B. \]

When computing geodetic coordinates with an accuracy up to \(0,''001\), using the table of rectangular coordinates of the corners of trapezia at the scale 1:10000, the final value may be taken as \(\delta l\) in the second approximation and \(\delta B\) in the third.

Below are examples of the solution of the direct and inverse problems.

In conclusion it should be noted that the solution of the problem of conversion from geodetic coordinates to rectangular coordinates in the Gauss projection and back by the proposed method, in comparison with the classical one, is very simple, clear, economical, and can provide sufficient accuracy. The computations may be performed both with the aid of an arithmometer and by tables of logarithms, which is especially important if the computations are carried out under field conditions. In view of the simplicity and clarity of the solution of the problem by this method, the computations may be performed by persons who do not have special training.

This method may be applied both for the independent solution of the problem of converting some coordinates into others, in particu-

DIRECT PROBLEM

Conversion from geodetic coordinates to rectangular coordinates for the corners of trapezium frames at scale 1:10000

Scheme X

\(B\) / \(l\) \(l_1 = 2^\circ 48' 45''\) \(l = 2^\circ 50' 21'',533\) \(l_2 = 2^\circ 52' 30''\) Correction terms
\(\delta l = 96'',533;\quad \delta l : \Delta l = 0,429\ 0536\)
\(B_2 = 55^\circ 35' 00''\) \(x_2 = 6\ 165\ 871,987\) \(x_2' = 6\ 166\ 033,429\) \(I_x^l = +0,434\)
\(II_x^l = 0,000\)
\(B = 55\ 33\ 54,375\) \(x_0 = 6\ 163\ 843,310\) \(x' = 6\ 163\ 912,592\)
\(\Delta_x = -0,437\)
\(x = 6\ 163\ 912,155\)
\(x_0' = 6\ 164\ 004,792\) \(\Delta_x^l = +0,434\)
\(B_1 = 55\ 32\ 30\) \(x_1 = 6\ 161\ 235,012\) \(x_1' = 6\ 161\ 396,545\) \(I_x^B = -0,004\)
\(II_x^B = +0,001\)
\(\delta B = 84'',375\)
\(\dfrac{\delta B}{\Delta B} = 0,5625\)
\((x_2 - x_1) = 4\ 636,975\)
\((x_2 - x_1)\dfrac{\delta B}{\Delta B} = 2608,298\)
\(x_0' - x_0 = 161,482\)
\((x_0' - x_0)\dfrac{\delta l}{\Delta l} = 69,282\)
\(x_2' - x_1' = 4\ 636,884\)
\((x_2' - x_1')\dfrac{\delta B}{\Delta B} = 2,608,247\)
\(\Delta_x^B = -0,003\)
\(\Delta_x = -0,437\)

Scheme Y

\(B\) \(l_1 = 2^\circ 48' 45''\) \(l = 2^\circ 50' 21'',533\) \(l_2 = 2^\circ 52' 30''\) Correction terms
\(B_2\) \(y_2 = 177\ 340,160\) \(y_2' = 181\ 269,869\)
\(B\) \(y_0 = 177\ 422,357\) \(y' = 179\ 113,417\)
\(\Delta_y = +0,021\)
\(y = 179\ 113,438\)
\(y_0' = 181\ 363,894\) \(II_y^l = +0,009\)
\(I_y^B = +0,012\)
\(\Delta_y = +0,021\)
\(B_1\) \(y_1 = 177\ 528,038\) \(y_1' = 181\ 471,926\)
\(y_2 - y_1 = -187,878\)
\((y_2 - y_1)\dfrac{\delta B}{\Delta B} = -105,681\)
\(y_0' - y_0 = +3\ 941,537\)
\((y_0' - y_0)\dfrac{\delta l}{\Delta l} = +1\ 691,060\)
\(y_2' - y_1' = -192,057\)
\((y_2' - y_1')\dfrac{\delta B}{\Delta B} = -108,032\)

INVERSE PROBLEM

Conversion from rectangular coordinates to geodetic coordinates using the coordinates of the corners of trapezoid frames at scale 1 : 10000

Scheme $\delta B$

$B$ $l_1=2^\circ48'45''$ $l=?$ $l_2=2^\circ52'30''$ $\dfrac{\delta x,\ \delta l}{\Delta_x,\ \Delta_y}$
$B_2=55^\circ35'00''$ $x_2=6\ 165\ 871{,}987$ $(x_{02})'=6\ 165\ 941{,}36$
$(x_{02})''=6\ 165\ 941{,}251$
$x_2'=6\ 166\ 033{,}429$ $(\delta B)'=86{,}6$
$(\delta B)''=84{,}371$
$(\delta B)'''=84{,}375$
$B=?$ $(x_0)'-(x_{01})'=4\ 636{,}93$
$x'-(x_{01})'=2\ 608{,}16$
$x=6\ 163\ 912{,}155$
$-\Delta_x=\quad +0{,}437$
$x'=6\ 163\ 912{,}592$
$(x_0)'-(x_{01})''=4\ 636{,}336$
$x'-(x_{01})''=2\ 608{,}277$
$I_x^{\,l}=+0{,}434$
$II_x^{\,l}=\ \ \ 0{,}00$

$\Delta_x^l=+0{,}434$

$I_x^{\,B}=-0{,}004$
$II_x^{\,B}=-0{,}001$

$\Delta_x^B=-0{,}006$
$\Delta_x=-0{,}437$
$B_1=55^\circ32'30''$ $x_1=6\ 161\ 235{,}012$ $(x_{01})'=6\ 161\ 304{,}43$
$(x_{01})''=6\ 161\ 304{,}315$
$x_1'=6\ 161\ 395{,}545$
$x_2-x_1=4\ 636{,}975$
$x-x_1=2\ 677{,}1$
$x_1'-x_1=161{,}533$
$x_2'-x_2=161{,}442$
$B=55^\circ33'54'',375$

Scheme $\delta l$

$B$ $l_1=2^\circ48'45''$ $l=?$ $l_2=2^\circ52'30''$ $\delta l$
$B_2$ $y_2=177\ 340{,}160$ $y_2'=181\ 279{,}863$ $(\delta l)'=96'',69$
$B$ $(y_0)'=177\ 419{,}6$
$(y_0)''=177\ 422{,}362$
$y=179\ 113{,}438$
$-\Delta_y=\quad -0{,}021$
$y'=179\ 113{,}417$
$(y_0')=181\ 361{,}1$
$(y_0'')=181\ 36[[unclear: digit]]{,}899$
$(\delta l)''=96533$
$B_1$ $y_1=177\ 528{,}038$ $y_1'=181\ 471{,}920$
$y_2-y_1=-187{,}878$ $(y_0')'-(y_0)'=3\ 941{,}5$
$y-(y_1)'=1\ 693{,}8$
$(y_0')''-(y_0)''=3\ 941{,}537$
$y'-(y_1)''=1\ 697{,}055$
$y_2'-y_1'=-192{,}057$ $\Delta_y^l=+0{,}009$
$\Delta_y^B=+0{,}012$

$\Delta_y=+0{,}021$

$l=2^\circ50'21'',533$

Schemes $\delta B$ and $\delta l$ are solved jointly.

when the initial points for substantiating a topographic survey are astronomical points, and also for checking the computation by formulas (3). In addition, this method of coordinate recomputation can also be applied for passing from one coordinate zone to another. The transition from zone to zone can be carried out in any combination: from a six-degree zone to a six-degree zone, from a six-degree zone to a three-degree zone and back, and from a three-degree zone to a three-degree zone.

In solving these problems, \(y\) and \(l\) should always be considered positive; if \(y\) or \(l\) is negative, the sign of the result will also be negative.

For the practical application of this method it is necessary to compile tables of rectangular coordinates of specified points of the degree grid, obtained with an accuracy of up to \(0.001\) m. Such tables may be tables of the coordinates of the corners of trapezia at the scale 1:10000, that is, at intervals of \(2'30''\) in latitude and \(3'45''\) in longitude. It is advisable to have tables compiled at one-minute intervals both in latitude and in longitude. If the latter are available, there is undoubtedly no need for other tables connected with the recalculation and transformation of coordinates.

REFERENCES

  1. Krasovskii, F. N. Manual of Higher Geodesy, Part II, Geodezizdat, 1942.
  2. Kavraisky, V. V. Tables of Rectangular Gauss–Krüger Coordinates for Plotting Kilometer Grids on Topographic Maps, ONTI, 1936.
  3. Tables for Computing Gauss–Krüger Coordinates, Geodezizdat, 1946.
  4. Tables of Gauss–Krüger Coordinates, Geodezizdat, 1947.
  5. Tables of Rectangular Coordinates of Frame Corners, Frame Dimensions, and Areas of Trapezia for Topographic Surveys at the Scale 1:5000, Geodezizdat, 1953.

Submission history

Conversion from Geodetic Coordinates to Rectangular Coordinates and Back Using Special Tables