INFINITY MARINER: Position Lines & Numericals

Position Lines by use of Sun

Latitude by meridian altitude:

Lat by mer alt Sun

On 23^rd Sept 2008, in DR 23° 40’N 161° 56’E, the sextant meridian altitude of the Sun’s lower limb was 66° 10.6’. If IE was 2.3’ on the arc and HE was 10.5m, find the latitude and the LOP.

Step 1: Obtain GMT mer pass sun

LMT mer pass sun 23d 11h 52m 00s

LIT (E) (-) 10 47 44

GMT mer pass Sept 23d 01h 04m 16s

Step 2: Obtain dec sun for GMT

Declination Sept 23d 01h 00° 09.0’S (See Nautical Almanac below)

d (+1.0) for 4m (+) 00.1’

Declination Sept 23d 01h 04m 00° 09.1’S

Step 3: Calculate lat & direction of LOP

Sext Alt	66° 10.6’
IE (on)	(-) 02.3’	Off is Plus; On is minus
Obs alt	66° 08.3’
Dip (10.5m)	(-) 05.7’	From Almanac first page
App alt	66° 02.6’
Tot corrn LL	(+) 15.5’	From Almanac first page
T alt	66° 18.1’S	See note 2 below
MZD	23° 41.9’N	See note 3 below
Dec	00° 09.1’S
Latitude	23° 32.8’N	See note 4 below
LOP	E - W	Always E-W at mer pass

Notes

Explanations in the box are for guidance only. They are not part of the working
Thumb rule 1: Name True alt N or S - same as azimuth at mer pass.
In this case, DR lat is 23° 40’N while dec is practically nil. Hence the sun’s Az at mer pass must be S and so the true altitude is named S.
Thumb rule 2: Name MZD opposite to that of true altitude. This naming system is a quick and sure method for you to decide whether MZD & dec are to be added or subtracted to obtain the latitude without the necessity of drawing a figure.
Thumb rule 3: If MZD and dec are of same name, add and retain name. If of opposite names, subtract the smaller one from the larger one and retain the name of the larger one
In many cases, it can be decided whether MZD and dec are to be added or subtracted by knowing that the answer must be very close to the DR lat. However, if either MZD or declination is very small, both addition and subtraction would give results reasonably close to DR lat, as in the foregoing worked example, and the unwary navigator would be puzzled as to what to do.
In some cases, you are not given the DR lat but then you would be given an indication of the Az at mer pass. By using the above naming system, you can get the solution easily and quickly.

Intercept:

Intercept Sun

On 29^th Nov 2008 in DR 26° 27’N 130° 27’W, the sextant altitude of the Sun’s UL East of the meridian was 28° 11’, “GMT 17h 47m 49s” . If HE was 10m and IE was 2.3’ off the arc, calculate the direction of the LOP and the intercept.

Step 1: Obtain the LHA & dec for the GMT (See the next page)

From the nautical almanac (NA), obtain the GHA & dec sun, & thence the LHA sun as shown in normal font in the table here.

For easy reference, insert the given lat in the empty cell below the dec.

GHA 29d 17h	077° 52.2’	dec	21° 37.2’S
Incr 47m 49s	011° 57.3’	d (+0.4)	(+) 00.3’
GHA for GMT	089° 49.5’	dec	21° 37.5’S
Long(W)	(-)130° 27.0’	lat	26° 27.0’N
LHA	319° 22.5’
P	040° 37.5’

The cells in bold font are needed to enter tables & calculate the CZD & later, the Az.

Step 2: Apply intercept formula & obtain CZD.

Cos CZD = Cos P $\times$ Cos Lat. $\times$ Cos Dec. ± Sin Lat. $\times$ Sin Dec.

Note: If Lat and dec are same name (+), contrary names (-). Hence in the formula, (+) has been put above the (-). In this case, lat & dec have contrary names so sign in formula becomes MINUS.

Position Lines by use of Stars

Time of meridian passage

Time of mer pass star

Sometimes, it may be necessary to find the GMT of meridian passage to the nearest second. In such a case, proceed as follows:

At mer pass of $\star$ , LHA $\star$ = 360°

Hence at mer pass of $\star$ , GHA $\gamma$ + E (or - W) long + SHA $\star$ = LHA $\star$ = 360°

SHA $\star$ is obtained from that day’s page in the Nautical Almanac. DR long is given.

In the worked example, GHA $\gamma$ + 75° 00’ + 245° 03.0 = 360° or GHA $\gamma$ = 039°57.0’

Hence, at mer pass $\star$ , GHA $\gamma$ =	039°57.0’
From almanac, for GMT December 01d 21h, GHA $\gamma$ =	026°05.2’
55m 18s $\leftarrow$ Increment $\leftarrow$	013°51.8’

Note: In referring to the almanac, it has been assumed that the date of GMT would be the same as that of ship i.e., 1st December. However, it may be possible that the date of GMT may be one day earlier or later than that assumed. It is therefore necessary to check up by applying LIT to the assumed GMT, as shown below, and alterations made, if necessary, to the GMT date and time.

GMT mer pass $\star$	Dec 01d 21h 55m 18s
LIT (E)	(+) 05h 00m 00s
LMT mer pass $\star$	02d 02h 55m 18s	But required date is 1st Dec
One sidereal day*	(-) 23h 56m 04s
Correct LMT mer pass $\star$	Dec 01d 02h 59m 14s
LIT (E)	(-) 05h 00m 00s
Correct GMT mer pass $\star$	Nov 30d 21h 59m 14s	Answer

The time interval between any two successive meridian passages of $\gamma$ is one sidereal day which is 23h 56m 04s. Hence one sidereal day may be added or subtracted as appropriate to ensure that the meridian passage is correct for ship’s date.

Latitude by meridian altitude

Lat by mer alt star

On 1^st Dec 2008, AM at ship in DR 45° 20’S 75°00’E, the sextant meridian altitude of the star PROCYON was 39° 28.8’. If IE was 1.5’ off the arc and HE was 25m, find the latitude and LOP.

Sext Alt	39° 28.8’
IE (off)	(+) 01.5’
Obs alt	39° 30.3’
Dip (25m)	(-) 08.8’
App alt	39° 21.5’
Tot corrn	(-) 01.2’
T alt	39° 20.3’N
MZD	50° 39.7’S
Dec	05° 12.2’N
Latitude	45° 27.5’S
LOP	E - W

Notes repeated for quick reference from mer alt sun

Name True alt N or S - same as azimuth at mer pass. In this case, DR lat is 45° 20’S while dec is 05° 14.6’N. Hence the sun’s Az at mer pass must be N and so the true altitude is named N.
Name MZD opposite to that of true altitude.
If MZD and dec are of same name, add and retain name. If of opposite names, subtract the smaller one from the larger one and retain the name of the larger one.
In many cases, you can decide whether MZD and dec are to added or subtracted by knowing that the answer must be very close to the DR lat. However, if either MZD or declination is very small, both addition and subtraction would give results reasonably close to DR lat and the unwary navigator would be puzzled as to what to do.
In some cases, you are not given the DR lat but then you would be given an indication of the Az at mer pass. By using the above naming system, you can get the solution easily and quickly.

Intercept

Intercept star

On 23^rd Aug 2008 in DR 34° 31’S 003° 30’W, at about 1800 hours at ship, the sextant altitude of the star SPICA was 45° 38.7’, “GMT 18h 17m 19s”. If HE was 11m & IE was 2.1’ on the arc, find the intercept and the direction of the LOP.

Step 1: Obtain the LHA & dec for the GMT (See Nautical Almanac below)

From the nautical almanac (NA), obtain the GHA & dec star, and thence the LHA star as shown in normal font in the table here. For easy reference, insert the given lat in the empty cell below the dec.

GHA $\gamma$ 23d 18h	242° 23.9’	dec $\star$	11° 12.4’S
Incr 17m 19s	004° 20.5’	lat	34° 31’S
GHA $\gamma$ for GMT	246° 44.4’
Long(W)	(-) 003° 30.0’
LHA $\gamma$	243° 14.4’
SHA $\star$	158° 35.3’
LHA $\star$	041° 49.7’
P	041° 49.7’

The cells in bold font are needed to enter tables & calculate the CZD & later, the Az.

Note: P is the angle inside the PZX triangle. If the LHA is between 0 &180°, then P = LHA. If LHA is more than 180°, then P = 360° - LHA.

Step 2: Apply intercept formula & obtain CZD.

Cos CZD = Cos P $\cdot$ Cos lat $\cdot$ Cos dec $\pm$ Sin lat $\cdot$ Sin dec

Note: If Lat and dec are same name (+), contrary names (-). Hence in the formula, (+) has been put above the (-). In this case, lat & dec have same names so sign in formula becomes PLUS.

Cos CZD = (Cos 41° 49.7’ $\cdot$ Cos 34° 31’ $\cdot$ Cos 11° 12.4’) + (Sin 34° 31’ $\cdot$ Sin 11° 12.4’) = 0.71239.

CZD = 44° 34.2’

Step 3: Correct sext alt & obtain TZD

Sext alt	45° 38.7’
IE (on )	(-) 02.1’
Obs alt	45° 36.6’
Dip (11m)	(-) 05.8’
App alt	45° 30.8’
Tot Corrn	(-) 01.0’
T alt	45° 29.8’
TZD	44° 30.2’

Step 4: Calculate the intercept

TZD	44° 30.2’
CZD	44° 34.2’
Intercept	04.0’	TOWARDS

Step 5: Calculate Az & thence the direction of LOP

By calculator or from ABC Tables:

A	0.768N
B	0.297S
C	0.471N
Az	N 68.8° W
Az	291.2°(T)
LOP	021.2-201.2°(T)

Step 6: Answer with a rough diagram

DR 34° 31’S 003° 30’W; Int 4.0M Towards, Az 291.2°(T); LOP 021.2-201.2°(T)

Longitude by chronometer

On 23^rd Aug 2008 PM at ship in DR 34° 31’S 003° 30’W, the sextant altitude of the star SPICA was 45° 27.2’, “GMT 18h 17m 19s”. If HE was 11m and IE was 2.1’ on the arc, calculate the direction of the LOP and the longitude where it cuts the DR latitude.

Step 1: Obtain the GHA & dec for the GMT

From the nautical almanac (NA), obtain the GHA & dec of star, and thence the LHA star as shown in normal font in the table here.

For easy reference, insert the given lat in the empty cell below the declination. (Use page of almanac under intercept star).

GHA $\gamma$ 23d 18h	242° 23.9’	dec $\star$	11° 12.4’S
Incr 17m 19s	004° 20.5’	lat	34° 31’S
GHA $\gamma$ for GMT	246° 44.4’
SHA $\star$	158° 35.3’
GHA $\star$	045° 19.7’

		See note below

Note: LHA $\star$ is needed here only to name the Azimuth as E or W in the quadrantal notation. Before mer pass, LHA $\star$ would be between 180 & 360° and Azimuth would be named E. After mer pass, LHA $\star$ would be between 000 & 180° and Azimuth would be named W.

The cells in bold font are needed to enter tables & calculate the LHA & later, the Az.

Step 2: Correct sext alt & obtain true alt

Sext alt	45° 38.7’
IE (on)	(-) 02.1’
Obs alt	45° 36.6’
Dip (11m)	(-) 05.8’
App alt	45° 30.8’
Tot Corrn	(-) 01.0’
T alt	45° 29.8’

Step 3: Apply ‘long by chron’ formula & obtain LHA.

$\cos \text{P} = \frac{\sin \ \text{T} \ \text{alt} \ \pm \sin \ \text{lat} \cdot \sin \ \text{dec}}{\cos \ \text{lat} \cdot \cos \text{dec}}$

Note: If Lat & dec are same name (-), contrary names (+). Hence in the formula, (-) has been put above the (+). In this case, lat and dec have same names so sign in formula becomes MINUS.

$\cos \text{P} = \frac{\sin \ 45^{\circ} \ 29.8' - \sin \ 34^{\circ} \ 31' \cdot \text{sin} \ 11^{\circ} \ 12.4'}{\cos \ 34^{\circ} \ 31' \cdot \cos \ 11^{\circ} \ 12.4'} = 0.74616$

P = $41.741^\circ$ = 41° 44.5'. Sight is after mer pass so LHA = P

Step 4: Calculate the obs long

LHA	041° 44.5’
GHA	045° 19.7’
Obs long	003°35.2’W

Step 5: Calculate Az & thence the direction of LOP

By calculator or from ABC Tables:

A	0.768N
B	0.297S
C	0.471N
Az	N 68.8° W
Az	291.2°(T)
LOP	021.2 - 201.2°(T)

Step 6: Answer with a rough diagram

DR lat 34° 31’S Obs long 003°35.2’W; Az 291.2°(T); LOP 021.2 - 201.2°(T)

Ex-Meridian

Ex meridian altitude star

On 2^nd March 2008, PM at ship in DR 16° 12’N 92° 10’E, the sextant altitude of the star CAPELLA near the meridian was 60° 29.4’ “GMT 12h 32m 18s”. HE was 48m and IE was 2.0’ off the arc, find the direction of the LOP and a position through which it passes.

Step 1: Obtain the LHA & dec for the GMT (See Nautical Almanac below)

From the nautical almanac (NA), obtain the GHA & dec star & then calculate the LHA star as shown in the table here.

For easy reference, insert the given DR at in the empty cell below the declination.

GHA $\gamma$ 02d 12h	340° 39.0’	dec $\star$	46° 00.6’N
Incr 32m 18s	008° 05.8’	lat	16° 12’N
GHA $\gamma$ for GMT	348° 44.8’
SHA $\star$	280° 40.3’
GHA $\star$	269° 25.1’
Long (E)	(+) 92° 10.0’
LHA $\star$	001° 35.1’
P	001° 35.1’

The cells in bold font are needed to calculate the Obs lat and the Az.

Step 2: Correct sext alt & obtain TZD

Sext alt	60° 29.4’
IE (off)	(+) 02.0’
Obs alt	60° 31.4’
Dip (48m)	(-) 12.2’
App alt	60° 19.2’
Tot Corrn	(-) 00.6’
T alt	60° 18.6’N
TZD	29° 41.4’S

Step 3: By calculator - Apply ‘ex-meridian’ formula & obtain MZD

Cos MZD = Cos TZD + [(1 - Cos P) . Cos DR lat . Cos dec]

Cos MZD = Cos 29° 41.4’ + [(1-Cos 001° 35.1’) . Cos 16° 12’ . Cos 46° 00.6’] = 0.86897.

MZD = 29.66° = 29° 39.6’

Step 4: By Nautical Tables - Obtain MZD

From Table I - Value A = 2.63

From Table II, First Correction	01.7’
From Table III, Second Correction	(-) 0.0’
Reduction to the TZD observed	01.7’

TZD	29° 41.4’
Reduction	01.7’
MZD	29° 39.7’

Step 5: Calculate the obs lat

MZD	29° 39.6’S
Dec	46° 00.6’N
Obs lat	16° 21.0’N

Step 6: Calculate Az & thence the direction of LOP

By calculator or from ABC Tables:

A	10.500S
B	37.451N
C	26.951N
Az	N2.0°W
Az	358°(T)
LOP	088 – 268°(T)

Step 7: Answer with a rough diagram

Obs lat 16° 21.0’N DR long 92° 10’E; Az 358°(T)); LOP 088 – 268°(T)

Position Lines by use of Moon

Longitude correction

Longitude correction for time of meridian passage of the moon

Each event of the moon – rising, culminating & setting – occurs about 40 to 50 minutes later every consecutive day. That is: for every 360° of rotation of the earth, there is a delay in the time of occurrence by about 40 to 50 minutes. So for part rotation, equal to longitude, a correction has to be applied to the time of the occurrence of each event tabulated in the Nautical Almanac. This correction is called longitude correction. The computation & use of the longitude correction is illustrated in the following worked example.

Time of meridian passage

Precise time of mer pass moon

Sometimes, it may be needed to find the precise GMT of meridian passage. Note the word precise or exact.

Calculation of precise time of mer pass

At mer pass, LHA = 360°

Hence at mer pass, GHA + E (or - W) long = LHA = 360°

In the worked example, GHA + 75° 00’ = 360° or GHA = 285° 00.0’

Hence, at mer pass, GHA moon =	285° 00.0’
From almanac, for GMT April 28d 00h, GHA =	272° 19.5’
53m 07s $\leftarrow$ Increment $\leftarrow$	012° 40.5’

Precise time of mer pass: April 28d 00h 53m 07s

Latitude by meridian altitude

On 28^th April 2008 in DR 25° 20’N 75°00’E, the sextant meridian altitude of the Moon’s LL was 42° 05.8’. If IE was 1.5’ off the arc and HE was 25m, calculate the latitude and LOP.

Notes

The LMT of mer pass of the Moon given in the almanac is approximate only and requires correction for the longitude of the observer.

Step 1: Find the daily difference of mer pass moon (See Nautical Almanac below)

Open the Nautical Almanac to the pages containing given date. From the bottom right corner of the right hand side page, extract the approximate LMT of the upper mer pass of the moon for the given date. Compare it with the earlier day if longitude is east and later day if longitude is west and obtain the daily difference.

Approx LMT mer pass moon	Apr 28d 06h 02m
Long E, earlier mer pass moon	Apr 27d 05h 14m
Daily difference	00h 48m = 48m

Step 2: Obtain longitude correction

$\text{Longitude correction} = \frac{\text{longitude} \times \text{daily difference}}{360} = \frac{75 \times 48}{360} = 10m$

Step 3: Compute the GMT of mer pass moon

Approx LMT mer pass Moon	Apr 28d 06h 02m
Longitude correction E is minus	(-) 10m
Corrected LMT mer pas moon	Apr 28d 05h 52m
LIT (E)	(-) 05h 00m
GMT mer pass Moon	Apr 28d 00h 52m

Step 4: Obtain the declination for time of mer pass

dec Apr 28d 00h	21º 49.0’S
d (-9.4) 52m	(-) 08.2’
dec for GMT	21º 40.8’S

Step 5: Calculate obs lat & state direction of LOP

Sext Alt	42° 05.8’
IE (off)	(+) 01.5’
Obs alt	42° 07.3’
Dip (25m)	(-) 08.8’
App alt	41° 58.5’
Main corrn	(+) 52.4’
HP (55.5)	(+) 02.8’
T altitude	42° 53.7’S
MZD	47° 06.3’N
dec	21º 40.8’S
Latitude	25° 25.5’N
LOP	E - W

Notes repeated for quick reference from previous lessons

Name True alt N or S - same as azimuth at mer pass.
Name MZD opposite to that of true altitude.
If MZD and dec are of same name, add and retain name. If of opposite names, subtract the smaller one from the larger one & retain the name of the larger one.
In many cases, you can decide whether MZD and dec are to added or subtracted by knowing that the answer must be close to the DR lat. However, if either MZD or dec is very small, both addition & subtraction would give results reasonably close to DR lat and the unwary navigator would be puzzled as to what to do.
In some cases, you are not given the DR lat but then you would be given the Az at mer pass. By the above naming system, you can get the solution easily.

Intercept

Intercept Moon

On 17^th Jan 2008 PM at ship in DR 34° 56’N 093° 30’W, the sextant altitude of the Moon’s UL was 48° 15.4’, “GMT 22h 47m 41s”. If HE was 16m and IE was 2.8’ on the arc, calculate the intercept and the direction of the LOP.

Step 1: Obtain the LHA & dec for the GMT (See page of Nautical Almanac earlier under intercept Moon).

From the nautical almanac (NA), obtain the GHA & dec moon, and thence calculate the LHA moon as shown in normal font in the table here.

For easy reference, insert the given lat in the empty cell below the dec.

GHA Jan 17d 22h	035° 33.1’	dec	24° 01.5’N
Incr 47m 41s	011° 22.7’	d(+9.1)	(+) 7.2’
v(+5.4)	(+) 04.3’	dec	24° 08.7’N
GHA for GMT	047° 00.1’	lat	34° 56’N
Long(W)	(-)093° 30.0’
LHA	313° 30.1’
P	046° 29.9’

Step 2: Apply intercept formula & obtain CZD.

Cos CZD = Cos P $\cdot$ Cos lat $\cdot$ Cos dec $\pm$ Sin lat $\cdot$ Sin dec

Note: If Lat and dec are same name (+), contrary names (-). Hence in the formula, (+) has been put above the (-). In this case, lat & dec have same names so sign in formula becomes PLUS.

Cos CZD = (Cos 046° 29.9’ . Cos 34°56’ . Cos 24° 08.7’) + (Sin 34°56’ . Sin 24° 08.7’) = 0.74921.

CZD = 41.48° = 41° 28.7’

Step 3: Correct sext alt & obtain TZD

Sext Alt	48° 15.4’
IE (on)	(-) 02.8’
Obs alt	48° 12.6’
Dip (16m)	(-) 07.0’
App alt	48° 05.6’
Main corrn	(+) 48.3’
HP (59.7)	(+) 04.4’
UL only	(-) 30.0’
T altitude	48° 28.3’
TZD	41° 31.7

Step 4: Calculate the intercept

TZD	41° 31.7’
CZD	41° 28.7’
Intercept	03.0’	Away

Step 5: Calculate Az & thence the direction of LOP

By calculator or from ABC Tables:

A	0.663S
B	0.618N
C	0.045S
Az	S87.9°E
Az	092.1°(T)
LOP	002.1-182.1°(T)

Step 6: Answer with a rough diagram

DR 34° 56’N 093° 30’W; Int 3.0M Away Az 092.1°(T); LOP 002.1-182.1°(T)

Longitude by chronometer

Long by chron Moon

On 17^thJan 2008 PM at ship in DR 34° 56’N 093° 30’W, the sextant altitude of the Moon’s UL was 48° 15.4’, “GMT 22h 47m 41s”. If HE was 16m and IE was 2.8’ on the arc, calculate the LOP and the longitude where it cuts the DR latitude.

Step 1: Obtain the LHA & dec for the GMT

(See page of Nautical Almanac earlier under intercept Moon).

From the nautical almanac (NA), obtain the GHA & dec moon, and thence calculate the LHA moon as shown in normal font in the table here.

For easy reference, insert the given lat in the empty cell below the dec.

GHA Jan 17d 22h	035° 33.1’	dec	24° 01.5’N
Incr 47m 41s	011° 22.7’	d(+9.1)	(+) 7.2’
v(+5.4)	(+) 04.3’	dec	24° 08.7’N
GHA for GMT	047° 00.1’	lat	34° 56’N

Step 2: Calculate the true alt

Sext Alt	48° 15.4’
IE (on)	(-) 02.8’
Obs alt	48° 12.6’
Dip (16m)	(-) 07.0’
App alt	48° 05.6’
Main corrn	(+) 48.3’
HP (59.7)	(+) 04.4’
UL only	(-) 30.0’
T altitude	48° 28.3’

Step 3: Apply long by chron formula & obtain LHA.

$\text{Cos\ P} =\frac{\text{Sin\ T \ alt} \pm \text{Sin\ lat} \cdot \text{Sin\ dec}}{\text{Cos\ lat} \cdot \text{ Cos\ dec}}$

Note: If Lat and dec are same name (-), contrary names (+). Hence in the formula, the (-) sign has been put above the (+). In this case, lat & dec have same names so sign in formula becomes MINUS.

$\text{Cos\ P}=\frac{\text{Sin}\ 48^\circ 28.3' - \text{Sin}\ 34^\circ 56' \cdot \text{Sin}\ 24^\circ \ 08.7'}{\text{Cos}\ 34^\circ 56' \cdot \text{Cos}\ 24^\circ\ 08.7'}=0.68761$

P = 46.559°. Since LMT of sight is 17d 16h & approx LMT of mer pass from the Nautical Almanac is 19h 32m, the LHA = 360° - 46.559° = 313.441° = 313° 26.5’

Step 4: Calculate the obs long

LHA	313° 26.5’
GHA	047° 00.1’
Obs long	093° 33.6’W

Step 5: Calculate Az & thence the direction of LOP

By calculator or from ABC Tables:

A	0.661S
B	0.617N
C	0.044S
Az	S87.9°E
Az	092.1°(T)
LOP	002.1-182.1°(T)

Step 6: Answer with a rough diagram

DR lat 34° 56’N Obs long 093° 33.6’W; Az 092.1°(T); LOP 002.1-182.1°(T)

Ex-Meridian:

Ex-meridian moon

On 28^th April 2008 AM at ship in DR 25° 20’N 75° 00’E, the sextant altitude of the Moon’s UL was 41° 49.0’ at 01h 32m 40s GMT by GPS clock. If IE was 2.5’ on the arc and HE was 15m, calculate the LOP and the latitude where it crosses the DR longitude.

Step 1: Obtain the LHA & dec for the GMT (See Nautical Almanac below)

From the nautical almanac (NA), obtain the GHA & dec moon & then calculate the LHA moon as shown in the table here.

For easy reference, insert the given DR at in the empty cell below the declination.

GHA Moon Apr 28d 01h	286° 50.2’	dec	21° 39.6’S
Incr 32m 40s	007° 47.7’	d(-9.4)	-05.1’
v(+11.8)	06.4’	dec	21° 34.5’S
GHA for GMT	294° 44.3’	lat	25° 20’N
Long (E)	(+) 75° 00.0’
LHA	009° 44.3’
P	009° 44.3’

Step 2: Correct sext alt & obtain TZD

Sext alt	41° 49.0’
IE (on)	(-) 02.5’
Obs alt	41° 46.5’
Dip (15m)	(-) 06.8’
App alt	41° 39.7’
Main Corrn	(+) 52.6’
HP (55.5)	(+) 02.6’
Only for UL	(-) 30.0’
T alt	42° 04.9’S
TZD	47° 55.1’N

Step 3: By calculator - Apply ‘ex-meridian’ formula & obtain MZD

Cos MZD = Cos TZD + [(1 - Cos P) . Cos DR lat . Cos dec]

Cos MZD = Cos 47° 55.1’ + [(1-Cos 009° 44.3’) . Cos 25° 20’ . Cos 21° 34.5’] = 0.68230.

MZD = 46.976° = 46° 58.6’

By Nautical Tables - Obtain MZD

From Table I - Value A = 2.32

From Table II, First Correction	56.8’
From Table III, Second Correction	(-) 0.4’
Reduction to the TZD observed	56.4’

TZD	47° 55.1’
Reduction	00° 56.4’
MZD	46° 58.7’

Step 4: Calculate the obs lat

MZD	46° 58.6’N
Dec	21° 34.5’S
Obs lat	25° 24.1’N

Step 5; Calculate Az & thence the direction of LOP

By calculator or from ABC Tables:

A	2.767 S
B	2.337 S
C	5.104 S
Az	S 012.2° W
W	192.2°
LOP	102.2° / 282.2°

Step 6: Answer with a rough diagram

Obs lat 25° 24.1’N DR long 75° 00’E; Az 012.2°(T); LOP 102.2 – 282.2°(T)

Position Lines by use of Planet

Time of meridian passage:

Note: The LMT of mer pass planet, given in the Nautical Almanac, is for the middle day of the three days listed in that page, in this case for the April 28th. The exact time will then be only a few seconds away from this time. Had the required date been the upper date or the lower one, 27th nor 29th April in this case, the precise time may differ by about 3 or 4 minutes. However, since the change of declination per hour (value of d) for planets is very small, no appreciable error creeps in by taking the LMT directly as tabulated in the almanac.

Latitude by meridian altitude:

Lat by meridian altitude of Planets

On 28^th April 2008, AM at ship in DR 25° 20’N 75°00’E, the sextant meridian altitude of JUPITER was 43° 04.5’. If IE was 1.5’ off the arc and HE was 25m, find the latitude and LOP.

Step 1: Find the approx GMT mer pass

Open the Nautical Almanac to the pages containing given date. (See Nautical Almanac below)

Look at the bottom right corner of the left hand side page.

Extract the approx LMT of the mer pass of the planet.

Approx LMT mer pass Jupiter	April 28^th 05h 10m 00s
LIT (E)	(-) 05h 00m 00s
Approx GMT mer pass Jupiter	April 28^th 00h 10m 00s

Step 2: Obtain the declination for time of mer pass

dec	21º 38.3’S
d 0.0	0.0’
dec	21º 38.3’S

Step 3: Calculate obs lat & state direction of LOP

Sext Alt	43° 04.5’
IE (off)	(+) 01.5’
Obs alt	43° 06.0’
Dip (25m)	(-) 08.8’
App alt	42° 57.2’
Tot corrn	(-) 01.0’
Addl corrn	-
T alt	42° 56.2’S
MZD	47° 03.8’N
Dec	21º 38.3’S
Latitude	25° 25.5’N
LOP	E - W

Notes repeated for quick reference from previous lessons

Name True alt N or S - same as azimuth at mer pass.
Name MZD opposite to that of true altitude.
If MZD and dec are of same name, add and retain name. If of opposite names, subtract the smaller one from the larger one & retain the name of the larger one.
In many cases, you can decide whether MZD and dec are to added or subtracted by knowing that the answer must be close to the DR lat. However, if either MZD or dec is very small, both addition & subtraction would give results reasonably close to DR lat and the unwary navigator would be puzzled as to what to do.
In some cases, you are not given the DR lat but then you would be given the Az at mer pass. By the above naming system, you can get the solution easily.

Intercept

Intercept planet

On 17^th Jan 2008 AM at ship in DR 34° 56’N 093° 30’W, the sextant altitude of the VENUS was 18° 06.4’, “GMT 12h 44m 19s”. If HE was 16m and IE was 2.8’ on the arc, calculate the intercept and the direction of the LOP.

Step1: Obtain the LHA & dec for the GMT (See attached page of almanac)

From the nautical almanac (NA), obtain the GHA & dec planet, and thence calculate the LHA planet as shown in normal font in the table here.

For easy reference, insert the given lat in the empty cell below the dec.

GHA 17d 12h	035° 19.4’	dec	21° 45.0’S
Incr 44m 19s	011° 04.8’	d(+0.3)	(+) 0.2’
v(-0.8)	(-) 0.6’	dec	21° 45.2’S
GHA for GMT	046° 23.6’	lat	34° 56’N
Long(W)	(-)093° 30.0’
LHA	312° 53.6’
P	047° 06.4’

The cells in bold font are needed to enter tables and calculate the CZD and later, the Az.

Step 2:Apply intercept formula & obtain CZD.

Cos CZD = Cos P $\cdot$ Cos lat $\cdot$ Cos dec $\pm$ Sin lat $\cdot$ Sin dec

Note: If Lat and dec are same name (+), contrary names (-). Hence in the formula, (+) has been put above the (-). In this case, lat & dec have contrary names so sign in formula becomes MINUS.

Cos CZD = (Cos 047°06.4’ . Cos 34°56’ . Cos 21°45.2’) - (Sin 34°56’ . Sin 21°45.2’) = 0.30604.

CZD = 72.18° = 72° 10.8’

Step 3: Correct sext alt & obtain TZD

Sext alt	18° 06.4’
IE (on)	(-) 02.8’
Obs alt	18° 03.6’
Dip (16m)	(-) 07.0’
App alt	17° 56.6’
Tot Corrn	(-) 03.0’
T alt	17° 53.6’
TZD	72° 06.4’

Step 4: Calculate the intercept

TZD	72° 06.4’
CZD	72° 10.8’
Intercept	04.4’	TOWARDS

Step 5: Calculate Az & thence the direction of LOP

By calculator or from ABC Tables:

A	0.649S
B	0.545S
C	1.194S
Az	S 45.6° E
Az	134.4°(T)
LOP	044.4-224.4°(T)

Step 6: Answer with a rough diagram

DR 34°56’N 093° 30’W; Int 4.4M Towards Az 134.4°(T); LOP 044.4-224.4°(T)

Longitude by chronometer planet

On 17^th Jan 2008 AM at ship in DR 34° 56’N 093° 30’W, the sextant altitude of the VENUS was 18° 06.4’, “GMT 12h 44m 19s”. If HE was 16m and IE was 2.8’ on the arc, calculate the direction of the LOP and the longitude where it cuts the DR latitude.

Step 1: Obtain the GHA & dec for the GMT

From the nautical almanac (NA), obtain the GHA & dec planet, (See page of almanac under Intercept planet).

For easy reference, insert the given lat in the empty cell below the dec.

GHA 17d 12h	035° 19.4’	dec	21° 45.0’S
Incr 44m 19s	011° 04.8’	d(+0.3)	(+) 0.2’
v(-0.8)	(-) 0.6’	dec	21° 45.2’S
GHA for GMT	046° 23.6’	lat	34° 56’N

The cells in bold font are needed for use in further calculations.

Step 2: Correct sext alt & obtain Talt

Sext alt	18° 06.4’
IE (on)	(-) 02.8’
Obs alt	18° 03.6’
Dip (16m)	(-) 07.0’
App alt	17° 56.6’
Tot Corrn	(-) 03.0’
T alt	17° 53.6’

Step 3: Apply long by chron formula & obtain LHA.

$\cos \text{P} = \frac{\sin \ \text{T} \ \text{alt} \ \pm \sin \ \text{lat} \ . \ \sin \ \text{dec}}{\cos \ \text{lat} \ . \ \cos \text{dec}}$

Note: If Lat and dec are same name (-), contrary names (+). Hence in the formula, the (-) sign has been put above the (+). In this case, lat & dec have contrary names so sign in formula becomes PLUS.

$\cos \text{P} = \frac{\sin \ 17^{\circ} \ 53.6' + \sin \ 34^{\circ} \ 56' \ . \ \sin \ 21^{\circ} \ 45.2'}{\cos \ 34^{\circ} \ 56' \ . \ \cos \ 21^{\circ} \ 45.2'}= 0.682217811$

P = $46.983^\circ$ . Since LMT of sight is 17d 06h & approx LMT of mer pass from the Nautical Almanac is 09h 39m, the LHA = $360^\circ$ - $46.983^\circ$ = $313.017^\circ$ =

$313^\circ \ 01.0'$

Step 4: Calculate the obs long

LHA	313° 01.0'
GHA	046° 23.6'
Obs long	093° 22.6' W

Step 5: Calculate Az & thence the direction of LOP

By calculator or from ABC Tables:

A	0.652 S
B	0.546 S
C	1.198 S
Az	S 45.5° E
Az	134.5°
LOP	044.5° / 224.5°

Step 6: Answer with a rough diagram

DR 34° 56’N 093° 22.6' W; Az 134.5° LOP 044.5° - 224.5°

Ex-Meridian

Ex meridian altitude planet

On 6th March 2008, PM at ship in in DR 20° 04’S 40° 37’E, the sextant altitude of MARS near the meridian was 43° 48.9’ at “GMT 16h 03m 42s”, IE was 1.4’ on the arc and HE was 18m, find the LOP and a position through which it passes.

Step1: Obtain the LHA & dec for the GMT (See Nautical Almanac below)

From the nautical almanac (NA), obtain the GHA & dec planet & then calculate the LHA planet as shown in the table here.

For easy reference, insert the given DR at in the empty cell below the declination.

GHA MARS Mar 06d 16h	313° 56.7’	dec	26° 09.6’N
Incr 03m 42s	000° 55.5’	d(0.0)	00.0’
V(+1.6)	00.1’	dec	26° 09.6’N
GHA for GMT	314° 52.3’	lat	20° 04’S
Long (E)	(+)40° 37.0’
LHA	355° 29.3’
P	004° 30.7’

The cells in bold font are needed to calculate the Obs lat and the Az.

Step2: Correct sext alt & obtain TZD

Sext alt	43° 48.9’
IE (on)	(-) 01.4’
Obs alt	43° 47.5’
Dip (18m)	(-) 07.5’
App alt	43° 40.0’
Tot Corrn	(-) 01.0’
Addl corrn	(+) 00.1’
T alt	43° 39.1’N
TZD	46° 20.9’S

Step 3: By calculator - Apply ‘ex-meridian’ formula & obtain MZD

Cos MZD = Cos TZD + [(1 - Cos P). Cos DR lat. Cos dec]

Cos MZD = Cos 46° 20.9’ + [(1-Cos 004° 30.7’) . Cos 20° 04’. Cos 26° 09.6’] = 0.69288.

MZD = 46.141° = 46° 08.4’

By Nautical Tables - Obtain MZD

From Table I - Value A = 2.31

From Table II, First Correction	12.5’
From Table III, Second Correction	(-) 0.0’
Reduction to the TZD observed	12.5’

TZD	46° 20.9’
Reduction	12.5’
MZD	46° 08.4’

Step 4: Calculate the obs lat

MZD	46° 08.4’S
Dec	26° 09.6’N
Obs lat	19° 58.8’S

Step 5: Calculate Az & thence the direction of LOP

By calculator or from ABC Tables:

A	4.608N
B	6.244N
C	10.852N
Az	N5.6°E
Az	005.6°(T)
LOP	095.6 – 275.6°(T)

Step 6: Answer with a rough diagram

Obs lat 19° 58.8’S DR long 40° 37’E; Az 005.6°(T); LOP 095.6 – 275.6°(T)

Latitude by Meridian Altitude

Introduction

Latitude by meridian altitude is the simplest celestial observation to capture and calculate. Meridian passage (commonly known as Mer Pass) happens when the object passes the observer's meridian, i.e. passes through south or north. This is usually done with the sun for a noon latitude though it can be done with any other celestial object also.

How to capture Meridian Passage using a Sextant?

First calculate the estimated time of the meridian altitude of the sun using the nautical almanac. A few minutes before this happens, start observing the altitude of the object with a sextant. The altitude will be increasing. Keep on adjusting the sextant so as to keep the reflected image of the object on the horizon. As the object passes through the meridian, a maximum altitude is observed. The sun will not rise any more. As soon as the sun starts going down, record the reading of the vertical angle.

Correct the sextant altitude thus observed for index error, semi-diameter, parallax and refraction.

Use the rational horizon diagram in case of a doubt. They are drawn differently for north and south latitudes.

The following media explains about the Meridian Passage

In the media, the second method shown for calculating the meridian passage time of a body, by using LHA of the body as $360^{\circ}$ , is more accurate, because it gives the answer in seconds. However in practice this method is not necessary because the value of Declination is obtained from the almanac by entering only the date, hours and minutes of GMT, and seconds are not required.

Ex-Meridian of Sun

Latitude by Ex-Meridian Altitude

It may not always be possible to obtain the altitude of a body, when on the meridian, owing to clouds, hazy horizon, rain or other causes.

This method is used to find only the latitude of the observer when the heavenly body is near the observer’s meridian and where the local hour angle is small. It can be computed either by Ex-Meridian tables, or by the formula.

In some places we cannot use this method to find the altitude of a body when on the meridian, due to clouds, unclear horizon, rain or other reasons. This method is used only to find the observer’s latitude when the heavenly body is close to the observer’s meridian and where the LHA is small.

Ex-Meridian table IV, in Norie’s tables, provides the limitations of hour angle such that at what range the angle must be obtained and worked by the Ex-Meridian method, without significant error. These tables are made according to the observer’s DR latitude and the body’s declination, for both same names and opposite names. The rule must be followed that the hour angle in minutes of time should be less than the approximate meridian zenith distance of the body in degrees.

The zenith distance of the body is less in quantity, if it is on the meridian and the zenith distance is little larger if it is near the meridian (before or after meridian passage).

Ex-Meridian Tables

Table I: Enter latitude and declination to find quantity of “A”. Value A can also be calculated by formula:

$A = \frac{1.9635 \times \cos\ Lat. \times \cos\ Dec.}{\sin(Lat. \pm Dec.)}$

For denominator, the latitude and declination are:

Subtracted if on upper meridian and same name
Added if on upper meridian and different name
Subtracted on lower meridian.

Table IV: The values of “A” and LHA “t” must be entered into Table IV to check whether the hour angle is within the limit of the table. From Table IV, find out the time limit for an hour angle; it should be within the limit to travel.

Table II: “A” and LHA must be entered to get first correction.

Table III: To obtain the 2^nd correction, first correction and True Altitude must be entered.

1^st Correction - 2^nd Correction = Reduction

TZD - Reduction = MZD

Procedure for solving Ex-Meridian problem

Calculation of and

Calculate these values as explained in Intercept problem.

Calculation of Meridian Zenith Distance and Latitude

Use formula: $\cos MZD = \cos TZD + {(1- \cos P) \times Cos \ Lat. \times \cos Decl.}$
Lat. $= MZD \pm Decl.$
Sign is decided by drawing the figure on plane of Rational horizon and locating the points and on it. is Lat., is and is Declination.

Calculation of True Azimuth and Position line

Calculate these values as explained in Long. by Chron. problem.

Example:

On 1st March 1992 at $DR \ Lat. 29^\circ 24' N$ Long. $165^\circ 00' W,$ the sextant altitude of the Sun’s lower limb near the observer’s meridian was $53^\circ 24'.$ GMT 12h 15m 56s. Index error 1.2'

off the arc. Height of eye was 14.6m. Find the direction of position line and a position through which it passed.

$GHA = 161^\circ 56.9' + 3^\circ 59' = 165^\circ 55.9' ; LHA = 165^\circ 55.9' - 165^\circ = 000^\circ 55.9'$

$Decl. = 7^\circ 12.3' - 0.3' = 07^\circ 12.0' S$

$T. Alt. = 53^\circ 24' + 1.2'$ (I.E. off the arc) - 6.7'

(Dip for H.E. 14.6m.) + 15.5'

(T.C. for L.L.)

$= 53^\circ 34' ; TZD = 36^\circ 26'$

$\cos MZD = \cos TZD + {(1 - \cos P) \times \cos Lat. \times \cos Decl.}$

$= \cos 36^\circ 26' + {(1 - \cos 0^\circ 55.9') \times \cos 29^\circ 24' \times \cos 7^\circ 12'}$

$MZD = 36^\circ 25.3' ;Lat. = 36^\circ 25.3' - 7^\circ 12' = 29^\circ 13.3' N$ $\Tan$

$A = \tan Lat. \div \tan P = \tan 29^\circ 13.3 \div \tan 0^\circ 55.9' = 34.398 S$

$B = \tan Decl. \div \sin P = \tan 7^\circ 12' \div \sin 0^\circ 55.9' = 7.769 S$

$\tan Az. = 1 \div (C \times \cos Lat.) = 1 \div (42.167 \times \cos 29^\circ 13.3') ; T. Az. = S \ 001.6^\circ$ $W = 181.6^\circ$

$PL = 091.6^\circ / 271.6^\circ$

Long by Chron and Intercept Methods

Latitude and longitude are measures of position vital for navigation. In the eighteenth century, there was no way to measure it and ships were meeting an adverse fate resulting in a huge loss of life and cargo.

The measurement of longitude was a problem that came into sharp focus as people started making transoceanic voyages. Determining latitude was relatively easy as this could be calculated using the altitude of the sun at noon with the aid of the almanac giving the sun's declination for the day. For longitude, early ocean navigators had to rely on dead reckoning. This DR was inaccurate on long voyages when out of sight of land leading to ships getting lost at sea. These ships sometimes came too close to land, leading to grounding.

In 1707 there was a death of 1550 British sailors when Admiral Shovell's fleet foundered in the Isles of Scilly in severe weather. The main cause of the disaster was the inability of the fleet to determine its position. Finding an adequate solution to determining longitude was of paramount importance. This led to the invention of the marine chronometer, a clock which measures time with great accuracy.

Use the DR latitude and calculate the observed longitude. The position line passes through the intersection of the

latitude and observed longitude. Combine the position line obtained through this method with another position line, using celestial or terrestrial means, to obtain a fix or the position of the observer on the earth.

The media below explains the Longitude by Chronometer method with a solved example

The Longitude by Chronometer method of calculating sights is more accurate when the azimuth of the object is towards east or west. The azimuth depending upon the hemisphere of the observer, changes towards the south or north. Hence, the cross of the position line with the assumed latitude becomes more and more oblique and the position less accurate. For this reason "Long by Chron" (as it is popularly called) is a less versatile method of calculating sights than the intercept method which can be used for all azimuths and is explained below.

Procedure for solving Longitude by Chronometer problem

Calculation of $\underline{GHA}$

Using the date and hour of given in the problem, obtain the of Sun, Moon or planets from the main pages of the almanac. For stars obtain the of Aries.
Obtain the increment for min. and sec. of from the Increment tables in the almanac and add it to the obtained above for all bodies.
Obtain the of star from the main pages of the almanac and add it to the of Aries calculated above, to get the final of star.
For Moon and planets, obtain the 'v' value for the respective body, given in the main pages. Use it to obtain the 'v' Corr. from the Increment tables. Add this Corr. to the obtained above to get the final For Venus, subtract the 'v' Corr. if so indicated in the main pages of the almanac.
For Sun the 'v' Corr. is not applicable.

Calculation of Declination

Obtain the of the Sun, Moon and planets for the date and hour of from the main pages of the almanac.
Obtain the 'd' value for the respective body from the main pages, and using it obtain the 'd' Corr. from the Increment tables. If the of the body is increasing every hour then add this 'd' Corr. to the obtained above, otherwise subtract it. This gives the final
For stars the is directly given in the main pages.

Calculation of True Altitude

Measure the Sextant altitude of the body as defined. For Sun and Moon take the measurement from their lower or upper limb.
If Index error (IE) of the sextant is “on the arc” then subtract it from the Sextant altitude to obtain Observed altitude. If the IE is “off the arc” then add it.
Estimate your Height of Eye (HE) from your standing position on the Bridge to the water level outside the ship.
Using the HE obtain the value of Dip from the almanac, and subtract it from the Observed altitude to obtain the Apparent altitude.
For Sun, ascertain the Total correction from the almanac, which depends on the month of the year, the limb being used and the Apparent altitude. This correction is actually a combination of Refraction, Semi-diameter and Parallax corrections. Apply the total correction to the Apparent altitude according to the given sign to get the final True altitude.
For all planets and stars, the Total correction is given in another table in the almanac. Actually this is the Refraction correction, and so it is subtracted from the Apparent altitude to get the True Altitude of stars, Jupiter and Saturn.
For Mars and Venus, an Additional correction is given in the almanac, which is actually the Parallax correction. This is added to the Apparent altitude to obtain their final True altitude.
For Moon the Total correction is obtained from the almanac as follows:

Enter the upper half of the table with the Apparent altitude to obtain the main correction.
Move down the column of the main correction to enter the lower half of the same table.
Obtain the value of Horizontal Parallax (HP) from the main pages of the almanac.
Using the HP and the limb used for measurement of sextant altitude, obtain the HP correction from the lower half of the table.
Add both the main and HP corrections to the Apparent Altitude, but if Upper limb is used for measurement of sextant altitude, then subtract 30'.

In case the atmospheric temperature and/or pressure is very different from the standard temperature of $10^\circ C$ and 1010millibars, then apply another correction obtained from the almanac as indicated in the table.

Calculation of LHA

Use the following formula to calculate $\angle \text{P}$ :
$\cos P = {\sin \ T. Alt. \pm (\sin \ Lat. \times \sin Decl.)} \div (\cos Lat. \times \cos Decl.)$
Sign is –ve if Lat. and have same names. Sign is +ve if they have opposite names.
If body is East of the meridian of observer, then $\text{LHA} = 360^{\circ} - \angle \text{P}$ . If body is West of meridian, then $\text{LHA} = \angle \text{P}$ .
Difference of and gives the Longitude. If then Long. Is West. If then Long. Is East.

Calculation of True Azimuth and Position Line

Use the following formulae to calculate values of and
$A= \tan Lat. \div \tan P ; B = \tan Decl. \div \sin P ; C = A \pm B$
Direction of is opposite to direction of if value of lies between $270^\circ$ and $090^\circ,$ otherwise it is same as Lat.
Direction of is same as direction of
is the algebraic sum of and and direction of is the direction of or whichever is greater.
Use the following formula to calculate $T. Az. --- \tan Az. = 1 \div (C \times \cos Lat.)$
Quadrant of is the direction of and whether the body is East or West of the observer.
Calculate Position line by adding and subtracting $90^\circ$ from

Intercept or Marcq St Hilaire Method

The intercept method or Marcq St Hilaire method is an astronomical method of calculating an observer's position on earth. It yields a line of position on which the observer is located and the intersection of two such lines gives the observer's position. Sights may be taken at short intervals, usually of stars during twilight. It can also be used for a fix from the sun, but the interval between sights will be larger because of the need for a relatively large change in azimuth - of at least thirty degrees - and the lines of position will be crossed by a running fix.

This method solves the PZX triangle to find the zenith distance and azimuth, which are the angular distance between the zenith and the celestial body and the bearing of the celestial body. The calculated zenith distance is compared to the true zenith distance which is obtained from the corrected true altitude of the body. The difference between them is call the intercept and is either toward or away from the body

Procedure for solving Intercept problem

Calculation of $\underline{GHA},$ Declination and True Altitude

Calculate all the values as explained in Long. by Chron. Problem.

Calculation of $\underline{LHA}$

If Long. is West, subtract it from to obtain If Long. Is East, add it to

Calculation of True Zenith Distance

$TZD = 90^\circ - T. Alt.$

Calculation of Calculated Zenith Distance and Intercept

Use formula: $\cos \ CZD = (\cos P \times \cos Lat. \times \cos Decl.) \pm (\sin Lat. \times \sin Decl.)$
Sign is +ve if and have same names. Sign is -ve if they have opposite names.

Intercept $= CZD \sim TZD$
If intercept is said to be “Towards” i.e. it is measured on the from the position towards the direction of the body.
If intercept is said to be “Away” i.e. it is measured on the from the position in a direction opposite to the direction of the body.

Calculation of True Azimuth and Position Line

Calculate all the values as explained in Long. by Chron. Problem.

Example

On 25^th February 1992, in $DR \ 20^\circ \ 04' S \ 090^\circ \ 04' \ W,$ the sextant altitude of Moon’s UL was $52^\circ \ 26.8'$ at GMT 14h 52m 16s. IE 0.6'

off the arc. HE 19m. Calculate the position line and the intercept.

$GHA = 117^\circ \ 43.5' + 12^\circ \ 28.3'$ (Increment) + 8.5'

(v Corr.) $= 130^\circ 20.3'$

$LHA = 130^\circ 20.3' - 90^\circ 4'$ (Long. W) $= 040^\circ 16.3'$

$Decl. = 24^\circ 16.3' + 2'$ ( d Corr.) $= 24^\circ 18.3' S$

$T. Alt. = 52^\circ 26.8' (S. Alt.) + 0.6'$ (IE off the arc) -7.7'

(Dip for HE 19m.) + 45.2'

(Main Corr.) + 2.9'

(2^nd Corr.) - 30'

(3^rd Corr.) $= 52^\circ 37.8'$

$TZD = 90^\circ - 52^\circ 37.8' = 37' 22.2^\circ$

$\cos CZD = (\cos P \times \cos Lat. \times \cos Decl.) \pm (\sin Lat. \times \sin Decl.)$

Sign is +ve because Lat.

and

have same names.

$\cos CZD = (\cos 40^\circ 16.3 \times \cos 20^\circ 04' \times \cos 24^\circ 18.3') + (\sin 20^\circ 04' \times \sin 24^\circ 18.3')$

$CZD = 37^\circ 24.2' ;$ Intercept $= 37^\circ 24.2' - 37^\circ 22.2' = 2'$ (Towards)

$A= \tan 20^\circ 04' \div \tan 40^\circ 16.3' = 0.431 N ; B$

$= \tan 24^\circ 18.3' \div \sin 40^\circ 16.3' = 0.699 S$

$C= 0.431 - 0.699 = 0.268 \ S ; \tan Az. = 1 \div (0.268 \times \cos 20^\circ 04')$

$T. Az. = S \ 75.9^\circ W = 255.9^\circ ; PL = 165.9^\circ/ 345.9^\circ$

Position Lines by use of Pole Star

The Pole star

The Pole star, also called Polaris, is a star of very high declination (over 89°) and hence its azimuth is very nearly North at all times, resulting in a LOP very nearly E-W. By applying special corrections (given in the Nautical Almanac) to the true altitude, you obtain the Obs lat (latitude where the LOP cuts the DR longitude). You obtain the true azimuth also from the same page of the Nautical Almanac. You obtain the direction of the LOP as usual, by applying ±90° to the azimuth. The solution to an altitude of Polaris should specify the Obs lat, the DR long and the direction of the LOP.

Note: The Obs lat obtained through Polaris is not the exact latitude of the ship as the LOP is nearly E-W not necessarily exactly E-W.

Limitations of observations of Polaris

Polaris cannot be seen from the Southern Hemisphere as it will be below the Rational Horizon.
Sextant observations of Polaris are not made in latitudes below about 10°N as heavy refraction would result in inaccuracy.
Sextant observations of Polaris are possible only during twilight when both, Polaris and the horizon, would be visible.

Advantages of observations of Polaris

Identification of Polaris in the sky is very easy.
The altitude of Polaris is very nearly the latitude of the observer.
Calculation of Obs latitude is very simple.
Calculation of true Azimuth to obtain compass error is very simple as compared to the use of ABC tables for other CBs.

Worked example

On 1^st Sept 2008, AM at ship in DR 18° 00’N 178° 11’E, the sextant altitude of the Pole star was 18° 47.4’ “GMT 31 Aug. 17h 22m 26s”. HE = 12.5m. IE = 1.6’ on the arc. Find the direction of the LOP and a position through which to draw it. If the azimuth was 001°(C), and variation was 1.3°E, find the deviation for that compass course.

Step 1: Obtain LHA for GMT (see part page of almanac herein)

From NA, extract values for GHA $\gamma$ , apply long & obtain LHA.

GHA 31d 17h	235° 14.6’
Incr 22m 26s	005° 37.4’
GHA	240° 52.0’
Long E	(+)178° 11.0’
LHA	059° 03.0’

Step 2: Obtain true altitude

Sext alt	18° 47.4’
IE (on)	(-) 01.6’
Obs alt	18° 45.8’
Dip (12.5m)	(-) 06.2’
App alt	18° 39.6’
Tot Corrn	(-) 02.8’
T alt	18° 36.8’

Step 3: Apply corrections & obtain Obs lat

		Corrections from Pole Star Tables in NA (See Nautical Almanac below)
T alt	18° 36.8’	Calculated by you in step 3
a₀	00° 19.2’	Enter with LHA 059° 03.0’ - mental interpolation only
a₁	00° 00.6’	Same column as a₀, against lat – no interpolation
a₂	00° 00.3’	Same column as a₀, against month – no interpolation
Sum	18° 56.9’	Sum of all rows herein
Minus 1°	- 1°	As instructed in Pole Star tables in the Nautical Almanac
Obs lat	17° 56.9’N	Always N as Polaris not visible in south latitude

Step 4: Find true Az, LOP & Deviation of compass

Using same column as step 4, obtain Az against lat to one decimal place only see diagram A. Thence compare with Compass Azimuth & obtain compass error and deviation. LOP is at right angles to Azimuth.

True Az	359.8°(T)
Comp Az	001.0°(C)
Comp Err	1.2°(W)
Variation	1.3°(E)
Deviation	2.5°(W)

Step 5: Answer with a sketch of the LOP

Obs lat 17° 56.9’N DR long 178° 11.0’E; LOP 089.8°- 269.8°; Deviation 2.5°(W).

Important note: Sometimes, the DR lat is not given in the question. Then use the true altitude in place of DR lat to enter tables to get a₁ correction.

Combination of two or more celestial observations

Position based on staggered observations:

Worked example 1 - Fix by staggered CLOPs

In DR 46° 20’N 118° 41’E, an Obs long of the Sun at 0800 was 118° 46’E bearing 150°(T). The ship then steamed 238°(T) at 15 knots. At 1150, lat by mer alt Sun was 45° 51.2’N. Find the ship’s fix at 1150 and the 1200 position for entry in the Bridge Logbook.

Step 1: Identify time, position & direction of CLOP1

Time: 0800 DR lat 46° 20’N Obs long 118° 46’E; LOP1 direction = 150°± 90° = 240°-060°(T).

Step 2:Identify time & position to transfer CLOP1

CLOP 1 at 0800. CLOP 2 at 1150.

Distance = Interval x speed = 3h 50m @ 15 Knots = 57.5M; Co = 238°(T) = S58°W.

From TT, Co S58°W dist 57.5M, d’lat = 30.5’S & dep = 48.8’W

LOP1 0800 DR lat	46° 20.0'N	Obs long	118° 46.0'E
d'lat	0° 30.5'S	d'long*	1° 10.4'W
LOP1 1150 lat	45° 49.5'N	long	117° 35.6'E

*Using m’lat 46.1° and dep of 48.8'W, d’long =70.4’W = 1° 10.4’W

Step 3: Plot both LOPs for 1150 & identify fix

Step 4: Obtain 1150 fix

The fix lat is that obtained by mer alt at 1150. Dep of fix is 3’E of LOP1 1150 long.

Calculate 1150 fix long as shown in the table.

		LOP1 1150 long	117° 35.6'E
		d'long*	0° 4.3'E
1150 fix lat	45° 51.2'N	1150 fix long	117° 39.9’E

*Using lat 45.9°N & dep of 3.0’E, d’long = 4.3’E

Step 5: Obtain 1200 fix for log book

Apply course of S58°W at 15 knots (2.5M) for run from 1150 to 1200

1150 fix	45° 51.2'N	long	117° 39.9'E
d'lat	0° 01.3'S	d'long*	0° 03.0'W
1200 lat	45° 49.9'N	1200 long	117° 36.9'E

*Using lat 45.8° and dep of 02.1’W, d’long = 3.0’W.

Step 6: Answer: 1150 fix: lat 45° 51.2'N long 117° 39.9'E

1200 position for log book: lat 45° 49.9'N 117° 36.9'E

Worked example 2 - Fix by staggered CLOPs

In DR 18° 41’S 179° 56’E at 0500 ship’s time, star Sirius gave an intercept of 8.4’ towards Az 083°(T). At 1210, lat by mer alt sun gave Obs lat of 19° 14.9’S.

During the interval:

The course steered was 121°(G), Gyro error nil, for 70M by log.
Drift of 224°(T) for 10M was experienced throughout.

Find:

The position at 1210
The position at 1200 for entry in the bridge log book.

Step 1: Identify time, position & direction of CLOP1

Time: 0500; DR lat 18° 41’S & DR long179° 56’E; Intercept 8.4’M Towards Az 083°(T). CLOP1 direction = 083°± 90° = 173°-353°(T)

Step 2: Identify time & position to transfer CLOP1

This procedure is the same as that in day’s work.

Time: 1210.To the 0500 DR, apply:

The first run as 8.4M Towards 083°(T) (from DR to ITP).

Note: If the Intercept was away, you must apply the first course as Az 180°(T).

The second run as 70M on course121°(T) (movement though the water).
The third run as 10M on course 224°(T) (current).
The position arrived, as shown in the table, would be the 1210 EP to transfer LOP1.

Course (T)	Dist	D’lat N	D’lat S	Dep E	Dep W
083° = N83°E	08.4	01.0		08.3
121° = S59°E	70.0		36.1	60.0
224° = S44°W	10.0		07.2		06.9
Total		01.0	43.3	68.3	06.9
			01.0	06.9
Resultant			42.3	61.4

0500	DR lat	18° 41.0'S	DR long	179° 56.0'E
	D’lat	0° 42.3'S	D’long*	1° 04.9'E
1210	EP lat	19° 23.3'S	EP long	178° 59.1'W

*Using m’lat 19.0° and dep of 61.4’E, d’long = 64.9’E = 1° 04.9’E

Step 3: Plot both CLOPs for 1210 & identify the fix

Step 4: Obtain 1210 fix

Thelat of the fix is that obtained by mer alt at 1210.

Dep of fix is 1.0’W of 1210 EP long.

Calculate 1210 fix long as shown in the table.

		1210 EP long	178° 59.1W
		d'long*	0° 01.1’W
1210 fix lat	19° 14.9’S	1210 fix long	179° 00.2W

*Using lat 19.3°S & dep of 1.0’W, d’long = 1.1’W

Step 4: Obtain 1200 fix for log book

From the 1210 fix, you have to go back to 1200. The run would is very small & the movement through the water reversed would suffice.

Direction: Apply course (121°(T) or S59°E) reversed i.e., N59°W.

Distance: Log distance = 70M in 7h 10m so for 10 minutes, distance = 1.6M

From TT, for Co of N59°W dist 1.6M, d’lat = 1.4’N & dep = 0.8’W

1210 fix	19° 14.9’S	long	179° 00.2’W
d'lat	0° 01.4'N	d'long*	0° 00.8’W
1200 lat	19° 13.5’S	1200 long	179° 01.0W

*Using lat 19.2° and dep of 0.8’W, d’long = 0.8’W.

Step 5: Answer

1210 fix: 19° 14.9’S 179° 00.2’W

1200 position for log book: 19° 13.5’S179° 01.0W

Determining position by simultaneous observations

Worked example 1– Two Long by Chrons

In DR 20° 36’N 146° 11’W, find the position of your ship using the following stellar observations:

Star	Obs long	Az (T)
Spica	146° 13.4’W	046°
Deneb	146° 19.3’W	130°

Step 1: Chose a suitable scale

A scale of 1 cm: 1M is suitable here.

Step 2: Draw first Celestial Line of Position (CLOP)

Choose a suitable spot on the paper & call it DR lat & Obs long 1.

LOP 1 = Az ± 90° = 046° ± 90° = 136° - 316°(T)

Draw CLOP 1 as shown in the accompanying diagram.

Step 3: Draw the second CLOP

The direction of CLOP 2 is Az ± 90° = 130° ± 90° = 220-040°(T). Draw it through DR lat & Obs long 2 for which you have to draw the meridian of Obs long 2.

Obslong 1	146° 13.4’W
Obs long 2	146° 19.3’W
D’long from Obs Long 1	000° 05.9’W

D’long from Obs long 1 to Obs long 2 = 5.9’W.

Using d’long 5.9’W & lat 20.6°, dep = 5.5’W.

Draw meridian of Obs long 2 5.5 M W of Obs long 1 and insert CLOP 2 as shown in the accompanying diagram.

Step 4: Identify the fix & obtain its lat & long

The point of intersection of the two CLOPs is the fix as shown in the accompanying diagram. Measure off the d’lat, apply it to DR lat & obtain the latitude of the fix as shown in the table. Measure off the dep from Obs long 1 (because the fix is closer to Obs long 1 than Obs long 2) and convert into d’long. Apply d’long to Obs long 1 and obtain the longitude of the fix as shown in the table.

DR lat	20° 36.0'N	DR lat	20° 36.0'N	Obs long 2	146° 19.3'W
D’lat	0° 03.1'N	Fix lat	20° 39.1'N	D’long*	0° 02.7'E
Fix lat	20° 39.1'N	M’lat	20° 37.5’N	Fix long	146° 16.6'W

*Using m’lat 20.6° & dep 2.5’E, d’long = 2.7’E

Step 5: Answer: Lat 20° 39.1'N long 146° 16.6'W

Worked example 2 – Three Intercepts

Find your ship’s position in DR 20° 11’S 140° 36’E if you obtained the following intercepts:

Intercept	Name	Azimuth
3.2 M	Away from	042°(T)
5.0 M	Towards	100°(T)
8.2 M	Towards	170°(T)

Step 1: Chose a suitable scale

Here one intercept is 8.2 M which is large signifying that a scale of 1 inch: 1 M is not suitable. Hence scale chosen is 1 cm: 1M.

Step 2: Draw CLOP 1

Choose a suitable spot on the middle of the paper & call it the DR position.

Lay off the Intercept Terminal Point (ITP) 3.2 M Away from Az 042°(T) as shown here in the accompanying diagram.

Step 3: Draw CLOPs 2 & 3

From the DR position lay off intercept 2 & 3 & LOPs 2 & 3 as shown in the accompanying diagram.

Step 4: Identify the fix & obtain its lat & long

The point of intersection of the three CLOPs is the fix as shown in the diagram.

Measure off the d’lat, apply it to DR lat & obtain the lat of the fix as shown in the table. Measure off dep, convert it into d’long, apply it to DR long and obtain the long of the fix as shown in the table.

DR lat	20° 11.0'S	DR lat	20° 11.0'S	DR long	140° 36.0'E
D’lat	0° 07.8'S	Fix lat	20° 18.8'S	D’long*	0° 04.3'E
Fix lat	20° 18.8'S	M’lat	20° 15.0’S	Fix long	140° 40.3'E

*Using m’lat 20.2° & dep 4.0’E, d’long = 4.3’E

Step 5: Answer: Lat 20° 18.8'S long 140° 40.3'E

Worked example 3 – One Intercept & One Long by Chron

In DR 48° 24’N 179° 59’E, find the position of your ship from the following two observations:

Intercept 0.5’ Towards Az 335°(T) and Obs long 179° 55.4’W Az 030°(T)

Step 1: Chose a suitable scale

The intercept for CLOP 1 & the d’long between DR long & Obs long for CLOP 2, in this question, are small. Hence a scale of 1 Inch: M is suitable. In case your ruler does not have inches on it, use 1 cm: 1M.

Step 2: Draw first CLOP

Choose a suitable spot on the middle of the paper & call it the DR position.

Lay off the Intercept Terminal Point (ITP) 0.5 M Towards Az 335°(T). Draw CLOP 1 as shown in the accompanying diagram.

Step 3: Draw the second CLOP

The direction of the second CLOP is Az ± 90° = 30° ± 90° = 120-300°(T). CLOP 2 has to be drawn through DR lat & Obs long for which you have to draw the meridian of the Obs long.

DR Long	179° 59.0’E
Obs long	179° 55.4’W
D’long from DR	000° 05.6’E

D’long from DR long & Obs long = 5.6’E.

The easterly distance between them is dep which you have to calculate.

Using d’long of 5.6’E &M’lat 48.4°, you find that dep = 3.7’E.

Draw meridian of Obs long 3.7 M E of the DR long & insert CLOP 2 as shown in the accompanying diagram.

Step 4: Identify the fix & obtain its lat & long

The point of intersection of the two CLOPs is the fix as shown in the said diagram.

DR lat	48° 24.0’N	DR lat	48° 24.0’N	DR long	179° 59.0’E
D’lat	0° 01.3’N	Fix lat	48° 25.3’N	D’long*	0° 02.3'E
Fix lat	48° 25.3’N	M’lat	48° 24.6’N	Fix long	180° 01.3'E
					179° 58.7’W

*Using m’lat 48.4° & dep 1.5’E, d’long = 2.3’E

Step 5: State your answer

Lat 48° 25.3’N long 179° 58.7’W

INFINITY MARINER

Sunday, June 9, 2024

Position Lines & Numericals

Position Lines by use of Sun

Position Lines by use of Stars

Position Lines by use of Moon

Position Lines by use of Planet

Latitude by Meridian Altitude

Ex-Meridian of Sun

Long by Chron and Intercept Methods

Position Lines by use of Pole Star

Combination of two or more celestial observations

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