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7/29/2019 Exercitii Nav Astro
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Date: 31.10.1998.
DR Latitude: 4804.0 S
DR Longitude: 12000.0E
Star: Achernar - Below the Pole
Declination: 5714.7N
Bearing: North of observer
IE: 1.5 ON the arc
Height of Eye: 8.5 m
NESW is the rational horizon
P is the pole
Z is the Zenith
X is the Star below the pole
WQE is the celestial equator
WZE is the prime verticalNZSis the observers meridianZX is the zenith distance
PX is the polar distance
QX is the declination
PS is the Lat.
PZ is the co-Lat.
NX the True Alt. Below the pole
NP is the Latitude
SX is the Altitude.
Declination: 5714.7N
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(-) 9000.0
Polar Distance: 3245.3
DR Latitude: 4804.0S
Polar Distance: 3245.3 (Explanation See figure)
Approx. True Altitude: 1518.7
Refraction: (+) 3.4Reversed signs since we are doing the calculation backward
Dip: (+) 5.1 [Dip (in minutes) = 1.76Height of Eye (in metres)]
IE: + 1.5 (ON) this is not reversed, since this a equipment fault
Approx. Sext. Alt.: 1528.7S
SUN
Latitude by Meridian Altitude
The sun is on the observers meridian at noon Local Apparent Time.
The Local Mean Time of the MP at GREENWICH is given in the AlmanacSun Mer Pass.
The above is the approximate time of MP at any other meridian.
Process:
Take down the time of MP of the Sun: Name it Local Mean Time of MP
Convert the Longitude into TimeLong. (W) (+) and Long. (E) (-)
Add or subtract the Longitude in time from the LMT to get the GMT of the MPLook up the Almanac to get the Suns declination correct it as required.
Find the TZD from the T.Alt.
Add or subtract the declination to the TZD to get the Latitude.
Date: 06.11.1998.
00h 12h
Equation Of time 06th------------------- 16m23s 16m22s
Equation of time 07th------------------- 16m21s 16m19s
Now interpolate for --------------------- 14h 11m 00s
LAT (ship)--------------------------------- 12 00 00
Long. WEST-------------------- (+) 02 11 00
Greenwich Apparent Time (GAT) 14 11 00
Equation of Time-------------- (+) 16 22
GMT (Noon at ship) 14 27 22
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Sext Altitude: 6350.0 North of Observer
Bearing: North of observer
IE: 3.3 OFF the arc
Height of Eye: 7 m
Latitude: ??
Longitude: 03245.0W
NESW is the rational horizon
P is the pole
Z is the Zenith
X is the StarWQE is the celestial equator
WZE is the prime vertical
NZSis the observers meridianZX is the zenith distanceNX is the true altitude
PX is the polar distance
PZ is the co-Lat
QX is the declination
ZQ is the Latitude.
QZ is the Latitude
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Correct the Altitude:
Sext. Alt.: 6350.0
IE: + 3.3 (OFF)
Observed Altitude: 6353.3
Dip: (-) 4.8 [Dip (in minutes) = 1.76Height of Eye (in metres)]
Convert the Longitude into Time.
Divide the Longitude by 15So, 03245.0 = 02h 11m
GMT = MP + Long. (W)
D H M
LMT 06 11 44
Long W + 02 11
GMT 06 13 55
6348.5
Refraction: (-) 0.5
6348.0
Parallax: (+) 0.1
Semi Diameter: (+) 16.2
True Altitude: 6404.3
(-) 9000.0
True Zenith Distance: 2555.7 (ZX)
Corrected Declination: S 1601.3 (QX)
Latitude: 4157.0 (ZQ)
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SUN
FINDING THE LONGITUDE
Date: 03.11.1998.
Time: ------------------ 0800DR Latitude: ---------- 3206.0NDR Longitude: -------- 14054.0EChron. Time: --------- 03d 22h 02m 07s
Chron. Error: SLOW----------- 00m 40sSext Altitude: --------- 3601.0 (sun LL)IE: --------------------- 3.6 ON the arcHeight of Eye: -------- 12.2 m
Chron. Time:--------------- 03 22 02 07
SLOW----------------------- (+) 00 40
GMT-------------------------03 22 02 47
GHA (sun) For22h------- 15406.3
Incr.For 02m 47s---------- (+) 041.8
GHA ------------------------- 15448.1
Decl-------------------------- 1512.7S
-------------------------------- (-) 9000.0Polar Distance ------------ 7447.3 (PX)
PZPX---------------------1653.3PZZX---------------------0359.9
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NESW is the rational horizon
P is the pole
Z is the Zenith
X is the Sun
WQE is the equinoctial
WZE is the prime verticalNZS is the observers meridianZX is the zenith distance
NX is the true altitude
PX is the polar distance
CX is the Declination
aX is the altitude
PZ is the co-Lat
Angle ZPX is the Hour Angle
Angle PZX is the Azimuth
Known: PZ, PX, AND ZX.
To find: angle P and angle Z
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Correct the Altitude:[Dip (in mts) = 1.76Height of Eye (in m)
Sext. Alt.(LL):------------ 3601.0IE:-------------------------- (-) 3.6 (ON)----------------------------- 3557.4Dip:------------------------- (-) 6.2----------------------------- 3551.2Corr. (Almanac)---------- (+) 14.7True Altitude:------------- 3605.9------------------------ (-) 9000.0True Zenith Distance:---- 5354.1 (ZX)Latitude:----------------- 3206.0N
------------------------ (-) 9000.0
Co-Lat. 5754.0 (PZ)
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Once the Longitude has been found out, the position line that has been obtained is plotted on theposition of DR Latitude and the Observed Longitude.
From the above calculation it is apparent that the DR Latitude plays an important part indetermining the observed Longitude, thus if the position of the DR Latitude is in error then
the obs, Longitude will be way out.
However the Azimuth will not be since a slight error in Latitude will not make much difference
in the Azimuth.
Longitude = LHA ~ GHA
LHA (sun):-------- 30148.7
GHA (sun):-------- 15448.1-----------------------14102.7 E
Obs. Longitude:----------- 14700.6 E
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WORKING ON BOARD:
In ancient timesbefore the advent of GPS, the sight of the sun was of paramount importance,since this was one of the primary means of locating the position of a ship. Observation of starswas not that fortunate because the horizon had to be good and the cloud cover negligible.
The Sun as a result was observed in the Morning at around 0800h and the Longitude obtained.At Noon, the MP of the sun was observed and the position line obtained from the morning sight
was transferred to the Noon position, thus the Observed Latitude and the transferred Position linegave the observed position of the ship. In the late afternoon, the sun was again observed to check
up on the accuracy of the position as well as for any extraordinary drift, the speed and drift beingtaken as estimated.
In this we would be using the figures as obtained in the above example.
After the Sun was observed in the morning the Ship sailed a distance of 33 NM on a course of
225T.
Principle of Position Lines
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Celestial Sphere Showing the Earth and the 2 stars. The projected spots of the stars on the
surface of the EarthIn the above we see two heavenly bodies as viewed from the earth, they are projected on theimaginary celestial sphere. The positions of the two bodies if projected on the earth would be at
their respective GPs as shown in the other figure (marked as X)
We may assume any latitude or longitude of the GPs of the two bodies.
Now with the True Zenith Distance as observed, we may draw a circle with radius equal to thatof the TZD on the surface of the Earth.
Then at every point on the circle the observed altitude (true) would be the same since it is a circleof equal TZD of that body.
However distant a ship may be as long as the location of the ship is on the circumference of thecircle the TZD as found would be the same.
Let us now assume the following that the body has a declination of 20N and a GHA of 2 hours,then this bodys GP would be at lat. 20N and Long. 030 W.
Let the body as observed have a true altitude of 70, then the TZD would be 20
A ship from which the bearing of the body is due North would be on the equator and on the same
meridian as the body. That is the ship would have a position of Lat. 00 and Long. 030W. Thisbecause the line of bearing, North cuts the circle at a definite position.
However from another ship if the same body is observed as bearing 180 then that ship would beon a parallel of latitude that is 20 due north of the GP of the body, or the Latitude of the ship
would be 40N. The longitude would be same at 030W. This because the line of bearing, South,cuts the circle at a definite position.
Again if another ship observes the body on a bearing of 116 True then, this line of bearingwould cut the circle at the ships position as shown in the figure.
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If this above ship takes another observation now of a different body whose declination is thesame at 20N but the GHA is at 4 hours, its GP would be would be lat. 20N and
Long. 060W. Let the TZD as obtained from the observed altitude to be 14 and the bearing fromthe ship as 225. Then the ships position as referenced to this body would lie on the dotted linewhere the dotted line cuts the circle.
Thus if a ship observes two or more bodies at the same time (nearly) the points of intersection of
the bearing lines would give the position of the ship with some degree of accuracy. However thepoints intersect at two opposite directions, thus the bearings of the bodies from the ship would
determine the exact position of the ship.
So far so good.
But we cannot draw such huge circles and accuracy would suffer. So instead of drawing the
circles what we do instead is we draw a part of the arc, this arc although a curved line may beassumed to be a straight line for a very short length.
The question is now of fixing the centre of the body, this is overcome by estimating the positionof the shipthat is using the DR position of the ship.
Now since the body/ bodies are observed from this DR position the line of bearing would pass
through this position.
The Zenith Distance of the body is calculated for the time of observation, and the ZenithDistance as obtained from the observation of the altitude is compared.
Or the CZD is compared with the TZD.
If the TZD and the CZD are the same then the Ship is at the position of the DR. The arc of the
circle (which is drawn as explained earlier with the radius as the TZD) would be drawn at right
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angles to the line of bearing (since the radius always cuts the circumference at right anglesthearc of the circumference being a straight line as assumed). If two or more bodies have been
observed and assuming that the ship was stationary during the observation (there beingabsolutely no movement of the ship) and also that the observer was a genius at sights then
theoretically the arcs (straight lines representing the arcs) would all intersect at exactly the DR
and the DR would be the observed position of the ship.However if the arcs do not intersect then the point at which they do intersect becomes theobserved position of the ship.
But when comparing the CZD and the TZD, the difference, called the Intercept may be +ve orve.
Or we may say that CZD is more than TZD by a value or the reverse. In that case how do wedraw the line of bearing and where do we place the arc of the circle of position?
First we mark the DR of the ship and the bearing lines are drawn since the DR position is closeto the actual position and the bearing was taken from that location the bearing line would pass
through the DR position.
Now the question where do we draw the arc (straight line) on the bearing line, obviously it will
be at right angles to the bearing line as a radius always touches the circle at right angles.
Modifying the figure for ease in drawing on the chart paper, we have:
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Here we see that the Body positions are very far in the distantthe dotted lines indicate this.
Also the arcs are shown as arcs, but when we draw on the ship we would use straight lines. Also
the arcs are drawn centered on the bodies. Whereas the arcs or the straight line we would bedrawing would be centered on the DR position.
What has been shown above is the observed position as plotted with the TZD, but this again isnot possible on the ship, since we still cannot plot the GP of the bodies.
Thus on the ship we first calculate the Zenith Distance for the time of observation using thedetails given in the Almanac.
Next we find the true altitude and obtain the True Zenith Distance.
Next we compare the CZD with the TZD. If we have the CZD is greater than the TZD then:
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The ship is somewhere on the approx. arc of CZD, the actual position is however on the ARC ofthe TZD circle.
In the above since the CZD (estimated) radius is more than the actual radius that is the TZD,
therefore the ship has to be on the arc TZD, so we term this difference between the CZD and theTZD as TOWARDS, that is we are moving towards the body for accuracy.
If however the TZD had been greater then the Ships position would have to be shifted on to the
TZD arc which being greater, meant that the ship was being moved AWAY from the body.
Summing up:
For plotting we first calculate the CZD using the time of sight and the Almanac, the ship is onthe arc drawn with the CZD as radiusthis is the assumption.
Next we calculate the TZD using the observed altitude.
We compare the two, and find which one is greater, if:
CZD is greater than TZD then the difference (Intercept) is termed TOWARDS
TZD is greater than CZD then the difference (Intercept) is termed AWAY.
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If two or more bodies were observed then the ship would lie at the intersection of the arcs Position Lines.
After the above position lines have been drawn, the intersection of the position line forms theposition at which the ship was at the time of observation.
This plotting may be done on any chart (old and cancelledpreferably of a region close to theequator)
The intersection position is now measured with reference to the DR.
The distance above or below the DR is measured using a fixed scale, which was used for the
initial marking of the Intercept. And the distance so measured is considered as the Difference ofLatitude to the DR latitude.
The distance east or west of the DR is considered as the departure from the DR Longitude.
Applying the Dlat we find the OBS. LATITUDE.
Using Traverse table and the mean of the two latitudes we can convert the Departure measuredinto Difference of Longitude and applying this to the DR Longitude we would get the OBS.
LONGITUDE.
In general working the scale used is the Longitude scale, since the Longitude scale remains
constant on a Mercator chart but not the Latitude scale.
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NOTE: The time interval between the sights of the bodies if too long then run between theindividual sights have to be applied, however if the entire time period is less than 3 minutes then
the run interval may be omitted due to the distance run being insignificant. Practicing theshooting of the stars, makes taking up to 4 stars complete, within 3-4 minutes.
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The Marcq St. Hilaire method of fixing the position of the Ship
(Also known as the Intercept method)
In this method the TZD is compared with the CZD and the Intercept is found out. The position of
the PL may have to be drawn Towards the body or Away from the body.
Let us now assume that another two stars were observed at an interval of time of 30ssaypractically the same time.
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The DR used for calculation remains the same.
We have Fomalhaut with an azimuth of 118T giving a PL of 028 - 208 and with an Interceptof 7.5 Towards
Say star A had an azimuth of 255 giving a PL of 165 - 345 and had an Intercept of 16.0 Away
And star B had an azimuth of 135 giving a PL of 045 - 225 and had an Intercept of24.0 Towards
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Apply the D.Lat to the DR Lat. To obtain the Observed Latitude
Use the Traverse Tables to get the D.Long for the measured Departure using the mean Lat.
Apply the D.Long to the DR Long. To get the Observed Longitude
IF THE STAR OBSERVATION CANNOT BE TAKEN AT THE SAME TIME AND THEYARE OBSERVED WITH A LARGE INTERVAL OF TIME THEN RUN HAS TO BE
APPLIED, ALL THE STARS HAVE TO BE WORKED WITH DIFFERENT DRS AND THEPL OBTAINED WOULD HAVE TO BE SHIFTED TO THE MEAN TIME OF SIGHT AS
SELECTED, TRANFEREING PLS AS SHOWN EARLIER ON,