CN106525026B - Project to the parsing astronomical positioning method of the first point of Aries equatorial system of coordinates - Google Patents
Project to the parsing astronomical positioning method of the first point of Aries equatorial system of coordinates Download PDFInfo
- Publication number
- CN106525026B CN106525026B CN201610931149.0A CN201610931149A CN106525026B CN 106525026 B CN106525026 B CN 106525026B CN 201610931149 A CN201610931149 A CN 201610931149A CN 106525026 B CN106525026 B CN 106525026B
- Authority
- CN
- China
- Prior art keywords
- cos
- sin
- triangle
- lat
- point
- Prior art date
- Legal status (The legal status is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the status listed.)
- Active
Links
Classifications
-
- G—PHYSICS
- G01—MEASURING; TESTING
- G01C—MEASURING DISTANCES, LEVELS OR BEARINGS; SURVEYING; NAVIGATION; GYROSCOPIC INSTRUMENTS; PHOTOGRAMMETRY OR VIDEOGRAMMETRY
- G01C21/00—Navigation; Navigational instruments not provided for in groups G01C1/00 - G01C19/00
- G01C21/02—Navigation; Navigational instruments not provided for in groups G01C1/00 - G01C19/00 by astronomical means
-
- G—PHYSICS
- G01—MEASURING; TESTING
- G01C—MEASURING DISTANCES, LEVELS OR BEARINGS; SURVEYING; NAVIGATION; GYROSCOPIC INSTRUMENTS; PHOTOGRAMMETRY OR VIDEOGRAMMETRY
- G01C21/00—Navigation; Navigational instruments not provided for in groups G01C1/00 - G01C19/00
- G01C21/20—Instruments for performing navigational calculations
Abstract
The invention discloses a kind of parsing astronomical positioning methods for projecting to the first point of Aries equatorial system of coordinates.This projects to the parsing astronomical positioning method of the first point of Aries equatorial system of coordinates, survey person position is projected into the first point of Aries equatorial system of coordinates, parallactic triangle is established in the first point of Aries equatorial system of coordinates, using observation celestial body azimuth, perhaps altitude information by resolving parallactic triangle with spherical triangle is associated with obtains position of heavenly body or survey person's position data.Compared with prior art, the beneficial effects of the present invention are: parsing astrofix calculates simplicity, it is easier especially repeatedly to observe same celestial body calculating；It observes two celestial body azimuths or elevation measurement position of heavenly body and is not strict with simultaneity, be ok whithin a period of time；It is continuously tracked conducive to realization, real-time resolving astrofix；The almanac that nautical almanac etc. provides celestial body Greenwich hour angle can significantly simplify；A kind of new astronomical positioning method is provided for measurement position of heavenly body or survey person position.
Description
Technical field
The present invention relates to the method for measurement position of heavenly body or survey person position, especially one kind to project to first point of Aries equator seat
Mark the parsing astronomical positioning method of system.
Background technique
The principle of astrofix acquires the subpoint of celestial body on the ground according to observation time, it is round to obtain astronomical fix
The heart, amendment observation altitude of the heavenly body to height obtain true zenith distance, obtain astronomical fix radius of circle, observe two celestial bodies and obtain two
Astronomical fix circle, generally there are two points for two astronomical fix circle intersections, and what it is close to estimated position is a little observation position；
The principle of altitude difference method is, very high with astronomical observation on calculated azimuth of celestial body direction to select accommodation as basic point
The difference of degree and celestial body height calculation is that intercept does vertical line, which is the tangent line of astronomical fix circle, uses cutting by section
Line is instead of astronomical fix circle.It is edited referring to " navigation " Guo Yu, Zhang Jiping, Dai Ran, Dalian publishing house, the Maritime Affairs University Of Dalian
.2014.8, P119-120,172-173.
The hour angle equatorial system of coordinates, also referred to as dependent equatorial coordinate system, the circle on the basis of celestial equator, with Green (or the survey person) noon
The intersection point of circle and celestial equator is origin, and geometry extremely north pole, coordinate is declination and hour angle.It is red in the hour angle equatorial system of coordinates
Latitude is unrelated with survey person, and hour angle is related with survey person.Due to earth rotation, hour angle of heavenly body changes constantly, position of heavenly body statement
Be a certain moment celestial body instantaneous position.
The first point of Aries equatorial system of coordinates, also referred to as second equatorial system of coordinates, the circle on the basis of celestial equator, using the first point of Aries as origin,
Geometry extremely north pole, coordinate are declination and right ascension.In the first point of Aries equatorial system of coordinates, the position of celestial body and earth rotation without
It closes, the position of the first point of Aries and fixed star is basically unchanged (variation is very slow), and the position of the sun, planet and the moon changes.Referring to
" navigation " Guo Yu, Zhang Jiping, Dai Ran chief editor, Dalian publishing house, the Maritime Affairs University Of Dalian .2014.8, P123-125.
The method of tradition measurement position of heavenly body is meridian method, is measured when lower transit on celestial body passes through observer's meridian.Son
Noon method measures position of heavenly body, is to measure celestial body apex distance on celestial body transit time, solves to obtain day according to celestial body apex distance and latitude of an observer
Body declination obtains right ascension of a celestial body according between celestial body transit time.Referring to " uranometry method " Li Dongming, inscription on ancient bronze objects is respected, the summer one fly etc.
It writes, Beijing China Science Tech Publishing House .2006.8, P7-8.
Meridian method is actually to measure in the special occasion that Local time angle of a heavenly body is 0 ° or 180 °.
Apply for a patent previously: Li Qinglin ZL201510626370.7 is " with the side for assuming the observation celestial body positioning of longitude and latitude method
Observation any time celestial body azimuth or height are given in method ", obtain the position line, cut by solving parallactic triangle
The method for measuring survey person position.
It is applied for a patent previously: Li Qinglin ZL201610725091.4 " parsing measurement position of heavenly body or survey person position
It gives in astronomical positioning method " and is marked and drawed without mapping, be associated with spherical triangle by resolving, solved with two orientation or height
Analysis obtains the astronomical positioning method of position of heavenly body or survey person's position data.
Method position of heavenly body before this will be converted to subcelestial point, which is time of measuring moment celestial body in the earth
Surface projection, parallactic triangle are built upon the hour angle equatorial system of coordinates；In the hour angle equatorial system of coordinates, celestial body has Sunday view fortune
It is dynamic, preferably meet people's daily habits, cut in of mapping previous asks survey person position to be also more convenient.
In the hour angle equatorial system of coordinates, the position of heavenly body moment is changing, and different time observes same celestial body, the position of celestial body
It is different；Simultaneity is needed for two celestial body azimuths of observation or height parsing positioning, non-concurrent observation data need to carry out
Altitude correction of zenith difference amendment；
Celestial body same for continuous observation, because the position of heavenly body moment is changing, each observation data calculation requires basis
Observation time first acquires position of heavenly body, although what this also can be achieved on, to be continuously tracked same celestial body or observation two days
Body orientation or height parsing, real-time resolving astrofix are clearly a kind of unfavorable factor for increasing calculation amount.
Summary of the invention
To solve the above-mentioned problems, the parsing astronomical positioning method for projecting to the first point of Aries equatorial system of coordinates is herein proposed.
The technical scheme adopted by the invention is that: survey person position is projected into the first point of Aries equatorial system of coordinates, it is red in the first point of Aries
Parallactic triangle is established in road coordinate system, using observation celestial body azimuth or altitude information, by resolving parallactic triangle and pass
Connection spherical triangle obtains position of heavenly body or survey person's position data.
Compared with prior art, the beneficial effects of the present invention are: parsing astrofix calculates simplicity, especially repeatedly observation is same
One day body calculates easier；It observes two celestial body azimuths or elevation measurement position of heavenly body and is not strict with simultaneity, at one section
It is ok in time；It is continuously tracked conducive to realization, real-time resolving astrofix；Nautical almanac etc. provides celestial body Greenwich hour angle
Almanac can significantly simplify；A kind of new astronomical positioning method is provided for measurement position of heavenly body or survey person position.
Existing nautical almanac provides the first point of Aries, the sun, the moon and planet Greenwich hour angle and declination and its variation per hour
Amount；Project to the first point of Aries equatorial system of coordinates, data need to be provided are as follows: Ge Lindian, the sun, the moon and the daily right ascension of planet and declination and
Its variable quantity, data volume are the 1/24 of existing nautical almanac；Green's point right ascension is identical as first point of Aries Greenwich hour angle, but because it is
Even variation provides daily data without overproof；Existing nautical almanac is every one page on the 3rd, total 366/3=122 pages annual,
Project to the first point of Aries equatorial system of coordinates be 122/24=5.08, every year page totally 5.08 can keep smart with existing nautical almanac etc.
Degree.
Detailed description of the invention
Fig. 1 are as follows: parallactic triangle figure in the first point of Aries equatorial system of coordinates.
Fig. 2 are as follows: ipsilateral two places observe same celestial body and are associated with spherical triangle corner relational graph.
Fig. 3 are as follows: heteropleural two places observe same celestial body and are associated with spherical triangle corner relational graph.
Fig. 4 an are as follows: ground observes ipsilateral two days bodies and is associated with spherical triangle corner relational graph.
Fig. 5 an are as follows: ground observes two days bodies of heteropleural and is associated with spherical triangle corner relational graph.
Specific embodiment
Survey person position is projected into the representation method that the first point of Aries equatorial system of coordinates is respectively measured:
For the representation method for being different from the hour angle equatorial system of coordinates, the intersection point of Green's noon line and equator is projected to the Spring Equinox by us
The point equatorial system of coordinates, the declination are 0 °, and right ascension is Green's point right ascension GRA, it is clear that Green's point right ascension GRA is numerically equal to hour angle
First point of Aries Greenwich hour angle GHA Υ in the equatorial system of coordinates, but it is to measure eastwards along celestial equator that it is opposite to that, which to measure direction, value model
Enclose 0 ° -360 °.
Survey person position is projected into the first point of Aries equatorial system of coordinates, survey person's declination is equal to latitude of an observer (Dec=Lat), surveys
Person's right ascension (RAZ) it is to be measured eastwards along celestial equator to the arc distance of survey person's projected position noon line from first point of Aries Υ, it is clear that survey person's right ascension
RAZDiffering between Green's point right ascension GRA is survey person's longitude Long,
Equally the arc distance that survey person's noon encloses and celestial body noon circle is folded on celestial equator is known as celestial body place right ascension by us
(LRA), it is expressed as being enclosed by survey person's noon with three figure method being measured eastwards along celestial equator to the celestial body noon and enclose, 0 ° -360 ° of value range；With half
Circule method expression is enclosed to be measured eastwards or westwards along celestial equator to the celestial body noon by survey person's noon and be enclosed, and 0 ° -180 ° of value range, needs to name
Mark east E or western W；Celestial body place right ascension is equal to the difference of right ascension of a celestial body and survey person's right ascension, LRA=RA-RAZ。
The foundation of parallactic triangle in the first point of Aries equatorial system of coordinates:
In the first point of Aries equatorial system of coordinates, enclosed by survey person's noon, celestial body noon circle and celestial body vertical circle constitute spherical triangle, title
For parallactic triangle, three vertex are survey person's projected position, position of heavenly body and pole of looking up to heaven, and three sides are colatitude (90 ° of-Lat), remaining
Away from (90 ° of-Dec) and apex distance (90 ° of-h), three angles are semicircle azimuth A, semicircle place right ascension LRA, position of heavenly body angle X, are such as schemed
Shown in 1.Obviously, the parallactic triangle in the first point of Aries equatorial system of coordinates is with the parallactic triangle difference in the hour angle equatorial system of coordinates
With semicircle place right ascension LRA instead of semicircle local hour angle LHA, other each sides and angle are all corresponding.
Survey person position is projected to the apparent motion situation of the first point of Aries equatorial system of coordinates:
The position of the first point of Aries and fixed star is basically unchanged (variation is very slow), and the sun does annual apparent motion, right ascension and declination
Using year as mechanical periodicity；The moon does all moon apparent motions, right ascension and declination using the moon as mechanical periodicity；Planet right ascension and declination are by respective
Period of motion variation.
Green's point and survey person's subpoint do diurnal motion, i.e. Green's point and survey person's subpoint right ascension becomes by the period of day
Change；It is considered as Green's point and survey person's subpoint inswept celestial sphere one week daily.
Because star place is basically unchanged (variation is very slow), and position of sun is with year in the first point of Aries equatorial system of coordinates
Mechanical periodicity, moon positions are using the moon as mechanical periodicity, and compared with the change rate in the hour angle equatorial system of coordinates, the sun is about 1/365,
The moon is about 1/30, and planet change rate is also much smaller.So continuous several times observe same celestial body in the first point of Aries equatorial system of coordinates
It calculates easier；It observes two celestial body azimuths or elevation measurement position of heavenly body and is not strict with simultaneity, whithin a period of time
It is ok；It is continuously tracked conducive to realization, real-time resolving astrofix.
The parsing astronomical positioning method for projecting to the first point of Aries equatorial system of coordinates can be realized the single celestial body azimuth of observation and height
Parse position of heavenly body or survey person position, it is known that survey person position and position of heavenly body resolve true azimuth of celestial body and height, two places observation
Perhaps two celestial body azimuths or height parsing survey person position are observed in height parsing one ground of position of heavenly body for same celestial body azimuth.
Specific embodiment is as follows.
Specific embodiment 1 observes the true bearing A and height h of unknown celestial body B, solution at known location Z (Lat/Long)
Calculate position of heavenly body；Green point right ascension GRA is obtained according to the time, survey person position Z is projected into the first point of Aries equatorial system of coordinatesIn parallactic triangle, known top is away from (90 ° of-h), colatitude as shown in Figure 1:
(90 ° of-Lat) and azimuth A three elements are solved remaining away from (90 ° of-Dec) using cosine formula:
Cos (90 ° of-Dec)=cos (90 ° of-h) * cos (90 ° of-Lat)
+sin(90°-h)*sin(90°-Lat)*cos(A)
=sin (h) * sin (Lat)+cos (h) * cos (Lat) * cos (A)
Away from (90 ° of-Dec) more than solving, declination Dec is obtained；It reuses cosine formula deformation and solves local right ascension LRA:
Resolving obtains local right ascension LRA, obtains right ascension of a celestial body RA=LRA+RAZ, obtain position of heavenly body (Dec, RA).
Specific embodiment 2 is resolved in the true bearing A and height h that unknown position Z observes known celestial body B (Dec/RA)
Survey person position；Green point right ascension GRA is obtained according to the time, survey person position Z is projected into the first point of Aries equatorial system of coordinates, in astronomy three
It is known remaining away from (90 ° of-Dec), apex distance (90 ° of-h) and azimuth A three elements as shown in Figure 1: in angular, it is asked using sine formula
Solve local right ascension LRA:
Local right ascension LRA is solved, longitude Long=RA-LRA-GRA is obtained；It reuses the deformation of Napier formula and solves colatitude
(90 ° of-Lat):
Resolving obtains colatitude (90 ° of-Lat), obtains latitude Lat；Obtain survey person position (Lat, Long).
Specific embodiment 3 observes known celestial body B (Dec/RA) at known location Z (Lat/Long), it is very square to resolve celestial body
Position A and height h；Green point right ascension GRA is obtained according to the time, survey person position Z is projected into the first point of Aries equatorial system of coordinatesBy right ascension of a celestial body RA and survey person's right ascension RAZObtain local right ascension LRA=RA-RAZ,
It is known remaining away from (90 ° of-Dec), colatitude (90 ° of-Lat) and place right ascension LRA three elements as shown in Figure 1: in parallactic triangle,
Apex distance (90 ° of-h) is solved using cosine formula:
Cos (90 ° of-h)=cos (90 ° of-Dec) * cos (90 ° of-Lat)
+sin(90°-Dec)*sin(90°-Lat)*cos(LRA)
=sin (Dec) * sin (Lat)+cos (Dec) * cos (Lat) * cos (LRA)
Apex distance (90 ° of-h) is solved, height h is obtained；It reuses cosine formula deformation and solves azimuth A:
Resolving obtains azimuth A；Obtain the true bearing and height (A, h) of celestial body.
Specific embodiment 4, in known location Z1(Lat1/Long1)、Z2(Lat2/Long2), same unknown celestial body is observed
The true bearing of B is respectively A1, A2, resolves position of heavenly body；Green point right ascension GRA1, GRA2 are obtained according to the time, by Z1And Z2Projection
To the first point of Aries equatorial system of coordinates,
Z1And Z2It is ipsilateral such as Fig. 2, heteropleural such as Fig. 3 positioned at celestial body B；Orthodrome tie point Z is used on spherical surface1And Z2Obtain two auxiliary three
Angular △ PZ1Z2With △ BZ1Z2With parallactic triangle △ PZ1B and △ PZ2B composition association spherical triangle, such as Fig. 2 Fig. 3 institute
Show；
In triangle △ PZ1Z2In, it is known that both sides (90 ° of-Lat1), (90 ° of-Lat2) and its angle (RAZ2-RAZ1), it uses
Cosine formula solves other one side Z1Z2:
cos(Z1Z2)=cos (90 ° of-Lat1) * cos (90 ° of-Lat2)+sin (90 ° of-Lat1)
*sin(90°-Lat2)*cos(RAZ2-RAZ1)
=sin (Lat1) * sin (Lat2)+cos (Lat1) * cos (Lat2)
*cos(RAZ2-RAZ1)
It reuses cosine formula deformation and solves other two corners ∠ PZ1Z2With ∠ PZ2Z1:
∠ BZ is solved by association spherical triangle corner relationship again1Z2With ∠ BZ2Z1；
Ipsilateral ∠ BZ1Z2=| ∠ PZ1Z2-A1| and ∠ BZ2Z1=| ∠ PZ2Z1+A2| (if resulting value is greater than 180 ° with 360 °
Subtract resulting value) or heteropleural ∠ BZ1Z2=| ∠ PZ1Z2-A1| and ∠ BZ2Z1=| ∠ PZ2Z1-A2|；
Under A1, A2 known case, two places are ipsilateral or heteropleural can only be one of situation in celestial body, therefore two places are observed
Ask position of heavenly body that can obtain one group of solution in same celestial body azimuth；
In triangle △ BZ1Z2In, it is known that two corners ∠ BZ1Z2、∠BZ2Z1And its contained side Z1Z2, deformed using four element equations
Solve other both sides (90 ° of-h1) and (90 ° of-h2):
In triangle △ PZ1In B, it is known that both sides (90 ° of-h1), (90 ° of-Lat1) and its included angle A 1 are asked using cosine formula
Solution is other on one side (90 ° of-Dec):
Cos (90 ° of-Dec)=cos (90 ° of-h1) * cos (90 ° of-Lat1)+sin (90 ° of-h1)
*sin(90°-Lat1)*cos(A1)
=sin (h1) * sin (Lat1)+cos (h1) * cos (Lat1) * cos (A1)
(90 ° of-Dec) is solved, declination Dec is obtained, cosine formula deformation is reused and solves other one jiao of LRA1:
Or in triangle △ PZ2In B, it is known that both sides (90 ° of-h2), (90 ° of-Lat2) and its included angle A 2 solve other one
Side (90 ° of-Dec) and LRA2 method are the same as solution △ PZ1B；
Right ascension of a celestial body RA=LRA+RA is obtained by LRAZ, obtain position of heavenly body B (Dec, RA).
Specific embodiment 5, in known location Z1(Lat1/Long1)、Z2(Lat2/Long2), same unknown celestial body is observed
The height of B is respectively h1, h2, resolves position of heavenly body；Green point right ascension GRA1, GRA2 are obtained according to the time, by Z1And Z2Projection
To the first point of Aries equatorial system of coordinates,
Z1And Z2Positioned at celestial body B ipsilateral such as Fig. 2, heteropleural such as Fig. 3, orthodrome tie point Z is used on spherical surface1And Z2Obtain two auxiliary three
Angular △ PZ1Z2With △ BZ1Z2With parallactic triangle △ PZ1B and △ PZ2B composition association spherical triangle, such as Fig. 2 and Fig. 3 institute
Show；
In triangle △ PZ1Z2In, it is known that both sides (90 ° of-Lat1), (90 ° of-Lat2) and its angle (RAZ2-RAZ1), it uses
Cosine formula solves other one side Z1Z2:
cos(Z1Z2)=cos (90 ° of-Lat1) * cos (90 ° of-Lat2)+sin (90 ° of-Lat1)
*sin(90°-Lat2)*cos(RAZ2-RAZ1)
=sin (Lat1) * sin (Lat2)+cos (Lat1) * cos (Lat2)
*cos(RAZ2-RAZ1)
It reuses cosine formula deformation and solves other two corners ∠ PZ1Z2With ∠ PZ2Z1:
In triangle △ BZ1Z2In, it is known that three sides (90 ° of-h1), (90 ° of-h2) and Z1Z2, deformed and solved using cosine formula
Two corners ∠ BZ1Z2With ∠ BZ2Z1:
A1 and A2 are solved by association spherical triangle corner relationship again；
A1=| ∠ PZ1Z2-∠BZ1Z2| and A2=| ∠ PZ2Z1-∠BZ2Z1|,
With A1=∠ PZ1Z2+∠BZ1Z2And A2=∠ PZ2Z1+∠BZ2Z1If (resulting value is greater than 180 ° and subtracts institute with 360 °
It must be worth)；
At the unknown orientation of known altitude, two groups of solutions can be existed simultaneously, therefore two places are observed same altitude of the heavenly body and asked
Position of heavenly body is generally possible to obtain two groups of solutions；
In triangle △ PZ1In B, it is known that both sides (90 ° of-h1), (90 ° of-Lat1) and its included angle A 1 are asked using cosine formula
Solution is other on one side (90 ° of-Dec):
Cos (90 ° of-Dec)=cos (90 ° of-h1) * cos (90 ° of-Lat1)+sin (90 ° of-h1)
*sin(90°-Lat1)*cos(A1)
=sin (h1) * sin (Lat1)+cos (h1) * cos (Lat1) * cos (A1)
(90 ° of-Dec) is solved, declination Dec is obtained, cosine formula deformation is reused and solves other one jiao of LRA1:
Or in triangle △ PZ2In B, it is known that both sides (90 ° of-h2), (90 ° of-Lat2) and its included angle A 2 solve other one
Side (90 ° of-Dec) and LRA2 method are the same as solution △ PZ1B；
Right ascension of a celestial body RA=LRA+RA is obtained by LRAZ, obtain position of heavenly body B (Dec, RA).
Specific embodiment 6 observes known celestial body B in unknown position Z1(Dec1/RA1)、B2(Dec2/RA2) true bearing
Respectively A1, A2 resolve survey person position；Green point right ascension GRA is obtained according to the time, survey person Z is projected into first point of Aries equator and is sat
Mark system, B1And B2It is ipsilateral such as Fig. 4, heteropleural such as Fig. 5 positioned at survey person Z；Orthodrome tie point B is used on spherical surface1And B2Obtain two it is auxiliary
Help triangle △ PB1B2With △ ZB1B2With parallactic triangle △ PZB1With △ PZB2Composition association spherical triangle, such as Fig. 4 and Fig. 5
It is shown；
In triangle △ PB1B2In, it is known that both sides (90 ° of-Dec1), (90 ° of-Dec2) and its angle (RA2-RA1) use
Cosine formula solves other one side B1B2:
cos(B1B2)=cos (90 ° of-Dec1) * cos (90 ° of-Dec2)+sin (90 ° of-Dec1)
*sin(90°-Dec2)*cos(RA2-RA1)
=sin (Dec1) * sin (Dec2)+cos (Dec1) * cos (Dec2)
*cos(RA2-RA1)
In triangle △ PZB1In, it is known that side (90 ° of-Dec1) and angle A1, unknown (90 ° of-h1) and (90 ° of-Lat) are used
Cosine formula:
Cos (90 ° of-Dec1)=cos (90 ° of-h1) * cos (90 ° of-Lat)+sin (90 ° of-h1) * sin (90 ° of-Lat) * cos
(A1)---(1)
In triangle △ PZB2In, it is known that side (90 ° of-Dec2) and angle A2, unknown (90 ° of-h2) and (90 ° of-Lat) are used
Cosine formula:
Cos (90 ° of-Dec2)=cos (90 ° of-h2) * cos (90 ° of-Lat)+sin (90 ° of-h2) * sin (90 ° of-Lat) * cos
(A2)---(2)
In triangle △ ZB1B2In, it is known that side B1B2With angle (A2-A1) (body is ipsilateral within two days) or angle (A1+A2) (two days
Body heteropleural) (if resulting value, which is greater than 180 °, subtracts resulting value with 360 °), unknown (90 ° of-h1) and (90 ° of-h2) use cosine public affairs
Formula:
cos(B1B2)=cos (90 ° of-h1) * cos (90 ° of-h2)+sin (90 ° of-h1) * sin (90 ° of-h2) * cos (A2-
A1)---(3)
Or
cos(B1B2)=cos (90 ° of-h1) * cos (90 ° of-h2)+sin (90 ° of-h1) * sin (90 ° of-h2) * cos (A1+
A2)---(4)
Under orientation known case, two days bodies are ipsilateral or heteropleural can only be one of situation in survey person, and body is same within two days
Totally 3 unknown numbers, solution obtain (90 ° of-h1), (90 ° of-h2) for side (1)-(3) or two days body heteropleural (1) (2) (4) three formula simultaneous
(90 ° of-Lat), obtains latitude Lat；
In triangle △ PZB1In, it is known that three sides (90 ° of-Dec1), (90 ° of-h1), (90 ° of-Lat) are become using cosine formula
Shape solves other one jiao of LRA1,
Or in triangle △ PZB2In, it is known that three sides (90 ° of-Dec2), (90 ° of-h2), (90 ° of-Lat) use cosine public affairs
Formula deformation solves other one jiao of LRA2 method with solution △ PZB1；
Longitude Long=RA-LRA-GRA is obtained by LRA；Obtain survey person position Z (Lat, Long).
Specific embodiment 7 observes known celestial body B in unknown position Z1(Dec1/RA1)、B2(Dec2/RA2) height
Respectively h1, h2 resolve survey person position；Green point right ascension GRA is obtained according to the time, survey person Z is projected into first point of Aries equator and is sat
Mark system, B1And B2It is ipsilateral such as Fig. 4, heteropleural such as Fig. 5 positioned at survey person Z；Orthodrome tie point B is used on spherical surface1And B2Obtain two it is auxiliary
Help triangle △ PB1B2With △ ZB1B2With parallactic triangle △ PZB1With △ PZB2Composition association spherical triangle, such as Fig. 4 and Fig. 5
It is shown；
In triangle △ PB1B2In, it is known that both sides (90 ° of-Dec1), (90 ° of-Dec2) and its angle (RA2-RA1) use
Cosine formula solves other one side B1B2:
cos(B1B2)=cos (90 ° of-Dec1) * cos (90 ° of-Dec2)+sin (90 ° of-Dec1)
*sin(90°-Dec2)*cos(RA2-RA1)
=sin (Dec1) * sin (Dec2)+cos (Dec1) * cos (Dec2)
*cos(RA2-RA1)
It reuses cosine formula deformation and solves other two corners ∠ PB1B2With ∠ PB2B1:
In triangle △ ZB1B2In, it is known that three sides (90 ° of-h1), (90 ° of-h2), B1B2, deformed and solved using cosine formula
In addition two corners ∠ ZB1B2With ∠ ZB2B1:
∠ PB is solved by association spherical triangle corner relationship again1Z and ∠ PB2Z；
∠PB1Z=| ∠ PB1B2-∠ZB1B2| and ∠ PB2Z=| ∠ PB2B1-∠ZB2B1|, with ∠ PB1Z=∠ PB1B2+∠
ZB1B2With ∠ PB2Z=∠ PB2B1+∠ZB2B1(if resulting value is greater than 180 ° and subtracts resulting value with 360 °)；
At the unknown orientation of known altitude, two groups of solutions can be existed simultaneously, therefore a ground observes two altitude of the heavenly bodies and asks survey
Person position is generally possible to obtain two groups of solutions；
In triangle △ PB1In Z, it is known that both sides (90 ° of-Dec1), (90 ° of-h1) and its angle ∠ PB1Z uses cosine public affairs
In addition formula solves on one side (90 ° of-Lat):
Cos (90 ° of-Lat)=cos (90 ° of-Dec1) * cos (90 ° of-h1)+sin (90 ° of-Dec1)
*sin(90°-h1)*cos(PB1Z)
=sin (Dec1) * sin (h1)+cos (Dec1) * cos (h1) * cos (PB1Z)
(90 ° of-Lat) is solved, latitude Lat is obtained, cosine formula deformation is reused and solves other one jiao of LRA1:
Or in triangle △ PB2In Z, it is known that both sides (90 ° of-Dec2), (90 ° of-h2) and its angle ∠ PB2Z, using remaining
String equations (90 ° of-Lat) and LRA2 method are the same as solution △ PB1Z；
Longitude Long=RA-LRA-GRA is obtained by LRA；Obtain survey person position Z (Lat, Long).
Numerical symbol conversion and naming rule in formula: above-mentioned association spherical triangle is Euler's spherical triangle, public
Calculated value value range [0-180 °] is participated in formula；
1) latitude Lat1 perseverance takes positive value+(no matter north latitude or south latitude), seeks Z1Z2When, latitude Lat2 and latitude Lat1 are of the same name
Take positive value+, different name takes negative value-, value range [N/S 0-90 °]；
2) latitude Lat is resolved to obtain, positive value+take is of the same name with declination Dec, and negative value-takes and declination different name, value model
Enclose [N/S 0-90 °]；
3) longitude Long, east longitude E take positive value+, west longitude W takes negative value-；Resolve to obtain longitude Long, as a result positive value+be east longitude E,
As a result negative value-is west longitude W, value range [E/W 0-180 °]；
4) B is sought1B2When, declination Dec1 takes positive value+(no matter north latitude or south latitude), declination Dec2 with declination Dec1 is of the same name takes
Positive value+, different name takes negative value-, value range [N/S 0-90 °]
5) declination Dec is resolved to obtain, positive value+take is of the same name with latitude of an observer Lat, and negative value-takes and latitude of an observer different name, value model
Enclose [N/S 0-90 °]；
6) Green point right ascension GRA and right ascension of a celestial body RA perseverance be positive value+, value range [0-360 °]；
7) place right ascension LRA takes semicircle right ascension, perseverance take positive value+, value range [E/W 0-180 °]；Three figure method is converted to half
Initial value is taken for east orientation E when circule method rule: less than 180 °, and greater than 180 ° Shi Weixi take (360 ° of-LRA) to E；
8) resolve local right ascension LRA is semicircle right ascension, perseverance for positive value+, value range [E/W 0-180 °], name weighs half
Circle the second title of orientation E/W；Semicircular method is converted to three figure method rule: east orientation E takes initial value, and west takes (360 ° of-LRA) to W；
9) observation true bearing A takes semicircle orientation, perseverance take positive value+, value range [N/S//E/W 0-180 °]；Circumferential orientation
Be converted to semicircle orientation rule: in north latitude, 0-180 ° is named as NE with initial value, and 180 ° -360 ° are named as NW with (360 ° of-A)；?
South latitude, 0-180 ° is named as SE with (180 ° of-A), and 180 ° -360 ° are named as SW with (A-180 °)；Same celestial body side is observed in two places
When position, two semicircular orientation is unified to be named according to Lat1；
10) resolve azimuth A be semicircle orientation, naming rule: for the first title with latitude of an observer Lat, the second title is same
Local right ascension LRA.
The 1800LT on December 20, of specific calculated example 1,1998 (1000GMT) survey person position Z (N30 ° 32 ' 06 "/E123 °
29 ° 53 ' 27 of 12 ' 18 A51 ° 22 ' 47 of the true bearing of observation Capella Capella ") " and height h ", resolve position of heavenly body；
Green's point right ascension GRA is obtained 238 ° 47 ' 12 according to the time ", survey person position Z is projected into first point of Aries equatorial coordinates
System
In parallactic triangle, it is known that apex distance (90 ° of-h)=90 ° -29 ° 53 ' 27 "=60 ° 06 ' 33 ", colatitude (90 ° of-Lat)=90 ° -
30 ° 32 ' 06 "=59 ° 27 ' 54 " and A=NE51 ° 22 ' 47 of azimuth " three elements, using cosine formula solve it is remaining away from (90 °-
Dec):
Cos (90 ° of-Dec)=cos (60 ° 06 ' 33 ") * cos (59 ° 27 ' 54 ")
+sin(60°06′33″)*sin(59°27′54″)*cos(51°22′47″)
Solve it is remaining away from (90 ° of-Dec)=44 ° 00 ' 18 ", obtain declination Dec=N45 ° 59 ' 42 "；Reuse cosine public affairs
Formula deformation solves local right ascension LRA:
Resolving obtains local right ascension LRA=77 ° 10 ' 00 ", known according to observed bearing NE for east orientation E, obtains right ascension of a celestial body RA
=LRA+RAZ=77 ° 10 ' 00 "+1 ° 59 ' 30 "=79 ° 09 ' 30 "；Obtain position (the Dec N45 ° 59 ' of Capella Capella
42 ", 79 ° 09 ' 30 of RA ").
The 1800LT on December 20, of specific calculated example 2,1998 (1000GMT) is in unknown survey person's position detection Capella
51 ° 22 ' 47 of the true bearing A " and height h 29 ° 53 ' 27 " of Capella (Dec N45 ° 59 ' 42 " 79 ° 09 ' 30 of/RA "), solution
Calculate survey person position；
Green's point right ascension GRA is obtained 238 ° 47 ' 12 according to the time ", survey person position Z is projected into first point of Aries equatorial coordinates
System, in parallactic triangle, it is known that it is remaining away from (90 ° of-Dec)=90 ° -45 ° 59 ' 42 "=44 ° 00 ' 18 ", apex distance (90 ° of-h)=
90 ° -29 ° 53 ' 27 "=60 ° 06 ' 33 " A=NE51 ° 22 ' 47 " three elements use sine formula to solve local right ascension with azimuth
LRA:
It solves local right ascension LRA=77 ° 10 ' 02 " or 102 ° 49 ' 58 ", is known according to observed bearing NE for east orientation E, obtained
Long=RA-LRA-GRA=79 ° 09 ' 30 of longitude " -77 ° 10 ' 02 " -238 ° 47 ' 12 "=E123 ° 12 ' 16 "；Or Long=
79 ° 09 ' 30 " -102 ° 49 ' 58 " -238 ° 47 ' 12 "=E97 ° 32 ' 20 "；Reuse Napier formula deformation solve colatitude (90 °-
Lat):
Or
Resolving obtains colatitude (90 ° of-Lat)=59 ° 27 ' 52 " or 35 ° 14 ' 56 ", obtains latitude Lat=N30 ° 32 ' 08 "
Or Lat=N54 ° 45 ' 04 "；Obtain survey person position (Lat N30 ° 32 ' 08 ", Long E123 ° 12 ' 16 ") or (Lat
N54 ° 45 ' 04 ", Long E97 ° 32 ' 20 ").
The 1800LT on December 20, of specific calculated example 3,1998 (1000GMT) survey person position Z (N30 ° 32 ' 06 "/E123 °
12 ' 18 79 ° 09 ' 30 of ") observation Capella Capella (Dec N45 ° 59 ' 42 "/RA "), resolve true azimuth of celestial body A and height
h；
Green's point right ascension GRA is obtained 238 ° 47 ' 12 according to observation time ", survey person position Z is projected into first point of Aries equator
Coordinate system
By RA=79 ° 09 ' 30 of right ascension of a celestial body " with survey person's right ascension RAZ" obtain local right ascension LRA=RA-RA for=1 ° 59 ' 30Z=79 ° 09 '
30 " -1 ° 59 ' 30 "=E77 ° 10 ' 00 ", in parallactic triangle, it is known that it is remaining away from (90 ° of-Dec)=90 ° -45 ° 59 ' 42 "=44 °
00 ' 18 ", LRA=E77 ° 10 ' 00 " three want for colatitude (90 ° of-Lat)=90 ° -30 ° 32 ' 06 "=59 ° 27 ' 54 " and local right ascension
Element solves apex distance (90 ° of-h) using cosine formula:
Cos (90 ° of-h)=cos (44 ° 00 ' 18 ") * cos (59 ° 27 ' 54 ")
+sin(44°00′18″)*sin(59°27′54″)*cos(77°10′00″)
Solve apex distance (90 ° of-h)=60 ° 06 ' 33 ", obtain height h=29 ° 53 ' 27 "；Cosine formula deformation is reused to ask
Solve azimuth A:
Resolving obtains azimuth A=NE 51 ° 22 ' 47 "=51 ° 22 ' 47 "；Obtain Capella Capella true bearing and
Height (29 ° 53 ' 27 of 51 ° 22 ' 47 of A ", h ").
In above-mentioned 3, positive and negative calculation result has about 2 rads of error, is to round up to make because of decimal after data rad
At, it is A=51 ° 22 ' 46 that 4 are resolved after celestial body azimuth and height decimal point in example 3 " .5656, h=29 ° 53 ' 27 " .0271,
Example 1,2 is carried out using this data to resolve, and is as a result consistent completely；Error producing cause is identical in subsequent calculations example.
The 1800LT on December 20, of specific calculated example 4,1998 (1000GMT) is in survey person position Z1(N18°18′12″/
E113 ° 11 ' 18 the true bearing of observation Capella Capella ") be 45 ° 54 ' 38 of A1 ", 1810LT (1010GMT) is in survey person position
Z2The true bearing of (N38 ° 28 ' 12 "/E121 ° 55 ' 30 ") observation Capella Capella is A2 55 ° 47 ' 25 ", resolve day position
It sets；
Green's point right ascension GRA1 is obtained 238 ° 47 ' 12 according to the time " and GRA2 241 ° 17 ' 36 ", by Z1And Z2It projects to
The first point of Aries equatorial system of coordinates, Name A1=in semicircle orientation
NE45 ° 54 ' 38 " and A2=NE55 ° 47 ' 25 "；Resolve Z1Z2:
cos(Z1Z2)=cos (90 ° -18 ° 18 ' 12 ") * cos (90 ° -38 ° 28 ' 12 ")
+sin(90°-18°18′12″)*sin(90°-38°28′12″)
*cos(3°13′06″-351°58′30″)
Solve Z1Z2=22 ° 25 ' 04 ", then solve two corners ∠ PZ1Z2With ∠ PZ2Z1:
Solve ∠ PZ1Z2=23 ° 35 ' 47 " and ∠ PZ2Z1=150 ° 57 ' 40 "；
It is solved again by association spherical triangle corner relationship:
∠BZ1Z2=| ∠ PZ1Z2-A1|=45 ° 54 ' 38 " -23 ° 35 ' 47 "=22 ° 18 ' 51 ",
∠BZ2Z1=| ∠ PZ2Z1+A2|=360 °-(150 ° 57 ' 40 "+55 ° 47 ' 25 ")=153 ° 14 ' 55 ",
(90 ° of-h1) is solved again:
Solve (90 ° of-h1)=75 ° 01 ' 51 ", then solve (90 ° of-Dec):
Cos (90 ° of-Dec)=cos (75 ° 01 ' 51 ") * cos (90 ° -18 ° 18 ' 12 ")
+sin(75°01′51″)*sin(90°-18°18′12″)*cos(45°54′38″)
Solve (90 ° of-Dec)=44 ° 00 ' 18 ", obtain declination Dec=45 ° 59 ' 42 ", then solve LRA1:
Solve LRA1=87 ° 11 ' 00 ", known according to observed bearing NE for east orientation E, obtains right ascension of a celestial body RA=LRA+RAZ=
87 ° 11 ' 00 "+351 ° 58 ' 30 "=79 ° 09 ' 30 "；
Resolving obtains the position (79 ° 09 ' 30 of Dec N45 ° 59 ' 42 ", RA ") of Capella Capella.
The 1800LT on December 20, of specific calculated example 5,1998 (1000GMT) is in survey person position Z1(N18°18′12″/
E113 ° 11 ' 18 the height of observation Capella Capella ") be 14 ° 58 ' 09 of h1 ", 1810LT (1010GMT) is in survey person position
Z2The height of (N38 ° 28 ' 12 "/E121 ° 55 ' 30 ") observation Capella Capella is h2 35 ° 25 ' 22 ", resolve day position
It sets；
Green's point right ascension GRA1 is obtained 238 ° 47 ' 12 according to the time " and GRA2 241 ° 17 ' 36 ", by Z1And Z2It projects to
The first point of Aries equatorial system of coordinates,
Resolve Z1Z2、∠PZ1Z2With ∠ PZ2Z1With calculated example 4, Z is solved1Z2=22 ° 25 ' 04 ", ∠ PZ1Z2=23 ° 35 '
47 " and ∠ PZ2Z1=150 ° 57 ' 40 "；∠ BZ is solved again1Z2With ∠ BZ2Z1:
Solve ∠ BZ1Z2=22 ° 18 ' 52 " and ∠ BZ2Z1=153 ° 14 ' 54 ",
It is solved again by association spherical triangle corner relationship:
A1=| ∠ PZ1Z2-∠BZ1Z2|=23 ° 35 ' 47 " -22 ° 18 ' 52 "=01 ° 16 ' 55 ",
Or A1=∠ PZ1Z2+∠BZ1Z2=23 ° 35 ' 47 "+22 ° 18 ' 52 "=45 ° 54 ' 39 ", then solve (90 °-
Dec):
Cos (90 ° of-Dec)=cos (90 ° -14 ° 58 ' 09 ") * cos (90 ° -18 ° 18 ' 12 ")
+sin(90°-14°58′09″)*sin(90°-18°18′12″)*cos(01°16′55″)
Or
Cos (90 ° of-Dec)=cos (90 ° -14 ° 58 ' 09 ") * cos (90 ° -18 ° 18 ' 12 ")
+sin(90°-14°58′09″)*sin(90°-18°18′12″)*cos(45°54′39″)
Solve (90 ° of-Dec)=03 ° 33 ' 11 " perhaps 44 ° 00 ' 19 " and obtain declination Dec=N86 ° 26 ' 49 " or Dec
=N45 ° 59 ' 41 ", then solve LRA1:
Or
LRA1=159 ° 35 ' 30 " or 87 ° 11 ' 00 " is solved, is directly named as known to celestial body azimuth, if orientation is unknown,
It is calculated separately according to E or W, RA=LRA+RAZ,
E:RA=159 ° 35 ' 30 "+351 ° 58 ' 30 "=511 ° 34 ' 00 " -360 °=151 ° 34 ' 00 ",
W:RA=360 ° -159 ° 35 ' 30 "+351 ° 58 ' 30 "=552 ° 23 ' 00 " -360 °=192 ° 23 ' 00 ",
Or E:RA=87 ° 11 ' 00 "+351 ° 58 ' 30 "=439 ° 09 ' 30 " -360 °=79 ° 09 ' 30 ",
W:RA=360 ° -87 ° 11 ' 00 "+351 ° 58 ' 30 "=624 ° 47 ' 30 " -360 °=264 ° 47 ' 30 "；
It is same to resolve LRA2:
Or
LHA2=148 ° 20 ' 54 " either 75 ° 56 ' 24 " is solved to be calculated separately with E or W,
E:RA=148 ° 20 ' 54 "+3 ° 13 ' 06 "=151 ° 34 ' 00 ",
W:RA=360 ° -148 ° 20 ' 54 "+3 ° 13 ' 06 "=214 ° 52 ' 12 ",
Or E:RA=75 ° 56 ' 24 "+3 ° 13 ' 06 "=79 ° 09 ' 30 ",
W:RA=360 ° -75 ° 56 ' 24 "+3 ° 13 ' 06 "=287 ° 16 ' 42 "；
LHA1 is identical as LHA2 calculated result for correctly as a result, obtaining right ascension of a celestial body RA=151 ° 34 ' 00 " or RA=
79 ° 09 ' 30 ",
Resolving obtains position (151 ° 34 ' 00 of Dec N86 ° 26 ' 49 ", RA ") or (Dec of Capella Capella
N45 ° 59 ' 41 ", 79 ° 09 ' 30 of RA ").
The 1800LT on December 20, of specific calculated example 6,1998 (1000GMT) observes Vega Vega (Dec in unknown position
N38 ° 47 ' 06 ", 279 ° 13 ' 00 of RA ") true bearing be A1 302 ° 20 ' 43 " and Capella Capella (Dec N45 ° 59 '
42 ", 79 ° 09 ' 30 of RA ") true bearing be A2 51 ° 22 ' 47 ", resolves survey person position；
Semicircle orientation is named as A1=NW57 ° 39 ' 17 " and A2=NE51 ° 22 ' 47 " on the Northern Hemisphere, obtains lattice according to the time
238 ° 47 ' 12 of woods point right ascension GRA ", projects to the first point of Aries equatorial system of coordinates for survey person Z；Resolve B1B2:
cos(B1B2)=cos (90 ° -38 ° 47 ' 06 ") * cos (90 ° -45 ° 59 ' 42 ")
+sin(90°-38°47′06″)*sin(90°-45°59′42″)
*cos(79°09′30″-279°13′00″)
Solve B1B2=93 ° 19 ' 57 " because two celestial body azimuth NW and NE are located at survey person's heteropleural, simultaneous (1) (2) (4) formula:
Cos (90 ° -38 ° 47 ' 06 ")=cos (90 ° of-h1) * cos (90 ° of-Lat)+sin (90 ° of-h1) * sin (90 ° -
Lat)*cos(57°39′17″)---(1)
Cos (90 ° -45 ° 59 ' 42 ")=cos (90 ° of-h2) * cos (90 ° of-Lat)+sin (90 ° of-h2) * sin (90 ° -
Lat)*cos(51°22′47″)---(2)
Cos (93 ° 19 ' 57 ")=cos (90 ° of-h1) * cos (90 ° of-h2)+sin (90 ° of-h1) * sin (90 ° of-h2) * cos
(57°39′17″+51°22′47″)---(4)
Solve (90 ° of-h1)=66 ° 15 ' 14 ", (90 ° of-h2)=60 ° 06 ' 32 ", (90 ° of-Lat)=59 ° 27 ' 54 ", obtain
Lat=N30 ° 32 ' 06 of latitude ", then solve LRA1:
Solve LRA1=82 ° 46 ' 32 ", known according to observed bearing NW for west to W, obtains longitude Long=RA-LRA-GRA
=279 ° 13 ' 00 " -238 ° 47 ' 12 "=- 236 ° 47 ' 40 "=E123 ° 12 ' 20 of-(360 ° -82 ° 46 ' 32 ") ",
It is named as A1=SW122 ° 20 ' 43 " and A2=SE128 ° 37 ' 13 " in Southern Hemisphere semicircle orientation, resolving obtains Z
(Lat=S51 ° 50 ' 56 ", Long=E121 ° 02 ' 04 "), in this position, two observed altitudes is negative value (h1=-23 ° of .7, h2
=-29 ° of .9), celestial body is actually invisible；
Resolving obtains survey person position Z (Lat N30 ° 32 ' 06 ", Long E123 ° 12 ' 20 ").
The 1800LT on December 20, of specific calculated example 7,1998 (1000GMT) observes Vega Vega (Dec in unknown position
N38 ° 47 ' 06 ", 279 ° 13 ' 00 of RA ") height be h1 23 ° 44 ' 47 " and Capella Capella (Dec N45 ° 59 '
42 ", 79 ° 09 ' 30 of RA ") height be h2 29 ° 53 ' 27 ", resolves survey person position；
Green's point right ascension GRA is obtained 238 ° 47 ' 12 according to the time ", survey person Z is projected into the first point of Aries equatorial system of coordinates；Solution
Calculate B1B2With calculated example 6, B is solved1B2=93 ° 19 ' 57 ", then solve ∠ PB1B2With ∠ PB2B1:
Solve ∠ PB1B2=13 ° 48 ' 30 " and ∠ PB2B1=15 ° 32 ' 01 ", then solve ∠ ZB1B2With ∠ ZB2B1:
Solve ∠ ZB1B2=55 ° 10 ' 52 " and ∠ ZB2B1=60 ° 05 ' 01 ",
∠ PB is solved by association spherical triangle corner relationship again1Z:
∠PB1Z=| ∠ PB1B2-∠ZB1B2|=55 ° 10 ' 52 " -13 ° 48 ' 30 "=41 ° 22 ' 22 ",
Or ∠ PB1Z=∠ PB1B2+∠ZB1B2=13 ° 48 ' 30 "+55 ° 10 ' 52 "=68 ° 59 ' 22 ", then solve (90 °-
Lat):
Cos (90 ° of-Lat)=cos (90 ° -38 ° 47 ' 06 ") * cos (90 ° -23 ° 44 ' 47 ")
+sin(90°-38°47′06″)*sin(90°-23°44′47″)*cos(41°22′22″)
Or
Cos (90 ° of-Lat)=cos (90 ° -38 ° 47 ' 06 ") * cos (90 ° -23 ° 44 ' 47 ")
+sin(90°-38°47′06″)*sin(90°-23°44′47″)*cos(68°59′22″)
Solve (90 ° of-Lat)=38 ° 01 ' 51 " perhaps 59 ° 27 ' 54 " and obtain latitude Lat=N51 ° 58 ' 09 " or Lat
=N30 ° 32 ' 06 ", then solve LRA1:
Or
LRA1=100 ° 53 ' 17 " or 82 ° 46 ' 30 " is solved, is directly named as known to celestial body azimuth, if orientation is unknown,
It is calculated separately according to E or W, Long=RA-LRA-GRA,
E:Long=279 ° 13 ' 00 " -100 ° 53 ' 17 " -238 ° 47 ' 12 "=W60 ° 27 ' 29 ",
W:Long=279 ° 13 ' 00 "-(360 ° -100 ° 53 ' 17 ") -238 ° 47 ' 12 "=E141 ° 19 ' 05 ",
Or E:Long=279 ° 13 ' 00 " -82 ° 46 ' 30 " -238 ° 47 ' 12 "=W42 ° 20 ' 42 ",
W:Long=279 ° 13 ' 00 "-(360 ° -82 ° 46 ' 30 ") -238 ° 47 ' 12 "=E123 ° 12 ' 18 ",
It is same to resolve LRA2:
Or
LHA2=99 ° 10 ' 13 " either 77 ° 10 ' 00 " is solved to be calculated separately with E or W,
E:Long=79 ° 09 ' 30 " -99 ° 10 ' 13 " -238 ° 47 ' 12 "=E101 ° 12 ' 05 ",
W:Long=79 ° 09 ' 30 "-(360 ° -99 ° 10 ' 13 ") -238 ° 47 ' 12 "=W60 ° 27 ' 29 ",
Or E:Long=79 ° 09 ' 30 " -77 ° 10 ' 00 " -238 ° 47 ' 12 "=E123 ° 12 ' 18 ",
W:Long=79 ° 09 ' 30 "-(360 ° -77 ° 10 ' 00 ") -238 ° 47 ' 12 "=W82 ° 27 ' 42 ",
LHA1 is identical as LHA2 calculated result be it is correct, obtain longitude Long=W60 ° 27 ' 29 " or Long=E123 °
12 ' 18 ",
Resolving obtain survey person position Z (Lat N51 ° 58 ' 09 ", Long W60 ° 27 ' 29 ") or (Lat N30 ° 32 ' 06 ",
Long E123°12′18″)。
The parsing astronomical positioning method for projecting to the first point of Aries equatorial system of coordinates is resolved actually and with mathematics to obtain spherical surface
The intersecting point coordinate of the height position line such as the isoazimuth position line or spherical surface, it is noted that following item when use:
1) using single celestial body azimuth and elevation measurement survey person position, the isoazimuth position line of same celestial body with etc. height position
Line is set according to the difference at azimuth and height, intersection point may be one or two, and the angle of cut may be 0 ° -90 ° of any angle,
I.e. two positions line may be vertically also possible to tangent, and in cusp position and its to be positioned about precision poor.
2) two orientation or elevation measurement position of heavenly body or survey person position solve declination according to data simultaneously
Or latitude of an observer is unrelated with observation time, i.e., observes data simultaneously and do not have the time, can also solve to obtain declination or
Latitude of an observer.
3) data are observed compared with altitude observations of a celestial body data in celestial body azimuth, and altitude observations of a celestial body data are by atmospheric density
The influence of vertical change, observation altitude of the heavenly body are preferably not below 30 °；Data are observed not by atmospheric density vertical change in celestial body azimuth
Influence；On the ground under height accuracy of observation identical with orientation, observation celestial body azimuth positioning accuracy is better than observation celestial body
Altitude location.
Claims (10)
1. a kind of parsing astronomical positioning method for projecting to the first point of Aries equatorial system of coordinates observes the side of unknown celestial body in known location
Perhaps height to measure position of heavenly body or observe orientation or the height of known celestial body in unknown position measures survey person position for position
It sets, it is characterised in that: survey person position is projected into the first point of Aries equatorial system of coordinates, astronomy three is established in the first point of Aries equatorial system of coordinates
It is angular, using observation celestial body azimuth or altitude information, celestial body is obtained with spherical triangle is associated with by resolving parallactic triangle
Position or survey person's position data.
2. a kind of parsing astronomical positioning method for projecting to the first point of Aries equatorial system of coordinates according to claim 1, feature
It is: using single celestial body azimuth and altitude information is observed, obtains position of heavenly body or survey person position by resolving parallactic triangle
Data；Known survey person position and position of heavenly body obtain true azimuth of celestial body and height data by resolving parallactic triangle.
3. a kind of parsing astronomical positioning method for projecting to the first point of Aries equatorial system of coordinates according to claim 2, feature
It is: observes the true bearing A and height h of unknown celestial body B at known location Z (Lat/Long), resolve position of heavenly body；According to when
Between obtain Green point right ascension GRA, survey person position Z is projected into the first point of Aries equatorial system of coordinatesIn parallactic triangle, it is known that apex distance (90 ° of-h), colatitude (90 ° of-Lat) and side
Parallactic angle A three elements are solved remaining away from (90 ° of-Dec) using cosine formula:
Cos (90 ° of-Dec)=cos (90 ° of-h) * cos (90 ° of-Lat)
+sin(90°-h)*sin(90°-Lat)*cos(A)
=sin (h) * sin (Lat)+cos (h) * cos (Lat) * cos (A)
Away from (90 ° of-Dec) more than solving, declination Dec is obtained；It reuses cosine formula deformation and solves local right ascension LRA:
Resolving obtains local right ascension LRA, obtains right ascension of a celestial body RA=LRA+RAZ, obtain position of heavenly body (Dec, RA).
4. a kind of parsing astronomical positioning method for projecting to the first point of Aries equatorial system of coordinates according to claim 2, feature
It is: in the true bearing A and height h that unknown position Z observes known celestial body B (Dec/RA), resolves survey person position；According to the time
Obtain Green point right ascension GRA, survey person position Z projected into the first point of Aries equatorial system of coordinates, in parallactic triangle, it is known that it is remaining away from
(90 ° of-Dec), apex distance (90 ° of-h) and azimuth A three elements solve local right ascension LRA using sine formula:
Local right ascension LRA is solved, longitude Long=RA-LRA-GRA is obtained；It reuses the deformation of Napier formula and solves colatitude
(90 ° of-Lat):
Resolving obtains colatitude (90 ° of-Lat), obtains latitude Lat；Obtain survey person position (Lat, Long).
5. a kind of parsing astronomical positioning method for projecting to the first point of Aries equatorial system of coordinates according to claim 2, feature
It is: observes known celestial body B (Dec/RA) at known location Z (Lat/Long), resolve true azimuth of celestial body A and height h；According to
Time obtains Green point right ascension GRA, and survey person position Z is projected to the first point of Aries equatorial system of coordinates
By right ascension of a celestial body RA and survey person's right ascension RAZObtain local right ascension LRA=RA-RAZ, in parallactic triangle, it is known that it is remaining away from (90 °-
Dec), colatitude (90 ° of-Lat) and local right ascension LRA three elements, solve apex distance (90 ° of-h) using cosine formula:
Cos (90 ° of-h)=cos (90 ° of-Dec) * cos (90 ° of-Lat)
+sin(90°-Dec)*sin(90°-Lat)*cos(LRA)
=sin (Dec) * sin (Lat)+cos (Dec) * cos (Lat) * cos (LRA)
Apex distance (90 ° of-h) is solved, height h is obtained；It reuses cosine formula deformation and solves azimuth A:
Resolving obtains azimuth A；Obtain astronomical observation true bearing and height (A, h).
6. a kind of parsing astronomical positioning method for projecting to the first point of Aries equatorial system of coordinates according to claim 1, feature
Be: observing same celestial body azimuth or altitude information using two places, by resolve parallactic triangle be associated with spherical triangle
Obtain position of heavenly body data；Two celestial body azimuths or altitude information are observed using one, by resolving parallactic triangle and association
Spherical triangle obtains survey person's position data.
7. a kind of parsing astronomical positioning method for projecting to the first point of Aries equatorial system of coordinates according to claim 6, feature
It is: in known location Z1(Lat1/Long1)、Z2(Lat2/Long2), the true bearing of the same unknown celestial body B of observation is respectively
A1, A2 resolve position of heavenly body；Green point right ascension GRA1, GRA2 are obtained according to the time, by Z1And Z2Project to first point of Aries equator seat
Mark system,In ball
Orthodrome tie point Z is used on face1And Z2Obtain two auxiliary triangle shape △ PZ1Z2With △ BZ1Z2With parallactic triangle △ PZ1B and
△PZ2B composition association spherical triangle；
In triangle △ PZ1Z2In, it is known that both sides (90 ° of-Lat1), (90 ° of-Lat2) and its angle (RAZ2-RAZ1), use cosine
The other one side Z of equations1Z2:
cos(Z1Z2)=cos (90 ° of-Lat1) * cos (90 ° of-Lat2)+sin (90 ° of-Lat1)
*sin(90°-Lat2)*cos(RAZ2-RAZ1)
=sin (Lat1) * sin (Lat2)+cos (Lat1) * cos (Lat2)
*cos(RAZ2-RAZ1)
It reuses cosine formula deformation and solves other two corners ∠ PZ1Z2With ∠ PZ2Z1:
∠ BZ is solved by association spherical triangle corner relationship again1Z2With ∠ BZ2Z1；
Ipsilateral ∠ BZ1Z2=| ∠ PZ1Z2- A1 | and ∠ BZ2Z1=| ∠ PZ2Z1+ A2 |, it is subtracted if resulting value is greater than 180 ° with 360 °
Resulting value or heteropleural ∠ BZ1Z2=| ∠ PZ1Z2- A1 | and ∠ BZ2Z1=| ∠ PZ2Z1-A2|；
Under A1, A2 known case, two places are ipsilateral or heteropleural can only be one of situation in celestial body, therefore two places observation is same
Ask position of heavenly body that can obtain one group of solution in celestial body azimuth；
In triangle △ BZ1Z2In, it is known that two corners ∠ BZ1Z2、∠BZ2Z1And its contained side Z1Z2, deformed and solved using four element equations
In addition both sides (90 ° of-h1) and (90 ° of-h2):
In triangle △ PZ1In B, it is known that both sides (90 ° of-h1), (90 ° of-Lat1) and its included angle A 1 are solved another using cosine formula
Outer one side (90 ° of-Dec):
Cos (90 ° of-Dec)=cos (90 ° of-h1) * cos (90 ° of-Lat1)+sin (90 ° of-h1)
*sin(90°-Lat1)*cos(A1)
=sin (h1) * sin (Lat1)+cos (h1) * cos (Lat1) * cos (A1)
(90 ° of-Dec) is solved, declination Dec is obtained, cosine formula deformation is reused and solves other one jiao of LRA1:
Or in triangle △ PZ2In B, it is known that in addition both sides (90 ° of-h2), (90 ° of-Lat2) and its included angle A 2 solve on one side
(90 ° of-Dec) and LRA2 method are the same as solution △ PZ1B；
Right ascension of a celestial body RA=LRA+RA is obtained by LRAZ, obtain position of heavenly body B (Dec, RA).
8. a kind of parsing astronomical positioning method for projecting to the first point of Aries equatorial system of coordinates according to claim 6, feature
It is: in known location Z1(Lat1/Long1)、Z2(Lat2/Long2), the height of the same unknown celestial body B of observation is respectively
H1, h2 resolve position of heavenly body；Green point right ascension GRA1, GRA2 are obtained according to the time, by Z1And Z2Project to first point of Aries equator seat
Mark system,In ball
Orthodrome tie point Z is used on face1And Z2Obtain two auxiliary triangle shape △ PZ1Z2With △ BZ1Z2With parallactic triangle △ PZ1B and
△PZ2B composition association spherical triangle；
In triangle △ PZ1Z2In, it is known that both sides (90 ° of-Lat1), (90 ° of-Lat2) and its angle (RAZ2-RAZ1), use cosine
The other one side Z of equations1Z2:
cos(Z1Z2)=cos (90 ° of-Lat1) * cos (90 ° of-Lat2)+sin (90 ° of-Lat1)
*sin(90°-Lat2)*cos(RAZ2-RAZ1)
=sin (Lat1) * sin (Lat2)+cos (Lat1) * cos (Lat2)
*cos(RAZ2-RAZ1)
It reuses cosine formula deformation and solves other two corners ∠ PZ1Z2With ∠ PZ2Z1:
In triangle △ BZ1Z2In, it is known that three sides (90 ° of-h1), (90 ° of-h2) and Z1Z2, deformed using cosine formula and solve two corners
∠BZ1Z2With ∠ BZ2Z1:
A1 and A2 are solved by association spherical triangle corner relationship again；
A1=| ∠ PZ1Z2-∠BZ1Z2| and A2=| ∠ PZ2Z1-∠BZ2Z1|, with A1=∠ PZ1Z2+∠BZ1Z2With A2=∠
PZ2Z1+∠BZ2Z1If resulting value, which is greater than 180 °, subtracts resulting value with 360 °；
At the unknown orientation of known altitude, two groups of solutions can be existed simultaneously, therefore two places observe same altitude of the heavenly body and seek celestial body
Position can obtain two groups of solutions；
In triangle △ PZ1In B, it is known that both sides (90 ° of-h1), (90 ° of-Lat1) and its included angle A 1 are solved another using cosine formula
Outer one side (90 ° of-Dec):
Cos (90 ° of-Dec)=cos (90 ° of-h1) * cos (90 ° of-Lat1)+sin (90 ° of-h1)
*sin(90°-Lat1)*cos(A1)
=sin (h1) * sin (Lat1)+cos (h1) * cos (Lat1) * cos (A1)
(90 ° of-Dec) is solved, declination Dec is obtained, cosine formula deformation is reused and solves other one jiao of LRA1:
Or in triangle △ PZ2In B, it is known that in addition both sides (90 ° of-h2), (90 ° of-Lat2) and its included angle A 2 solve on one side
(90 ° of-Dec) and LRA2 method are the same as solution △ PZ1B；
Right ascension of a celestial body RA=LRA+RA is obtained by LRAZ, obtain position of heavenly body B (Dec, RA).
9. a kind of parsing astronomical positioning method for projecting to the first point of Aries equatorial system of coordinates according to claim 6, feature
It is: observes known celestial body B in unknown position Z1(Dec1/RA1)、B2(Dec2/RA2) true bearing is respectively A1, A2, is resolved
Survey person position；Green point right ascension GRA is obtained according to the time, survey person Z is projected into the first point of Aries equatorial system of coordinates, with big on spherical surface
Circular sliding slopes point B1And B2Obtain two auxiliary triangle shape △ PB1B2With △ ZB1B2With parallactic triangle △ PZB1With △ PZB2Composition
It is associated with spherical triangle；
In triangle △ PB1B2In, it is known that both sides (90 ° of-Dec1), (90 ° of-Dec2) and its angle (RA2-RA1) use cosine
The other one side B of equations1B2:
cos(B1B2)=cos (90 ° of-Dec1) * cos (90 ° of-Dec2)+sin (90 ° of-Dec1)
*sin(90°-Dec2)*cos(RA2-RA1)
=sin (Dec1) * sin (Dec2)+cos (Dec1) * cos (Dec2)
*cos(RA2-RA1)
In triangle △ PZB1In, it is known that side (90 ° of-Dec1) and angle A1, unknown (90 ° of-h1) and (90 ° of-Lat) use cosine
Formula:
Cos (90 ° of-Dec1)=cos (90 ° of-h1) * cos (90 ° of-Lat)+sin (90 ° of-h1) * sin (90 ° of-Lat) * cos
(A1)---(1)
In triangle △ PZB2In, it is known that side (90 ° of-Dec2) and angle A2, unknown (90 ° of-h2) and (90 ° of-Lat) use cosine
Formula:
Cos (90 ° of-Dec2)=cos (90 ° of-h2) * cos (90 ° of-Lat)+sin (90 ° of-h2) * sin (90 ° of-Lat) * cos
(A2)---(2)
In triangle △ ZB1B2In, it is known that side B1B2With two days same side angles of body (A2-A1) or two days body heteropleural angles (A1+A2), if
Resulting value is greater than 180 ° and subtracts resulting value with 360 °, and unknown (90 ° of-h1) and (90 ° of-h2) uses cosine formula:
cos(B1B2)=cos (90 ° of-h1) * cos (90 ° of-h2)+sin (90 ° of-h1) * sin (90 ° of-h2) * cos (A2-A1) ---
(3)
Or
cos(B1B2)=cos (90 ° of-h1) * cos (90 ° of-h2)+sin (90 ° of-h1) * sin (90 ° of-h2) * cos (A1+A2) ---
(4)
Under orientation known case, two days bodies are ipsilateral or heteropleural can only be one of situation in survey person, and body is ipsilateral within two days
(1)-(3) or two days body heteropleural (1) (2) (4) three formula simultaneous totally 3 unknown numbers, solution obtain (90 ° of-h1), (90 ° of-h2) and
(90 ° of-Lat), obtains latitude Lat；
In triangle △ PZB1In, it is known that three sides (90 ° of-Dec1), (90 ° of-h1), (90 ° of-Lat) are asked using cosine formula deformation
Other one jiao of LRA1 is solved,
Or in triangle △ PZB2In, it is known that three sides (90 ° of-Dec2), (90 ° of-h2), (90 ° of-Lat) are become using cosine formula
Shape solves other one jiao of LRA2 method with solution △ PZB1；
Longitude Long=RA-LRA-GRA is obtained by LRA；Obtain survey person position Z (Lat, Long).
10. a kind of parsing astronomical positioning method for projecting to the first point of Aries equatorial system of coordinates according to claim 6, feature
It is: observes known celestial body B in unknown position Z1(Dec1/RA1)、B2(Dec2/RA2) height is respectively h1, h2, is resolved
Survey person position；Green point right ascension GRA is obtained according to the time, survey person Z is projected into the first point of Aries equatorial system of coordinates, with big on spherical surface
Circular sliding slopes point B1And B2Obtain two auxiliary triangle shape △ PB1B2With △ ZB1B2With parallactic triangle △ PZB1With △ PZB2Composition
It is associated with spherical triangle；
In triangle △ PB1B2In, it is known that both sides (90 ° of-Dec1), (90 ° of-Dec2) and its angle (RA2-RA1) use cosine
The other one side B of equations1B2:
cos(B1B2)=cos (90 ° of-Dec1) * cos (90 ° of-Dec2)+sin (90 ° of-Dec1)
*sin(90°-Dec2)*cos(RA2-RA1)
=sin (Dec1) * sin (Dec2)+cos (Dec1) * cos (Dec2)
*cos(RA2-PA1)
It reuses cosine formula deformation and solves other two corners ∠ PB1B2With ∠ PB2B1:
In triangle △ ZB1B2In, it is known that three sides (90 ° of-h1), (90 ° of-h2), B1B2, other two are solved using cosine formula deformation
Angle ∠ ZB1B2With ∠ ZB2B1:
∠ PB is solved by association spherical triangle corner relationship again1Z and ∠ PB2Z；
∠PB1Z=| ∠ PB1B2-∠ZB1B2| and ∠ PB2Z=| ∠ PB2B1-∠ZB2B1|, with ∠ PB1Z=∠ PB1B2+∠ZB1B2
With ∠ PB2Z=∠ PB2B1+∠ZB2B1If resulting value, which is greater than 180 °, subtracts resulting value with 360 °；
At the unknown orientation of known altitude, two groups of solutions can be existed simultaneously, therefore a ground observes two altitude of the heavenly bodies and asks survey person position
Two groups of solutions can be obtained by setting；
In triangle △ PB1In Z, it is known that both sides (90 ° of-Dec1), (90 ° of-h1) and its angle ∠ PB1Z is asked using cosine formula
Solution is other on one side (90 ° of-Lat):
Cos (90 ° of-Lat)=cos (90 ° of-Dec1) * cos (90 ° of-h1)+sin (90 ° of-Dec1)
*sin(90°-h1)*cos(PB1Z)
=sin (Dec1) * sin (h1)+cos (Dec1) * cos (h1) * cos (PB1Z)
(90 ° of-Lat) is solved, latitude Lat is obtained, cosine formula deformation is reused and solves other one jiao of LRA1:
Or in triangle △ PB2In Z, it is known that both sides (90 ° of-Dec2), (90 ° of-h2) and its angle ∠ PB2Z uses cosine public affairs
Formula solves (90 ° of-Lat) and LRA2 method with solution △ PB1Z；
Longitude Long=RA-LRA-GRA is obtained by LRA；Obtain survey person position Z (Lat, Long).
Priority Applications (1)
| Application Number | Priority Date | Filing Date | Title |
|---|---|---|---|
| CN201610931149.0A CN106525026B (en) | 2016-11-01 | 2016-11-01 | Project to the parsing astronomical positioning method of the first point of Aries equatorial system of coordinates |
Applications Claiming Priority (1)
| Application Number | Priority Date | Filing Date | Title |
|---|---|---|---|
| CN201610931149.0A CN106525026B (en) | 2016-11-01 | 2016-11-01 | Project to the parsing astronomical positioning method of the first point of Aries equatorial system of coordinates |
Publications (2)
| Publication Number | Publication Date |
|---|---|
| CN106525026A CN106525026A (en) | 2017-03-22 |
| CN106525026B true CN106525026B (en) | 2019-09-03 |
Family
ID=58292473
Family Applications (1)
| Application Number | Title | Priority Date | Filing Date |
|---|---|---|---|
| CN201610931149.0A Active CN106525026B (en) | 2016-11-01 | 2016-11-01 | Project to the parsing astronomical positioning method of the first point of Aries equatorial system of coordinates |
Country Status (1)
| Country | Link |
|---|---|
| CN (1) | CN106525026B (en) |
Families Citing this family (1)
| Publication number | Priority date | Publication date | Assignee | Title |
|---|---|---|---|---|
| CN108645410B (en) * | 2018-05-15 | 2022-04-15 | 天津大学 | Ship astronomical positioning method without elevation angle observation |
Citations (4)
| Publication number | Priority date | Publication date | Assignee | Title |
|---|---|---|---|---|
| US5344325A (en) * | 1993-05-17 | 1994-09-06 | Wang Zn Hu | Automatic tracking astronomical globe |
| CN102162731A (en) * | 2011-01-12 | 2011-08-24 | 哈尔滨工业大学 | High-precision satellite independent navigation method based on pulse data of sun, earth and moon integrated sensor |
| CN105222777A (en) * | 2015-09-29 | 2016-01-06 | 李清林 | By the method for hypothesis longitude and latitude method observation celestial body location |
| CN105760688A (en) * | 2016-03-04 | 2016-07-13 | 李清林 | Method for determining celestial body position or spherical target position by longitude and latitude assumption method |
-
2016
- 2016-11-01 CN CN201610931149.0A patent/CN106525026B/en active Active
Patent Citations (4)
| Publication number | Priority date | Publication date | Assignee | Title |
|---|---|---|---|---|
| US5344325A (en) * | 1993-05-17 | 1994-09-06 | Wang Zn Hu | Automatic tracking astronomical globe |
| CN102162731A (en) * | 2011-01-12 | 2011-08-24 | 哈尔滨工业大学 | High-precision satellite independent navigation method based on pulse data of sun, earth and moon integrated sensor |
| CN105222777A (en) * | 2015-09-29 | 2016-01-06 | 李清林 | By the method for hypothesis longitude and latitude method observation celestial body location |
| CN105760688A (en) * | 2016-03-04 | 2016-07-13 | 李清林 | Method for determining celestial body position or spherical target position by longitude and latitude assumption method |
Non-Patent Citations (1)
| Title |
|---|
| 用空间解析几何法进行天文定位定向的误差分析;王昊京等;《红外与激光工程》;20150831;第44卷(第8期);第2364-2374页 |
Also Published As
| Publication number | Publication date |
|---|---|
| CN106525026A (en) | 2017-03-22 |
Similar Documents
| Publication | Publication Date | Title |
|---|---|---|
| Gerkema | An introduction to tides | |
| CN105222777B (en) | With the method for assuming the observation celestial body positioning of longitude and latitude method | |
| de Grijs | A (not so) brief history of lunar distances: Lunar longitude determination at sea before the chronometer | |
| CN106441297B (en) | The gravity error vector acquisition methods and device of inertial navigation system | |
| CN106525026B (en) | Project to the parsing astronomical positioning method of the first point of Aries equatorial system of coordinates | |
| Brosche et al. | Earth’s Rotation from Eons to Days: Proceedings of a Workshop Held at the Centre for Interdisciplinary Research (ZiF) of the University of Bielefeld, FRG. September 26–30, 1988 | |
| Van Allen | Basic principles of celestial navigation | |
| CN106383982B (en) | The astronomical positioning method of parsing measurement position of heavenly body or survey person position | |
| Chapman | The elements of astronomy for surveyors | |
| Howell | Practical Celestial Navigation | |
| Bennett | Instruments, mathematics, and natural knowledge: Thomas Harriot's place on the map of learning | |
| Nicolai | The map projection of portolan charts | |
| Rogers et al. | Working group report: a brief history of trigonometry for mathematics educators | |
| Deakin | A guide to the mathematics of map projections | |
| Izraelevitz | The Cultural Adaptation of the Astrolabe | |
| Penprase | Exploring the Earth and Sky | |
| Newton | The historical acceleration of the Earth | |
| ISTIWA’AINI | COMPARATIVE STUDY OF MIZWALA AND | |
| Lapaine | Geodesy: determining the Earth’s shape and size | |
| Newcomb et al. | Astronomy for high schools and colleges | |
| Leahy | William John Wills as scientist | |
| Barlow et al. | Elementary Mathematical Astronomy: With Examples and Examination Papers | |
| Evers | Elementary Treatise on Nautical Astronomy: For the Use of Science Classes and Seamen | |
| Deakin | Satellite orbits | |
| Dutta | Āryabhata and axial rotation of Earth: 1. Khagola (The celestial sphere) |
Legal Events
| Date | Code | Title | Description |
|---|---|---|---|
| C06 | Publication | ||
| PB01 | Publication | ||
| SE01 | Entry into force of request for substantive examination | ||
| SE01 | Entry into force of request for substantive examination | ||
| GR01 | Patent grant | ||
| GR01 | Patent grant |