celestial navigation

5/20/2018 Celestial Navigation

1/25

Celestial Navigation

Exercises

Latitude by Meridian Altitude STAR

Date: 31.10.1998.

Sext Altitude: 5725

Star: Acrux

Declination: 6305.3S

Bearing: South of observer

IE: 3.3 OFF the arc

Height of Eye: 6 m

Latitude: ??

NESWis the rational horizon

Pis the pole

Zis the Zenith

Xis the Star

WQEis the celestial equator

WZEis the prime vertical

NZSis the observers meridianZXis the zenith distance

SXis the true altitude

PXis the polar distance

QXis the declination

PZis the co-Lat.

QZis the Latitude


2/25

Correct the Altitude:

Sext. Alt.: 5725.00

IE: + 3.3 (OFF)

Observed Altitude: 5728.3

Dip: (-) 4.3 [Dip (in minutes) = 1.76Height of Eye (in metres)]

5724.0

Refraction: (-) 0.6

True Altitude: 5723.4

(-) 9000.0

True Zenith Distance: 3236.6N

Declination: 6305.3S [Samenames add, different names subtract]

Latitude: 3028.7S


3/25

Calculating the Meridian Altitude STAR

Date: 01.11.1998.

Sext Altitude: 5725

Star: Altair

Declination: 0852.1N

Bearing: South of observer

IE: 2.0 ON the arc

Height of Eye: 7 m

DR Latitude: 4414.0N (approximate)

DR Longitude: 15800.0E


Pis the pole

Zis the Zenith

Xis the Star




SXis the true altitude



PZis the co-Lat.

QZis the Latitude


DR Latitude: 4414.0N

Declination: 0852.1N [Same names subtract, different names add]


4/25

{Since this is reverse calculation to find the altitude, the add / subtract rules change}

Approx. MZD 3521.9N

(-) 9000.0

Approx. True Altitude: 5438.1 S

Refraction: (+) 0.7 Reversed signs since we are doing the calculation backward

Dip: (+) 4.6 [Dip (in minutes) = 1.76Height of Eye (in metres)]

IE: + 2.0 (ON) this is not reversed, since this a equipment fault

Approx. Sext. Alt.: 5445.4S


Below the Pole

FINDING THE LATITUDE

Date: 06.11.1998

Sext Altitude: 1508North of the observer

Longitude: 15800.0E

Star: Alioth


Bearing: North of observer

IE: 1.6 OFF the arc

Height of Eye: 10 m


Pis the pole

Zis the Zenith

Xis the Star below the pole






PZis the co-Lat.

NXthe True Alt. Below the pole

QX1 is the declination

NPis the Latitude


5/25


(-) 9000.0

Polar Distance: 3402.0

Sext. Alt.: 5725.00 N

IE: + 1.6 (OFF)

1509.6


1504.0

Refraction: (-) 3.6True Alt.: 1500.4 N

Polar Distance:(+) 3402.0

Latitude: 4902.4N (Explanation See figure)

Important: When the object is on the meridian below the pole it always bears North in North

Lat., and South in South Lat. Or you may say that the Latitude always has the same name as the

bearing of the object.


6/25


Below the Pole

FINDING THE ALTITUDE

Date: 31.10.1998.

DR Latitude: 4804.0 S

DR Longitude: 12000.0E

Star: Achernar - Below the Pole



IE: 1.5 ON the arc

Height of Eye: 8.5 m


Pis the pole

Zis the Zenith

Xis the Star below the pole





QXis the declinationPSis the Lat.

PZis the co-Lat.

NXthe True Alt. Below the pole

NPis the Latitude

SXis the Altitude.


7/25


(-) 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 atnoon 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


8/25

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 MP

Look 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

Sext Altitude: 6350.0 North of Observer


IE: 3.3 OFF the arc

Height of Eye: 7 m

Latitude: ??

Longitude: 03245.0W


9/25


Pis the pole

Zis the Zenith

Xis the Star




NXis the true altitude


PZis the co-Lat


ZQis the Latitude.

QZis the Latitude


10/25


Sext. Alt.: 6350.0

IE: + 3.3 (OFF)

Observed Altitude: 6353.3


Convert the Longitude into Time.

Divide the Longitude by 15

So, 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)


11/25

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) For 22h------- 15406.3

Incr.For 02m 47s---------- (+) 041.8

GHA ------------------------- 15448.1

Decl-------------------------- 1512.7S

-------------------------------- (-) 9000.0Polar Distance ------------ 7447.3 (PX)

PZPX--------------------- 1653.3

PZZX--------------------- 0359.9


12/25


Pis the pole

Zis the Zenith

Xis the Sun

WQEis the equinoctial



NXis the true altitude


CXis the Declination

aX is the altitude

PZis 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


13/25

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)


14/25

Once the Longitude has been found out, the position line that has been obtained is plotted on the

position of DR Latitude and the Observed Longitude.

From the above calculation it is apparent that the DR Latitude plays an important part in

determining 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


15/25

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 stars

was 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 line

gave 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 being

taken 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


16/25

Celestial Sphere Showing the Earth and the 2 stars. The projected spots of the stars on the

surface of the Earth

In the above we see two heavenly bodies as viewed from the earth, they are projected on the

imaginary 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 that

of 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 circle

of equal TZD of that body.

However distant a ship may be as long as the location of the ship is on the circumference of the

circle 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. This

because 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 be

on 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 bearing

would cut the circle at the ships position as shown in the figure.


17/25

If this above ship takes another observation now of a different body whose declination is the

same 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 from

the 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 the

points 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 be

assumed 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 position

of 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 Zenith

Distance as obtained from the observation of the altitude is compared.

Or the CZD is compared with the TZD.


18/25

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

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 being

absolutely no movement of the ship) and also that the observer was a genius at sights thentheoretically 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 the

observed position of the ship.

But when comparing the CZD and the TZD, the difference, called the Intercept may be +veorve.

Or we may say that CZD is more than TZD by a value or the reverse. In that case how do we

draw 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 close

to 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:


19/25

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 be

drawing would be centered on the DR position.

What has been shown above is the observed position as plotted with the TZD, but this again is

not 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 the

details 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:


20/25

The ship is somewhere on the approx. arc of CZD, the actual position is however on the ARC of

the 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 on

the 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


21/25

TZD is greater than CZD then the difference (Intercept) is termed AWAY.

If two or more bodies were observed then the ship would lie at the intersection of the arcsPosition Lines.

After the above position lines have been drawn, the intersection of the position line forms the

position at which the ship was at the time of observation.

This plotting may be done on any chart (old and cancelledpreferably of a regionclose to the equator)

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 Dlatwe find the OBS. LATITUDE.

Using Traverse table and the mean of the two latitudes we can convert the Departure measured

into Difference of Longitude and applying this to the DR Longitude we would get the OBS.

LONGITUDE.


22/25

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.

NOTE: The time interval between the sights of the bodies if too long then run between the

individual 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.


23/25

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.


24/25

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 7.5Towards

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 of24.0Towards


25/25

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 THEY

ARE OBSERVED WITH A LARGE INTERVAL OF TIME THEN RUN HAS TO BE

APPLIED, ALL THE STARS HAVE TO BE WORKED WITH DIFFERENT DRS AND THE

PL OBTAINED WOULD HAVE TO BE SHIFTED TO THE MEAN TIME OF SIGHT ASSELECTED, TRANFEREING PLS AS SHOWN EARLIER ON,

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