CN102750457A - Millimeter-scale high-precision measurement transforming method - Google Patents

Abstract

The invention discloses a millimeter-scale high-precision measurement transforming method and belongs to the technical field of measurement methods. The millimeter-scale high-precision measurement transforming method is characterized by including the following steps of firstly, deducing a calculation formula of the curvature radius M of the meridian; secondly, deducing a calculation formula of the curvature radius N of the prime vertical; thirdly, deducing a calculation formula of the arc length X of the meridian; fourthly, deducing a calculation formula, required by high precision, of transforming geodetic coordinates into planar coordinates by a gauss projection forward calculation method; and fifthly, deducing a formula, required by high precision, of transforming the planar coordinates into geodetic coordinates by a gauss projection back calculation method. The millimeter-scale high-precision measurement transforming method is high in data precision and small in errors and is used for measurement transformation.

Description

A kind of grade high-acruracy survey conversion method

Technical field

The present invention relates to a kind of measurement conversion methods, more specifically more particularly to a kind of grade high-acruracy survey conversion method.

Background technique

Previous coordinate transformation method, usually using the Gauss projection formula that derives decades ago, centre expansion item only has 3 to 4, and precision can only achieve centimetre even decimeter grade.And geodetic coordinates and plane coordinates, (overall situation) plane coordinates and local coordinate, calculating before needs separate computations, inconvenient.Layman not necessarily knows the data such as the affiliated reel number in somewhere, central meridian value.

Summary of the invention

The purpose of the present invention is to provide the small grade high-acruracy survey conversion methods of a kind of data precision height, error.

The technical scheme of the present invention is realized as follows: a kind of grade high-acruracy survey conversion method, wherein this method includes the following steps: that (1) derives the calculation formula of radius of curvature of meridian M are as follows:

M=m₀+m₂sin²B+m₄sin⁴B+m₆sin⁶B+m₈sin⁸B+m₁₀sin¹⁰B；

(2) calculation formula of radius of curvature in prime vertical N is derived are as follows:

N=n₀+n₂sin²B+n₄sin⁴B+n₆sin⁶B+n₈sin⁸B+n₁₀sin¹⁰B；

(3) calculation formula of Meridian arc length X is derived:

$X=a_{0}B-\frac{a_2}{2}\sin 2B+\frac{a_4}{4}\sin 4B-\frac{a_6}{6}\sin 6B+\frac{a_8}{8}\sin 8B-\frac{a_{10}}{10}\sin 10B,$ Reach 10 grades of expansion items needed for high-precision；

In above-mentioned formula, B indicates the latitude of point to be measured； $a_0={\mathrm{<figure-callout\; id="0"\; label="m"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">m</figure-callout>}}_0+\frac{m_2}{2}+\frac{3}{8}{m}_4+\frac{5}{16}{m}_6+\frac{35}{128}{m}_8+\frac{63}{256}{m}_{10},$ $a_2=\frac{m_2}{2}+\frac{m_4}{2}+\frac{15}{31}{m}_6+\frac{7}{16}{m}_8+\frac{105}{256}{m}_{10},$ $a_4=\frac{m_4}{8}+\frac{3}{16}{m}_6+\frac{7}{32}{m}_8+\frac{15}{64}{m}_{10},$ $a_6=\frac{m_6}{32}+\frac{m_8}{16}+\frac{45}{512}{m}_{10},$ $a_8=\frac{m_8}{128}+\frac{5}{256}{m}_{10},$ $a_{10}=\frac{1}{512}{m}_{10};$

In above-mentioned formula, m₀=a (1-e²)、 $m_2=\frac{3}{2}{e}^2{\mathrm{<figure-callout\; id="0"\; label="m"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">m</figure-callout>}}_0,$ $m_4=\frac{5}{4}{e}^2{m}_2,$ $m_6=\frac{7}{6}{e}^2{m}_4,$

E is the first eccentricity of ellipsoid, is formulated asA in above-mentioned formula is semimajor axis of ellipsoid, f flattening of ellipsoid；B is semiminor axis of ellipsoid, b=(1-f) * a；

(4) geodetic coordinates needed for method derives high-precision is just being calculated using gauss projection and is turning plane coordinates calculation formula:

$x=X+\frac{N}{2{\mathrm{\&rho;}}^{\&prime;\&prime;2}}\sin B\cos B\&CenterDot;l^{\&prime;\&prime;2}+\frac{N}{24{\mathrm{\&rho;}}^{\&prime;\&prime;4}}\sin B\mathrm{co}{s}^{3}B(5-t^2+9{\mathrm{\&eta;}}^2+4{\mathrm{\&eta;}}^4)l^{\&prime;\&prime;4};$

$y=\frac{N}{{\mathrm{\&rho;}}^{\&prime;\&prime;}}\cos B\&CenterDot;l^{\&prime;\&prime;}+\frac{N}{6{\mathrm{\&rho;}}^{\&prime;\&prime;3}}{\cos }^{3}B(1-t^2+{\mathrm{\&eta;}}^2)l^{\&prime;\&prime;3}+\frac{N}{120{\mathrm{\&rho;}}^{\&prime;\&prime;5}}{\cos }^{5}B(5-18t^2+t^4+14{\mathrm{\&eta;}}^2-58{\mathrm{\&eta;}}^2{t}^2)l^{\&prime;\&prime;5};$

Wherein x, y are the plane transverse and longitudinal coordinate value of measurement point；In formula:

ρ ": 1 radian=ρ " second is constant: 57.295779513082320876798154814105；

B: the latitude of point to be measured；L ": central meridian longitude-subpoint longitude of band where projection (difference of longitude, unit are the seconds)；

t:tanB；η²=e '² cos²B；A is semimajor axis of ellipsoid, f flattening of ellipsoid；B is semiminor axis of ellipsoid, b=(1-f) * a；

Elliptical first eccentricityElliptical second eccentricity

η²=e '²cos²B；(5)

Plane coordinates needed for anti-method derives high-precision is just being calculated using gauss projection turns geodetic coordinates formula:

$B=B_f-\frac{t_f}{2M_f{N}_f}{y}^2+\frac{t_f}{24M_f{N}_f^3}(5+3t_f^2+{\mathrm{\&eta;}}_f^2-9{\mathrm{\&eta;}}_f^2{t}_f^2-4{\mathrm{\&eta;}}_f^4)y$

$l=\frac{y}{N_f\cos B_f}-\frac{y^3}{6N_f^3\cos B_f}(1+2t_f^2+{\mathrm{\&eta;}}_f^2)+\frac{y^5}{120N_f^5\cos B_f}(5+28t_f^2+24t_f^4+6{\mathrm{\&eta;}}_f^2+8{\mathrm{\&eta;}}_f^2{t}_f^2),$ Wherein B, l are the earth transverse and longitudinal coordinate value of measurement point；In formula: t=tanB；η²=e '²cos²B；

In a kind of above-mentioned grade high-acruracy survey conversion method, the derivation process of the step (1) specifically: (1) differential arc length DK=ds is taken in a part of meridional ellipse, correspondingly there is increment of coordinate dx, point n is the center of curvature of differential arc dS, and then line segment Dn and Kn is radius of curvature of meridian M；

The defined formula of the radius of curvature of arbitrary plane curve are as follows:

∠ KBx, that is, latitude can be acquired from differential triangle DKE:

Above formula is substituted into (7-53), is obtained: $M=-\frac{\mathrm{dx}}{\mathrm{dB}}\&CenterDot;\frac{1}{\sin B}---(7-54)$

It is the normal of P point as Pn, is easy to get:

Elliptic equation: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1---(7-12)$

X is taken and is led, is obtained $\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{a^2}{b^2}\&CenterDot;\frac{x}{y}---(7-13)$

(7-13) compares and can obtain with (7-11):

Therefore, y=x (1-e2) tanB (7-14) is acquired

(7-14) is substituted into (7-12), both sides are shifted with a2cos2B is multiplied:

Above formula is substituted into (7-14) to obtain: $Y=\frac{a}{W}(1-e^2)\sin B=\frac{b\sin B}{V},$

It is obtained by (7-16): $\frac{\mathrm{dx}}{\mathrm{dB}}=a\left(\frac{-\sin \mathrm{BW}-\cos B\frac{\mathrm{dW}}{\mathrm{dB}}}{W^2}\right)(7-55),$

Due to $\frac{\mathrm{dW}}{\mathrm{dB}}=\frac{d\sqrt{1-e^2{\sin }^{2}B}}{\mathrm{dB}}=\frac{-e^2\sin B\cos B}{W}(7-56),$

Above formula is substituted into (7-55), and because of W²=1-e²sin²B,

$\frac{\mathrm{dx}}{\mathrm{dB}}=-\frac{a\sin B}{W^3}(1-e^2{\sin }^{2}B-e^2{\cos }^{2}B)=-\frac{a\sin B}{W^3}(1-e^2),$

Then obtain radius of curvature of meridian formula are as follows:

Series is unfolded by Newton binomial theorem, takes to 10 items, obtains: M=m₀+m₂sin²B+m₄sin⁴B+m₆sin⁶B+m₈sin⁸B+m₁₀sin¹⁰B,

In formula: $\begin{array}{c}{\mathrm{<figure-callout\; id="0"\; label="m"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">m</figure-callout>}}_0=a(1-e^2),m_2=\frac{3}{2}{e}^2{\mathrm{<figure-callout\; id="0"\; label="m"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">m</figure-callout>}}_0,m_4=\frac{5}{4}{e}^2{m}_2,m_6=\frac{7}{6}{e}^2{m}_4,m_8=\frac{9}{8}{e}^2{m}_6\\ m_{10}=\frac{11}{10}{e}^2{m}_8\end{array}$

It increases with the increase of B in relation to by M and latitude B.

In a kind of above-mentioned grade high-acruracy survey conversion method, the derivation process of the step (2) specifically: cross the tangent line PT for the parallel circle PHK that P point is made centered on O ', which is located at perpendicular in the parallel circle plane of meridian plane；Because prime vertical is also perpendicularly to meridian plane, therefore PT is also tangent line of the prime vertical at P point；I.e. PT is perpendicular to Pn；So PT is common tangent of the parallel circle PHK and prime vertical PEE ' at P point；Known by wheat Neil theorem, assuming that cutting arc by a little drawing two on curved surface, one is bevel arc, and this two sections of arcs have common tangential at that point, and at this moment the radius of curvature of bevel arc at this point is equal to the radius of curvature of method section arc multiplied by two-section arc plane holder cosine of an angle；Thus scheme the angle it is found that between parallel circle plane and prime vertical plane, as geodetic latitude B, if the radius of parallel circle indicates have with r

R=NcosB (7-62)

Again because parallel circle radius r is equal to the abscissa x of P point, that is,

Therefore, radius of curvature in prime vertical can be indicated with following two formula:

Series is unfolded by Newton binomial theorem in above formula, takes to 10 items, obtains:

N=n₀+n₂sin²B+n₄sin⁴B+n₆sin⁶B+n₈sin⁸B+n₁₀sin¹⁰B  (7-70)

In formula: ${\mathrm{<figure-callout\; id="0"\; label="n"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">n</figure-callout>}}_0=a,n_2=\frac{1}{2}{e}^2{\mathrm{<figure-callout\; id="0"\; label="n"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">n</figure-callout>}}_0,$ $n_4=\frac{3}{4}{e}^2{n}_2,$ $n_6=\frac{5}{6}{e}^2{n}_4,n_8=\frac{7}{8}{e}^2{n}_6,$ $n_{10}=\frac{8}{9}{e}^2{n}_8.$

In a kind of above-mentioned grade high-acruracy survey conversion method, the derivation process of the step (3) specifically: certain differential arc PP '=dx on meridian is taken, enabling P point latitude is B, and P ' latitudes are B+dB, and the radius of curvature of meridian of P point is M, then have:

Dx=MdB

It can be found out to the arc length the parallel circle of any latitude B by following integral since equator:

$X={\&Integral;}_0^B\mathrm{MdB}$

M can be expressed with following formula in formula:

M=a₀-a₂cos2B+a₄cos4B-a₆cos6B+a₈cos8B-a₁₀cos10B

Wherein: $\left.\begin{array}{c}{a}_0={\mathrm{<figure-callout\; id="0"\; label="m"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">m</figure-callout>}}_0+\frac{m_2}{2}+\frac{3}{8}{m}_4+\frac{5}{16}{m}_6+\frac{35}{128}{m}_8+\frac{63}{256}{m}_{10}\\ a_2=\frac{m_2}{2}+\frac{m_4}{2}+\frac{15}{31}{m}_6+\frac{7}{16}{m}_8+\frac{105}{256}{m}_{10}\\ a_4=\frac{m_4}{8}+\frac{3}{16}{m}_6+\frac{7}{32}{m}_8+\frac{15}{64}{m}_{10}\\ a_6=\frac{m_6}{32}+\frac{m_8}{16}+\frac{45}{512}{m}_{10}\\ a_8=\frac{m_8}{128}+\frac{5}{256}{m}_{10}\\ a_{10}=\frac{1}{512}{m}_{10}\end{array}\right\}$

It is integrated, Meridian arc length calculating formula is obtained after being arranged:

$X=a_{0}B-\frac{a_2}{2}\sin 2B+\frac{a_4}{4}\sin 4B-\frac{a_6}{6}\sin 6B+\frac{a_8}{8}\sin 8B-\frac{a_{10}}{10}\sin 10B.$

The precision that the present invention converts after adopting the above method, by improving coordinate, to meet the demand of the user required to high-precision geographical location information.The multiple cumulative calculation of another point is calculated especially by a point, whole error is reduced, and generated economic benefit, social benefit are especially considerable.

Detailed description of the invention

The present invention is described in further detail for embodiment in reference to the accompanying drawing, but does not constitute any limitation of the invention.

One of schematic diagram when Fig. 1 is present invention derivation half calculation formula diameter of meridian circle curvature；

Two of schematic diagram when Fig. 2 is present invention derivation radius of curvature of meridian calculation formula；

Fig. 3 is the schematic diagram when present invention derives radius of curvature in prime vertical calculation formula；

Fig. 4 is the schematic diagram when present invention derives Meridian arc length calculation formula.

Specific embodiment

Referring to FIG. 1 to 4, a kind of grade high-acruracy survey conversion method of the invention, this method include the following steps:

(1) calculation formula of radius of curvature of meridian M is derived:

As depicted in figs. 1 and 2, a differential arc length DK=ds is taken in a part of meridional ellipse, correspondingly having increment of coordinate dx, point n is the center of curvature of differential arc dS, and then line segment Dn and Kn is radius of curvature of meridian M；

The defined formula of the radius of curvature of arbitrary plane curve are as follows:

∠ KBx, that is, latitude can be acquired from differential triangle DKE:

Above formula is substituted into (7-53), is obtained: $M=-\frac{\mathrm{dx}}{\mathrm{dB}}\&CenterDot;\frac{1}{\sin B}---(7-54)$

It is the normal of P point as Pn, is easy to get:

Elliptic equation: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(7-12)$

X is taken and is led, is obtained $\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{a^2}{b^2}\&CenterDot;\frac{x}{y}---(7-13)$

(7-13) compares and can obtain with (7-11):

Therefore, y=x (1-e2) tanB (7-14) is acquired

(7-14) is substituted into (7-12), both sides are shifted with a2cos2B is multiplied:

Above formula is substituted into (7-14) to obtain: $y=\frac{a}{W}(1-e^2)\sin B=\frac{b\sin B}{V},$

It is obtained by (7-16): $\frac{\mathrm{dx}}{\mathrm{dB}}=a\left(\frac{-\sin \mathrm{BW}-\cos B\frac{\mathrm{dW}}{\mathrm{dB}}}{W^2}\right)---(7-55),$

Due to $\frac{\mathrm{dW}}{\mathrm{dB}}=\frac{d\sqrt{1-e^2{\sin }^{2}B}}{\mathrm{dB}}=\frac{-e^2\sin B\cos B}{W}---(7-56),$

Above formula is substituted into (7-55), and because of W²=1-e²sin²B,

$\frac{\mathrm{dx}}{\mathrm{dB}}=-\frac{a\sin B}{W^3}(1-e^2{\sin }^{2}B-e^2{\cos }^{2}B)=-\frac{a\sin B}{W^3}(1-e^2),$

Then obtain radius of curvature of meridian formula are as follows:

Series is unfolded by Newton binomial theorem, takes to 10 items, obtains: M=m₀+m₂sin²B+m₄sin⁴B+m₆sin⁶B+m₈sin⁸B+m₁₀sin¹⁰B,

In formula: ${\mathrm{<figure-callout\; id="0"\; label="m"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">m</figure-callout>}}_0=a(1-e^2),$ $m_2=\frac{3}{2}{e}^2{\mathrm{<figure-callout\; id="0"\; label="m"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">m</figure-callout>}}_0,$ $m_4=\frac{5}{4}{e}^2{m}_2,$ $m_6=\frac{7}{6}{e}^2{m}_4,$ $m_8=\frac{9}{8}{e}^2{m}_6$ $m_{10}=\frac{11}{10}{e}^2{m}_8,$

It increases with the increase of B in relation to by M and latitude B, and changing rule is as shown in the table:

(2) calculation formula of radius of curvature in prime vertical N is derived:

As shown in figure 3, crossing the tangent line PT for the parallel circle PHK that P point is made centered on O ', which is located at perpendicular in the parallel circle plane of meridian plane；Because prime vertical is also perpendicularly to meridian plane, therefore PT is also tangent line of the prime vertical at P point；I.e. PT is perpendicular to Pn；So PT is common tangent of the parallel circle PHK and prime vertical PEE ' at P point；Known by wheat Neil theorem, assuming that cutting arc by a little drawing two on curved surface, one is bevel arc, and this two sections of arcs have common tangential at that point, and at this moment the radius of curvature of bevel arc at this point is equal to the radius of curvature of method section arc multiplied by two-section arc plane holder cosine of an angle；Thus figure is it is found that angle between parallel circle plane and prime vertical plane, as geodetic latitude B, if the radius of parallel circle indicates have with r:

R=NcosB (7-62)

Again because parallel circle radius r is equal to the abscissa x of P point, that is,

Therefore, radius of curvature in prime vertical can be indicated with following two formula:

Series is unfolded by Newton binomial theorem in above formula, takes to 10 items, obtains:

N=n₀+n₂sin²B+n₄sin⁴B+n₆sin⁶B+n₈sin⁸B+n₁₀sin¹⁰B    (7-70)

In formula: ${\mathrm{<figure-callout\; id="0"\; label="n"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">n</figure-callout>}}_0=a,n_2=\frac{1}{2}{e}^2{\mathrm{<figure-callout\; id="0"\; label="n"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">n</figure-callout>}}_0,$ $n_4=\frac{3}{4}{e}^2{n}_2,$ $n_6=\frac{5}{6}{e}^2{n}_4,$ $n_8=\frac{7}{8}{e}^2{n}_6,$ $n_{10}=\frac{8}{9}{e}^2{n}_8.$

(3) calculation formula of Meridian arc length X is derived:

As shown in figure 4, taking certain differential arc PP '=dx on meridian, enabling P point latitude is B, and P ' latitudes are B+dB, and the radius of curvature of meridian of P point is M, is then had:

Dx=MdB

It can be found out to the arc length the parallel circle of any latitude B by following integral since equator:

$X={\&Integral;}_0^B\mathrm{MdB}$

M can be expressed with following formula in formula:

M=a₀-a₂cos2B+a₄cos4B-a₆cos6B+a₈cos8B-a₁₀cos10B

Wherein: $\left.\begin{array}{c}{a}_0={\mathrm{<figure-callout\; id="0"\; label="m"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">m</figure-callout>}}_0+\frac{m_2}{2}+\frac{3}{8}{m}_4+\frac{5}{16}{m}_6+\frac{35}{128}{m}_8+\frac{63}{256}{m}_{10}\\ a_2=\frac{m_2}{2}+\frac{m_4}{2}+\frac{15}{32}{m}_6+\frac{7}{16}{m}_8+\frac{105}{256}{m}_{10}\\ a_4=\frac{m_4}{8}+\frac{3}{16}{m}_6+\frac{7}{32}{m}_8+\frac{15}{64}{m}_{10}\\ a_6=\frac{m_6}{31}+\frac{m_8}{16}+\frac{45}{512}{m}_{10}\\ a_8=\frac{m_8}{128}+\frac{5}{256}{m}_{10}\\ a_{10}=\frac{1}{512}{m}_{10}\end{array}\right\}$

It is integrated, Meridian arc length calculating formula is obtained after being arranged:

$X=a_{0}B-\frac{a_2}{2}\sin 2B+\frac{a_4}{4}\sin 4B-\frac{a_6}{6}\sin 6B+\frac{a_8}{8}\sin 8B-\frac{a_{10}}{10}\sin 10B;$ To reach 10 grades of expansion items needed for high-precision；In above-mentioned formula, B indicates the latitude of point to be measured； $a_0={\mathrm{<figure-callout\; id="0"\; label="m"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">m</figure-callout>}}_0+\frac{m_2}{2}+\frac{3}{8}{m}_4+\frac{5}{16}{m}_6+\frac{35}{128}{m}_8+\frac{63}{256}{m}_{10},$ $a_2=\frac{m_2}{2}+\frac{m_4}{2}+\frac{15}{32}{m}_6+\frac{7}{16}{m}_8+\frac{105}{256}{m}_{10},$ $a_4=\frac{m_4}{8}+\frac{3}{16}{m}_6+\frac{7}{32}{m}_8+\frac{15}{64}{m}_{10},$ $a_6=\frac{m_6}{32}+\frac{m_8}{16}+\frac{45}{512}{m}_{10},$ $a_8=\frac{m_8}{128}+\frac{5}{256}{m}_{10},$ $a_{10}=\frac{1}{512}{m}_{10};$

In above-mentioned formula, m₀=a (1-e²)、 $m_2=\frac{3}{2}{e}^2{\mathrm{<figure-callout\; id="0"\; label="m"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">m</figure-callout>}}_0,$ $m_4=\frac{5}{4}{e}^2{m}_2,$ $m_6=\frac{7}{6}{e}^2{m}_4,$

E is the first eccentricity of ellipsoid, is formulated asA in above-mentioned formula is semimajor axis of ellipsoid, f flattening of ellipsoid；B is semiminor axis of ellipsoid, b=(1-f) * a；

(4) geodetic coordinates needed for method derives high-precision is just being calculated using gauss projection and is turning plane coordinates calculation formula:

$x=X+\frac{N}{2{\mathrm{\&rho;}}^{\&prime;\&prime;2}}\sin B\cos B\&CenterDot;l^{\&prime;\&prime;2}+\frac{N}{24{\mathrm{\&rho;}}^{\&prime;\&prime;4}}\sin B\mathrm{co}{s}^{3}B(5-t^2+9{\mathrm{\&eta;}}^2+4{\mathrm{\&eta;}}^4)l^{\&prime;\&prime;4};$

$y=\frac{N}{{\mathrm{\&rho;}}^{\&prime;\&prime;}}\cos B\&CenterDot;l^{\&prime;\&prime;}+\frac{N}{6{\mathrm{\&rho;}}^{\&prime;\&prime;3}}{\cos }^{3}B(1-t^2+{\mathrm{\&eta;}}^2)l^{\&prime;\&prime;3}+\frac{N}{120{\mathrm{\&rho;}}^{\&prime;\&prime;5}}{\cos }^{5}B(5-18t^2+t^4+14{\mathrm{\&eta;}}^2-58{\mathrm{\&eta;}}^2{t}^2)l^{\&prime;\&prime;5};$

Wherein x, y are the plane transverse and longitudinal coordinate value of measurement point；In formula:

ρ ": 1 radian=ρ " second is constant: 57.295779513082320876798154814105；

B: the latitude of point to be measured；L ": central meridian longitude-subpoint longitude of band where projection (difference of longitude, unit are the seconds)；

t:tanB；η²=e '²cos²B；A is semimajor axis of ellipsoid, f flattening of ellipsoid；B is semiminor axis of ellipsoid, b=(1-f) * a；

Elliptical first eccentricity

Elliptical second eccentricity

η²=e '²cos²B；

Its process specifically:

Formula is learned in projection according to the map:

X=F₁(L,B)

Y=F₂(L, B) (8-1)

And ellipsoid in Differential Geometry, claims Cauchy-Riemann condition to the orthomorphic general formulae of plane:

$\left.\begin{array}{c}\frac{\&PartialD;x}{\&PartialD;q}=\frac{\&PartialD;y}{\&PartialD;l}\\ \frac{\&PartialD;x}{\&PartialD;l}=-\frac{\&PartialD;y}{\&PartialD;q}\end{array}\right\}---(8-51)$

It can be obtained by gauss projection:

$\left.\begin{array}{c}x={\mathrm{<figure-callout\; id="0"\; label="m"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">m</figure-callout>}}_0+m_2{l}^2+m_4{l}^4+m_6{l}^6\\ y=m_{1}l+m_3{l}^3+m_5{l}^5\end{array}\right\}---(8-64)$

M0 in formula, m1, m2 are undetermined coefficients, they are all the functions of latitude B.

By third condition:

With

It asks partial derivative to substitute into l and q respectively (8-64) formula, obtains

$\left.\begin{array}{c}{m}_1+3m_3{l}^2+5m_5{l}^4=\frac{{\mathrm{<figure-callout\; id="0"\; label="dm"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">dm</figure-callout>}}_0}{\mathrm{dq}}+\frac{{\mathrm{dm}}_2}{\mathrm{dq}}{l}^2+\frac{{\mathrm{dm}}_4}{\mathrm{dq}}{l}^4\\ 2m_{2}l+4m_4{l}^3+6m_6{l}^5=-\frac{{\mathrm{dm}}_1}{\mathrm{dq}}l-\frac{{\mathrm{dm}}_3}{\mathrm{dq}}{l}^3-\frac{{\mathrm{dm}}_5}{\mathrm{dq}}{l}^5\end{array}\right\}---(8-65)$

For make above two formula both sides it is equal, necessary and sufficient condition is that the coefficient of the same power of l is equal, thus has

$\left.\begin{array}{c}{m}_1=\frac{{\mathrm{<figure-callout\; id="0"\; label="dm"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">dm</figure-callout>}}_0}{\mathrm{dq}}\\ m_2=-\frac{1}{2}\&CenterDot;\frac{{\mathrm{dm}}_1}{\mathrm{dq}}\\ m_3=\frac{1}{3}\&CenterDot;\frac{{\mathrm{dm}}_2}{\mathrm{dq}}\\ m_4=-\frac{1}{4}\&CenterDot;\frac{{\mathrm{dm}}_3}{\mathrm{dq}}\\ m_5=\frac{1}{5}\&CenterDot;\frac{{\mathrm{dm}}_4}{\mathrm{dq}}\\ m_6=-\frac{1}{6}\&CenterDot;\frac{{\mathrm{dm}}_5}{\mathrm{dq}}\end{array}\right\}---(8-66)$

(8-66) is a kind of recurrence formula, as long as m has been determined₀It can successively determine remaining each coefficient.Known by second condition: the point on centrally located meridian, the ordinate x after projection, which should be equal to preceding measure from equator to the Meridian arc length X of the point of projection, to be had that is, in the first formula of (8-64) formula as l=0:

X=X=m₀                                    (8-67)

Take into account (for central meridian)

$\frac{\mathrm{dX}}{\mathrm{dB}}=M$

$\frac{\mathrm{dB}}{\mathrm{dq}}=\frac{N\cos B}{M}=\frac{r}{M}=V^2\cos B$

:

$m_1=\frac{{\mathrm{<figure-callout\; id="0"\; label="dm"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">dm</figure-callout>}}_0}{\mathrm{dq}}=\frac{\mathrm{dX}}{\mathrm{dq}}=\frac{\mathrm{dX}}{\mathrm{dB}}\&CenterDot;\frac{\mathrm{dB}}{\mathrm{dq}}=r=N\cos B=\frac{c}{V}\cos B---(8-\mathrm{68,69})$

$m_2=-\frac{1}{2}\&CenterDot;\frac{{\mathrm{dm}}_1}{\mathrm{dq}}=-\frac{1}{2}\&CenterDot;\frac{{\mathrm{dm}}_1}{\mathrm{dB}}\frac{\mathrm{dB}}{\mathrm{dq}}=\frac{N}{2}\sin B\cos B---(8-70)$

Successively derivation, and the left and right (8-66) Ke get m3, m4, m5 are successively substituted into, m6 is respectively worth:

$\left.\begin{array}{c}{m}_3=\frac{N}{6}{\cos }^{3}B(1-t^2+{\mathrm{\&eta;}}^2)\\ m_4=\frac{N}{24}\sin B{\cos }^{3}B(5-t^2+9{\mathrm{\&eta;}}^2+4{\mathrm{\&eta;}}^4)\\ m_5=\frac{N}{120}{\cos }^{5}B(5-18t^2+t^4+14{\mathrm{\&eta;}}^2-58{\mathrm{\&eta;}}^2{t}^2)\\ m_6=\frac{N}{720}\sin B{\cos }^{5}B(61-58t^2+t^4+270{\mathrm{\&eta;}}^2-330{\mathrm{\&eta;}}^2{t}^2+200{\mathrm{\&eta;}}^4-232{\mathrm{\&eta;}}^4{t}^2)\end{array}\right\}---(8-71)$

M4- > m5- > m6- > m7 specific derivation process

$m_4=\frac{N}{24}\sin B{\cos }^{3}B(5-t^2+9{\mathrm{\&eta;}}^2+4{\mathrm{\&eta;}}^4)$

$m_5=\frac{1}{5}\&CenterDot;\frac{{\mathrm{dm}}_4}{\mathrm{dq}}=\frac{1}{5}\&CenterDot;\frac{{\mathrm{dm}}_4}{\mathrm{dB}}\&CenterDot;\frac{\mathrm{dB}}{\mathrm{dq}}=\frac{1}{5}\&CenterDot;\frac{{\mathrm{dm}}_4}{\mathrm{dB}}\&CenterDot;V^2\cos B$

Enable (5-t²+9η²+4η⁴)=Temp,

It is expressed as (...) '

$[\frac{N}{24}\sin B{\cos }^{3}B(5-t^2+9{\mathrm{\&eta;}}^2+4{\mathrm{\&eta;}}^4){]}^{\&prime;}\&CenterDot;V^2\cos B$

$=\frac{1}{24}\{\frac{N}{V^2}{\mathrm{\&eta;}}^{2}t\sin B{\cos }^{3}B\&CenterDot;\mathrm{Temp}+N{[\left(\sin B{\cos }^{3}B\right)\&CenterDot;\mathrm{Temp}]}^{\&prime;}\}{V}^2\cos B$

$=\frac{N}{24}\left\{\begin{array}{c}{\mathrm{\&eta;}}^{2}t\sin B{\cos }^{4}B\&CenterDot;\mathrm{Temp}+[(-3{\sin }^{2}B{\cos }^{2}B+{\cos }^{4}B)\&CenterDot;\mathrm{Temp}\\ -({16\mathrm{\&eta;}}^4+18{\mathrm{\&eta;}}^2+2\mathrm{se}{c}^{2}B){\sin }^{2}B{\cos }^{2}B]V^2\cos B\end{array}\right\}$

$=\frac{N}{24}{\cos }^{5}B\{{\mathrm{\&eta;}}^2{t}^2\&CenterDot;\mathrm{Temp}+[(-3t^2+1)\&CenterDot;\mathrm{Temp}-(16{\mathrm{\&eta;}}^4+18{\mathrm{\&eta;}}^2+2s{\mathrm{ec}}^{2}B)t^2\left]V^2\right\}$

①η²t²Temp=η²t²·(5-t²+9η²+4η⁴)

=5 η²t²-η²t⁴+9η⁴t²+4η⁶t²

②(3t²+ 1) Temp=(- 3t²+1)·(5-t²+9η²+4η⁴)

=5-16t²+3t⁴+9η²-27η²t²+4η⁴-12η⁴t²

③②-(16η⁴+18η²+2sec² B)t²

=5-16t²+3t⁴+9η²-45η²t²+4η⁴-28η⁴t²-2t²sec²B

  ③×(η²+1)

4.=5-16t²+3t⁴+14η²-61η²t²+3η²t⁴+13η⁴-73η⁴t²+4η⁶

  -28η⁶t²-2η²t²sec²B-2t²sec²B

Take sec into account²B=1+t²Above formula is substituted into, then

⑤①+④

=5-18t²+t⁴+14η²-58η²t²+13η⁴-64η⁴t²+4η⁶-24η⁶t²

To sum up

$m_5=\frac{N}{120}{\cos }^{5}B(5-18t^2+t^4+14{\mathrm{\&eta;}}^2-58{\mathrm{\&eta;}}^2{t}^2+13{\mathrm{\&eta;}}^4-64{\mathrm{\&eta;}}^4{t}^2+4{\mathrm{\&eta;}}^6-24{\mathrm{\&eta;}}^6{t}^2)$

The item of item 6 times of η 4 times is all smaller, smaller to whole value effect, so omitting in following derivation process

$m_5=\frac{N}{120}{\cos }^{5}B(5-18t^2+t^4+14{\mathrm{\&eta;}}^2-58{\mathrm{\&eta;}}^2{t}^2)$

$m_6=-\frac{1}{6}\&CenterDot;\frac{{\mathrm{dm}}_5}{\mathrm{dq}}=-\frac{1}{6}\&CenterDot;\frac{{\mathrm{dm}}_5}{\mathrm{dB}}\&CenterDot;\frac{\mathrm{dB}}{\mathrm{dq}}=-\frac{1}{6}\&CenterDot;\frac{{\mathrm{dm}}_5}{\mathrm{dB}}\&CenterDot;V^2\cos B$

Enable (5-18t²+t⁴+14η²-58η²t²)=Temp,

It is expressed as (...) '

${\left[\frac{N}{120}{\cos }^{5}B(5-18t^2+t^4+14{\mathrm{\&eta;}}^2-58{\mathrm{\&eta;}}^2{t}^2)\right]}^{\&prime;}\&CenterDot;V^2\cos B$

$=\frac{1}{120}[\frac{N}{V^2}{\mathrm{\&eta;}}^{2}t{\cos }^{5}B\&CenterDot;\mathrm{Temp}+N{({\cos }^{5}B\&CenterDot;\mathrm{Temp})}^{\&prime;}]\&CenterDot;V^2\cos B$

$=\frac{N}{120}\sin B{\cos }^{5}B[{\mathrm{\&eta;}}^2\mathrm{Temp}+(-5\mathrm{Temp}+(\frac{80}{{\cos }^{2}B}+\frac{{4t}^2}{{\cos }^{2}B}-28{\mathrm{\&eta;}}^{2}t+116({\mathrm{\&eta;}}^2{t}^2-\frac{V^2}{{\cos }^{2}B})))V^2]$

${\mathrm{\&eta;}}^2\mathrm{Temp}+(-5\mathrm{Temp}+(\frac{80}{{\cos }^{2}B}+\frac{4t^2}{{\cos }^{2}B}-28{\mathrm{\&eta;}}^{2}t+116({\mathrm{\&eta;}}^2{t}^2-\frac{V^2}{{\cos }^{2}B})))V^2$

$={\mathrm{\&eta;}}^2(5-18t^2+t^4+14{\mathrm{\&eta;}}^2-58{\mathrm{\&eta;}}^2{t}^2)-5\&times;(5-18t^2+t^4+14{\mathrm{\&eta;}}^2-58{\mathrm{\&eta;}}^2{t}^2)V^2$

$+\frac{80}{{\cos }^{2}B}({\mathrm{\&eta;}}^2+1)+\frac{{4t}^2}{{\cos }^{2}B}({\mathrm{\&eta;}}^2+1)-28{\mathrm{\&eta;}}^{2}t({\mathrm{\&eta;}}^2+1)+116{\mathrm{\&eta;}}^2{t}^2({\mathrm{\&eta;}}^2+1)$

$-\frac{116\&times;{({\mathrm{\&eta;}}^2+1)}^2}{{\cos }^{2}B}$

$=5{\mathrm{\&eta;}}^2-18{\mathrm{\&eta;}}^2{t}^2+{\mathrm{\&eta;}}^2{t}^4+14{\mathrm{\&eta;}}^4-58{\mathrm{\&eta;}}^4{t}^2+(-25+90t^2-5t^4-70{\mathrm{\&eta;}}^2+290{\mathrm{\&eta;}}^2{t}^2)({\mathrm{\&eta;}}^2+1)$

$+(80{\mathrm{\&eta;}}^2+80)(t^2+1)+(4{\mathrm{\&eta;}}^2{t}^2+4t^2)(t^2+1)-28{\mathrm{\&eta;}}^{4}t-28{\mathrm{\&eta;}}^{2}t$

$+116{\mathrm{\&eta;}}^4{t}^2+116{\mathrm{\&eta;}}^2{t}^2-116({\mathrm{\&eta;}}^4+2{\mathrm{\&eta;}}^2+1)(t^2+1)$

$=5{\mathrm{\&eta;}}^2-18{\mathrm{\&eta;}}^2{t}^2+{\mathrm{\&eta;}}^2{t}^4+14{\mathrm{\&eta;}}^4-58{\mathrm{\&eta;}}^4{t}^2$

$-25{\mathrm{\&eta;}}^2+90{\mathrm{\&eta;}}^2{t}^2-5{\mathrm{\&eta;}}^2{t}^4-70{\mathrm{\&eta;}}^4+290{\mathrm{\&eta;}}^4{t}^2-25+90t^2-5t^4-70{\mathrm{\&eta;}}^2+290{\mathrm{\&eta;}}^2{t}^2$

$+80{\mathrm{\&eta;}}^2{t}^2+80{\mathrm{\&eta;}}^2+80t^2+80+4{\mathrm{\&eta;}}^2{t}^4+4t^4+4{\mathrm{\&eta;}}^2{t}^2+4t^2-28{\mathrm{\&eta;}}^{4}t-28{\mathrm{\&eta;}}^{2}t$

$+126{\mathrm{\&eta;}}^4{t}^2+116{\mathrm{\&eta;}}^2{t}^2-116{\mathrm{\&eta;}}^4{t}^2-232{\mathrm{\&eta;}}^2{t}^2-116t^2-116{\mathrm{\&eta;}}^4-232{\mathrm{\&eta;}}^2-116$

$=-61+58t^2-t^4-270{\mathrm{\&eta;}}^2+330{\mathrm{\&eta;}}^2{t}^2-200{\mathrm{\&eta;}}^4+232{\mathrm{\&eta;}}^4{t}^2$

$=-61+58t^2-t^4-242{\mathrm{\&eta;}}^2+330{\mathrm{\&eta;}}^2{t}^2-172{\mathrm{\&eta;}}^4+232{\mathrm{\&eta;}}^4{t}^2-28{\mathrm{\&eta;}}^{4}t-28{\mathrm{\&eta;}}^{2}t$

To sum up obtain m6:

$m_6=\frac{N}{720}\sin b{\cos }^{5}B(61-58t^2+t^4+242{\mathrm{\&eta;}}^2-330{\mathrm{\&eta;}}^2{t}^2+172{\mathrm{\&eta;}}^4-232{\mathrm{\&eta;}}^4{t}^2+28{\mathrm{\&eta;}}^{4}t+28{\mathrm{\&eta;}}^{2}t)$

$m_6=\frac{N}{720}\sin B{\cos }^{5}B(61-58t^2+t^4+242{\mathrm{\&eta;}}^2-330{\mathrm{\&eta;}}^2{t}^2)$

$m_7=\frac{1}{7}\&CenterDot;\frac{{\mathrm{dm}}_6}{\mathrm{dq}}=\frac{1}{7}\&CenterDot;\frac{{\mathrm{dm}}_6}{\mathrm{dB}}\&CenterDot;\frac{\mathrm{dB}}{\mathrm{dq}}=\frac{1}{7}\&CenterDot;\frac{{\mathrm{dm}}_6}{\mathrm{dB}}\&CenterDot;V^2\cos B$

To sum up:

(8-64) formula is substituted into, gauss projection is obtained and just calculates formula,

$x=X+\frac{N}{2{\mathrm{\&rho;}}^{\&prime;\&prime;2}}\sin B\cos B\&CenterDot;l^{\&prime;\&prime;2}+\frac{N}{24{\mathrm{\&rho;}}^{\&prime;\&prime;4}}\sin B{\cos }^{3}B(5-t^2+9{\mathrm{\&eta;}}^2+4{\mathrm{\&eta;}}^4)l^{\&prime;\&prime;4}$

$y=\frac{N}{{\mathrm{\&rho;}}^{\&prime;\&prime;}}\cos B\&CenterDot;l^{\&prime;\&prime;}+\frac{N}{6{\mathrm{\&rho;}}^{\&prime;\&prime;3}}{\cos }^{3}B(1-t^2+{\mathrm{\&eta;}}^2)l^{\&prime;\&prime;3}$

$+\frac{N}{120{\mathrm{\&rho;}}^{\&prime;\&prime;5}}{\cos }^{5}B(5-18t^2+t^4+14{\mathrm{\&eta;}}^2-58{\mathrm{\&eta;}}^2{t}^2)l^{\&prime;\&prime;5}---(8-73)$

(5) plane coordinates needed for deriving high-precision using gauss projection reverse calculation algorithms turns geodetic coordinates formula:

$B=B_f-\frac{t_f}{2M_f{N}_f}{y}^2+\frac{t_f}{24M_f{N}_f^3}(5+3t_f^2+{\mathrm{\&eta;}}_f^2-9{\mathrm{\&eta;}}_f^2{t}_f^2-4{\mathrm{\&eta;}}_f^4)y$

$l=\frac{y}{N_f\cos B_f}-\frac{y^3}{6N_f^3\cos B_f}(1+2t_f^2+{\mathrm{\&eta;}}_f^2)+\frac{y^5}{120N_f^5\cos B_f}(5+28t_f^2+24t_f^4+6{\mathrm{\&eta;}}_f^2+8{\mathrm{\&eta;}}_f^2{t}_f^2),$ Wherein B, l

The as the earth transverse and longitudinal coordinate value of measurement point；In formula: t=tanB；η²=e '²cos²B；Its process specifically: projection equation:

Meet following three conditions:

1. being the symmetry axis that central meridian is projection after x coordinate axial projection；2. length is constant after x coordinate axial projection；3. projection has conformality property, i.e. orthomorphic condition.

The derivation method of inversion formula is similar with positive formula of calculating, its basic thought is to calculate the intersection point dimension B of projection of the ordinate on ellipsoid according to x first_f, then press B_fCalculate (B_f- B) and through poor l, finally obtain

$\left.\begin{array}{c}B=B_f-(B_f-B)\\ L={\mathrm{<figure-callout\; id="0"\; label="L"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">L</figure-callout>}}_0+l\end{array}\right\}---(8-75)$

In order to simplify difference (B_f- B) and l calculation formula derivation, we are still by means of isometric latitude q.Since gauss projection region is little, it is also little that wherein y value is compared with ellipsoid radius, therefore (q, l) can be expanded into the power series of y.With respect to above-mentioned first condition, i.e., projection is to the symmetrical requirement of roller noon, and q should be the even function of y, and l should be the odd function of y, therefore

$\left.\begin{array}{c}q={\mathrm{<figure-callout\; id="0"\; label="n"\; filenames="HDA00001904011800012.png"\; state="\{\{state\}\}">n</figure-callout>}}_0+n_2{y}^2+n_4{y}^4\\ l=n_{1}y+n_3{y}^3+n_5{y}^5\end{array}\right\}---(8-76)$

Known by third condition:

$\frac{\&PartialD;q}{\&PartialD;x}=\frac{\&PartialD;l}{\&PartialD;y},$

$\frac{\&PartialD;l}{\&PartialD;x}=-\frac{\&PartialD;q}{\&PartialD;y},$

(8-76) formula is sought into partial derivative to x and y respectively;And consider above-mentioned second condition, as y=0, point is in central meridian, that is, l=0, x=X,

Then undetermined coefficient n can be found out, they are substituted into above formula, then

$\left.\begin{array}{c}q=q_f-\frac{y^2}{2}\left(\frac{d^{2}q}{{\mathrm{dX}}^2}\right)+\frac{y^4}{24}\left(\frac{d^{4}q}{{\mathrm{dX}}^4}\right)\\ l=y{\left(\frac{\mathrm{dq}}{\mathrm{dX}}\right)}_f-\frac{y^3}{6}{\left(\frac{d^{3}q}{{\mathrm{dX}}^3}\right)}_f\end{array}\right\}---(8-77)$

Q in formula_fWith

Correspond to intersection point latitude B_fNumerical value.

By (8-38) formula it is found that q must have fixed functional relation with B, the present is set

B=f (q), B_f=f(q_f)

Again

B=f(q)=f(q_f+dq)

dq=q-q_f

Then have by Taylor series expansion

$B=f\left(q_f\right)+\left(\frac{\mathrm{dB}}{\mathrm{dq}}\right)\mathrm{dq}+\frac{1}{2}\left(\frac{d^{2}B}{{\mathrm{dq}}^2}\right){\mathrm{dq}}^2+\frac{1}{6}\left(\frac{d^{3}B}{{\mathrm{dq}}^3}\right){\mathrm{dq}}^3---(8-80)$

Therefore

$-(B_f-B)=B-B_f=\left(\frac{\mathrm{dB}}{\mathrm{dq}}\right)(q-q_f)+\frac{1}{2}\left(\frac{d^{2}B}{{\mathrm{dq}}^2}\right)\&CenterDot;{(q-q_f)}^2+\frac{1}{6}\left(\frac{d^{3}B}{{\mathrm{dq}}^3}\right)\&CenterDot;{(q-q_f)}^3---(8-81)$

In order to acquire the practical inversion formula of Gauss coordinate, we must also find out the all-order derivative value in (8-77) formula and (8-81) formula $\frac{d^{n}q}{{\mathrm{dX}}^n}$ And $\frac{d^{n}B}{{\mathrm{dq}}^n}.$

Derivation numerical value first

First derivative is known by (8-68) formula

$\frac{\mathrm{dq}}{\mathrm{dx}}=\frac{\mathrm{dq}}{\mathrm{dX}}=\frac{1}{N_f\cos B_f}$

Other all-order derivatives can be according to above formula gradually to X derivation:

It asks $\frac{d^{4}q}{{\mathrm{dX}}^4}$

Subscript f is omitted for variable following in easy calculating process

${\left[\frac{1+2t^2+{\mathrm{\&eta;}}^2}{N^3\cos B}\right]}^{\&prime;}\&CenterDot;\frac{V^2}{N}$

$=\frac{{(1+2t^2+{\mathrm{\&eta;}}^2)}^{\&prime;}{N}^3\cos B-{\left(N^3\cos B\right)}^{\&prime;}(1+2t^2+{\mathrm{\&eta;}}^2)}{N^6{\cos }^{2}B}\&CenterDot;\frac{V^2}{N}$

$=\frac{{(1+{2t}^2+{\mathrm{\&eta;}}^2)}^{\&prime;}{N}^3\cos B-(\frac{{3N}^3}{V^2}{\mathrm{\&eta;}}^{2}t\&CenterDot;\cos B-\sin B\&CenterDot;N^3)(1+2t^2+{\mathrm{\&eta;}}^2)}{N^6{\cos }^{2}B}\&CenterDot;\frac{V^2}{N}$

$=\frac{(4t\&CenterDot;{\sec }^{2}B-2{\mathrm{\&eta;}}^{2}t)({\mathrm{\&eta;}}^2+1)-({3\mathrm{\&eta;}}^{2}t-t({\mathrm{\&eta;}}^2+1))(1+2t^2+{\mathrm{\&eta;}}^2)}{N^4\cos B}$

(4t·sec²B-2η²t)(η²+1)-(3η²t-t(η²+1))(1+2t²+η²)

=t [(4sec²B-2η²)(η²+1)-(2η²-1)(1+2t²+η²)]

=t [(4 η²sec²B-2η⁴+4·sec²B-2η²)-(2η²+4η²t²+2η⁴-1-2t²-η²)]

=t (4 η²sec²B-2η⁴+4·sec²B-2η²-η²-4η²t²-2η⁴+1+2t²)

=t (4 η²sec²B-4η⁴+4·sec²B-3η²-4η²t²+1+2t²)

=t (4 η²(t²+1)-4η⁴+4·(t²+1)-3η²-4η²t²+1+2t²)

=t (η²-4η⁴+6t²+5)

$\frac{d^{4}q}{{\mathrm{dX}}^4}=\frac{t_f}{{N_f}^4\cos B_f}(5+6t_f^2+{\mathrm{\&eta;}}_f^2-4{\mathrm{\&eta;}}_f^4)$

It asks $\frac{d^{5}q}{{\mathrm{dX}}^5}$

${\left\{\frac{t}{N^4\cos B}(5+{6t}^2+{\mathrm{\&eta;}}^2-4{\mathrm{\&eta;}}^4)\right\}}^{\&prime;}\&CenterDot;\frac{V^2}{N}$

$=\left\{\frac{{\left[t(5+6t^2+{\mathrm{\&eta;}}^2-4{\mathrm{\&eta;}}^4)\right]}^{\&prime;}{N}^4\cos B-{\left(N^4\cos B\right)}^{\&prime;}\left[t(5+6t^2+{\mathrm{\&eta;}}^2-4{\mathrm{\&eta;}}^4)\right]}{N^8{\cos }^{2}B}\right\}\&CenterDot;\frac{V^2}{N}$

$=\left\{\frac{(5+23t^2+18t^4+{\mathrm{\&eta;}}^2-{\mathrm{\&eta;}}^2{t}^2-4{\mathrm{\&eta;}}^4+12{\mathrm{\&eta;}}^4{t}^2)-(\frac{4}{V^2}{\mathrm{\&eta;}}^2{t}^2-t^2)(5+{6t}^2+{\mathrm{\&eta;}}^2-4{\mathrm{\&eta;}}^4)}{N^5\cos B}\right\}\&CenterDot;V^2$

$=\frac{5+28t^2+24t^4+6{\mathrm{\&eta;}}^2+8{\mathrm{\&eta;}}^2{t}^2-3{\mathrm{\&eta;}}^4+4{\mathrm{\&eta;}}^4{t}^2-4{\mathrm{\&eta;}}^6+24{\mathrm{\&eta;}}^6{t}^2}{N^5\cos B}$

[t(5+6t²+η²-4η⁴)]'

=sec²B(5+6t²+η²-4η⁴)+12t²sec²B-2η²t²+16η⁴t²

=5sec²B+18t²sec²B+η²sec²B-4η⁴sec²B-2η²t²+16η⁴t²

=5 (t²+1)+18t²(t²+1)+η²(t²+1)-4η⁴(t²+1)-2η²t²+16η⁴t²

=5t²+5+18t⁴+18t²+η²t²+η²-4η⁴t²-4η⁴-2η²t²+16η⁴t²

=5+23t²+18t⁴+η²-η²t²-4η⁴+12η⁴t²

${\left(N^4\cos B\right)}^{\&prime;}$

$=\frac{4N^4}{V^2}{\mathrm{\&eta;}}^{2}t\cos B-\sin B\&CenterDot;N^4$

$(\frac{4}{V^2}{\mathrm{\&eta;}}^2{t}^2-t^2)(5+6t^2+{\mathrm{\&eta;}}^2-4{\mathrm{\&eta;}}^4)\&CenterDot;V^2$

$=(3{\mathrm{\&eta;}}^2{t}^2-t^2)(5+6t^2+{\mathrm{\&eta;}}^2-4{\mathrm{\&eta;}}^4)$

$=-5t^2-{6t}^4+14{\mathrm{\&eta;}}^2{t}^2+18{\mathrm{\&eta;}}^2{t}^4+7{\mathrm{\&eta;}}^4{t}^2-12{\mathrm{\&eta;}}^6{t}^2$

(5+23t²+18t⁴+η²-η²t²-4η⁴+12η⁴t²)·V²

=5 η²+5+23t²η²+23t²+18t⁴η²+18t⁴+η⁴+η²

-η⁴t²-η²t²-4η⁶-4η⁴+12η⁶t²+12η⁴t²

=6 η²+5+22η²t²+23t²+18η²t⁴+18t⁴-4η⁶-3η⁴+12η⁶t²+11η⁴t²

=5+23t²+18t⁴+6η²+22η²t²+18η²t⁴+11η⁴t²-4η⁶-3η⁴+12η⁶t²

5+23t²+18t⁴+6η²+22η²t²+18η²t⁴+11η⁴t²-4η⁶-3η⁴+12η⁶t²

-(-5t²-6t⁴+14η²t²+18η²t⁴+7η⁴t²-12η⁶t²)

=5+28t²+24t⁴+6η²+8η²t²-3η⁴+4η⁴t²-4η⁶+24η⁶t²

: $\frac{d^{5}q}{{\mathrm{dX}}^5}=\frac{1}{N_f^5\cos B_f}(5+28t_f^2+24t_f^4+6{\mathrm{\&eta;}}_f^2+8{\mathrm{\&eta;}}_f^2{t}_f^2-3{\mathrm{\&eta;}}_f^4+4{\mathrm{\&eta;}}_f^4{t}_f^2-4{\mathrm{\&eta;}}_f^6+24{\mathrm{\&eta;}}_f^6{t}_f^2)$

So that

$\left.\begin{array}{c}\frac{d^{2}q}{{\mathrm{dX}}^2}=\frac{t_f}{{N_f}^2\cos B_f}\\ \frac{d^{3}q}{{\mathrm{dX}}^3}=\frac{1}{{N_f}^3\cos B_f}(1+2{t_f}^2+{{\mathrm{\&eta;}}_f}^2)\\ \frac{d^{4}q}{{\mathrm{dX}}^4}=\frac{t_f}{{N_f}^4\cos B_f}(5+6{t_f}^2+{{\mathrm{\&eta;}}_f}^2-4{{\mathrm{\&eta;}}_f}^4)\\ \frac{d^{5}q}{{\mathrm{dX}}^4}=\frac{1}{N_f^5\cos B_f}(5+{28t}_f^2+24t_f^4+6{\mathrm{\&eta;}}_f^2+8{\mathrm{\&eta;}}_f^2{t}_f^2-3{\mathrm{\&eta;}}_f^4+4{\mathrm{\&eta;}}_f^4{t}_f^2-4{\mathrm{\&eta;}}_f^6+24{\mathrm{\&eta;}}_f^6{t}_f^2)\end{array}\right\}---(8-83)$

Next is differentiated

First derivative is apparent from by (8-38) formula

${\left(\frac{\mathrm{dB}}{\mathrm{dq}}\right)}_f={\left(\frac{N\cos B}{M}\right)}_f={V_f}^2\cos B_f---(8-84)$

Other all-order derivatives can gradually obtain q according to above formula,

$\frac{d^{2}B}{{\mathrm{dq}}^2}=\frac{V^2\cos B}{\mathrm{dB}}\&CenterDot;\frac{\mathrm{dB}}{\mathrm{dq}}$

$=(-2{\mathrm{\&eta;}}^{2}t\cos B-\sin B\&CenterDot;V^2)\&CenterDot;V^2\cos B$

$=-\sin B({3\mathrm{\&eta;}}^2+1)\&CenterDot;V^2\cos B$

$\frac{d^{3}B}{{\mathrm{dq}}^3}=\frac{-\sin B_f\cos B_f(1+4{{\mathrm{\&eta;}}_f}^2+3{{\mathrm{\&eta;}}_f}^4)}{\mathrm{dB}}\&CenterDot;\frac{\mathrm{dB}}{\mathrm{dq}}$

$=[{\left({-\sin B}_f\cos B_f\right)}^{\&prime;}(1+4{{\mathrm{\&eta;}}_f}^2+3{{\mathrm{\&eta;}}_f}^4)+{(1+4{{\mathrm{\&eta;}}_f}^2+3{{\mathrm{\&eta;}}_f}^4)}^{\&prime;}\left({-\sin B}_f\cos B_f\right)]\&CenterDot;V^2\cos B$

$=[{\cos }^{2}B(t^2-1)(1+4{{\mathrm{\&eta;}}_f}^2+3{{\mathrm{\&eta;}}_f}^4)+(-8{\mathrm{\&eta;}}^{2}t-12{\mathrm{\&eta;}}^{4}t)(-\sin B_f\cos B_f)]\&CenterDot;V^2\cos B$

$={\cos }^{2}B[(t^2-1)(1+4{{\mathrm{\&eta;}}_f}^2+3{{\mathrm{\&eta;}}_f}^4)+({8\mathrm{\&eta;}}^2+12{\mathrm{\&eta;}}^4)t^2]\&CenterDot;V^2\cos B$

$={\cos }^{2}B[(t^2-1)(1+4{{\mathrm{\&eta;}}_f}^2+3{{\mathrm{\&eta;}}_f}^4)+{8\mathrm{\&eta;}}^2{t}^2+12{\mathrm{\&eta;}}^4{t}^2]\&CenterDot;V^2\cos B$

$={\cos }^{2}B(t^2+4{{\mathrm{\&eta;}}_f}^2{t}^2+3{{\mathrm{\&eta;}}_f}^4{t}^2-1-4{{\mathrm{\&eta;}}_f}^2-3{{\mathrm{\&eta;}}_f}^4+8{\mathrm{\&eta;}}^2{t}^2+12{\mathrm{\&eta;}}^4{t}^2)\&CenterDot;V^2\cos B$

$={\cos }^{3}B[-1+t^2-4{{\mathrm{\&eta;}}_f}^2+12{{\mathrm{\&eta;}}_f}^2{t}^2-3{{\mathrm{\&eta;}}_f}^4+15{{\mathrm{\&eta;}}_f}^4{t}^2]\&CenterDot;V^2$

(-sinBcosB)'

=-(cos²B-sin²B)=(t²-1)cos²B

(1+4η_f ²+3η_f ⁴)'

=-8 η²t-12η⁴t

As a result,

$\left.\begin{array}{c}{\left(\frac{d^{2B}}{{\mathrm{dq}}^2}\right)}_f=-\sin B_f\cos B_f(1+4{\mathrm{\&eta;}}_f^2+3{\mathrm{\&eta;}}_f^4)\\ =-\sin B_f\cos B_f(3{\mathrm{\&eta;}}_f^2+1)\&CenterDot;V_f^2\\ {\left(\frac{d^{3}B}{{\mathrm{dq}}^3}\right)}_f=-{\cos }^3{B}_f(1-t_f^2+5{\mathrm{\&eta;}}_f^2-13{\mathrm{\&eta;}}_f^2{t}_f^2)\\ ={\cos }^3{B}_f(-1+t_f^2-4{\mathrm{\&eta;}}_f^2+12{\mathrm{\&eta;}}_f^2{t}_f^2-3{\mathrm{\&eta;}}_f^4+15{\mathrm{\&eta;}}_f^4{t}_f^2)\&CenterDot;V_f^2\end{array}\right\}$

After by related formula is updated to (8-77) first formula in (8-83) formula, but it is available

$q-q_f=-\frac{t_f\sec B_f}{2N_f^2}{y}^2+\frac{t_f\sec B_f}{{24N}_f^4}(5+6t_f^2+{\mathrm{\&eta;}}_f^2-4{\mathrm{\&eta;}}_f^4)y^4$

$-\frac{t_f\sec B_f}{720N_f^6}(61+180t_f^2+120t_f^4+46{\mathrm{\&eta;}}_f^2+48{\mathrm{\&eta;}}_f^2{t}_f^2)y^6$

To above formula, quadratic sum cube is obtained respectively,

$\left.\begin{array}{c}{(q-q_f)}^2=\frac{t_f^2{\sec }^2{B}_f}{{4N}_f^4}{y}^4-\frac{t_f^2{\sec }^2{B}_f}{{24N}_f^6}(5+{6t}_f^2+{\mathrm{\&eta;}}_f^2-{4\mathrm{\&eta;}}_f^4)y^6\\ {(q-q_f)}^3=-\frac{t_f^3{\sec }^3{B}_f}{{8N}_f^6}{y}^6\end{array}\right\}---(8-86)$

(8-84) ~ (8-86) formula is substituted into (8-81) formula, by (8-82), (8-83) formula substitutes into second formula of (8-77) formula, and omits

WithSecondary item, it is collated

$-(B_f-B)=B-B_f=$

$\left({V_f}^2\cos B_f\right)(q-q_f)$

$+\frac{1}{2}(-\sin B_f\cos B_f(1+{{4\mathrm{\&eta;}}_f}^2+{{3\mathrm{\&eta;}}_f}^4))\&CenterDot;{(q-q_f)}^2$

$+\frac{1}{6}(-{\cos }^3{B}_f(1-{t_f}^2+{{5\mathrm{\&eta;}}_f}^2-13{{\mathrm{\&eta;}}_f}^2{t_f}^2))\&CenterDot;{(q-q_f)}^3$

$\left({V_f}^2\cos B_f\right)(-\frac{t_f\sec B_f}{2N_f^2}{y}^2+\frac{t_f\sec B_f}{{24N}_f^4})(5+{6t}_f^2+{\mathrm{\&eta;}}_f^2-4{\mathrm{\&eta;}}_f^4)y^4$

$-\frac{t_f\sec B_f}{720N_f^6}(61+180t_f^2+120t_f^4+46{\mathrm{\&eta;}}_f^2+48{\mathrm{\&eta;}}_f^2{t}_f^2)y^6)$

$=-\frac{t_f}{2M_f{N}_f}{y}^2+\frac{t_f}{24M_f{N}_f^3}(5+{6t}_f^2+{\mathrm{\&eta;}}_f^2-4{\mathrm{\&eta;}}_f^4)y^4$

$-\frac{t_f}{720M_f{N}_f^5}(61+180t_f^2+120t_f^4+46{\mathrm{\&eta;}}_f^2+48{\mathrm{\&eta;}}_f^2{t}_f^2)y^6$

$\frac{1}{2}\left({-\sin B}_f{\cos B}_f(1+{{4\mathrm{\&eta;}}_f}^2+3{{\mathrm{\&eta;}}_f}^4)\right)\&CenterDot;{(q-q_f)}^2$

$=\frac{1}{2}(-\sin B({3\mathrm{\&eta;}}^2+1)\&CenterDot;V^2\cos B)(\frac{t_f^2{\sec }^2{B}_f}{{4N}_f^4}{y}^4-\frac{t_f^2{\sec }^2{B}_f}{24N_f^6})(5+6t_f^2+{\mathrm{\&eta;}}_f^2-4{\mathrm{\&eta;}}_f^4)y^6)$

$=\frac{-({3\mathrm{\&eta;}}^2+1)t_f^3}{{8\mathrm{MN}}_f^3}{y}^4+\frac{({3\mathrm{\&eta;}}^2+1)t_f^3}{{48\mathrm{MN}}_f^5}(5+6t_f^2+{\mathrm{\&eta;}}_f^2-4{\mathrm{\&eta;}}_f^4)y^6$

$\frac{1}{6}{\cos }^3{B}_f(-1+t_f^2-{4\mathrm{\&eta;}}_f^2+{12\mathrm{\&eta;}}_f^2{t}_f^2-{3\mathrm{\&eta;}}_f^4+{15\mathrm{\&eta;}}_f^4{t}_f^2)\&CenterDot;V_f^2\&CenterDot;{(q-q_f)}^3$

$=\frac{1}{6}{\cos }^{3}B[-1+t^2-{{4\mathrm{\&eta;}}_f}^2+{{12\mathrm{\&eta;}}_f}^2{t}^2-{{3\mathrm{\&eta;}}_f}^4+15{{\mathrm{\&eta;}}_f}^4{t}^2]\&CenterDot;V^2(-\frac{t_f^3{\sec }^3{B}_f}{8N_f^6}{y}^6)$

$=\frac{1}{6}[-1+t^2-4{{\mathrm{\&eta;}}_f}^2+12{{\mathrm{\&eta;}}_f}^2{t}^2-3{{\mathrm{\&eta;}}_f}^4+15{{\mathrm{\&eta;}}_f}^4{t}^2](-\frac{t_f^3}{8M{N}_f^5}{y}^6)$

$\left({V_f}^2\cos B_f\right)(q-q_f)+\frac{1}{2}(-\sin B_f\cos B_f(1+4{{\mathrm{\&eta;}}_f}^2+3{{\mathrm{\&eta;}}_f}^4))\&CenterDot;{(q-q_f)}^2$

$+\frac{1}{6}(-{\cos }^3{B}_f(1-{t_f}^2+{{5\mathrm{\&eta;}}_f}^2-13{{\mathrm{\&eta;}}_f}^2{t_f}^2))\&CenterDot;{(q-q_f)}^3$

$=-\frac{t_f}{{2M}_f{N}_f}{y}^2+\frac{t_f}{24M_f{N}_f^3}(5+{6t}_f^2+{\mathrm{\&eta;}}_f^2-4{\mathrm{\&eta;}}_f^4)y^4$

$-\frac{t_f}{720M_f{N}_f^5}(61+180t_f^2+120t_f^4+46{\mathrm{\&eta;}}_f^2+48{\mathrm{\&eta;}}_f^2{t}_f^2)y^6$

$-\frac{({3\mathrm{\&eta;}}^2+1)t_f^3}{{8\mathrm{MN}}_f^3}{y}^4+\frac{({3\mathrm{\&eta;}}^2+1)t_f^3}{48{\mathrm{MN}}_f^5}(5+6t_f^2+{\mathrm{\&eta;}}_f^2-{4\mathrm{\&eta;}}_f^4)y^6$

$+\frac{1}{6}[-1+t^2-4{{\mathrm{\&eta;}}_f}^2+12{{\mathrm{\&eta;}}_f}^2{t}^2-3{{\mathrm{\&eta;}}_f}^4+15{{\mathrm{\&eta;}}_f}^4{t}^2](-\frac{t_f^3}{8{\mathrm{MN}}_f^5}{y}^6)$

$=-\frac{t_f}{2M_f{N}_f}{y}^2+\frac{t_f}{24M_f{N}_f^3}(5+3t_f^2+{\mathrm{\&eta;}}_f^2-9{\mathrm{\&eta;}}^2{t}_f^2-4{\mathrm{\&eta;}}_f^4)y^4$

$-\frac{t_f}{720M_f{N}_f^5}[(61+180t_f^2+120t_f^4+46{\mathrm{\&eta;}}_f^2+48{\mathrm{\&eta;}}_f^2{t}_f^2)$

$+{15t}^2(-6-{5t}^2-20{{\mathrm{\&eta;}}_f}^2-6{{\mathrm{\&eta;}}_f}^2{t}^2-2{{\mathrm{\&eta;}}_f}^4+15{{\mathrm{\&eta;}}_f}^4{t}^2+12{\mathrm{\&eta;}}_f^6)]y^6$

$=-\frac{t_f}{2M_f{N}_f}{y}^2+\frac{t_f}{24M_f{N}_f^3}(5+{3t}_f^2+{\mathrm{\&eta;}}_f^2-{9\mathrm{\&eta;}}^2{t}_f^2-{4\mathrm{\&eta;}}_f^4)y^4$

,

$\left.\begin{array}{c}B=B_f-\frac{t_f}{{2M}_f{N}_f}{y}^2+\frac{t_f}{24M_f{N}_f^3}(5+3t_f^2+{\mathrm{\&eta;}}_f^2-9{\mathrm{\&eta;}}_f^2{t}_f^2-4{\mathrm{\&eta;}}_f^4)y^4\\ l=\frac{y}{N_f\cos B_f}-\frac{y^3}{{6N}_f^3\cos B_f}(1+{2f}_t^2+{\mathrm{\&eta;}}_f^2)\\ +\frac{y^5}{120N_f^5\cos B_f}(5+{28t}_f^2+{24t}_f^4+{6\mathrm{\&eta;}}_f^2+{8\mathrm{\&eta;}}_f^2{t}_f^2)\end{array}\right\}---(8-87)$

Experimental example 1

It calculates 3 degree of band x just to calculate, calculates B for 40 degree

Calculation of curvature radius N	6387083.0473111
		Calculate square of η	0.00395432903655276

Angular transition in seconds is radian

For 5400 seconds (1.5 degree), p_11 is ρ * 3600

double l=l_11/p_11;

0.0261799387799149

It is that unit l/ ρ result is identical with degree

0.0261799387799149

The formula items for calculating abscissa x are X, a1, a2, a3

Corresponding Meridian arc length X is calculated by dimension B	4429607.36780102
		a1	1077.78288797253
a2	0.156473173757087
		a3	1.00688009324594E-05

Calculate final result abscissa x

4430685.30717223

Experimental example 2

6 degree of band y values are calculated, obtain following result now by taking maximum 3 degree of the longitude (10800 seconds, 0.052359877559829887307710723054658 radian) of 6 degree of bands as an example for the precision for examining error maximum:

a1	256206.456439453
		a2	-0.040251526038781
a3	-4.06069611771454E-05
		y	256206.41614732

Experimental example 3

Inverse is carried out by taking WGS84 as an example,

a=6378137

b=6356752.3142451794975639665996337

m0	6335552.71700043
		m2	63617.7573920752
m4	532.351802628968
		m6	4.15772613110754
m8	0.0313125734436458
		m10	0.00023058009161107

A0 unit (degree)	6367449.14582304
		A0 unit (radian)	111132.952547912
a2/2	16038.5086626504
		a4/4	16.8326131614634
a6/6	0.0219843738776171
		a8/8	3.11416246803452E-05
a10/10	4.50351741427872E-08

When calculating intersection point latitude Bf, used recurrence (iterative calculation) is needed.

By taking X=4433842.59393608 as an example

I is the number of iterations

Bf(X,i)

Bf (5)-Bf (4)=9.744107619769639495e-11 degree < (1e-6 seconds=2.7777777777777777777777777777778e-10 degree)

Bf6-Bf5=8.5552019681364349743458447721599e-14 degree < < (1e-6 seconds=2.7777777777777777777777777777778e-10 degree)

Bf (5)-Bf (4)=40.0388486815572-40.0388486814598=9.74e-11 < (1e-6 seconds=2.7777777777777777777777777777778e-10 degree)

Bf6-Bf5=40.0388486815573-40.0388486815572=1.e-13 degree < < (1e-6 seconds=2.7777777777777777777777777777778e-10 degree)

So using Bf6 for iterative value.

Claims (4)

1. a kind of grade high-acruracy survey conversion method, which is characterized in that this method includes the following steps:

(1) calculation formula of radius of curvature of meridian M is derived are as follows:

M=m₀+m₂sin²B+m₄sin⁴B+m₆sin⁶B+m₈sin⁸B+m₁₀sin¹⁰B；

(2) calculation formula of radius of curvature in prime vertical N is derived are as follows:

N=n₀+n₂sin²B+n₄sin⁴B+n₆sin⁶B+n₈sin⁸B+n₁₀sin¹⁰B；

(3) calculation formula of Meridian arc length X is derived:

In above-mentioned formula, B indicates the latitude of point to be measured；

In above-mentioned formula, m₀=a (1-e²)、

E is the first eccentricity of ellipsoid, is formulated as

A in above-mentioned formula is semimajor axis of ellipsoid, f flattening of ellipsoid；B is semiminor axis of ellipsoid, b=(1-f) * a；

(4) geodetic coordinates needed for method derives high-precision is just being calculated using gauss projection and is turning plane coordinates calculation formula:

Wherein x, y are the plane transverse and longitudinal coordinate value of measurement point；In formula:

ρ ": 1 radian=ρ " second is constant: 57.295779513082320876798154814105；

B: the latitude of point to be measured；L ": central meridian longitude-subpoint longitude of band where projection (difference of longitude, unit are the seconds)；

t:tanB；η²=e '²cos²B；A is semimajor axis of ellipsoid, f flattening of ellipsoid；B is semiminor axis of ellipsoid, b=(1-f) * a；

Elliptical first eccentricityElliptical second eccentricity

η²=e '²cos²B；(5) plane coordinates needed for deriving high-precision using gauss projection reverse calculation algorithms turns geodetic coordinates formula:

Wherein B, l are the earth transverse and longitudinal coordinate value of measurement point；In formula: t=tanB；η²=e '²cos²B。

2. a kind of grade high-acruracy survey conversion method according to requiring 1, it is characterized in that, the derivation process of the step (1) specifically: (1) differential arc length DK=ds is taken in a part of meridional ellipse, correspondingly there is increment of coordinate dx, point n is the center of curvature of differential arc dS, and then line segment Dn and Kn is radius of curvature of meridian M；

The defined formula of the radius of curvature of arbitrary plane curve are as follows:

∠ KBx, that is, latitude can be acquired from differential triangle DKE:

Above formula is substituted into (7-53), is obtained:

It is the normal of P point as Pn, is easy to get:

Elliptic equation:

X is taken and is led, is obtained

(7-13) compares and can obtain with (7-11):

Therefore, y=x (1-e2) tanB (7-14) is acquired

(7-14) is substituted into (7-12), both sides are shifted with a2cos2B is multiplied:

Above formula is substituted into (7-14) to obtain:

It is obtained by (7-16):

Due to

Above formula is substituted into (7-55), and because of W²=1-e²sin²B,

Then obtain radius of curvature of meridian formula are as follows:

Series is unfolded by Newton binomial theorem, takes to 10 items, obtains: M=m₀+m₂sin²B+m₄sin⁴B+m₆sin⁶B+m₈sin⁸B+m₁₀sin¹⁰B,

m₀=a (1-e²),

In formula:

It increases with the increase of B in relation to by M and latitude B.

3. a kind of grade high-acruracy survey conversion method according to requiring 1, it is characterized in that, the derivation process of the step (2) specifically: cross the tangent line PT for the parallel circle PHK that P point is made centered on O ', which is located at perpendicular in the parallel circle plane of meridian plane；Because prime vertical is also perpendicularly to meridian plane, therefore PT is also tangent line of the prime vertical at P point；I.e. PT is perpendicular to Pn；So PT is common tangent of the parallel circle PHK and prime vertical PEE ' at P point；Known by wheat Neil theorem, assuming that cutting arc by a little drawing two on curved surface, one is bevel arc, and this two sections of arcs have common tangential at that point, and at this moment the radius of curvature of bevel arc at this point is equal to the radius of curvature of method section arc multiplied by two-section arc plane holder cosine of an angle；Thus scheme the angle it is found that between parallel circle plane and prime vertical plane, as geodetic latitude B, if the radius of parallel circle indicates have with r

R=NcosB (7-62)

Again because parallel circle radius r is equal to the abscissa x of P point, that is,

Therefore, radius of curvature in prime vertical can be indicated with following two formula:

Series is unfolded by Newton binomial theorem in above formula, takes to 10 items, obtains:

N=n₀+n₂sin²B+n₄sin⁴B+n₆sin⁶B+n₈sin⁸B+n₁₀sin¹⁰B(7-70)

In formula: n₀=a,

4. a kind of grade high-acruracy survey conversion method according to requiring 1, which is characterized in that the derivation process of the step (3) specifically: take certain differential arc PP '=dx on meridian, enabling P point latitude is B, P ' latitudes are B+dB, and the radius of curvature of meridian of P point is M, are then had:

Dx=MdB

It can be found out to the arc length the parallel circle of any latitude B by following integral since equator:

M can be expressed with following formula in formula:

M=a₀-a₂cos2B+a₄cos4B-a₆cos6B+a₈cos8B-a₁₀cos10B 

Wherein:

It is integrated, Meridian arc length calculating formula is obtained after being arranged:

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