Charles Brent. # Ex-meridian altitude tables, declination (0 -70 ), to which is added an explanation of maximum & minimum altitude, longitude as well as latitude from two observations of a heavenly body when near and on opposite sides of the meridian. Also a solution of the new navigation method online

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Online Library → Charles Brent → Ex-meridian altitude tables, declination (0 -70 ), to which is added an explanation of maximum & minimum altitude, longitude as well as latitude from two observations of a heavenly body when near and on opposite sides of the meridian. Also a solution of the new navigation method → online text (page 1 of 18)

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ASTRONOMY

LIBRARY

EX-MERIDIAN

ALTITUDE TABLES

DECLINATION (0-70)

TO WHICH IS ADDED AN EXPLANATION OP

MAXIMUM AND MINIMUM ALTITUDE,

LONGITUDE AS WELL AS LATITUDE FROM TWO

OBSERVATIONS OF A HEAVENLY BODY WHEN

NEAR AND ON OPPOSITE SIDES OF

THE MERIDIAN.

ALSO

A SOLUTION OF THE NEW NAVIGATION METHOD

BY

CHARLES BRENT, CAPTAIN, R.N.

ALBERT F. WALTER, M.A., NAVAL INSTRUCTOR, R.N.

GEORGE WILLIAMS, NAVAL INSTRUCTOR, R.N.

(Late Assistant to Director of Naval Education).

SEVENTH EDITION

LONDON

GEORGE PHILIP & SON, LTD., 32 FLEET STREET, E.C

Liverpool : PHILIP, SON & NEPHEW, Ltd., 20 Church Street

1914

(All rights reserved)

:RONOMY

LIBRARY

,

- "I!.- 1

PREFACE TO SEVENTH EDITION.

SINCE the publication of the Ex-Meridian Altitude Tables, the

authors have received a large number of communications from

officers who have used the Tables under almost every con-

dition of time and other circumstances, all testifying in the

most unqualified manner as to their practical utility, both in

giving excellent results and in saving the labour of calculation,

often a matter of consideration to the navigator. This gratify-

ing reception of the book has proved an incentive to the

publishing of a new edition.

April, 1914.

380031

THE

EX-MERIDIAN ALTITUDE PROBLEM;

WITH EXPLANATION AND USE OF TABLES.

It often happens that, although an observation of a celestial body can

be well taken a few minutes before or after its meridian passage, it may

be totally obscured by clouds when on the meridian.

Hence, to secure the latitude from an altitude near the meridian is

of the highest importance.

The great practical value of the Ex-meridian problem can scarcely be

over estimated in these days of quick passages.

Captain Lecky, in his valuable work entitled " Wrinkles in Practical

Navigation," says, " In addition to the extreme simplicity of the Ex-

" meridian problem, it has this to recommend it, that neither is the

" patience taxed, the eye fatigued, nor the instrument unnecessarily

" exposed by the usual weary waiting for the meridian altitude."

"Another point wherein the Ex-meridian altitude has

" a pull over the altitude on the meridian, is, that during twilight (the

" best time for observing) it may so happen that there is no star then

" culminating ; whereas it would be hard lines indeed if one or two

" could not be found, whose hour-angle east or west permitted the use of

" this method."

The method for deducing the latitude from observations near the

meridian, exhibited in this work, is one of great precision, and is applic-

able with equal facility to observations of all heavenly bodies.

In the spherical triangle formed by the elevated pole, the zenith of

the observer, and the celestial body, let

h be the easterly or westerly hour-angle.

p the polar distance, supposed to remain constant while the celestial

body describes the angle h.

c the co-latitude of the observer.

Zi the zenith distance of the celestial body, ana

z the zenith distance, when on the meridian of the observer.

Then for the superior meridian we have

Z = p (S> C.

Cos. z = Cos. p. Cos. c 4- Sin. p. Sin. c.

Cos. z^ = Cos. p. Cos. c + Sin. p. Sin. c. Cos. L

/I

6 THE EX-MERIDIAN ALTITUDE PROBLEM.

By subtraction

Cos.0 Cos.0! = Sin. p. Sin. c (1 Cos. h).

o Q; i t~ \ 2 - Sin. p. Sin. c Q . , ,

2. Sin. J (0! - 0) = ^ Sin.* &.

bin. (! +0) 2"

Now, the observation being made when the celestial body is near the

meridian, J (zi 0) and h are small ; thus, expressing the former in

seconds of arc and the latter in minutes of time, we may write

(0, - 0) Sin. 1" = 2.Cos.fe Cos to. /h

Sm. (0! + 0) \2

Also we may for Sin. \ (0,. + z] substitute Sin. z, or Sin. (lat. + dec.)

Cos, cfec. Cos, to. Sin. 2 15 X

"

, 2

2. Sin. (to. + dec.). Sin. 1

If we represent the co-efficient of h* by G we have

3 x _ = a A 2

or = ^ 0. A 2

For the inferior meridian we have

z p + a

Cos. = Cos. p. Cos. c Sin. >. Sin. c.

Cos. X = Cos. p. Cos. c Sin. p. Sin. c. Cos. IT h.

By subtraction

Cos. L Cos. = Sin. p. Sin. c. (1 Cos. tr &).

Thence as before

Cos, cfec. Cos, to. Sin. 2 15 X

" 01 " = 2. Sin. (to. + dec.). Sin. 1" *" "" h

and T = G. TT h

2

or = 0! + (7. TT

2

Table I. gives the corrections to be applied to observed altitudes of

the sun's lower limb, and of the stars, involving for the sun dip

refraction mean semidiameter 16' and parallax ; for the stars dip

and refraction.

Table II. indicates the intervals of time before and after the meridian

passage of the celestial body, during which observations may be taken.

Table III. contains the value of (7, which is the change in altitude

during the minute preceding or succeeding the meridian transit.

When the altitude is taken near the inferior meridian, the value of G

will be found in that part of Table III. where the declination is of a

contrary name to the latitude.

G is tabulated for the correct latitude. In practice the navigator

would use his dead reckoning latitude as an argument to find (7, and

generally that will be sufficient Should the latitude obtained from the

THE EX-MERIDIAN ALTITUDE PROBLEM.

observation differ considerably from the dead reckoning one, re-enter the

table for another value of C and re- work the observation.

Table IV. gives the product of C and h 2 or C and TT h. This product

is to be added to the altitude of the body when the observation is taken

near the superior meridian otherwise subtracted from the altitude, or,

which is the same thing in the end, added to the declination.

Table V. gives the approximate apparent time of the meridian passage

of the principal fixed stars at the superior transit. By adding or sub-

tracting 12 hours to these times we can obtain with sufficient accuracy

the meridian passage at the inferior transit. This Table, in conjunction

with Table II., will enable the navigator to select suitable stars for an

ex-meridian observation during twilight.

The following Examples, taken on board H. M. S. " Orlando" will

illustrate and explain the use of the Tables.

EXAMPLE I.

July 20th, 1889, at about llh. 15m. A.M., in lat. D.R. 28 S., long. D.R.

177 30' W., the observed altitude of the sun's lower limb was 39 53' 30",

when a chronometer, which was fast on G.M.T. Oh. 27m. 44s., showed

llh. 40m. 47s.; height of the eye 16 feet. Find the latitude and the

direction of the " Sumner Line."

TimebyChron li

Error fast

m.

40

27

S.

47

44

G.M.T 11 13

Equation of time - 6

G. A. T 11 7

Long. D.R a 50

Westerly hour angle 23 17

Easterly hour angle 43

Sun's obs. alt

Correction Table I.

53

10

30

48

Sun's true alt 40 4 IS

r 2"'00 1 1 38

C.{ -10 3 5

06 1 51

Mer. alt. at place of obs 41 10 52

Mer.zen.dist 48 49 8

Decimation 20 41 47 N.

Latitude 28 7 21 S.

The azimuth can be obtained from tables, diagrams, or by calculation,

thus :

In spherical triangle of position

Sin. Az. : Sin. hour angle : : Sin. pol. dist. : Sin. zen. dist.

Or Sin. Az. = Sin. h. Cos. dec. Sec. alt.

Sin. h. (Oh. 43m. Os.) ............................. 9'270735

Cos. dec. (20 42') ................................. 9'971018

Sec. alt (40 4') .................................... 10-116171

N. 1311'E .................................. Sin. 9'357924

Or, M,o h f i,r ili n'f

On chart through point

C 2"-

lon g- D.R * 77 30 ' W - \draw " Sumner Line "

^ 2go ^ g j w igo R

* Entering Table III., with lat. D.R. 28 S., and dec. 20' 40' N., contrary names gives

Table IV., with hour angle 43m. Os., and C 2" 16, we have the above corrections.

8

THE EX-MERIDIAN ALTITUDE PROBLEM.

EXAMPLE II.

July 16th, 1888, at about 8h. P.M., in lat. D.E. 8 30' N., long. D.R.

72 30' E., when a chronometer which was fast on G.M.T. 3m. 40s. showed

3h. 35m. Os., the observed altitude of Jupiter near the meridian was

62 24' 0"; height of the eye 16 feet. Find the latitude and direction of

" Sumner Line."

Time by Chron 3 35

Errorfast 3 40

G. M. T 3 31 20

Long. D.R 4 50

Ship mean time 8 21 20

Sid. time at U. M. noon 7 39 1

/ 30

I 5

Acceleration for 3h. 31m.

Sidereal time of obs 16 56

R. A. of Jupiter 15 37 53

Westerly hour angle

23 3

Jupiter's obs. alt 62 24

Correction Table I - 4 27

Jupiter's true alt 62 19 33

*Cf4"00 35 16

\ -06 32

Mer. alt. at place of obs 62 55 21

Mer. zen. dist 27 4 39

Decimation 18 36 24S.

Latitude 8 28 15N.

Sin. Az. = Sin. h. Cos. dec. Sec. alt.

Sin. h. (Oh. 23m. 3s.) ........................... 9'001756

Cos. dec. (18 36') ........................................... 99,6702

Sec. alt. (62 20') .......................................... 10'333176

S. 11 49' W .............................................. Sin. 9'3 11634

On chart through point

?* N!

draw " Sumner Line " W 12

* Entering Table III., with lat. D.R. 8 30' N., and dec. 18* 36' S., contrary names gives C. 4" '06.

From Table IV., with hour angle 23m. Os., and C. 4" '06, we have the above corrections.

tThe line of position may at once be projected on the chart without calculating the

azimuth as follows :

From the point A on the chart lay off a space A B, taken from the scale of longitude, equal

to i h (in arc) to the right if hour angle is easterly, and to the left if westerly.

On the meridian through this point B lay off B C, the value of C% 9 , taken from the scale of

latitude, in a direction away from the heavenly body.

Join A C, which is the line of position required. This method of projecting the line of

position is correct as long as the hour angle limit shown in Table II. is not exceeded.

tFor this idea we are indebted to Mr. Niven.

EXAMPLE III.

July 2nd, 1889, at about 6h. 10m. P.M., in lat. D.R 30 40' S., long.

D.R. 162 45 ' E., the observed altitude of a 1 Crucis near the meridian was

57 41' 30", when a chronometer, which was fast on G.M.T. Oh. 27m. Os.,

THE EX-MERIDIAN ALTITUDE PROBLEM.

showed 7h. 51m. 30s.; height of the eye 16 feet. Find the latitude and

direction of " Sumner Line."

h. m. s.

Time bv Chron 7 51 30

Error fast 27

G. M. T 19 24

Long. D.R. .. 10 51

30

Ship mean time 6 15 30

Sid. time at G. M. noon 6 3S 55

I q f

Acceleration for 1 9h. 24m.

Sid. time of obs ...................... 12 57

11. A. of Star ..................... 12 20

36

27

Westerly hour angle

37 9

Obs. alt. of Star 57 41 30

Correction Table I - 4 35

True alt. of Star 57 36 55

roo 22 49

'40 9 8

-07 . 1 36

True ait.

fit

* C\ '4

I <

Mer. alt. at place of obs 58 10 28

Her. zen.dist ... 31 49 32

Decimation 62 29 21 S.

Latitude .. .. 30 39 49 S.

Sin. Az. = Sin. h. Cos. dec. Sec. alt.

Sin. h. (Oh. 37m. 9s.) 9*207873

Cos. dec. (62 y 29') 9'664648

Sec. alt. (57 37') 10'27H75

S. 8 W... ..Sin. 9-143696

On chart through poi.t {ft* *

* Entering Table III., with lat. D.R. 30* 40' S., dec. 62* 30' S., same name gives C l"-47

From Table IV., with hour angle 37m. 9s., and C 1*47, we have the above corrections.

EXAMPLE IV.

July 6th, 1889, at about 5h. A.M., in latitude D.R. 33 15' S., longitude

D.R. 172 E., when a chronometer, which was fast on G.M.T. Oh. 27m. 11s.,

showed 6h. 6m. 18sec., the observed altitude of a 1 Crucis near the meri-

dian below the pole was 6 2' 0"; height of the eye 16 feet. Find the

latitude and direction of " Sumner Line."

Time by Chron ..................... 6

Error fast ..............

m. B.

6 18

27 11

G. M. T 5 39

Long. D R 11 28

Ship mean time 17 7 7

Sid. time at G. M. noon 6 54 41

49

6

Acceleration for 5h. 39m.

Sid. time of obs 2 43

R. A. a 1 Crucis .. .. 12 20 26

Westerly hour angle from

Superior Meridian ... 11 42 17

Westerly hour angle from In-

ferior Meridian .

17 43

Star's obs. alt 620

Dip 3 56

Refraction .. 8 35

Star's true alt. . 5 49 29

Star's declination 62 29 21 S.

*C7"-70 3 40

\ -06 18

62 33 19

27 26 41

Star's true alt. . ; 5 49 29

Latitude ... .. 33 16 108.

* Entering Table III., with lat. D.R. 33 15' S., dec. 62 29' S., contrary names (see

explanation of Table III.) gives C. 0"76.

From Table IV., with hour angle 17h. 43s., and C. 0"76, we have the above corrections.

10

THE EX-MERIDIAN ALTITUDE PROBLEM

Sin. Az. = Sin. h. Cos. dec. Sec. alt.

Sin. A. (Oh. 17m 43s.) 8'887767

Cos. dec. (62 29') ... 9'664648

Sec. alt. (5 49') 10-002242

S. 2*3' W... ... Sin. 8'554657

On chart throu

gh points

long. D.K 172 E.

lat. obs. 33 16'

2E. \

.6'-2 S.j

draw " Sumner Line "

E. 2 S.

MAXIMUM AND MINIMUM ALTITUDES.

If the observer be changing his latitude, or the heavenly body its

decimation, the meridian altitude at the superior transit takes place after

the maximum altitude, if the combined effect of the two movements be

towards separation, and before if towards approachment.

This can be shown in a simple manner by supposing the two motions

united and wholly performed by one say the heavenly body as in the

accompanying figures on the plane of the celestial horizon.

Separation. Approachmetit.

Let K be the united action of the observer and a heavenly body, in

the plane of the meridian, towards separation or approach-

ment in knots or ' per hour or * per minute.

C the change in altitude per minute, in the plane of the meridian

due to the earth's rotation. Table III.

t expressed in minutes of time, the easterly or westerly hour

angle of the heavenly body, at the instant of maximum

altitude.

R the correction to be applied to the maximum altitude to reduce

it to the meridian.

Then K t will be the change in altitude produced by the movements

in latitude and declination.

G t* will be the change of altitude due to the earth's rotation.

THE EX-MERIDIAN ALTITUDE PROBLEM. 11

We shall have

R = K t * C t* or R = K t + C t>

OR = CKt ^ C 3 * CR ^= CKt + C* t*

: 1 u

It is evident that .R or its multiple C R will have its maximum or

minimum value when the quantities within the brackets vanish, that is

when

C t IjT ~ = or

and then R = + ^*- = + C t z

40

t will be an easterly hour angle when the observer and the

heavenly body are separating, and a westerly one when they

are approaching. It will be a portion of apparent solar,

lunar, planetary or sidereal time, according to the celestial

body observed, and should be converted into its mean time

equivalent, when mean time is required.

RULE A. If the ship be changing her longitude, an easterly t must

be increased by a westward, and decreased by an eastward

change in longitude, to obtain the correct hour angle.

RULE B. The converse for a westerly t.

Again, the meridian altitude at the inferior transit takes

place before the minimum altitude, if the tendency of the two

movements be towards separation, and after, if towards

approachment. Consequently, for the inferior meridian

RULES A and B must be reversed.

EXAMPLE.

February 1st, 1894, in lat. D.R. 47 45' N., long. D.R. 12 3 10' W. the

maximum altitude of the sun's lower limb was 24 48' 40", index

error + 2' 10"; height of the eye 15 feet; the ship steaming S. 22 J E.

(true) 25 knots. Find the latitude at ship apparent noon.

Green. Date. Declination. Change.

h. m. a o / / //

S. A. T. 000 S. 17 57 42'94

Long. DR. 48 40 W. Cor. 35

48 40 17 22

From Traverse Table.

S. 22i E. 25' d. lat. 23 ''1 dep. 9''57 = d. long. H' '3 = 57'2 sec.

Observer's change in latitude 23 '1 South per hour, or 23 '"1 South per minute.

Sun's declination 42 "'94 North 0"7 North

Observer and Sun approaching each other at K 23 "'8

12 THE EX-MERTDIAN ALTITUDE PROBLEM.

t.=K. R.= tt. 2

= = l"'39(8m. 34s.) *

2 x l"-39

= 8m. 34s. = 1 ' -41 ' ' Table IV.

As the observer and the sun are approaching each other, t. is a westerly hour angle.

Since the ship is changing her longitude to the eastward at the rate of 57'2 sec. per hour,

->r 8 sec. during an approximate interval of 8m. 34s. the corrected hour angle is 8m. 42s.

KULE B N

Therefore the maximum altitude took place 8m. 42s. after the meridian transit.

Obs. max. alt. 24 48 40

Index-error + 2 10

Correction. Table I + 10 27

R. always added for supr. mer + 1 41

25 2 58

64 57 2

Declination 17 22 S.

Latitude at instant of max. alt 47 56 40 N

Lat. run in 8m. 42s 3 20

Latitude at 8. A. noon ....48 ON.

LONGITUDE AND LATITUDE FROM EX-MERIDIAN

ALTITUDES.

The following method of finding the longitude as well as the latitude

from ex-meridian altitudes, founded on Art. 172, Vol. I. of Spherical and

Practical Astronomy, by Chauvenet, is inserted in the hope that it may

be of use to the navigator when from various causes he has been unable

to obtain a reliable longitude by other means.

Suppose two altitudes of the sun to have been taken on opposite sides

of the meridian, within the limits of Table II., from a ship in motion, and

the times noted by a chronometer.

Let z^ 2 be the zenith distances derived from the two altitudes.

h^ h^ the easterly and westerly hour-angles from the instants of

maximum altitude for each observation.

the value obtained from Table III. for the latitude and

declination at noon.

1 the chronometer or mean time interval between the

observations.

g the change in longitude during this interval.

Then if z represent the zenith distance of the maximum altitude we

have

*:.. ^' - .,0 Aj

z = z n - C h*

THE EX-MERIDIAN ALTITUDE PROBLEM. 13

By subtraction

z^ 2 = C h* ~

or *'-*- z$n3

Now the ship being in motion

Very approximately

, 7

h, +h 2

From addition of (1) and (2)

h

+ easterly ,~

................ (2).

2 2 C a 9)

and if o^ and a 2 represent the two altitudes

, ,- .....

2 2C(Ig)

* pZus, i^en ^e second altitude is the greater.

This ean be simplified by taking the condition of equal altitudes, for

then (3) becomes

As far as the altitudes are concerned (3) determines the " degree of

dependance " the navigator can place on the resulting longitude, which

will be shown by - - , when e is an assumed error committed

on the difference of the altitudes.

The preceding is equally applicable to the stars, but it must be re-

membered that &j + ^ 2 will then be a portion of sidereal time, or

lt l + h z = 1 (in sid. time) g.

It may not be out of place to call the attention of the navigator to

the following precautions :

I. The altitudes should be taken, when practicable, under identical

conditions, i.e., from the same part of the ship, by the same

observer, with the same instrument, shades and telescope.

II. Preference should be given to equal or nearly equal altitudes, with

as large an interval between the observations as is possible,

keeping in view the limits of Table II.

To show the detail of calculation, the following examples of " equal "

and "unequal altitudes" have been computed to place the observer at

time of meridian transit in lat. 48 O x N., long. 12 (X W., at different times

of the year, and the ship proceeding at high and moderate speeds.

Ex. I. October 17th, 1894, in lat. D.R. 47 50' N., long. D.R 11 50'

W., the following sights were taken near the meridian to find the longi-

tude and latitude at apparent noon.

H

Chron. times.

h. m.

3 5

3 10

THE EX-MERIDIAN ALTITUDE PROBLEM.

Obs. alts, nun's I. I.

sec*.

35

59

25

31

22

11

31

31

31

31

44

54

54

43

20

20

10

Index error + 1' 10" ; height of the eye 28 feet. The chronometer being

fast 3h. 17m. 36s. on G.M.T., and the ship steaming N. 67J W. (true) at

the rate of 19 knots.

From Traverse Table. N. 67i W. 19' d.lat. 7 ''3 dep. 17 '-6 = d.long. 26' '2 = 104'8

seconds.

Observer's change in lat. 7 ''3 N. per hour or 7 "'3 per minute.

Sun's dec. 54"'84S. 0"'9

Towards separation K 8"'2

t- K

' 2 C R = Ct*

= 8"'2 = i"54 ( 2m . 39s . )2

2 x r ' 54 = 11" Table IV

= 2m. 39'4s.

The ship's run in longitude in 2m. 39s. is about 46 sec. westerly. Therefore the

corrected hour angle is 2m. 44s. (See Rule A. under Max. and Min. Alts.)

EQUAL ALTITUDES. ,THE 2ND AND 3RD OBSERVATIONS.

Green, date. Sun's decn. Equation of time.

Oct. ... 17d- Oh. Qm- S. 9 21' 14"'l 14m. 36'84s.

Long. D.R. 47 Cor. 44 Cor. "4

Oct 17 47 9 21 58 14 37 '2- to app. time.

Chron. times.

Green, date of 1st obsn.

3h-

10m. 593.

3h.

10m. 593.

4

25 22

Error 3

17 36 fast.

7. 1

14 23

23

53 23

g (in lh. 14m. 23s. )

2 11 W.

csaH

;

I a .1

12 12

Ji^ % (1 -g)

36 6

S A T of merid alt

W

Oh. Om. Os- (a)

Observed altitude .

. 31 54'

20"

t (separation)

2 44

Index-error

Correction Table I.

. + 1

+ 9

10

28

SAT of max alt

23 57 16

36 6

CVorl"'54(36m. 6s. ) a

Table IV.

( 21

10

43

51

52

h t (always subtract! ve)

S. A. T. of 1st observation

.. 23 21 10 (b)

R (always additive) .

11

Equation of time

.. 14 37

32 38

35

S. M. T. of 1st observation

.. 23 6 33

G. M. T.

... 23 53 23

57 21

25

Declination

9 21

588.

Longitude at

Run between (6) and (a)

46 50 W.

1 10 W.

Latitude at max. alt.

Run in 2 1 "* 44 S >

... 47 59

27 N.

20 N.

~8 OW.

Latitude at noon

47 59

47 N.

or

12 0' W.

=====

THE EX-MERIDIAN ALTITUDE PROBLEM.

15

UNEQUAL ALTITUDES.

Chron. times.

THE IST AND SRD OBSERVATIONS.

Green, date of 1st obsn. Altitudes.

3h. 5m. 35s.

4 25 22

3h. 5m. 35s. a, 31 44' 0"

Error3 17 56 fast o a 31 54 20

/. 1 19 47 23

g (in lh. 19m. 47 s - ) 2 19 W.

47 59 a a fl! 10 20

= 620*

______

38m 44<? 4-

620

S.

S.

s.

s.

G.

Lc

Lc

I-g l n

(I-g) 38

A. T. of mer. alt

28

44

in.

Oh- Om.

- 2

0s. (a)

44

2 x

= 41m. 20s.

Observed altitude

Index-error

l"-54 x 77

31 44'

+ 1

+ 9

\ 28

m.-5

0'

10

28

28

14

8

11

S.

N

N.

N

t (separation)... ....

A T. of max. alt.

23 57

41

23 15

14

23 1

23 47

16

20

56 (b)

37

19

59

Chi 3 or T54s. (41m. 20s. ) "

Table IV.

R (always additive)...

Declination

A^ (always sub tractive)

A. T. of 1st observation

EG nation of time

M. T. of 1st observation...

M. T.

ngitude at

Run between (6) and (a)

nffitude at noon

32

38

39

57

g

21

21

21

58

46

40 W.

19 W.

Latitude at max . alt

. 47

59

23

20

47

,

11

59 W.

Latitude at noon

or

47

59

43

__

!

" Degree of Dependance." Supposing an error of 30" to have been

committed on the difference of the altitudes. The error in the longitude

will be

15 x 30"

For the equal altitudes = 2'

For the unequal altitudes

2 x 1 '-54 x 72m. -2

15 x 30"

2 x 1-54 x 77- '5

= r-9.

Ex. II December llth, 1894, in lat. D.R. 48 10' N., long. D.R. 12 15*

W., the following observations were taken near the meridian, to find the

longitude and latitude at apparent noon.

Obs. Alts, sun's I 1.

Chron. times.

h. m. s.

7 23

7 38

11

44

17

17

45

57

10

9 17

9 23

17 58

17 45 10

Index error + 2' 10" ; height of the eye 31 feet. The chronometer being

slow 4h. 18m. 23s. on G.M.T., and the ship steaming S. 22 W. (true) at 14

knots.

16

THE EX-MERIDIAN ALTITUDE PROBLEM.

From Traverse Table. S. 22 W. 14' d.lat.13' dep. 5' "24 = d.Iong. 7 ''83 = 31 '32 seconds.

Observer's change in lat. 13' S. per hour or 13" per minute.

Sun's ,,dec. 12" S. 0"'2

Towards approachment K 12" - 8 ,,

t - K

~ 2(7 R= Ct*

12"-8 = l"-28 (5mO a

2 x l "' 28 = 32" Table IV.

= 5 min.

The ship's run in longitude during 5m. is about 2'5s. westerly. Therefore the corrected

hour angle is 4m. 57*5s. (See Rule B. under Max. and Min. Alts.}

EQUAL ALTITUDES.

Green, date.

(1. h. m.

Deer. ..11

Long. D.R. 49

THE IST AND 4TH OBSERVATIONS.

Sun's decn. J quation of time.

S. 23 2 16'6 ' 29 S; 98

Cor. 9-6 Cor. -92

Deer. ... 11 49

Chron.

h. m.

7 33

9 23

23

^^

times.

s.

11

8

2 26-2 6 29-0 - to app. time.

Green, date of 1st Obs.

h. m. s.

7 33 11

Error 4 18 23 slow.

7

1

1

49

48

57

57'5 W.

59-5

g (in Ih. 50m.) .

7 - q .

23 51

^^ssssssss

Observed altitude

34

___

1*7 4*5

10

10

58

30

54

58

32

A, i (I - a) .

h.

54

m.

1

4

54

297

la\

WtJ

57-5

297

S, A. T. of mer. alt

t (ariproachment) , .,

S. A. T. of max. alt

A! (always subtractive)

S. A. T. of 1st observation

Equation of time

S. M. T. of 1st observation

ASTRONOMY

LIBRARY

EX-MERIDIAN

ALTITUDE TABLES

DECLINATION (0-70)

TO WHICH IS ADDED AN EXPLANATION OP

MAXIMUM AND MINIMUM ALTITUDE,

LONGITUDE AS WELL AS LATITUDE FROM TWO

OBSERVATIONS OF A HEAVENLY BODY WHEN

NEAR AND ON OPPOSITE SIDES OF

THE MERIDIAN.

ALSO

A SOLUTION OF THE NEW NAVIGATION METHOD

BY

CHARLES BRENT, CAPTAIN, R.N.

ALBERT F. WALTER, M.A., NAVAL INSTRUCTOR, R.N.

GEORGE WILLIAMS, NAVAL INSTRUCTOR, R.N.

(Late Assistant to Director of Naval Education).

SEVENTH EDITION

LONDON

GEORGE PHILIP & SON, LTD., 32 FLEET STREET, E.C

Liverpool : PHILIP, SON & NEPHEW, Ltd., 20 Church Street

1914

(All rights reserved)

:RONOMY

LIBRARY

,

- "I!.- 1

PREFACE TO SEVENTH EDITION.

SINCE the publication of the Ex-Meridian Altitude Tables, the

authors have received a large number of communications from

officers who have used the Tables under almost every con-

dition of time and other circumstances, all testifying in the

most unqualified manner as to their practical utility, both in

giving excellent results and in saving the labour of calculation,

often a matter of consideration to the navigator. This gratify-

ing reception of the book has proved an incentive to the

publishing of a new edition.

April, 1914.

380031

THE

EX-MERIDIAN ALTITUDE PROBLEM;

WITH EXPLANATION AND USE OF TABLES.

It often happens that, although an observation of a celestial body can

be well taken a few minutes before or after its meridian passage, it may

be totally obscured by clouds when on the meridian.

Hence, to secure the latitude from an altitude near the meridian is

of the highest importance.

The great practical value of the Ex-meridian problem can scarcely be

over estimated in these days of quick passages.

Captain Lecky, in his valuable work entitled " Wrinkles in Practical

Navigation," says, " In addition to the extreme simplicity of the Ex-

" meridian problem, it has this to recommend it, that neither is the

" patience taxed, the eye fatigued, nor the instrument unnecessarily

" exposed by the usual weary waiting for the meridian altitude."

"Another point wherein the Ex-meridian altitude has

" a pull over the altitude on the meridian, is, that during twilight (the

" best time for observing) it may so happen that there is no star then

" culminating ; whereas it would be hard lines indeed if one or two

" could not be found, whose hour-angle east or west permitted the use of

" this method."

The method for deducing the latitude from observations near the

meridian, exhibited in this work, is one of great precision, and is applic-

able with equal facility to observations of all heavenly bodies.

In the spherical triangle formed by the elevated pole, the zenith of

the observer, and the celestial body, let

h be the easterly or westerly hour-angle.

p the polar distance, supposed to remain constant while the celestial

body describes the angle h.

c the co-latitude of the observer.

Zi the zenith distance of the celestial body, ana

z the zenith distance, when on the meridian of the observer.

Then for the superior meridian we have

Z = p (S> C.

Cos. z = Cos. p. Cos. c 4- Sin. p. Sin. c.

Cos. z^ = Cos. p. Cos. c + Sin. p. Sin. c. Cos. L

/I

6 THE EX-MERIDIAN ALTITUDE PROBLEM.

By subtraction

Cos.0 Cos.0! = Sin. p. Sin. c (1 Cos. h).

o Q; i t~ \ 2 - Sin. p. Sin. c Q . , ,

2. Sin. J (0! - 0) = ^ Sin.* &.

bin. (! +0) 2"

Now, the observation being made when the celestial body is near the

meridian, J (zi 0) and h are small ; thus, expressing the former in

seconds of arc and the latter in minutes of time, we may write

(0, - 0) Sin. 1" = 2.Cos.fe Cos to. /h

Sm. (0! + 0) \2

Also we may for Sin. \ (0,. + z] substitute Sin. z, or Sin. (lat. + dec.)

Cos, cfec. Cos, to. Sin. 2 15 X

"

, 2

2. Sin. (to. + dec.). Sin. 1

If we represent the co-efficient of h* by G we have

3 x _ = a A 2

or = ^ 0. A 2

For the inferior meridian we have

z p + a

Cos. = Cos. p. Cos. c Sin. >. Sin. c.

Cos. X = Cos. p. Cos. c Sin. p. Sin. c. Cos. IT h.

By subtraction

Cos. L Cos. = Sin. p. Sin. c. (1 Cos. tr &).

Thence as before

Cos, cfec. Cos, to. Sin. 2 15 X

" 01 " = 2. Sin. (to. + dec.). Sin. 1" *" "" h

and T = G. TT h

2

or = 0! + (7. TT

2

Table I. gives the corrections to be applied to observed altitudes of

the sun's lower limb, and of the stars, involving for the sun dip

refraction mean semidiameter 16' and parallax ; for the stars dip

and refraction.

Table II. indicates the intervals of time before and after the meridian

passage of the celestial body, during which observations may be taken.

Table III. contains the value of (7, which is the change in altitude

during the minute preceding or succeeding the meridian transit.

When the altitude is taken near the inferior meridian, the value of G

will be found in that part of Table III. where the declination is of a

contrary name to the latitude.

G is tabulated for the correct latitude. In practice the navigator

would use his dead reckoning latitude as an argument to find (7, and

generally that will be sufficient Should the latitude obtained from the

THE EX-MERIDIAN ALTITUDE PROBLEM.

observation differ considerably from the dead reckoning one, re-enter the

table for another value of C and re- work the observation.

Table IV. gives the product of C and h 2 or C and TT h. This product

is to be added to the altitude of the body when the observation is taken

near the superior meridian otherwise subtracted from the altitude, or,

which is the same thing in the end, added to the declination.

Table V. gives the approximate apparent time of the meridian passage

of the principal fixed stars at the superior transit. By adding or sub-

tracting 12 hours to these times we can obtain with sufficient accuracy

the meridian passage at the inferior transit. This Table, in conjunction

with Table II., will enable the navigator to select suitable stars for an

ex-meridian observation during twilight.

The following Examples, taken on board H. M. S. " Orlando" will

illustrate and explain the use of the Tables.

EXAMPLE I.

July 20th, 1889, at about llh. 15m. A.M., in lat. D.R. 28 S., long. D.R.

177 30' W., the observed altitude of the sun's lower limb was 39 53' 30",

when a chronometer, which was fast on G.M.T. Oh. 27m. 44s., showed

llh. 40m. 47s.; height of the eye 16 feet. Find the latitude and the

direction of the " Sumner Line."

TimebyChron li

Error fast

m.

40

27

S.

47

44

G.M.T 11 13

Equation of time - 6

G. A. T 11 7

Long. D.R a 50

Westerly hour angle 23 17

Easterly hour angle 43

Sun's obs. alt

Correction Table I.

53

10

30

48

Sun's true alt 40 4 IS

r 2"'00 1 1 38

C.{ -10 3 5

06 1 51

Mer. alt. at place of obs 41 10 52

Mer.zen.dist 48 49 8

Decimation 20 41 47 N.

Latitude 28 7 21 S.

The azimuth can be obtained from tables, diagrams, or by calculation,

thus :

In spherical triangle of position

Sin. Az. : Sin. hour angle : : Sin. pol. dist. : Sin. zen. dist.

Or Sin. Az. = Sin. h. Cos. dec. Sec. alt.

Sin. h. (Oh. 43m. Os.) ............................. 9'270735

Cos. dec. (20 42') ................................. 9'971018

Sec. alt (40 4') .................................... 10-116171

N. 1311'E .................................. Sin. 9'357924

Or, M,o h f i,r ili n'f

On chart through point

C 2"-

lon g- D.R * 77 30 ' W - \draw " Sumner Line "

^ 2go ^ g j w igo R

* Entering Table III., with lat. D.R. 28 S., and dec. 20' 40' N., contrary names gives

Table IV., with hour angle 43m. Os., and C 2" 16, we have the above corrections.

8

THE EX-MERIDIAN ALTITUDE PROBLEM.

EXAMPLE II.

July 16th, 1888, at about 8h. P.M., in lat. D.E. 8 30' N., long. D.R.

72 30' E., when a chronometer which was fast on G.M.T. 3m. 40s. showed

3h. 35m. Os., the observed altitude of Jupiter near the meridian was

62 24' 0"; height of the eye 16 feet. Find the latitude and direction of

" Sumner Line."

Time by Chron 3 35

Errorfast 3 40

G. M. T 3 31 20

Long. D.R 4 50

Ship mean time 8 21 20

Sid. time at U. M. noon 7 39 1

/ 30

I 5

Acceleration for 3h. 31m.

Sidereal time of obs 16 56

R. A. of Jupiter 15 37 53

Westerly hour angle

23 3

Jupiter's obs. alt 62 24

Correction Table I - 4 27

Jupiter's true alt 62 19 33

*Cf4"00 35 16

\ -06 32

Mer. alt. at place of obs 62 55 21

Mer. zen. dist 27 4 39

Decimation 18 36 24S.

Latitude 8 28 15N.

Sin. Az. = Sin. h. Cos. dec. Sec. alt.

Sin. h. (Oh. 23m. 3s.) ........................... 9'001756

Cos. dec. (18 36') ........................................... 99,6702

Sec. alt. (62 20') .......................................... 10'333176

S. 11 49' W .............................................. Sin. 9'3 11634

On chart through point

?* N!

draw " Sumner Line " W 12

* Entering Table III., with lat. D.R. 8 30' N., and dec. 18* 36' S., contrary names gives C. 4" '06.

From Table IV., with hour angle 23m. Os., and C. 4" '06, we have the above corrections.

tThe line of position may at once be projected on the chart without calculating the

azimuth as follows :

From the point A on the chart lay off a space A B, taken from the scale of longitude, equal

to i h (in arc) to the right if hour angle is easterly, and to the left if westerly.

On the meridian through this point B lay off B C, the value of C% 9 , taken from the scale of

latitude, in a direction away from the heavenly body.

Join A C, which is the line of position required. This method of projecting the line of

position is correct as long as the hour angle limit shown in Table II. is not exceeded.

tFor this idea we are indebted to Mr. Niven.

EXAMPLE III.

July 2nd, 1889, at about 6h. 10m. P.M., in lat. D.R 30 40' S., long.

D.R. 162 45 ' E., the observed altitude of a 1 Crucis near the meridian was

57 41' 30", when a chronometer, which was fast on G.M.T. Oh. 27m. Os.,

THE EX-MERIDIAN ALTITUDE PROBLEM.

showed 7h. 51m. 30s.; height of the eye 16 feet. Find the latitude and

direction of " Sumner Line."

h. m. s.

Time bv Chron 7 51 30

Error fast 27

G. M. T 19 24

Long. D.R. .. 10 51

30

Ship mean time 6 15 30

Sid. time at G. M. noon 6 3S 55

I q f

Acceleration for 1 9h. 24m.

Sid. time of obs ...................... 12 57

11. A. of Star ..................... 12 20

36

27

Westerly hour angle

37 9

Obs. alt. of Star 57 41 30

Correction Table I - 4 35

True alt. of Star 57 36 55

roo 22 49

'40 9 8

-07 . 1 36

True ait.

fit

* C\ '4

I <

Mer. alt. at place of obs 58 10 28

Her. zen.dist ... 31 49 32

Decimation 62 29 21 S.

Latitude .. .. 30 39 49 S.

Sin. Az. = Sin. h. Cos. dec. Sec. alt.

Sin. h. (Oh. 37m. 9s.) 9*207873

Cos. dec. (62 y 29') 9'664648

Sec. alt. (57 37') 10'27H75

S. 8 W... ..Sin. 9-143696

On chart through poi.t {ft* *

* Entering Table III., with lat. D.R. 30* 40' S., dec. 62* 30' S., same name gives C l"-47

From Table IV., with hour angle 37m. 9s., and C 1*47, we have the above corrections.

EXAMPLE IV.

July 6th, 1889, at about 5h. A.M., in latitude D.R. 33 15' S., longitude

D.R. 172 E., when a chronometer, which was fast on G.M.T. Oh. 27m. 11s.,

showed 6h. 6m. 18sec., the observed altitude of a 1 Crucis near the meri-

dian below the pole was 6 2' 0"; height of the eye 16 feet. Find the

latitude and direction of " Sumner Line."

Time by Chron ..................... 6

Error fast ..............

m. B.

6 18

27 11

G. M. T 5 39

Long. D R 11 28

Ship mean time 17 7 7

Sid. time at G. M. noon 6 54 41

49

6

Acceleration for 5h. 39m.

Sid. time of obs 2 43

R. A. a 1 Crucis .. .. 12 20 26

Westerly hour angle from

Superior Meridian ... 11 42 17

Westerly hour angle from In-

ferior Meridian .

17 43

Star's obs. alt 620

Dip 3 56

Refraction .. 8 35

Star's true alt. . 5 49 29

Star's declination 62 29 21 S.

*C7"-70 3 40

\ -06 18

62 33 19

27 26 41

Star's true alt. . ; 5 49 29

Latitude ... .. 33 16 108.

* Entering Table III., with lat. D.R. 33 15' S., dec. 62 29' S., contrary names (see

explanation of Table III.) gives C. 0"76.

From Table IV., with hour angle 17h. 43s., and C. 0"76, we have the above corrections.

10

THE EX-MERIDIAN ALTITUDE PROBLEM

Sin. Az. = Sin. h. Cos. dec. Sec. alt.

Sin. A. (Oh. 17m 43s.) 8'887767

Cos. dec. (62 29') ... 9'664648

Sec. alt. (5 49') 10-002242

S. 2*3' W... ... Sin. 8'554657

On chart throu

gh points

long. D.K 172 E.

lat. obs. 33 16'

2E. \

.6'-2 S.j

draw " Sumner Line "

E. 2 S.

MAXIMUM AND MINIMUM ALTITUDES.

If the observer be changing his latitude, or the heavenly body its

decimation, the meridian altitude at the superior transit takes place after

the maximum altitude, if the combined effect of the two movements be

towards separation, and before if towards approachment.

This can be shown in a simple manner by supposing the two motions

united and wholly performed by one say the heavenly body as in the

accompanying figures on the plane of the celestial horizon.

Separation. Approachmetit.

Let K be the united action of the observer and a heavenly body, in

the plane of the meridian, towards separation or approach-

ment in knots or ' per hour or * per minute.

C the change in altitude per minute, in the plane of the meridian

due to the earth's rotation. Table III.

t expressed in minutes of time, the easterly or westerly hour

angle of the heavenly body, at the instant of maximum

altitude.

R the correction to be applied to the maximum altitude to reduce

it to the meridian.

Then K t will be the change in altitude produced by the movements

in latitude and declination.

G t* will be the change of altitude due to the earth's rotation.

THE EX-MERIDIAN ALTITUDE PROBLEM. 11

We shall have

R = K t * C t* or R = K t + C t>

OR = CKt ^ C 3 * CR ^= CKt + C* t*

: 1 u

It is evident that .R or its multiple C R will have its maximum or

minimum value when the quantities within the brackets vanish, that is

when

C t IjT ~ = or

and then R = + ^*- = + C t z

40

t will be an easterly hour angle when the observer and the

heavenly body are separating, and a westerly one when they

are approaching. It will be a portion of apparent solar,

lunar, planetary or sidereal time, according to the celestial

body observed, and should be converted into its mean time

equivalent, when mean time is required.

RULE A. If the ship be changing her longitude, an easterly t must

be increased by a westward, and decreased by an eastward

change in longitude, to obtain the correct hour angle.

RULE B. The converse for a westerly t.

Again, the meridian altitude at the inferior transit takes

place before the minimum altitude, if the tendency of the two

movements be towards separation, and after, if towards

approachment. Consequently, for the inferior meridian

RULES A and B must be reversed.

EXAMPLE.

February 1st, 1894, in lat. D.R. 47 45' N., long. D.R. 12 3 10' W. the

maximum altitude of the sun's lower limb was 24 48' 40", index

error + 2' 10"; height of the eye 15 feet; the ship steaming S. 22 J E.

(true) 25 knots. Find the latitude at ship apparent noon.

Green. Date. Declination. Change.

h. m. a o / / //

S. A. T. 000 S. 17 57 42'94

Long. DR. 48 40 W. Cor. 35

48 40 17 22

From Traverse Table.

S. 22i E. 25' d. lat. 23 ''1 dep. 9''57 = d. long. H' '3 = 57'2 sec.

Observer's change in latitude 23 '1 South per hour, or 23 '"1 South per minute.

Sun's declination 42 "'94 North 0"7 North

Observer and Sun approaching each other at K 23 "'8

12 THE EX-MERTDIAN ALTITUDE PROBLEM.

t.=K. R.= tt. 2

= = l"'39(8m. 34s.) *

2 x l"-39

= 8m. 34s. = 1 ' -41 ' ' Table IV.

As the observer and the sun are approaching each other, t. is a westerly hour angle.

Since the ship is changing her longitude to the eastward at the rate of 57'2 sec. per hour,

->r 8 sec. during an approximate interval of 8m. 34s. the corrected hour angle is 8m. 42s.

KULE B N

Therefore the maximum altitude took place 8m. 42s. after the meridian transit.

Obs. max. alt. 24 48 40

Index-error + 2 10

Correction. Table I + 10 27

R. always added for supr. mer + 1 41

25 2 58

64 57 2

Declination 17 22 S.

Latitude at instant of max. alt 47 56 40 N

Lat. run in 8m. 42s 3 20

Latitude at 8. A. noon ....48 ON.

LONGITUDE AND LATITUDE FROM EX-MERIDIAN

ALTITUDES.

The following method of finding the longitude as well as the latitude

from ex-meridian altitudes, founded on Art. 172, Vol. I. of Spherical and

Practical Astronomy, by Chauvenet, is inserted in the hope that it may

be of use to the navigator when from various causes he has been unable

to obtain a reliable longitude by other means.

Suppose two altitudes of the sun to have been taken on opposite sides

of the meridian, within the limits of Table II., from a ship in motion, and

the times noted by a chronometer.

Let z^ 2 be the zenith distances derived from the two altitudes.

h^ h^ the easterly and westerly hour-angles from the instants of

maximum altitude for each observation.

the value obtained from Table III. for the latitude and

declination at noon.

1 the chronometer or mean time interval between the

observations.

g the change in longitude during this interval.

Then if z represent the zenith distance of the maximum altitude we

have

*:.. ^' - .,0 Aj

z = z n - C h*

THE EX-MERIDIAN ALTITUDE PROBLEM. 13

By subtraction

z^ 2 = C h* ~

or *'-*- z$n3

Now the ship being in motion

Very approximately

, 7

h, +h 2

From addition of (1) and (2)

h

+ easterly ,~

................ (2).

2 2 C a 9)

and if o^ and a 2 represent the two altitudes

, ,- .....

2 2C(Ig)

* pZus, i^en ^e second altitude is the greater.

This ean be simplified by taking the condition of equal altitudes, for

then (3) becomes

As far as the altitudes are concerned (3) determines the " degree of

dependance " the navigator can place on the resulting longitude, which

will be shown by - - , when e is an assumed error committed

on the difference of the altitudes.

The preceding is equally applicable to the stars, but it must be re-

membered that &j + ^ 2 will then be a portion of sidereal time, or

lt l + h z = 1 (in sid. time) g.

It may not be out of place to call the attention of the navigator to

the following precautions :

I. The altitudes should be taken, when practicable, under identical

conditions, i.e., from the same part of the ship, by the same

observer, with the same instrument, shades and telescope.

II. Preference should be given to equal or nearly equal altitudes, with

as large an interval between the observations as is possible,

keeping in view the limits of Table II.

To show the detail of calculation, the following examples of " equal "

and "unequal altitudes" have been computed to place the observer at

time of meridian transit in lat. 48 O x N., long. 12 (X W., at different times

of the year, and the ship proceeding at high and moderate speeds.

Ex. I. October 17th, 1894, in lat. D.R. 47 50' N., long. D.R 11 50'

W., the following sights were taken near the meridian to find the longi-

tude and latitude at apparent noon.

H

Chron. times.

h. m.

3 5

3 10

THE EX-MERIDIAN ALTITUDE PROBLEM.

Obs. alts, nun's I. I.

sec*.

35

59

25

31

22

11

31

31

31

31

44

54

54

43

20

20

10

Index error + 1' 10" ; height of the eye 28 feet. The chronometer being

fast 3h. 17m. 36s. on G.M.T., and the ship steaming N. 67J W. (true) at

the rate of 19 knots.

From Traverse Table. N. 67i W. 19' d.lat. 7 ''3 dep. 17 '-6 = d.long. 26' '2 = 104'8

seconds.

Observer's change in lat. 7 ''3 N. per hour or 7 "'3 per minute.

Sun's dec. 54"'84S. 0"'9

Towards separation K 8"'2

t- K

' 2 C R = Ct*

= 8"'2 = i"54 ( 2m . 39s . )2

2 x r ' 54 = 11" Table IV

= 2m. 39'4s.

The ship's run in longitude in 2m. 39s. is about 46 sec. westerly. Therefore the

corrected hour angle is 2m. 44s. (See Rule A. under Max. and Min. Alts.)

EQUAL ALTITUDES. ,THE 2ND AND 3RD OBSERVATIONS.

Green, date. Sun's decn. Equation of time.

Oct. ... 17d- Oh. Qm- S. 9 21' 14"'l 14m. 36'84s.

Long. D.R. 47 Cor. 44 Cor. "4

Oct 17 47 9 21 58 14 37 '2- to app. time.

Chron. times.

Green, date of 1st obsn.

3h-

10m. 593.

3h.

10m. 593.

4

25 22

Error 3

17 36 fast.

7. 1

14 23

23

53 23

g (in lh. 14m. 23s. )

2 11 W.

csaH

;

I a .1

12 12

Ji^ % (1 -g)

36 6

S A T of merid alt

W

Oh. Om. Os- (a)

Observed altitude .

. 31 54'

20"

t (separation)

2 44

Index-error

Correction Table I.

. + 1

+ 9

10

28

SAT of max alt

23 57 16

36 6

CVorl"'54(36m. 6s. ) a

Table IV.

( 21

10

43

51

52

h t (always subtract! ve)

S. A. T. of 1st observation

.. 23 21 10 (b)

R (always additive) .

11

Equation of time

.. 14 37

32 38

35

S. M. T. of 1st observation

.. 23 6 33

G. M. T.

... 23 53 23

57 21

25

Declination

9 21

588.

Longitude at

Run between (6) and (a)

46 50 W.

1 10 W.

Latitude at max. alt.

Run in 2 1 "* 44 S >

... 47 59

27 N.

20 N.

~8 OW.

Latitude at noon

47 59

47 N.

or

12 0' W.

=====

THE EX-MERIDIAN ALTITUDE PROBLEM.

15

UNEQUAL ALTITUDES.

Chron. times.

THE IST AND SRD OBSERVATIONS.

Green, date of 1st obsn. Altitudes.

3h. 5m. 35s.

4 25 22

3h. 5m. 35s. a, 31 44' 0"

Error3 17 56 fast o a 31 54 20

/. 1 19 47 23

g (in lh. 19m. 47 s - ) 2 19 W.

47 59 a a fl! 10 20

= 620*

______

38m 44<? 4-

620

S.

S.

s.

s.

G.

Lc

Lc

I-g l n

(I-g) 38

A. T. of mer. alt

28

44

in.

Oh- Om.

- 2

0s. (a)

44

2 x

= 41m. 20s.

Observed altitude

Index-error

l"-54 x 77

31 44'

+ 1

+ 9

\ 28

m.-5

0'

10

28

28

14

8

11

S.

N

N.

N

t (separation)... ....

A T. of max. alt.

23 57

41

23 15

14

23 1

23 47

16

20

56 (b)

37

19

59

Chi 3 or T54s. (41m. 20s. ) "

Table IV.

R (always additive)...

Declination

A^ (always sub tractive)

A. T. of 1st observation

EG nation of time

M. T. of 1st observation...

M. T.

ngitude at

Run between (6) and (a)

nffitude at noon

32

38

39

57

g

21

21

21

58

46

40 W.

19 W.

Latitude at max . alt

. 47

59

23

20

47

,

11

59 W.

Latitude at noon

or

47

59

43

__

!

" Degree of Dependance." Supposing an error of 30" to have been

committed on the difference of the altitudes. The error in the longitude

will be

15 x 30"

For the equal altitudes = 2'

For the unequal altitudes

2 x 1 '-54 x 72m. -2

15 x 30"

2 x 1-54 x 77- '5

= r-9.

Ex. II December llth, 1894, in lat. D.R. 48 10' N., long. D.R. 12 15*

W., the following observations were taken near the meridian, to find the

longitude and latitude at apparent noon.

Obs. Alts, sun's I 1.

Chron. times.

h. m. s.

7 23

7 38

11

44

17

17

45

57

10

9 17

9 23

17 58

17 45 10

Index error + 2' 10" ; height of the eye 31 feet. The chronometer being

slow 4h. 18m. 23s. on G.M.T., and the ship steaming S. 22 W. (true) at 14

knots.

16

THE EX-MERIDIAN ALTITUDE PROBLEM.

From Traverse Table. S. 22 W. 14' d.lat.13' dep. 5' "24 = d.Iong. 7 ''83 = 31 '32 seconds.

Observer's change in lat. 13' S. per hour or 13" per minute.

Sun's ,,dec. 12" S. 0"'2

Towards approachment K 12" - 8 ,,

t - K

~ 2(7 R= Ct*

12"-8 = l"-28 (5mO a

2 x l "' 28 = 32" Table IV.

= 5 min.

The ship's run in longitude during 5m. is about 2'5s. westerly. Therefore the corrected

hour angle is 4m. 57*5s. (See Rule B. under Max. and Min. Alts.}

EQUAL ALTITUDES.

Green, date.

(1. h. m.

Deer. ..11

Long. D.R. 49

THE IST AND 4TH OBSERVATIONS.

Sun's decn. J quation of time.

S. 23 2 16'6 ' 29 S; 98

Cor. 9-6 Cor. -92

Deer. ... 11 49

Chron.

h. m.

7 33

9 23

23

^^

times.

s.

11

8

2 26-2 6 29-0 - to app. time.

Green, date of 1st Obs.

h. m. s.

7 33 11

Error 4 18 23 slow.

7

1

1

49

48

57

57'5 W.

59-5

g (in Ih. 50m.) .

7 - q .

23 51

^^ssssssss

Observed altitude

34

___

1*7 4*5

10

10

58

30

54

58

32

A, i (I - a) .

h.

54

m.

1

4

54

297

la\

WtJ

57-5

297

S, A. T. of mer. alt

t (ariproachment) , .,

S. A. T. of max. alt

A! (always subtractive)

S. A. T. of 1st observation

Equation of time

S. M. T. of 1st observation

Online Library → Charles Brent → Ex-meridian altitude tables, declination (0 -70 ), to which is added an explanation of maximum & minimum altitude, longitude as well as latitude from two observations of a heavenly body when near and on opposite sides of the meridian. Also a solution of the new navigation method → online text (page 1 of 18)