Navigation and Nautical Astronomy, Stebbing

NAVIGATION

AND NAUTICAL ASTRONOMY

NAVIGATION AND

NAUTICAL ASTRONOMY

BY F.

C.

M.A. STEBBING, u

CHAPLAIN AND NAVAL INSTRUCTOR, ROYAL NAVY; FORMERLY EXAMINER IN NAVIGATION AND NAUTICAL ASTRONOMY TO THE ROYAL NAVAL COLLEGE, GREENWICH

MACMILLAN AND

CO., LIMITED NEW YOKK: THE MACMILLAN COMPANY

1903 All rights reserved

Astron.

First Edition, 1896.

Second Edition,

OLA40OW

:

il

1903.

FMVTED AT THE UHIVERHITY PREM IUOU MM K vsi. 00. '

i

DP*

PREFACE. THE

earlier portion of this book deals with that division of the subject which does not require a knowledge of astronomical definitions, nor, with the exception of Great Circle Sailing, of

Spherical Trigonometry -the investigation of Eaper's Rules for finding the distance from a Mountain Peak being inserted in ;

"

the chapter on Fixing a ship's position on a chart." The second portion the various necessary definitions having been stated and an explanation given of the method of constructing Nautical Astronomy diagrams consists of the discussion of Time, Greenwich Date, Sextant, Altitudes and the rules for their correction Longitude, Latitude, Chronometer and ;

Compass Errors concludes

the

;

Day's

Work and

practical

really

Sumner's Method. This and follows then a portion,

chapter containing the investigation of the various corrections,

and of the errors produced in Longitude, small errors in time or observation. These Latitude, etc., by errors are treated geometrically, and are thus shortly and simply explained without the use of Differential Calculus (not Dip, Parallax,

etc.;

in an elementary work) or the long analytical methods which usually take its place. The last chapter contains the investigation of Latitude by

admissible

observation of the Pole Star

Equation of Equal Altitudes (the method a short and simple one, much preferable to the ordinary method, which is, however, inserted as being familiar to many) Time at which the Sun will dip in a ship steamand a few other ing at a high rate of speed Interpolations ;

first

;

;

;

problems of interest in

A hoped

JSTautical

Astronomy. number of Miscellaneous Examples is added, which will

ingenuity.

provide, in

The

many

examples

cases,

exercises

throughout

367212

the

for

it

is

the student's

book

are

either

PREFACE.

vi

(ID

original

so)

such

or

the case of the Practical examples almost entirely as have been set in the various examinations

connected with the Royal Naval College, Greenwich. It must add considerably to the utility of a work if

the student

Navigation Nautical Almanac

is

able to actually

make

on

use of the

when working out examples, instead of the necessary data provided at the end of the questions. having As, however, Nautical Almanacs go out of print, and it would make too bulky and expensive a book if the whole or a large Almanac were

bound

up with it, all the composed for three examples requiring months of the year 1895, viz. April, June, and December,, which provide a considerable North and South Declination, and a medium Declination of the Sun, a sufficient variety the

of

portion

use

its

been

have

,

for all practical purposes.

Those

pages, therefore, or to which reference

elements Sun's

Noon

;

pages which contain the examples, viz.

of

parts

made

is

in

Declination, Equation of Time, Sidereal Time at Mean the Right Ascension, Declination, Semi -diameter, Hori-

and Time of Meridian Passage of the Moon Lunar Distances the Right Ascension and the one Planet to which reference is made together with the Pole Star Tables are bound up at the

zontal

Parallax,

;

the

requisite Declination of

;

;.

end of the work. The Right Ascension those

as

included, sufficiently

near

and

recorded

for

all

Declination in

any

Stars

of

Nautical

are

not

Almanac

are

nor are any practical purposes Differences in Lunar Observations, :

such as Second which do not vary with the year. Kvtiv care has been taken to ensure the correctness of the answers to the examples, but the notification of ;m\ errors that may be found will be gratefully received. It will be noticed that no definite rules are given for the tables,

solution etc.

of

The

the various problems principle of

in

finding latitude, longitude,

any problem is explained, the ne< -es.-su examples fully worked in illustration, -\

tommlaj given, and which should be sufficient, as

all

who have

to

make the

calculations requin-d in Navigation problems must be familiar with the use of Logarithms, and there would appear to be

more education ni'j

rale*

in

u..rkin^

thus,

than

in

mechanically

PKEFACE.

vii

For convenience of reference, the necessary definitions are printed in italics and numbered consecutively, and the principal formulae are collected at the end of their respective chapters. The Tables referred to are Inman's, which are in general use in the Eoyal

Navy

;

but any set of Tables will

suffice

in almost

remembered that the Log Haversine of every case, Inman is exactly the same as the Log Sine Square of other Tables, while the Half Log Haversine is half the Log Sine if

it

is

Square.

have also tables of Natural Sines, Cosines, the Natural Ver sines may be obtained by etc., from unity. the Cosines subtracting N.B, The word Log is omitted in the working of the examples, as it is easily understood that the numbers placed

Many

Tables

from which

against the various Sines, Cosines, etc., or the natural numbers, are the requisite logarithms from the Tables. The multiplication of methods of solving the various prob-

lems

has

been

book being which will

avoided

as

far

as

possible,

the

aim

of

the

Navigation with methods understand the straightforward

to provide the student of

principles of

enable

him

to

Navigation, leaving

it

to himself

to

select

after-

he pleases, any of the various plans for shortening his calculations, of which so many are to be obtained. The Author's thanks are especially due to Mr. H. B. Goodwin, of the Eoyal Naval College, Greenwich, and Mr. J. M. Pask, " of H.M.S. Britannia," from whom he has received valuable suggestions and advice. wards,

if

F. C. S.

H.M.S. "Cambridge,' June, 1896.

PREFACE TO THE SECOND EDITION. A

FEW alterations have Work has been practically Method

The chapter on Day's the examples of Sumner's have been re- worked so as to bring them into accord-

ance with present

made.

been

r^- written

day methods

;

and mention

;

is

made

of

a

determining the position line when the Sun's Altitude This is due to Mr. near the Meridian is greater than 60.

method

of

W.

D. Niveh, C.B., Director of Studies at the Eoyal Naval College, Greenwich, who also supplies a method of determining

formula

the

based

for

a

Kinematics, Author.

Interpolation, modification of

on

method

a

the

of

principles

by

proposed

the

v

A

few examples have been added at the end of Chapter " XVII. to show how " errors may be readily determined by the aid of

"

position lines."

The Double Altitude method been retained, as observations

of

of

finding

this

the

latitude

description are

has

still

on

those required at the yearly examination of Midof the shipmen Eoyal Navy. Moreover, it is advisable to have a method of finding latitude which is independent of the

the

list

of

error of the chronometer.

Every effort has been made to correct misprints and errors, and it is hoped that few now remain. A number of miscellaneous examples has been added, selected from those recently set at the Eoyal Naval College. F. C. S.

November,

1902.

CONTENTS. c

CHAPTER

I.

PAGE

DEFINITIONS IN NAVIGATION,

INTRODUCTION.

CHAPTER

CHAPTER THE

DEAD RECKONING.

THE

8

III.

SAILINGS.

MISCELLANEOUS EXAMPLES

CONSTRUCTION

OF

MERCATOR'S 25

CHAPTER METHODS OF FIXING A

GREAT CIRCLE

SAILING.

FIGURES,

ON A CHART,

58

-

V.

COMPOSITE SAILING,

CHAPTER ASTRONOMY

IV.

SHIP'S POSITION

CHAPTER

NAUTICAL

-

-

CHART,

68

VI.

DEFINITIONS.

CONSTRUCTION

OF

-

76

CHAPTER SEXTANT.

LOG LINE,

CONSTRUCTION AND USE OF TRAVERSE

"SAILINGS."

TABLE. IN

II.

CORRECTION OF COURSES.

COMPASS.

1

VII.

VERNIER.

CORRECTIONS IN ALTITUDE. ARTIFICIAL

-

-

HORIZON,

-

86

CONTENTS.

xii

CHAPTER

VIII. PAGE -

-

TI.MK,

CHAPTER To CONVERT ARC INTO TIME,

IX.

GREENWICH DATE.

ETC.

CORRECTION OF ELEMENTS FROM "NAUTICAL ALMANAC,"

CHAPTER PROBLEMS ON TIME.

ERROR OF

LONGITUDE.

ALTITUDE;

(2)

LATITUDE.

XI.

CHRONOMETER;

(1)

BY SINGLE -

138

XII.

MERIDIAN AND EX-MERIDIAN ALTITUDES.

TION TO MERIDIAN.

REDUC-

DOUBLE ALTITUDES. POLAR ANGLES.

CORRECTION FOR RUN.

POLE STAR,

CHAPTER ('(MI'ASS

HOUR ANGLES,

BY EQUAL ALTITUDES,

CHAPTER

112

X.

MERIDIAN PASSAGES.

CHAPTER

100

151

XIII.

TRUE BEARING OF TERRESTRIAL OBJECT.

Ki:i;ni;>.

MAXIMUM AZIMUTH,

-

CHAPTER

XIV.

CHAPTER

XV.

BUHNER'S METHOD,

WnuK.

_'!:;

<

LONGITUDE BY

l.i

\\K,

M \ITI.I;

\\

i

CONTENTS.

CHAPTER

xiii

XVII. PAGE.

OF CORRECTIONS (Dip, PARALLAX, REFRACINVESTIGATION OF ERRORS IN LATITUDE, TION, ETC.).

INVESTIGATION

HOUR ANGLE, COMPASS, ETC., ARISING FROM SMALL ERRORS IN LATITUDE, ALTITUDE, ETC., -

CHAPTER MISCELLANEOUS PROBLEMS, TUDES.

241

XVIII.

POLE STAR.

EQUAL ALTITABULAR VALUE OF AUXILIARY ANGLE A.

MINIMUM TWILIGHT.

ETC.

INTERPOLATIONS,

ETC.,

-

-

268

MISCELLANEOUS EXAMPLES,

286-

NAUTICAL ALMANAC,

305-

ANSWERS TO EXAMPLES,

323

/

INDEX,

-/

331

NAVIGATION AND NAUTICAL ASTRONOMY. c

CHAPTER

I.

INTRODUCTION. DEFINITIONS IN NAVIGATION.

means

Navigation, as a general term, denotes that science by of which the place of a ship on the sea, and her course

to or

from any given

1.

iv

includes

the

point, are determined.

This, of course,

of

winds, currents, tides, local magnetic The necessary attraction, etc., which have to be considered. information on these heads is to be obtained from the sailing directions,

tide tables,

Navigation

and

is

compass manuals,

etc.

;

and the term

usually restricted to the finding a ship's position

direct course.

This the

effect

may

common

be done in two ways (1) by the application of rules of Plane Trigonometry, the necessary; angles :

and distances being supplied by means of the compass arid lo^' line; (2) by means of astronomical observations, treated by the rules of Spherical Trigonometry. It has been proposed to call

the

first

of

these

methods

geo-navigation (y?j, the earth), and the second cselo-navigation (ccelum, the heavens), but the terms Navigation and Nautical

Astronomy have become sanctioned by long

use,

and are there-

fore retained. 2. The earth is very nearly a sphere, the length of the longest diameter (about the equator) being 7,926 miles and that of the shortest (that which joins the poles) being 7,899 miles. For all the practical purposes of Navigation the earth is con;

sidered as a sphere. S.

N.

A

NAVIGATION AND NAUTICAL ASTRONOMY.

2

N

it

The Axis of the earth is that diameter about which revolves, with uniform motion, from west to east.

Def.

1.

Def.

2.

meets

The Poles of the earth are the points where surface; as N, S in the figure.

its

Def.

The Earth's Equator

3.

equidistant

from

its

poles; as

is

a great

circle

on

its

its

axis

surface

EW.

N.B. The plane of a great circle passes through the centre of a sphere, the plane of a small circle does not.

Def. pole

4

>'

Meridians are great ///'!

circles

which pass through

the

earth.

The Latitude of a place is the arc of a meridian intercepted between the equator and the place, measured from Def.

5.

N

A place north of the equator is said to be a place south of the equator in south latitude. Thus AL is the north latitude of the place A, is the south latitude of the place D. to

90,

or S.

in north latitude,

DK

Def.

thus

6.

Tfie

Co-latitude is the complement is the co-latitude of A.

NA =90-4Z

Def.

7.

of the latitude;

Parallels of Latitude are small circles ivhose planes

are parallel

to

tfte

plane of

the equator, all places

on

a parallel

DEFINITIONS IN NAVIGATION. having the same

Thus

latitude.

tude, AL = BK = CM. Def. 8. A Nautical

be a parallel of

lati-

is equal to the mean length of a reckoned as 6080 feet.

Mile

and

minute of latitude,

ABC

if

3

is

Def

9. The True Difference of Latitude between two places arc of a meridian intercepted between their parallels, expressed in minutes of arc or in nautical miles.

the

is

Thus the true arc

ED

difference of latitude

between their respective

B

between

and

P

is

the

parallels.

evident that, if the places are on the same side of the equator, their true difference of latitude is obtained by subIt is

tracting the less latitude from the greater, while, if they are sides, it is obtained by adding the two latitudes

on opposite together.

the

is

Thus the true

sum

latitude

difference of latitude

as the place arrived at or in place left or from. Examples. 51

BK

(of

B),

KD (of P).

The true

lat.

between B and P and the south

difference of latitude

of the north latitude

26'

is

N

marked

S according

or

to the north or south of the

is

Find the true difference of latitude between Cape Clear, N, and Cape Finisterre, lat. 42 54' N. Lat. from, Lat. in,

-

51

26'

42

54

8

32

N N

60 true

diff. lat.,

-

-

-

512 miles S

-

Find the true difference of latitude between Cape and Cape Verd, lat. 14 43' N. Lnt. from, Lat. in,

St.

Roque,

-

5

28'

S

-

14

43

N

20

11

lat.

60 true

A

-

diff. lat.,

ship sails from 42

14'

N, 215 miles north,

Lat. from

true

diff.

lat, 215'

Lat. in,

N=

1211 miles

N

find the latitude in.

42

14'

3

35

45

49

N N N

5 28' S.

NAVIGATION AND NAUTICAL ASTRONOMY.

A

ship sails from latitude 27 Lat. from, true diff. lat., 320'

Find the true difference

N N

(1)

27

(2)

47 25

(3)

18 25 S

16'

320 miles north, find the latitude

N=

15'

S

5

20

N

39 41' 37 43

N N

Lat. from.

36 17 S

diff. lat.

(4)

35

(5)

12 27

(6)

19'

in.

55 S

of latitude in the following cases

Lat. in.

true

27

21

-

Lat. in,

Lat. from.

15' S,

S

N

3 19 S

:

Lat. in.

18 45' S 5

16 S

22 47

N

DEFINITIONS IN NAVIGATION. Examples.

Find the difference

of longitude

5

between Ushant and the

east point of Madeira.

diff.

5

-

Long, from, Long, in,

-

long.

W W

3'

16

39

11

36

696'

W

Find the difference of longitude between the Cape of Good Hope and Tristan d'Acunha. -

Long, from,

Long,

in,

18

29'

12

2

30

31

E

W

60 -

diff. long.,

Find the difference of longitude between 120

W

1831'

and 79

-

Long, from, Long, in,

-

-

W E.

120

W

79

E

199

E

360 161

W

60 9660'

diff. long.,

W

W

A

and changes her longitude 236 ship sails from longitude 1 20' miles to the eastward. What is her longitude in ? -

Long, from, diff.

long., 236' E,

Long, Examples.

2

in,

Find the difference

Long. from.

Long.

in.

20'

W

56

E

36

E

of longitude in the following cases

:

NAVIGATION AND NAUTICAL ASTRONOMY.

6

Rhumb Line or Loxodromic Spiral is a curve meridians at the same angle. generally an equiangular spiral on the surface of the

Def. 14. The ivhich cuts all It is

sphere, always approaching the pole, but never actually reaching it; the pole must always bear due north, and therefore cannot

be reached on any course except due north. Def. 15. The Course Steered is the angle between a meridian

and

and

the ship's fore

Def. 16. The

meridian and

aft line.

Made Good is rhumb line joining

Course the

the

angle between the

the place left

and

the

place arrived at Def. 17. The Distance between two places, or the Distance

Run

by a ship on any course, is the length of the arc of the rhumb line expressed in nautical miles. The Departure is the distance, in nautical miles, made good, east or west, by a ship sailing on a' rhumb line or it is the distance, in nautical miles, between two places on the same It is marked E or parallel of latitude. according as the ;

W

si

course has been towards the east or the west.

lip's

Def. 18. // an infinite number of points be taken on the line joining two places, the meridian through each of these points cuts off an arc of the parallel of latitude through the

rhumb

next point.

The

sum

of all these arcs is called the Departure.

(a;

I'!...

P

is

on

the place

"t

an

tl..

'I.

..mils poles; A is the place left, and h I'M-:. PBQ be meridians ;mrof a nautical mile (6080 feet) as tinto the length in which the sand-glass runs out bear- b0 of seconds

number number

the

:

of a knot

of seconds in an hour.

standard glass kimt should be

is

a 28-seconds glass;

$$Jx28 = 47'.S

fWt

In the Royal Navy tin limn- tin- length of a

nearly.

When

the ship's

greater than about six knots, a 14-seconds glas^ jx used, and tin- numlx-i- 0

S

ditf.

cos mid. lat,

9*818030

sec course,

diff.

2-371068

diff.

long.,

2-217484 2-354272

2-217484

lat,

tan course,

distance = 226'!'.

9'971614

Course, S 43 (2)

W

10'136788

lat.,

12-189098 diff.

long., 235'

7'

45"

W.

Find the latitude and longitude Lat. from, 27

15'

S

;

having given

in,

long, from, 93 21' distance, 325' miles.

cos course,

9'685571

distance,

2

^ 11^8 83

E

course,

;

tan course, diff lat

10'256248

sec mid.

1Q-Q56376

W

;

2'1 97454

-

_

2-197454

S 61

lat.,

'

_2

o

,

diff.

long.

=323-7'

Lat. from,

27

g 15 S

Lat. in,

29

53 S

Long, from, 93__21

27

15

Long,

2

Mid. lat, Examples.

(1)

2 510078

=157-6'

diff. lat.

)

3g

57

8

28

34

=5

Find the course and distance from

in,

87

A

to B,

24'

57

W W E E

having given

W W

NAVIGATION AND NAUTICAL ASTRONOMY.

38 This

obtains

(A.D.

its

name from Gerard

Mercator, a

Fleming

who

1569), chart in which

appears to have been the first to construct a the length of the degrees of latitude increased

as they receded from the equator. His E. Mr. of Caius upon by Wright, College,

method was improved Cambridge (A.D. 1599).

Suppose a globe or sphere, with meridians, parallels of rhumb lines, etc., drawn on it, to be circumscribed hollow a cylinder, which touches it all round the equator, by and suppose that this globe is continually expanded until each 22.

latitude,

point of it, in succession, touches the cylinder, the expansion at that point then ceasing, while that of the portion of the sphere not in contact with the cylinder still goes on. The meridians will become straight lines, and the parallels circles

on the surface of the cylinder, the former in the direction of the axes, the latter parallel to the base. If now the cylinder be unrolled and spread out into a plane, the surface thus produced will represent a Mercator's chart,

which gives a true representation of the form of each small portion of the earth's surface but varies greatly as to the scale on which these portions are represented in different latitudes. ;

The polar

regions, in particular, are very

much

distorted.

23. From the above description it appears that the distance between the meridians, on a Mercator's chart, is everywhere

made equal

to the

difference of

longitude, the

meridians becoming straight lines at right angles to the equator.

Each

parallel of latitude, therefore,

been increased in the ratio of sec of

the

parallel sailing principle which cuts all the meridians at the ;

becomes a straight

line

;

must have

1 by the rhumb line,

lat.

:

,

same angle, and the arcs of the

meridians must, therefore, to preserve the relative /proportions of the small figures, such as

ABC, CDE(&g.

18),

be also increased in the same

proportion as the parallels of latitude, i.e. sec lat.:l. Hence the degrees of latitude on a Mercator's chart increase as they recede from the equator, as the latitude increases; and all the parallel- f latitude, and every part of them, are larger than they are on the secant of

the globe, in .the proportion sec

lat.

:

I

THE

SAILINGS.

39

Hence also, though the latitudes and longitudes and bearings of places are accurately represented on a Mercator's chart, the distances are distorted in various proportions. Def. 37. The

Meridional Parts of a certain latitude give in nautical miles, of the distance on a

the length, expressed

Mercator's

chart

from

the

equator

to

the

parallel

of that

latitude.

Def. 38. The Meridional Difference of Latitude between two places is the length, expressed in nautical miles, of the line on

a Mercator's

chart,

To calculate 24.

Let

P

of the earth

of latitude.

the

Meridional Parts for any latitude.

PV

represent two meridians on the surface the equator, AB, CD, etc., arcs of parallels Let these arcs be expanded in tig. 19 (2) into

7 and

;

which represents the difference of latitude.

UV

the distances ab, cd, etc., on a Mercator's chart, to the difference of longitude.

uv being equal

UV

(2)

FIG. 19.

Then panded

the differences of latitude similarly into bd, df,

etc.

=

etc.,

etc.

BD, DF, where

etc.,

must be

ex-

NAVIGATION AND NAUTICAL ASTRONOMY.

40

BD = DF=FH... and of B being Then

= l'

of

bd + df -\-fh- meridional difference of latitude between

B

Now

suppose that

latitude, the latitude

)

that each

I.

and# = sec lat. B + sec lat. D +

Now

suppose

.B to

. . .

be a place on the equator, then

bd + df+jh = meridional parts for

= sec + sec 1 In the same

way

it

may

3'

'

-f-

sec

2'.

be shown that meridional parts for

I*

= secO+secl'+sec2'+...+sec(r-l').

A

closer

tances

approximation

BD, DF,

etc.,

may

each

= I",

be obtained by taking the disbut the above is sufficiently

For accurate calculation practical purposes. parts a formula obtained by means of the and the fact that the earth is Integral Calculus is required not a perfect sphere must be taken into account.

accurate for of

all

meridional

;

When

the

calculated,

parts for any latitude have been any other latitude may be found by

meridional

those

for

multiplying the true difference of latitude by the secant of the true middle latitude, which can be obtained by means of

Table

6,

Inmcins

Tables,

new

edition

(cf.

2G),

and adding

the result to the given meridional parts. To obtain the meridional difference of latitude between

two from the Table the meridional parts for each places take latitude, and then take the difference or the sum, according as the latitudes are of the same or different names.

To show that tan course = iner.

25.

chart,

lat.

20) represent a portion of a Mercator's the positions of two places on it. Completing

Let the figure

A and B

dm.

(fig.

the right-angled triangle ABC, EG will represent the difference f the meridional difference of latitude. longitude and

AC

Suppose that the construction

BC = departure x m, then by AC must = diff. lat. x m.

the principle of

THE SAILINGS. XT

Now

tan course

dep. = dep. x ra. = diff. long. = ,.* --= .-~ry-r ,.^ dm. lat. dm. lat. x m mer. din. lat.

CAB If

AD

41

-

represents the course from

be the true

diff.

A

between the two

lat,

.

r

,

,

AE

to B.

places,

and

DE

be drawn parallel to CB, will represent the true distance between them, because the relative proportions of distances, etc.,

on the globe are preserved on the chart.

AE = ADsecEAD,

.'.

distance = true

or

These two formulae solve

all

diff. lat.

x sec course.

problems in Mercator's

sailing.

FIG. 20.

Considering the two expressions dep. tan course = ,.~.^ diff. lat.

diff'. long. tan course = - - , r^-P mer. diff. lat.

-

we

i,

j-rr

.

obtain

diff.

by means

of

departure,

if

-

,

= dep. mer. i

i

long.

,

lat.

diff'. ,

,

which the

diff. long, may be obtained from the without the use of the middle latitude. desired,

26. Also, since

dep. cos true mid. lat.

= diff. long,

x cos true mid.

lat.,

diff. lat.

=

mer.

,..

,

,

.

diff. lat.'

which gives the latitude in which the distance between the made good between two

meridians is equal to the departure places. If the course is

error in

it,

the

nearly 90, and there is any probability of long, obtained from the expression

diff.

diff.

long.

= mer. diff. lat.

x tan course

NAVIGATION AND NAUTICAL ASTRONOMY.

42

may

very rapidly. In this case

it

method, which

is

from

tangents of angles near 90

be incorrect, as the

change

be preferable to use the middle-latitude used, in practice, for obtaining the diff. long. the departure in the ordinary daily reckoning.

Examples.

may

Find the course and distance from

(1)

Latitude.

A,

50

B,

47 30

15'

A

to B, having given

Longitude.

N N

A,

27

19'

B,

31

14

W W

Her. parts.

Lat. from, 50

15'

47 30

Lat. in,

N N

2 45

165'

diff.

mer.

Long, from, 27

19'

3246 '91

Long,

31

14

W W

3 55

AN'

in,

250-96

60 diff. lat.,

3497'87'

Mer.

60

diff. lat.

S

long., 235'

diff.

long., diff. lat.,

tan course,

2-371068

sec course,

2-399605

diff. lat.,

10-136728 2-217484

9*971463

Course, S 43

7'

15"

W

2*354212

W

226'5'

distance,

the course differing by about 30" and the distance by less than half a mile from the results obtained by middle-latitude method. (2)

Find the latitude and longitude in, having given lat. from 27 15' S, long, from 93 21' E, course S 61

W,

distance 325 miles. Mer. parts.

cos course,

9*685571

27

15'

1700*37

distance,

2-511883

29 53

1880*30 179-93

2-197454 diff. lat.

=157-6

=2 Lat. from, 27 I.

.

>

38'

mer.

S

2'255104

diff. lat.,

tan course,

10*256248

15 :.:

_':>

long.

=324-6 25'

W

21

E

87 56

K

=5 Long, from, 93 Long,

from

1352

dill',

wl.i.-h differs

1

tin- iiii/, margin if it is to represent south latitude; or at a suitable position towards the centre if both north and south latitudes 30.

pa*per if

are to be represented; or it may be found convenient to take, as the line of reference, a parallel about the middle of the Divide this line into equal parts to represent longit lullpaper. according to the scale on which the chart is to be constructed,

MERCATOR'S CHART.

53

At each of degrees of longitude required. Take out the meridional erect parts extremity perpendiculars. for each convenient degree of latitude for the limits between and the number

be drawn, and take the difference between each successive pair, thus obtaining the meridional differences

which the chart

is to

of latitude.

As the given scale represents CO miles of longitude, multiply by the number of miles in the meridional differences and divide by 00 the results will be the lengths, on the given scale, between the chosen degrees of latitude. Lay off these the and on lengths successively through the perpendicular lines, the scale

;

points thus obtained (but not through half degrees, unless the scale is very large) draw straight lines parallel to the original

represent parallels of latitude. Draw also straight lines at convenient intervals parallel to the perpendiculars to represent line, to

meridians.

The frame of the chart is thus completed. Its accuracy should be tested by seeing whether the two diagonals of the rectangle thus formed are equal, and whether the intervals representing longitude are of the same length at each end of the chart.

The intervals representing latitude and longitude should then be divided conveniently, the principal divisions numbered, all lines inked in, a neat margin added, and pencil lines rubbed

when the chart may be required,

out,

as

is ready for use, such positions, land, being inserted.

etc.,

31. (1) To fix a position on the chart. Place the edge of a parallel ruler along the parallel of latitude nearest to the given latitude move it until it passes :

on the graduated edge of the chart. through Measure the distance of the given longitude from the nearest meridian, and lay off this along the edge of the ruler. The position thus obtained is the position required, which should this

latitude

be marked by a small

The

cross.

and longitude from the chart by the reverse latitude

of any position of this method.

may

be taken

To lay off courses or bearings. charts for actual use compasses are drawn; those for use near the land, on the magnetic meridian, marked to quarter points, and, in recent charts, to degrees. (2)

On

NAVIGATION AND NAUTICAL ASTRONOMY.

54

On

those

made

more straight

for practice

sufficient

is

it

draw one or

to

representing the magnetic meridian and of a meridian and a parallel. the intersection passing through Correct the given compass course or bearing for the deviation lines,

due to the direction of the ship'* head, so as to obtain the magnetic course or bearing. Lay this off' from the magnetic meridian a line parallel to it through the given position will ;

be the required course or bearing. The intersection of the bearings of two or more points thus laid off fixes the position of the ship as she must be somewhere ;

on each

and, therefore, at their intersection. The converse of this method will obtain the course between line,

any two

positions, or the bearing of

any

point, etc.

To find the distance between two point*. The distance is found (nearly) by transferring the interval between the two positions to the graduated meridian, as nearly (3)

as

possible opposite to the

i.e.

positions,

as

much below

more southern latitude as above the more northern

:

the

this space

the distance required. If the two places have the same latitude, half the distance should be measured

turned into minutes

is

on the graduated meridian on each side of the parallel of latitude the total space measured, expressed in minutes, is If the places have the same (nearly) the distance required. longitude, the algebraic difference of their latitudes will be ;

the distance between them. off a given distance, take it, similarly, from the meridian and apply it to the line representing the graduated

(4)

To lay

course or bearing.

The whole subject may be ample, worked in

illustrated

by the following ex-

full.

Example. Construct a Mercator's chart on a scale of 1'5 inches to a to 55 30' N, and from 8 degree of longitude, extending from 53 30' to 11 W. Place on it the following positions (fig. 21):

W

N

Clare Island Light,

-

f>3

Achill Head,

-

53 58 :

H.-a.l,

Carrigan

RathliuO'Birii.- Ma.,,1,

Home

Head,

NOTE. hijjh

F

i

37

54 40 55 14 55 17

Tory Island Li-ht, in

1

50'

notion of charts on the

latitudes, see

Great

Ciivl.-

Sailin-.

N N N N N N <

9

5!

\\

Ifi

\V

8 41

\V \\

10

8

f>

8

o

W

8

15

W

Jnomonic projection, fur

DM

MERCATOB'S CHART.

A

ship's position 20' W.

was

fixed

55

by Cross Bearings, the variation allowed

being 25

^A

/Clare Island Light, A UMI TT lAchill Head,

Afterwards sailed as follows oi/aiiuaiu

Compass Course.

SEbE) ^TXTT^ NNE

1

:

fi J

... T, no KA Deviation, 2 50' ,

^ W.T

56

NAVIGATION AND NAUTICAL ASTRONOMY.

Tory Island

Light/

Home

\Rathlin O'Birne Island

^

Achill

\^Carrigan Head

B

Head

\ k-

A

Clare Island Light

8V

10 JO

Fio.

DR

Head*

MERCATOR'S CHART.

57

same over the whole chart, draw a line making an angle of 25 20' with a meridian to represent the magnetic meridian, from which to lay off the courses and bearings. (If desired, the courses, etc., may be corrected for variation as well as deviation, and the results laid off from the true to be the

meridian.) (2) (3)

Fix the given positions. Correct and lay off the bearings and coursesClare Island Light.

S 56

15'

2 50

E

W

Achill Head.

N

22 30' 2

50

E

W

1st Course.

N

28

8'

2 50

W W

CHAPTER METHODS OF FIXING A To lay

off

on a chart

IV.

SHIP'S POSITION

the

bearing of

any

ON A CHART.

object.

32. Correct

the given compass bearing for the deviation due to the direction of the ship's head. Measure this corrected

which will be the magnetic bearing, on the nearest compass drawn on the chart (such compasses being constructed on the magnetic meridian), by placing the parallel rulers so bearing,

that the edge passes through the centre of the compass and the requisite degree on the circumference. Then move the ruler

till

the edge passes through the object, when the line If will represent the bearing required.

drawn along the edge

the compass on the chart is not marked to degrees, correct the magnetic bearing for the variation, and measure off this true bearing from the true meridian. A line parallel to this through the object will represent the bearing required. (1) Cross Bearings. 33. If the bearing be taken of a headland, lighthouse, or other well-defined object, and this bearing laid off on a chart, it is manifest that the position of the ship must be somewlinv

on this

line.

Similarly, after,

on

if

be laid

tliis

line.

the bearing of another object, taken immediately the position of the ship must be somewhere And, as the only point common to two lines is

off,

the point where they meet, the position of the ship on tinchart will In- at tin* point of intersect ion of the two lin.-^ .

>!'

bearing,

The objects should be chos.-n so that tin- lines ilo not intersect f intersection, in such at a very acute angle, as tin- point In practice tin- I>.-;irin-- of a tliinl Case, is s.me\\ hat -loul.t t'ul. always taken as a check. -,\

i

FIXING A SHIPS POSITION ON A CHART. (2)

By

bearing

and

sextant angle.

two well-defined

objects cannot be seen at the same time from the standard compass, the position may be fixed by taking the bearing of one object, and the sextant angle between 34. If

and another

it

object.

bearing of the first object, and from any point of it lay off the sextant angle. A line through the second to the line thus obtained, will cut the bearing object, parallel

Lay

off the

of the first object in a point

which

is

the position of the ship.

(3) Doubling the angle on the bow. 35. The angle between the direction

of

the ship's head

and the bearing of an object is called " the angle on the bow." Let this angle be observed and the time noted; and, when the angle on the

bow

doubled, let the time be again distance of the ship from the object is equal to the distance run by the ship in the interval. Lay off the second noted.

is

Then the

corrected bearing and on the line measure the distance run, the point obtained will be the position of the ship.

Proof. Suppose that the ship is proceeding in the direction ABC (fig. 22), and, when she is at A, let be the "angle on the A bow," i.e. the difference between the ship's

BAD

FIG. 22.

course and the bearing of the object D. Let B be the position of when the angle on the bow is doubled, i.e. Magn. N.

the ship

when CBD = 2BAD. Therefore the side ship

But CBD = BAD + BDA. must = BAD, and the side BD = BA, i.e. the distance of D from the

= the

BDA

distance run.

A

ship steering S 36 E by compass observes that a lightship bears S 8 E after running 2 miles the angle on the bow is observed to be doubled.

Example

(fig. 23).

;

Required the magnetic bearing of the lightship at the the deviation being 2

second observation

W

;

;

.4^=38

LAS=]0

IQ-

J3AL = 28 = angle on the bow,

CBS = 38

and

'

:.

i.e.

Q

LBS' = I8

magnetic bearing of lightship at second observation

FIG. 23. is

S 18

W.

NAVIGATION AND NAUTICAL ASTKONOMY.

60

off

Laying

N

E

18

2'

from the lightship the position

of the ship

is

The compass bearing will be S 20 W. In this manner the required compass bearing, when the angle has been doubled, may be calculated, and all that has to be done is to note the time when the obtained.

object has this calculated compass bearing. Or, since ALB=BAL, the position of the ship may be obtained by an angle equal to the angle drawing from L a line LB, making with

AL

on the bow, and on

this line

measuring the distance run.

Four-point bearing. This is a particular case of the previous method. The object is observed when its bearing is four points or 45 from the direction of the ship's head, and again when on the beam the distance from the object at the second observation (4)

36.

:

being clearly the same as the distance run in the interval. These observations are easily made by means of a brass plate, with radiating lines, let into the bridge, the compass therefore not being required. (5)

in

Two

bearings of the same

object,

and

the distance

run

the interval. 37.

Lay

off,

and the course

from the

object, the

second bearing reversed,

of the ship; on the latter lay off the distance

run in the interval, and from the point thus obtained draw a The point in which this parallel to the first bearing. meets the second bearing will be the position of the ship at line

the second observation.

A is

Table based on trigonometrical calculation of these positions inserted as Table 15 in the new edition of Inman's Table*.

In these cases, in which there is an interval between the observations, the results can only be approximate, as the ship's course will be affected by tides, etc. But, if the interval is not large, the results will be sufficiently accurate for the purposes of navigation. ((>)

Angles observed with the w.rfdnf only. When the view i'rom thr standard compass

38.

is

obstructed,

the position of the ship may be obtained by sextant angles. Three objects are selected, and the angle subtended between tin

are

small,

writing 2a+0ftin

etc.,

(2a

+ 0)0 = (2/3 + + 0X0 + #)>

FIXING A SHIPS POSITION ON A CHART.

65

Hence, multiply the distance run towards the object by the of the first altitude (in minutes) and half the distance,

sum

and divide the product by the difference between the second altitude (in minutes) and the sum of the first altitude (in minutes) and the distance run. The quotient will be the distance, in miles, at the time of the second observation.

and 42 are the investigation

41

the practical Raper's Practice of Navigation, 10th Edit.,

in

given

of

rules

355,

359, 363. c

43.

If

Danger Angle.

it

desired to pass outside a danger which does not

is

show above water,

this result

may

be obtained by calculating

the "danger angle," on the following principles:

A

and

R

B

be two well-defined objects on the shore, the position of the submerged rock, shoal, etc. If the angle be measured by a protractor, the same angle will be

Let

ARE

subtended between

A

and

B

at

any point on the segment

of

a circle passing through A, R, B, angles in the same segment of a circle being equal to each other. If, therefore, an angle less is

ARB

than

be placed on a sextant, so long as this angle A and B, the ship must be outside the

subtended between

segment

ARB, and

called the

therefore clear of the danger.

ARB

is

"danger angle."

If it is desired to pass at any given distance outside R, a circle must be described with as centre and radius equal

R

to the given distance.

A

must then be described through A, B, touching this and the angle subtended between A and B at the point circle, of contact will be the "danger angle." circle

S.

N.

E

NAVIGATION AND NAUTICAL ASTRONOMY.

66

whose height

If there be a point

vertical

accurately known, a

is

be calculated

may

danger angle

tan P JIB = -gig,

and so long as the observed than

less

is

PRB,

altitude of

P

the ship must be out-

side R.

The chart used in the following examples West Coast of Ireland, which provides a great number of points suitable for observation. N.B.

is

No examples have been given on the use of the Station Pointer or on the Danger Angle, as they belong to the subject of Pilotage rather than to

FIG. 31.

"

Navigation

"

as defined at the

was thought advisable Examples.

18246,

N.B.

commencement

to state the principles

of this book, though on which they depend.

it

All bearings are by compass. Cross Bearings.

(1)

Daunt's Rock Lightship,

S 87

-

(2)

(3)

Sheep Head, Three Castle Head, Mizen Head, Great Skellig Light,

Hog

Island Centre,

W"|V

N 22 W N29 E N 78 E

Roche Point, Poor Head,

S

-

-

50JJ

E

Deviation, 2 E.

J

]

[Deviation,

S2fi|E

J

N 2 E N 81 E

] 1

3

Deviation, 3

W.

W.

N51 E

Bolus Head,

Bearing and Sextant Angle. (4)

(5)

Old Head of Kinsale, Sextant angle between Old Head of Kinsale and Seven Heads, -

Kerry Head,

N

4:

E

Sextant angle to Loop Head, (6) Clare Island Light,

Deviation, 2

W.

Deviation, 2

E.

59

S 61

-

Sextant angle to Achill Head,

E.

73 50'

S 28 E

-

l

Deviation,

-

E

78

Doubling the angle on the bow. (7)

A ship steaming S 72 E 16 knots, observes the Fastnet Light 24 on A quarter of an hour after the angle on the bow was doubled.

the bow.

Deviation, 4

A

W.

N

84 E 15 knots, observes the Old Head of Kinsale ship steaming on the bow. After 20 minutes the angle on the bow was doubled. Deviation, 3 E. (8)

Light

16*

FIXING A SHIP'S POSITION ON A CHAET.

67

W

A ship steaming N 9 10 knots, observes the Great Skellig Light on the bow. Half an hour after the angle on the bow was doubled.

(9)

24

Deviation, 2

W. Four-point bearing.

A ship steaming N 8 E 12 knots, observes Black Eock Light 4 points the bow. A quarter of an hour afterwards the Light was abeam.

(10)

on

Deviation, 5 E.

N82W

15 knots, observes the Fastnet Light (11) A ship steaming 4 points on the bow. Twenty minutes afterwards the Light was abeam. Deviation, 5 E. (12) A ship steaming N 50 E 12 knots, observes Daunt's Eock Lightship 4 points on the bow. Ten minutes afterwards the Lightship was abeam.

Deviation, 4 E.

Two bearings of an (13)

5 41

E

A ;

object,

and

the distance

run in

N

46 E, observes Achill (14) A ship steaming after she has run 4 miles the bearing is S 18E. (15)

S 88

4W.

A

W

;

the interval.

ship steaming S 3E, observes the Great Skellig Light bearing after she has run 5 miles the bearing is N 57 E. Deviation, 7 W.

ship steaming after she has

Head bearing S Deviation,

85

E

;

4E.

N 60 W, observes Tory Island Light bearing run 3 miles the bearing is S 55 W. Deviation,

CHAPTEE GREAT CIRCLE SAILING.

V.

COMPOSITE SAILING.

Great Circle Sailing. 44.

The introduction

of steam

power having rendered steam-

ships practically independent of the direction of the wind, they are able to steer towards their port by the most direct route. This most direct route is the smaller arc of the great circle

passing through the position of the ship and that of the port. The larger the diameter of the circle drawn through any

two

points, the more nearly will the arc coincide with the chord joining them. And, as a great circle is the circle of largest diameter that can be drawn on a sphere, its arc will

be the shortest distance on any circle joining the two points. On a Mercator's chart the rhumb line is represented by a straight line,

and the great

circle

by a curve

;

so that, at first

may appear that the distance on the rhumb line is less than that on the great circle. But the rhumb line appears

sight, it

as a straight line because of the distortion of the chart; and, a series of points on it be taken, and their positions marked on a globe, it will be found that the length of a piece of string

if

which passes through these points will be longer than that which measures the arc of the great circle. If, then, a ship could be kept on a great circle, her head would be always pointing directly towards her port, and all her distance would be made good. This, however, would necessitate a continual alteration of

the course (unless, as

not very likely, the ship's track was due E or equator, or due N r S on a meridian), unlike the

W

is

on

the

rhumb

line

course, which remains the same, although the ship's head does not actually point towards the port until it is in si^ht. In practice, therefore, points an taken mi the ,

{ ''"./,'.'

} '.'

Equation of Time

Def. 109. The

is

the

,'

,>

:.">',

between

difference

mean and apparent time at any angle of the real and mean suns;

instant; or it is the polar or it is the arc of the equinoctial (expressed in time) between the circles of declination of the real and mean suns. Def. 110. The Greenwich Date

is

the

Greenwich astronomical

time of any observation.

Time. 60.

In order to obtain a distinct notion of that to which

we

give the standards of

between

the

name

of

time or duration,

measurement, successive

celestial

to

the

occurrences

we must

intervals of

certain

refer,

as

which

elapse well-defined

phenomena. Those which most naturally occur to us as most universally suitable, are, the rotation of the earth on its axis, and its revolution in

its

orbit.

NAVIGATION AND NAUTICAL ASTRONOMY.

102

The former provides the measure of a " day," the latter of a "year"; the interval between two successive transits of some point in the heavens, or some heavenly body, over the same celestial meridian being called a "day"; the interval between two successive arrivals of the sun, in its orbit, at some point in the heavens being called a "year." The interval between the sun's leaving a fixed point in the heavens and returning to it again is called a "sidereal year." The interval between two successive transits of the first point of Aries (which is practically a fixed point as regards daily motion) over the same meridian is called a "sidereal day."

These standards are invariable, and are therefore of in astronomy; but they are not adapted to the ordinary purposes of life, for which the sun provides the standard of measurement the interval between two successive 61.

great

use

:

the same celestial meridian being a "solar day": the interval between two successive passages of the % e.artb in its orbit (or the sun, if considered to motap^ thrpVgk "tkfi* first point of Aries being called a "solar of

transits

the sun over

called

year." ,",

..

f ',%

:

-\

* .Th*Q ."iengih or" "tlie

solar year

is

not, however, quite invariable,

to various irregularities in the motion of the sun and of the first "point of Aries. Observations of the sun's longitude,

owing

extending over long periods, have given as the mean length of the solar year, and called a "mean solar or tropical year," a period of 365-242242 days. 62. The length of the mean solar year differs from that of the sidereal year, because the first point of Aries is not fixed, the equinoctial points moving back along the equinoctial to

meet the sun 50'2" each year. This

movement

"

Precession of the Equinoxes," caused by the attraction of the sun and moon on tl itprotuberant parts of the earth, the earth being a spheroid and not a perfect sphere. A result is that the longitudes of the fixed stars incrmx-

and

is

called the

is

.">

signs of the zodiac have moved back about 30 since the time of Hipjwn-luis, and tin- first point of Aries, is not in the constellation of Aries though retaining the na

a year.

The

,

at

all.

TIME.

103

The equinoctial points make a complete revolution

in about

26,000 years, the poles of the heavens describing circles of about 47 diameter round the poles of the ecliptic. On account of this motion of the first point of Aries, the it again before he has described a complete the heavens, and his arrival at that point will precede his arrival at some fixed point by the time he takes to describe

sun will arrive at circle of

50-2",

about 20 m. 23

i.e.

s.

Hence a mean A.

solar year contains 365 d. oh. 365 d. 6 h. sidereal year contains

Length of Solar

Day

48m.

47s.

9 m. 10

s.

not constant.

the sun's motion in Eight Ascension were uniform, days would be of the same length. But this is not

63. If all solar

The sun does not move uniformly in its orbit, which is an ellipse and not a circle and even if it did move uniformly, the corresponding motion in RA would not be uniform, as the the case.

;

ecliptic is inclined to the equinoctial.

In order, therefore, to obtain a uniform measure of time depending on the sun, an imaginary body, called the "Mean

move along

the equinoctial, with the of the real sun. The days measured average angular velocity this sun will be their by equal, length being the average of all the length x)f the apparent solar days throughout the year ; is

Sun,"

supposed to

and a clock which goes uniformly may be regulated to the time shown by this mean sun, such a clock showing " mean time." It

that

is

necessary to fix a starting point for the

mean and apparent time may never

mean

sun, so

a

differ

by large Hence a body, which may be termed an imaginary sun, is supposed to move in the ecliptic with the average angular velocity of the true sun, and to start with the true

interval.

sun at perigee

(i.e.

when

the sun

is

nearest to

the earth).

Then the mean SUD is supposed to start from the first point of Aries at the same time as this imaginary sun. Hence the Right Ascension of the mean sun is equal to the mean longitude of the real sun, which is the longitude of the supposed imaginary sun. 64.

Mean noon

the meridian, and

is

the instant

mean

time

is

when

the

mean sun

reckoned by the

is

on

westerly

NAVIGATION AND NAUTICAL ASTRONOMY.

104

hour angle of the mean sun, measured from

hours to 24

hours.

These 24 hours of mean time constitute the astronomical

mean

day, which commences at mean noon. civil day, however, commences at midnight and ends at the next midnight, being divided into two periods of 12 hours

The

each.

Hence astronomical and number of hours

the same

civil

time are only expressed by day thus

in the afternoon of each

Jan. 10th at 3 P.M. civil time

;

is

Jan. 10th 3 hours

is

Jan. 9th 15 hours

astronomical time. Jan. 10th at 3 A.M. civil time

astronomical time,

15 hours having elapsed since the previous astronomical noon

on Jan.

9th.

From

this is

we

obtain the practical rule: "Civil time P.M. on represented in astronomical time by the same

any day number of hours on that day

but in order to represent

civil

time A.M. in astronomical time, 12 hours are added to the time and the date is put one day back."

civil

;

65. The equation of time is the difference between apparent and mean time at any instant, or is the angle at the pole or

the arc of the equinoctial, expressed in time, between the circles of declination of the real and mean suns.

The equation

of time arises

from the

fact that the ecliptic moves in its

inclined to the equinoctial, and that the sun orbit with varying velocity. is

separately, and the algebraic of the effect will be equal to that due to their combined

The two causes may be examined

sum

action. (1) Neglecting the fact that the sun's orbit is 66. Suppose the sun to describe the orbit AS'L

an

ellipse.

with uniform

angular velocity, while the mean sun describes the equinoctial ASL with the same velocity. Suppose them to start together from A. Then when the true sun is at B the mean sun will be at

C

t

where

CD

AB = AC.

If

PBD

be the sun's

circle

of

measure the equation of time, and it is declination, clear that C and I) will only coincide at the equinoxes and solstices, when the real and mean suns are on the same circle will

LENGTH OF SOLAR DAY NOT CONSTANT. of declination.

From equinox

D, and behind

it

from

to solstice

G

solstice to equinox.

of the earth (which revolves on

its

will be in

105

advance of

Hence a meridian

axis in the direction of the

PN

arrow) will arrive at

noon

D

before

it

arrives

at'

C,

and apparent

mean noon,

or the equation of time will be subtractive from equinox to solstice, and vice versa from solstice to equinox. will precede

To illustrate this practically let Af represent the ecliptic and Af the equinoctial between the vernal equinox and summer solstice.

NAVIGATION AND NAUTICAL ASTRONOMY.

106 Let Aa',

Af

and

Af

Through PcD, etc. It is

a, b,

...

c...draw the

circles of declination

PaB, PbC,

Aa is greater than AB, ef the sun's longitude at first changes more right ascension, but as it approaches the

then evident that though

than Ff, rapidly than its less

is

be divided into six equal portions Aa, ab

a'b' ....

i.e.

longitude changes more slowly than the right Hence the numerical value of the equation of time will increase to a maximum value and then decrease to at the solstice, where the real and mean suns are again on the same circle of declination.

equinox

the

ascension.

Solving the equation tan Aa = tan

by

giving to

AB

.

sec 23

successively the value

obtain b.

m.

s.

27' 1

h.,

2

h.,

...6

h.

we

.LENGTH OF SOLAE DAY NOT CONSTANT. Hence from perigee

107

apogee the meridians of the earth will before they pass over the real sun, mean noon will take place before apparent noon, and the

pass over the

to

mean sun

equation of time will be additive

;

and

vice versa

from apogee

to perigee.

The above explanation

is sufficient

as

an

actual calculation of the equation of time, must be consulted.

For the works on astronomy

illustration.

found that the greatest value of the equation of due to the obliquity of the ecliptic, is 10 minutes in time very nearly, while that due to the unequal motion in the orbit has 7 minutes as its greatest value. 68. It is

time,

Hence a simple graphical representation will show the value of the equation at different periods and also that it vanishes four times a year. Draw a horizontal line

XT

to represent the time, equal Draw a curve periods. equal representing so that its ordinates, i.e. the perpendicular distances of points

AAAA

intervals

B

A

20-

on

FIG. 52.

above and below XY, may represent that part of the equation of time for each day which depends on the obliquity of the ecliptic while the curve BBBB similarly represents that part due to the unequal motion in the orbit. will it

;

AAAA

cross

the line

XY

at

the

equinoxes and

solstices,

and the

NAVIGATION AND NAUTICAL ASTRONOMY.

108

greatest value of the ordinates will be 10 m. at intermediate times. will cross the line at perigee and apogee,

XY

BBBB

with 7 m. as the greatest value of the intermediate ordinates, positive values being measured above XY, negative values below.

The value of the equation of time due to the combined two causes may be represented by a curve whose ordinates are the algebraic sum of the ordinates of the two

action of the

former curves. Let

X'Y

be the curve

it

(parallel, equal,

confusing the figure. time vanishes about

December

24th,

and

CCCC drawn

with respect to the line to XY), to avoid

and similarly divided It

will

be seen that the equation of

June 15th, August 31st, April has maximum values 4-14*5 m. about 15th,

4m. about May 14th; February llth; m. about -16-5 November 3rd. 2Gth; Difference in length between 69.

Owing

to the

+6

m. about July

a solar and a sidereal day.

immense distance

of the fixed stars, the

by comparison becomes a mere point. Hence meridian revolves from a fixed star to the same any given star again in the same time that the earth takes to revolve on its axis, i.e. a sidereal day. This would be the case with earth's

orbit

the sun

if

the earth had no annual motion.

But

as the earth

revolves round the sun, it advances almost a degree eastward in its orbit while it revolves once on its axis. If, then, the on a sun be on the meridian of any place given day, the earth

must perform rather more than a complete revolution on its axis before the sun is again on the meridian on the day followIn other words, the earth must perform one complete ing. revolution and as much in proportion of another as it has advanced in its orbit in that time, viz. about A $ f part of a -.

revolution at a

mean

rate.

The earth, therefore, must perform about 366 revolutions in about 365 days, and the period of each revolution being a -i-lereal day, there must be about 366 sidereal days in a mean .solar year.

And, generally, since the time of the rotation of a planet its axis is its sidereal day, the number of sidereal days will always exceed by one the number of solar days, whatever it

on

may

be.

DIFFERENCE BETWEEN SOLAR AND SIDEREAL DAY. This

may

E

109-

be illustrated by a figure. be the positions of the earth in

E

f

its orbit on two successive days, 8 the sun, F the direction of a fixed point, as a fixed star, which is on a meridian PP' of the earth at the same time as the sun when the earth is at E.

Let

and

When

the earth is at E' next day, and the fixed point is on the meridian, PP' will have revolved through 360, again but it will have to revolve in addition through the angle FE'S, or E'SE, before it again passes over the sun (PF is parallel to

PF

owing

to the

The value of the arc EE' Its

average value 70.

of F). varies from about 61' to about 57'.

be thus found.

The mean sun

describes

the

equinoctial

360

and therefore describes

solar year,

mean

may

immense distance

in

a

mean

C

= 59'

366-242342

8-33" in a

day that is, a meridian of the earth has to revolve 360 59' 8'33" in a mean solar day. through solar

;

Equivalent sidereal and 71.

A

mean

solar

mean

solar intervals.

year contains 365'242242

days

;

there

are therefore 366*242242 sidereal days in the same time. Hence if and 8 be the measures of the same interval

M

expressed in

mean and

sidereal time respectively,

M

365-242242 one mean solar year

Interval

and

8 Interval

366-242242

~~

one

mean

solar year'

NAVIGATION AND NAUTICAL ASTRONOMY.

110

365-242242

S

366-242242

S

366-242242

x=

where

and

=

-00273791,

M=

-99726957

x' /.

'00273043

and

S

=1-00273791 M.

From

these expressions the tables of time equivalents in Nautical Almanac are calculated also the tables for acceleration of sidereal on mean time, and retardation of mean

the

on

;

sidereal time.

72. The same results may. be obtained by considering that a meridian of the earth revolves through 360 in a sidereal day, and through 360 59' 8'33" in a mean solar day.

_ 360

M~

59' 8-38"

360

which expressions give exactly the same results and S in terms of each other. obtained for

as

before

M

Convert 17

Examples.

h.

27 m. 47

s.

mean time

into the equivalent in

sidereal time.

By Time h.

17 h.

mean time

=17

27m.

Equivalents. m.

2

27

s.

47'56 sidereal time. 4-44

47-13

47s. 17

/;.//

Correction for 17

30

Tables.

\\.

m.

17

27 2

h.,

39-13

s.

47 47'56 4-44

27 m., 47 s.,

-13

17

30

39-13

EXAMPLES. Convert 19

27m. 18s.

h.

Ill

sidereal time into the equivalent

By Time

m.

19 h. sidereal time = 18

56

53*24

27 m.

26

55-58

=

s.

s.

By

24

time.

6-77

Tables.

m.

h,

s.

3

h.,

6'76 4-42

"27 m.,

18

mean

17-95

19

Correction for 19

time.

Equivalents.

h.

18

mean

-05

s..

3

Total correction,

19

27

19

24

11 '23 to

be subtracted.

18 6-77

Leap Year.

As the mean

73.

while

it

year consists of 365*242242 days, the for ordinary purposes of life that necessary of an exact number of days, a shall consist year solar

is

the civil

plan has been devised to render the error in the thus produced as small as possible.

calendar

The mean solar year containing very nearly 365J days, three consecutive years are considered to consist of 365 days, and each fourth year (with the exceptions to be mentioned) to contain 366 days. This is named "Leap Year."

The in

error

thus produced would amount to about 3'1 days The extra day is therefore not added in the

400 years.

error to about

*1

"

"

years, thus reducing the century of a day in 400 years, or one day in about

case of three out of four

4000 years.

Thus If

to find leap year, divide the

there

of a

is

no remainder

it

"century" year, when was not a leap

the year 1900

number

of the year

by

4.

leap year, except in the case the divisor must be 400. Thus

is

year.

CHAPTER

IX.

TO CONVERT ARC INTO TIME, ETC. GREENWICH DATE. CORRECTION OF ELEMENTS FROM "NAUTICAL ALMANAC." To convert arc into time and

the converse.

74. The longitude of a place is defined to be the smaller arc of the equator intercepted between the first meridian and that of the place. It may also be considered as a difference

Thus the interval between two consecutive passages mean sun over any merifliart is 24 mean solar hours.

in time.

of the

Hence 24 hours

we

in time correspond to

360

of arc

;

from which

obtain that C

1

hour of time corresponds to 15 of 15' minute

1

second

1

arc.

15"

Also that 15 degrees of arc 1 degree 1

minute

4

1

second

^

by means

of

of time.

h.

1

4 m.

which

arc

s. s.

turned into time and the

can be

converse. Example*.

Convert 115

17' 45" into time. h.

115 17' 1

.

=Wh.

=71(11.

=

7

= HID. =l^m. = = ff s. =3 s. 115

17' 45"

-

m.

a.

40 1

8

3 7

41

11

TO CONVERT ARC INTO TIME. Convert 9

h.

47 m. 23 9 h.

s.

113

into arc.

=9x15=

47m. = V2'

135

=11|= = 5f =

11

9h. 47m. 23s. = 146

0"

0'

45 5

45

50

45

75. Hence we obtain the practical rules To convert arc into time, "Divide the degrees, etc., by 15 and multiply the remainder by 4." This will give hours and minutes, minutes and seconds, seconds and fractions of seconds :

respectively.

To convert time

into arc, "Multiply the hours

by

15, divide

the minutes, etc., by 4, and multiply the remainders by 15." This will give degrees, degrees and minutes, minutes and seconds respectively. Examples.

Convert the following arcs into time

:

(1)

47

19'

15"

(4)

163

46'

50"

(2)

98

47

30

(5)

147

29

42

(3)

119

26

45

16

(9)

10

19

47'5

(10)

8

16

29-3

Convert the following times into

-arc

j

(6)

h.

m.

3

25

h.

s.

(7)

4

17

19

(8)

9

27

27

m.

s.

In a similar manner hours, etc., of right ascension, or sidereal may be turned into the corresponding degrees, etc., of arc. In some Nautical Tables the tables of log sines, etc., and

time,

log haversines are calculated for time as well as arc, and the conversion of time into arc, or the converse, can be effected

by

inspection.

76. From the above considerations it appears that the longitude of a place shows the difference in time between that place and Greenwich, from which longitude is reckoned; for the longitude converted into time shows the interval which

elapse between the passage of the mean sun over the meridian of Greenwich and that of the given place, if in west or which has elapsed between its passage over a longitude meridian in east longitude and that over the meridian of

will

;

As the earth

Greenwich.

revolves from west to east, easterly mean sun earlier than that of

meridians will pass over the

Greenwich s.

N.

;

westerly meridia'ns will pass

H

later.

Hence the

NAVIGATION AND NAUTICAL ASTRONOMY.

114

time at a place east of Greenwich is before Greenwich time, and after it at a place west of Greenwich. Further, the difference of longitude between any two places converted into time will give the difference between the local times at those places at any given instant.

The Greenwich then, we wish to corresponding to the time at

date.

obtain the time at

77. If,

any other place,

Greenwich

we must apply

to the local time the longitude turned into time, adding

the longitude

is

west, subtracting

if

the longitude

is

it

if

east.

time at any place corresponding to a given Greenwich time, the longitude in time must be subtracted from the Greenwich time if the longitude is west, If it is required to obtain the

and added if the longitude is east. The Greenwich time thus found, to the nearest minute, is The knowledge of it is required called the Greenwich date. in almost every nautical problem, because the right ascensions, etc., of the various heavenly bodies are tabulated

declinations,

Almanac for certain therefore, we can make

in the Nautical

instants of Greenwich

time.

use of these elements

Before,

in the solution of Nautical

Astronomy problems, it is necessary the Greenwich time corresponding to the times of observation, in order that the proper values of that

we

should

the elements

know

may

be obtained.

The Greenwich time may be found more accurately by means of a chronometer whose error is known. The known error on Greenwich mean time being applied to the chronometer time It is, howwill produce the Greenwich mean time required. ever, necessary to know the local time and the longitude, at any rate approximately, as chronometers are only marked up to 12 hours, and it could not be decided whether or no the correct Greenwich mean time was more or less than 12 hours without a knowledge of the local time and the longitude. Examples.

Feb.

15th in longitude 45

W,

and 3

h. 15

corresponding to local times 8 h. A.M. h.

Long, in time,

Greenwich

-

h.

Feb. 15th,

-SOW

date, Feb. 14th, 23

Greenwich dates

P.M.

in.

20

Feb. 14th,

find the

m.

Long, in time, .

Greenwich

m.

3 15

-

-

SOW

date, Feb. 16th, 6 15

THE GREENWICH DATE. Feb. 15th in longitude 117 to local times 8 h. A.M.

and 3

h.

Long, in time,

Greenwich

-

-

-

-

E, find the Greenwich dates corresponding 15 m. P.M.

m.

h. -

Feb. 14th,

Feb. 15th,

7 48

m.

h.

20

date, Feb. 14th,

115

E

3 15

-

Long, in time,

1212

7

-

48

E

Greenwich date, Feb. 14th, 19 27

In the second part of this example, as the longitude to be subtracted greater than the local time, it is necessary to add 24 hours, 27 h. 15 m. on Feb. 14th corresponding to 3 h. 15 m. on Feb. 15th.

is

Feb. 15th in longitude 45 W, at about 7 A.M. local time a chronometer h. 47 m. 45 s., its error on Greenwich mean time being 13 m. 15 s.

showed 9 slow.

Required the Greenwich mean time. m.

h.

Chronometer time,

h.

s.

Feb. 14th,

9 47 45

-

13 15 +

Error, slow,

10

-

-

Long, in time,

3

-

Greenwich date, Feb. 14th,

1

m.

19

22

12

Greenwich mean time, 22

1

In this case the Greenwich date shows that 12 hours must be added to the 10 h. 1 m. obtained

by applying the

error of the chronometer to

the chronometer time.

To take out 78.

The

Almanac

sun's

for each

the

declination

day

at

Sun's Declination. is

tabulated in

the

Nautical

noon at Greenwich.

If it is required to find its value at the time

any observation

taken, the Greenwich date must first be obtained, and then the hourly variation multiplied by the hours and of the sun

is

fractions of

the

an hour in the Greenwich

correction to be

Greenwich, added

if

to

date. This will give the declination at noon at

applied the declination

is

increasing, subtracted if

decreasing.

This is the general principle of the rule, but there are modifications which require notice. The hourly variation represents the rate of change of the

noon of the given day, but this rate is not constant throughout the 24 hours, as may be seen from the fact that the hourly variation varies from day to day. declination at

When, therefore, the Greenwich date is more than 12 hours, the declination should be taken out for the nearer Greenwich

NAVIGATION AND NAUTICAL ASTRONOMY.

116

noon, and corrected for the difference between the Greenwich date and 24 hours, the correction obtained being subtracted if the declination is increasing, and vice versa. If great accuracy is desired, the hourly variation used should be that for the time midway between the Greenwich date and

the nearer Greenwich noon; but for the ordinary purposes of navigation, the hourly variation at the nearer noon is usually sufficient.

Find the sun's declination on Feb. 16th

Example.

10 h.

at

A.M. in

W.

longitude 79

Variation

m.

h.

Feb. 15th, _ -

Long.,

Greenwich

-

22

-

5

16+

-

in one hour.

Feb. 16th, at Greenwich) __ G f 12 lo DO C5 -J mean noon,

52-1"

.

date, \

J-

Feb. 16th,

Declination,

declination required,

-

12 16

3

171-93 2'

52"

The hourly

variation is multiplied by 3'3, as 16 m. is very an hour (6 minutes is or '1 .of an hour, so that nearly the number of minutes of the Greenwich date divided by 6 gives the decimal part of an hour). The change is subtracted from the noon declination as the declination

Next

^

of

*3

let

is

decreasing.

us take an example in which the Greenwich date

is

over 12 hours. Example.

Find the sun's declination on June 19th at 11 A.M.

in longi-

tude 33 E. Hourly m.

h.

June

18th,

Long. E,

-

-

23

-

2

Greenwich date^ June 18th, /

12-

~

Declination,

June 18th, at Greenwich j -J mean noon,

^

^ ^^ N g. g

variation.

3'31"

20'8

"

_

2'648

^"

declination required,

66'20

68-848 1'

8-8"

The hourly variation, having been taken for noon on June and multiplied by the hours, etc., of the Greenwich date

18th,

produces a result which is greater than the declination for noon June 19th, i.e. 23 26' 19", showing that the principle used must be incorrect.

TO TAKE OUT THE SUN'S DECLINATION.

H7

Taking, therefore, 20 h. 48 m. from 24 h., we obtain 3 h. 12 m. as the interval that will elapse before noon June 19th when the hourly variation is 2*28". 2-28"x3'2 = 7'3", the correction which must be subtracted from

23

26'

19" to obtain a

more correct value

of the declination

at the given time.

The same method applies to the sun's apparent right ascension, which is not, however, often required in navigation. 79. The sun's declination is tabulated for both apparent and mean noon at Greenwich. As the time kept on board ship at sea is apparent time, if the Greenwich mean time of an observation, such as latitude by meridian altitude, deviation by altitude azimuth, etc., where

not required to be known accurately, is wanted, the ship time would have to be corrected for the equation of time before finding the Greenwich date. But by applying the the G. M. T.

is

longitude directly to the ship time, the Greenwich apparent time can be found at once, and the declination for apparent noon corrected, so as to find the declination at the time of observation.

The practical difference will not be great, but, as a matter of principle and showing an understanding of what is being done, the point is worth notice.

Examples.

Eequired the sun's declination

Mean Day.

h.

time.

m.

:

NAVIGATION AND NAUTICAL ASTRONOMY.

118

To take out 81.

the

Equation of Time.

Obtain the Greenwich date, take out the equation of

Greenwich mean noon, multiply the by the hours and parts of an hour of the Greenwich date, or of the interval between the Greenwich date and 24 h., when the Greenwich date is over 12 h., and time

the

for

nearest

variation in one hour

apply the correction thus obtained; adding of time

is

the

if

is

if

equation in the

it decreasing, increasing, subtracting first case, and vice versa in the second case.

Required the equation of time at 6

Examples.

in longitude 53

20'

h.

18

tn. P.M.

on June 6th

W. Equation Variation

of time,

m.

h.

Jure

6th,

-

-

6 18.

Long.

W,

-

-

3 33

Greenwich

date,

June

m.

s.

Noon, June

6th,

1

in 1 hour.

s.

"445

37'41

4 -38-

20+

6th, 9 51 20

1

33-03

9'85

2225

4005 4-38325

Required the equation longitude 53

20'

of time

at 6 h.

18

in.

on June 6th

A.M.

in

W. Equation of time. h.

June Long.

5th,

-

-

W,

-

-

Greenwich

date,

June

m.

m.

s.

Noon, June 6th,

18 18

1

3 33 20

5th, 21 51

1

Variation,

s.

37 '41

445

96

2-15

38-37

2225 445

24

890

290

95675

Here the correction is + because the equation of time is decreasing, but the Greenwich date is before noon on June 6th. Care must be taken to mark the equation of time as additive to or subtractive from mean or apparent time, as the (as.-

may

In-,

at

tin*

time

tliat

it

is

taken nut.

should always be taken out for mean noon. When it is necessary to reduce ship apparent time to ship nn an time for the purpose of finding the Greenwich date, it is sufficient, in It

practice, to

apply the equation of time to the nearest minute

EIGHT ASCENSION OF THE MEAN SUN. Examples.

(1)

longitude 117 29'

Eequired the equation on June 8th.

time at 11

of

h.

13 m. A.M. in

E

(2) Required the equation of on April 15th. 104 23'

time at 9 h. 14 m. A.M. in

longitude

W

(3)

165

Eequired the equation of time at 4 E on December 1st.

h.

45 m.

P.M.

in

longitude

19'

To

take out the Right Ascension of the

Mean Sun.

The right ascension of the mean sun at Greenwich mean noon is found in the column headed "Sidereal Time" on page II. of each month in the Nautical Almanac. 82.

Since sidereal time

is

the hour angle of the

first

point of

Aries, or the right ascension of the meridian when the mean sun is on the meridian of Greenwich, the mean sun's right ascension is that of the meridian, and is therefore the same

as sidereal time at Greenwich

mean

noon.

the right ascension of the mean sun increases regularly 3 m. 55*55 s. in every 24 hours, its value at any other time

As

than Greenwich mean noon can be found by simple proportion, or more easily by the tables of time equivalents in the Nautical Almanac, or by the table "Acceleration of Sidereal on Mean

Time"

in Nautical Tables.

Example.

Eequired the right ascension on June 15th. 49 h.

June. 15th,

Long.

W,

-

Greenwich

of the

mean sun

m.

h.

-

6 15

Sidereal time at Greenwich

-

3 16

Increase for

9 hours, 31 minutes,

mean noon, -

-

-

-

-

m.

s.

5 34

4'20

1

28 '71 5'09

date, 9 31

Eight ascension of the mean sun, Examples. (1) Eequired the right ascension of 3 h. 57 m. P.M. in longitude 94 25' on April 6th.

the

W

(2)

m,

at 6 h. 15

W

P.M. in longitude

Eequired the right ascension of the mean sun at 6 on June 14th.

in longitude 114 33'

E

the

mean sun h.

at

18 m. A.M.

W

(3) Eequired the right in longitude 76 21' on

To take out

5 35 38

ascension of the

December

mean sun

at 2 h. 15 m. P.M.

19th.

Moon's Right Ascension and Declination.

83. The moon's R.A. and declination are tabulated for each hour of the day, and the change in 10 minutes of time at the commencement of each hour is also given.

120

NAVIGATION AND NAUTICAL ASTRONOMY.

Therefore, find the Greenwich date, take out the R.A. or declination for the hour of the Greenwich date, multiply the

change in 10 m. by the number of minutes in the Greenwich and mark off one more decimal place in the result. Add the correction so found, in the case of the R.A., and add or

date,

subtract in the case of the declination according as the declination is increasing or decreasing. If the

change in 10 m.

is

changing rapidly, take out the

RA.

or the declination for the nearest hour, and apply the correction accordingly, as in the case of the sun's declination.

5

Examples. Required the moon's 25 m. A.M. in longitude 41 27'

right ascension

W

h.

on April

April 5th,

Long.

W,

Greenwich

h.

At20h.,

date, April 5th,

Moon's R.A. s. m.

10

19

11-26

24-28

Variation in 10 m.

and declination at

6th. h.

m.

-

17

25

-

2

45

48 +

-

20

10

48

s.

SEMI-DIAMETER AND HORIZONTAL PARALLAX. To take out

the

121

Moons Semi-diameter and Horizontal Parallax.

These are tabulated for noon and midnight of each

84.

day.

The Greenwich date having been found, the correction is obtained by simple proportion for the hours that have elapsed since Greenwich noon or midnight. Example. Required the moon's semi-diameter and horizontal parallax 10' E on June 4th. h. 15 m. A.M. in long. 116

at 9

h.

June

3rd,

Long.

E.,

Greenwich

m.

-7

-

date,

June

13

3rd,

Moon's semi-diameter.

s.

15

21

-

Midnight, June 3rd,

15'

52 '3"

15

48'8

44

40

Noon, June 4th,

-

30

20

Change

in 12 h.,

-

IJh.,

Semi-diameter required,

3'5

-

"44

15' 51*86".

Moon's horizontal parallax.

Midnight, June 3rd, Noon, June 4th,

Change

8'9"

58'

-

57

56'1

12 '8

in 12h.,

1-6

lib.,

Horizontal parallax required, 58'

The horizontal parallax inserted

7'3".

in the Nautical

Almanac

the equatorial horizontal parallax, which is greatest at the equator and decreases as the latitude increases; but as the is

greatest value of the difference

is

about

12",

for almost all

purposes of Nautical Astronomy the parallax thus found be considered as the value at the place of observation. Examples.

(1)

Required

the

moon's

parallax at 7 h. 15 m. P.M. in long. 87 (2)

3

h.

9

h.

semi-diameter

and horizontal

on December 17th.

Required the moon's semi-diameter and horizontal on June llth. 27'

W

14 m. A.M. in long. 116

(3)

E

Required

the moon's semi-diameter

17 m. A.M. in kmg. 43

E

may

and

parallax

at

horizontal parallax

at

on April 10th.

To take out a Planet's Right Ascension and Declination. 85.

These are tabulated for Mean Noon and for Time of

Transit each day at Greenwich. The Greenwich date having been found, the proportional part of the change in 24 hours will give the required correction

NAVIGATION AND NAUTICAL ASTRONOMY.

122

Mean Noon. The correction may from the variation of R.A. or dec. in one

to the values inserted for also be obtained

hour of longitude.

To

take out the Right Ascension Star.

and Declination

of a fixed

"

86. Look out the R.A. of the star in the Table Mean places of stars," turn to the corresponding R.A. in the Table " Apparent places of stars," and take out the R.A. and dec. for the nearest

day. Example. its

mean

The RA. of Capella on December 27th differs by Hence using the mean value would cause an

value.

nearly two miles of longitude,

determine the longitude.

if

7

s.

from

error of

the star were observed in order

to

CHAPTER

X.

MERIDIAN PASSAGES.

PROBLEMS ON TIME.

HOUR ANGLES. Given apparent time and the equation of time, to find time; or given mean time and the equation of time, to find apparent time.

mean

87. Since the equation of

mean and apparent date,

time,

its

time

is

the difference between

value, corrected for the

and applied with the proper sign

time, will produce

mean

Greenwich

to apparent or

mean

or apparent time respectively.

Given mean time,

to

find sidereal time.

88. Let QAQ' represent the celestial equator, QPQ' being the meridian, ^1 the first point of Aries, the mean sun, west of the meridian in (1), east in (2). The sidereal time

m

FIG. 54.

APQ

measured by the angle or the arc AQ, as it is the westerly hour angle of the first point of Aries, or the time that has elapsed since the first point of Aries was on the meridian. is

NAVIGATION AND NAUTICAL ASTRONOMY.

124 In (1)

Sidereal time =

AQ = Am-\-mQ = Right ascension

of the

AQ = Am mQ = R.A. mean sun

(corrected for

mean sun (corrected the Greenwich date) + mean time.

for

In (2) Sidereal time =

mean time) = R.A. mean sun + mean time

Greenwich date)

h.

(24

24

h.

Hence, generally, = R.A. mean sun + mean time,

Sidereal time

24 hours being subtracted from the result

if

over 24 hours.

Examples. April 18th, in long. 49 E, the mean time was 5 required the sidereal time.

P.M.

6 m. 18

h.

s.

;

h.

S.M.T., April 18th, Long.,

-

Greenwich

m.

6 18

-

5

-

3 16

date, April 18th,

R.A.M.S. m. s.

h.

s.

1

0-

9'86

50m.,

8-21

1

50 18

1

1

h.

Corrected R.A.M.S., S.M.T.,

-

-

Sidereal time,-

December

W,

19th, in long. 77

the

-

m.

45 23'89

h.,

Correction for

45 41-96

s.

1

45 42

5

6 18

6 52

mean time was 2h. 16m.

15s. A.M.

;

required the sidereal time. h. -

S.M.T., Dec. 18th, Long.,

-

-

-

Greenwich date, Dec. 18th,

-

m.

R.A.M.S. m. s.

h.

P.

14 16 15 5

17 47 23'67

+

8

Correction for 19

h.,

24 m., -

3

7 '27 3'94

19 24 15 17 40 34-88

h.

m.

8.

Corrected R.A.M.S.,

17 40 35

S.M.T.,-

14 16 15 31 56 50 24

Sidereal time,

-

7 56 50

PEOBLEMS ON TIME.

125

If the time given is apparent, as in the case when it is required to find the sidereal time in observation of the pole star at sea, where the time kept is apparent, it must be

reduced to

mean by

which may for

all

the application of the equation of time, practical purposes of navigation be taken

out to the nearest minute. Examples. (1) April 24th, in long. 119 14 m. 21 s. P.M. required sidereal time.

E, given ship

mean time 8

h.

;

June

(2)

llth, in long. 57

W,

given ship

mean time

7 h. 19 m. 16

s.

A.M.

;

required sidereal time. (3)

June

(4)

December

28th, in long. 115 time. sidereal required

19 m. P.M. (5)

A.M.

;

;

50' E,

16th, in long. 73

given ship apparent time 5 20'

W,

h. 21

m. A.M.

;

given ship apparent time llh.

required sidereal time.

December

27th, in long. 23

E, given ship apparent time 6

h.

19 m.

required sidereal time.

mean

Given

time or apparent time, to find what heavenly meridian next after that time.

body will pass the 89.

AQ represent the celestial equator, PQ the celestial A the first point of Aries, m the mean sun, X a

Let

meridian,

heavenly body passing the meridian. Then mQ is the given mean time, or the given apparent time corrected for the equation of time, P and the R.A. mean sun (corrected for the Greenwich date).

Am

= R.A. mean

+ mean

sun time.

Hence that star in the catalogue in the Nautical Almanac whose right ascension is equal to that found will be on the meridian at the given time, or if there is no star with that R.A., the one whose R.A. is the next greater will be the first to pass the meridian after the given time. It is often convenient to know what bright stars will pass the meridian between two given times. The R.A. of the

meridian must be found, as above, corresponding to each of the given times, and the stars whose R.A's. lie between the sidereal times thus determined will be the stars required.

NAVIGATION AND NAUTICAL ASTRONOMY.

126

Stars that are thus on the meridian will not necessarily be This depends on the latitude, and

available for observation.

be easily deduced from the method of finding the latitude by observations on the meridian, that the zenith distance of a heavenly body is equal to the difference between it

will

the latitude and declination when they are of the same name, and the sum when they are of different names. Stars, therefore, whose declination of opposite name is greater

than the complement of the latitude, will be below the horizon when on the meridian. And further, in practice, stars cannot as a rule be observed unless their altitude is greater than 5. Examples. On April 25th, what bright star will be the the meridian after 11 P.M. apparent time, in longitude 15

first

to pass

W?

m.

h.

2 12 59-8

11

April 25th,

Equation of time, S.M.T.,

-

Long.,

-

-

211

-

-

Correction for 11

48'4

1

h.,

9-5

58m.,

10 57 49

-

100

Greenwich date, April 25th,

2 14 57

S.M.T.,

10 57 49

R.A. of meridian,

13 12 46

11 57 49

a Virginia (Spica), whose E.A.

Which

R.A.M.S. m. s.

h.

s.

is

13 h. 19 m.,

is

the star required.

bright stars will be available for observation on the meridian A.M., Feb. llth (apparent

between the hours of midnight, Feb. 10th, and 6 ? time), in latitude 47 N, longitude 124

W

h.

Feb. 10th,

-

Greenwich

20 16

date, Feb. 10th,

Equation of time. 14 m. 26 s. -f apparent time.

18 8 16

Greenwich

date, Feb. llth,

14m. 26s. R.A.M.S. m. s.

h.

21 21 14-8 h.,

3 17'1

16m,

2-6

-

21 25 11-3

Correction for 2

197

h.,

2-6

16m.,

33

21 25 33

12 14 26

18 14 26

21 24 .S.M.T.,

2 16

Equation of time.

R.A.M.S. h. m. 8.

Correction for 20

m.

h.

Feb. 10th,

8 16 +

-

Long.,

m.

12

-

9 38 69

ir.

PEOBLEMS ON TIME.

127

The bright stars whose Right Ascensions lie between 9 h. 38 m. 59 s. and 15 h. 39 m. 59 s. are a Leonis (Regulus), a Crucis, a Virginis (Spica), but of these, a Crucis, ft Centauri, a Bootes (Arcturus), and a Centauri declination 62 S, ft Centauri, 60 S, and a Centauri, 60 S, will be below ;

the horizon.

Hence Regulus,

Spica,

and Arctums are the principal

stars required.

In the 1896 Nautical Almanac the magnitudes of the stars are arranged on a new system. Hence by "bright stars" is meant "stars of magnitude not lower than the second."

N.B.

Examples. (1) Which bright stars will pass the meridian between the hours of 8 P.M. and midnight, apparent time, longitude 62 on April 18th?

W

(2) Which bright stars will pass the meridian between the hours of midnight, June 12th, and 4 A.M., June 13th, apparent time, longitude 73 E? (3) Which bright stars will pass the meridian between the hours of midnight, December 14th, and 6 A.M., December 15th, apparent time, longitude 145 E ?

(4)

Which

bright stars will be available for observation on the meridian

between the hours of 8

P.M.

and midnight, apparent time, December

in latitude 35 N, longitude 136

Given sidereal time,

to

W

1st,

?

find

mean

time or apparent time.

Let

QmAQ' represent the celestial equator, QQ' the the pole, A the first point of Aries, the mean sun, west of the meridian in (1), east in (2). 90.

meridian,

m

P

FIG. 56.

Then

AQ is

the right ascension of the meridian, or the given sidereal time, the right ascension of the mean sun, in (1) and 24 h. Qm in (2) the mean time required.

Am

mQ

NAVIGATION AND NAUTICAL ASTRONOMY.

128

-Am,

fig.

mean time = sidereal time

Qm = Am-AQ 24

t

(1),

R.A.

fig.

mean

sun.

(2),

mean time = R.A mean sun sidereal time, mean time = 24 h. + sidereal time R.A. mean

h.

.

or

Hence we have,

sun.

generally,

mean time = sidereal time

R.A. mean sun,

24 hours being added, if necessary, to the sidereal time. But as we cannot obtain the R.A. mean sun until

we know mean time, we must obtain an approximate value of the mean time by using in the above formula the R.A. mean sun To this for the day as tabulated in the Nautical Almanac. approximate mean time apply the longitude, and get a Greenwich date with which to obtain a more correct value of the R.A. mean the

This being subtracted from the given sidereal time will a more correct value of the mean time, by means of which give a still more correct value of the R.A. mean sun may be obtained; sun.

and so on being

any desired degree

to

of Approximation, the second

usually sufficient for navigation purposes.

Example. Given sidereal time 9 h. 47 m. 10 s., same instant on December 18th in longitude 67 W. h.

RA. mean

mean time

b.

s.

94710

-

Sidereal time,

m.

find

sun at noon, 18th, 17 47 24

Correction for 20

-

4 28

Long.,

3 17'13

h.,

27 m., 4g g

Ship mean time, nearly, 17th, 15 59 46

R.A.M.S. m. s.

17 43 27'11

-

Noon, 17th,

at the

4'436 .j

28

17 46 48-804

Greenwich date, 17th,

20 27 46

-

Second Approximation. R.A.M.S. m. a. 17 43 27'11

m. s. 9 47 10

h.

Sidereal time,

-

R.A. mean

-

.

suii,

Ship mean time,

-

-

174648-8 16

Correction for 20

*?

21'2

428 Greenwich

h.

"

317*13

h.,

m

'

J

4

2l8 " 17 46 48'898

date,

-

20 28 21

The new value of the RA. mean sun differs, therefore, only by '094 of a second from that already found. Hence 16 h. m. 21 s. is the mean time required, or 4 h. m. 21 s. A.M. If apparent time is required, the corrected equation of time is to be applied to the

mean time thus

found.

PEOBLEMS ON TIME.

129

14 h. 16 m. 29 s. on June 28th in Find the apparent time. (2) Given sidereal time 1 h. 14 m. 37 s. on April 26th in longitude 114 E. Find the apparent time.

Examples.

(1)

longitude 74 20'

To find

Gi\7 en sidereal time

W.

time when a heavenly body will be on the meridian.

the

This is a particular case of the preceding problem. a heavenly body is on the meridian its right ascension equal to the right ascension of the meridian or sidereal time. 91.

When is

Hence the mean time

of meridian passage of a heavenly found body by subtracting the right ascension of the mean sun from that of the body, increased, if necessary, by 24 hours, the right ascension of the mean sun being corrected to any is

desired degree of accuracy.

The apparent time required meridian passage at sea

is

to

find

the time

of

a star's

obtained by application of the equa-

tion of time.

Example.

Find the apparent time of the meridian passage of Eegulus on June 12th.

(a Leonis) in longitude 73

W

K.A.M.S. h.

Star's E.A.,

-

E.A. mean sun, approximately,

10

m.

h. in.

s.

2 48

5 22 14

Correction for

9

1

h.,

33 m.,

4 40 34

4 52

-

Long.,

Greenwich date,

h.

m.

s.

2 48

Equation of time. s.

5 23 49

Mean

4 38 59

time of passage,

23

92. It is

5-42

5 23 48-65

E.A. mean sun,

apparent time,

28'71

9 32 34

10

Star's E.A.,

s.

5 22 14'52

28

523 + to

mean

time.

4 39 22

necessary to remember that the ship's clock shows

correct apparent time only at noon, or, in ships of high speed, at such other times as may be found convenient, and that it will be

the ship has run to the westward from noon or such till the time of observation, and too slow if to the eastward, four minutes of time for each degree of longitude.

too fast

if

other time

Examples. (1) Find the apparent times of the meridian passages of a Bootes, a Scorpii, a Aquilae on June 14th in longitude 27 W. S.

N.

I

NAVIGATION AND NAUTICAL ASTRONOMY.

130

(2) Find the apparent times of the meridian passages of a Tauri, a Orionis, a Argus on December 16th in longitude 39 E. (3) Find the time which will be shown by a clock, put right at apparent noon, when the star a Leonis (Regulus) is on the meridian, on April 6th

in longitude

97 E, the ship having changed her longitude 67 miles to the eastward at the time of the observation.

To find

the time of the

Moon's Meridian -Passage in any longitude.

The time

of the moon's meridian passage at Greenwich is tabulated for each day in the Nautical Almanac', and, if 93.

the

moon were

like a

star with a constant

right ascension,

the Greenwich time of passage would give the local time of passage at any other meridian, allowing for a small change in the right ascension of the mean sun. If the time of a

passage were, say, 8 P.M., the local time at that instant, at a meridian 45 west of Greenwich, would be 5 P.M. by the time the meridian came to the star, 3 hours due to the star's

;

of longitude would have elapsed, and the star's meridian passage would be 8 P.M. in local time at that place

difference

and so for any other meridian. But the moon's R.A. increases much more rapidly than that of the sun, on account of the motion of the moon in her also

;

Therefore the time of the moon's meridian passage is considerably later each day, the amount, 40 m. to 66 in., orbit.

depending on the number of minutes by which the increase of R.A. of the

m

moon

exceeds that of the

mean

sun.

M

If be the mean sun on the meridian, A and the com spending positions of the first point of Aries and of the moon when ni is on the m.-i-i-lian next day, A will be at A

;

PKOBLEMS ON TIME. '

=

M

m. nearly), while

4<

will be

131

at M',

nearly) the moon's change of E.A. and This retardation of the meridian passage.

RR

being (very nearly) the to say, the moon

(very is

eastward each day with regard to the meridians, which revolve from west to east with the earth. Hence the interval between the moon's meridian passage over an easterly meridian and that of Greenwich will be is

farther

to

the

greater than that due to the difference of longitude, because the meridian will have to revolve through the difference of longitude, and, in addition, the while moved to the eastward.

amount the moon has mean-

In other words, the local time of passage over an easterly meridian is earlier than the time tabulated for the meridian and, similarly, the local time will be passage at Greenwich ;

later at a westerly meridian.

To find Let

94.

L

amount

the

be the longitude,

of the correction.

D

the difference between the

tabulated times of the moon's meridian passage, between which the required time of passage lies.

D

Then, as the retardation of longitude,

takes place in 24 hours or 360

by proportion 306 *

.

.

:

L

:

:D: correction.

correction =

^^ obu

X D,

and is subtractive for east longitude, additive for west longitude. Hence the rule " Take out the times of the moon's meridian passage for the given day and the day before for east longitude; or for the given day and the day after for west calculate or take from the tables the correction, longitude subtract it from the time for the given day for east longitude, or add it for west longitude," The result will be the local :

;

(not the Greenwich) meridian.

mean time

of

passage over the given

The longitude in time must be applied in the usual obtain the Greenwich date.

way

to

It is to be noticed that the day, and day before or after, are to be understood as astronomical day, etc., a point often overlooked in the working of examples. " Lower Meridian Passage is used in finding" O O the time in and not the latitude. high water, finding

The of

"

NAVIGATION AND NAUTICAL ASTRONOMY.

132

of the moon's meridian passage on April 5th

Find the time

Examples.

in longitude 47

W. h.

m.

8 58 '5

Mer. passage, April 5th,

9 50-2

-

-

6th,

517 Correction from table,

6

S.M.T. of passage,

-

9

Long.,

-

38 +

Greenwich date,

4'5

1212

-

April 5th.

Find the time of the moon's meridian passage on April 16th in longitude 53 E. h.

Mer. passage, April 16th, 15th,

-

-

-

m.

17 43'6

April 15th.

16 50-8

14th.

52-8 -

Correction,

8

.

S.M.T. of passage,

Greenwich date,

3

32-

14

3'6

Find the ship mean time and the Greenwich mean time meridian passage

longitude 73 W. longitude 49 E.

April 6th,

(2)

April 8th,

(3)

April 20th, longitude 89 E. the

15th. of the

moon's

:

(1)

To find

15th.

17 35'6

Long.,

longitude 116

W.

27th, longitude 127 (6) Dec. 5th, longitude 81

W.

(4) (5)

June June

8th,

E.

time of a Planet's Meridian Passage in any longitude.

the same as in the case of the moon, principle of the time that passage on any day is not always except The moon's right on the that than later preceding day. 94.*

The

is

h. to 24 h., and the constantly increases from daily change of R.A. is always greater than that of the mean sun; hence the time of the meridian passage is always later

ascension

day by day. But, as the apparent motion of a planet, as seen from the earth, is a combination of the actual motions of the two bodies in their respective orbits, the right ascension of a planet is sometimes increasing and sometimes decreasing. So long, therefore, as the planet's R.A. is increasing more rapidly than that of the

mean

sun, the time of its meridian

PEOBLEMS ON TIME.

133

passage will be later day by day. When the rate of increase is equal to that of the mean sun the time of passage will be practically the same on successive days; when it is less, or

when

the planet's R.A.

decreasing, the time of passage will These changes can be easily traced by

is

be earlier day by day. inspection of a planet's elements in the Nautical Almanac. Thus, in 1895, the R.A. of Venus increased at a gradually decreasing rate from January 1st until about August 26th, after which it decreased until about October 8th, and then

The time of meridian until June was later 28, when it remained day by day passage the same for three days it was then earlier day by day until November 29th, when it remained the same for five days, after which it was later again day by day until the end of the year. increased until the end of the year.

;

It is therefore necessary, in finding the

time of a planet's

meridian passage, to notice whether the time of passage

is

accelerated or retarded.

N.B.Fov

the mathematical investigation of a planet's stationary points see Godfray's Astronomy, 4th edition, pp. 276, seqq.

and retrograde motion,

Given the latitude of the place, and the zenith distance declination of a heavenly body, to find its hour angle, thence the 95.

P

mean

Let

the pole,

meridian in

and and

time.

NEWS Z

represent the horizon, NS the meridian, the heavenly body, west of the the zenith,

X

(1), east in (2).

E

w

FIG. 58.

Then

ZPX

hour angle in

is

the body's hour angle in

(2).

(1),

and 24 h.-the

NAVIGATION AND NAUTICAL ASTRONOMY.

134

ZPX = h, =90-

Let

In the triangle angle

ZPX

lat.

the three sides are known, to find the

ZPX. by the well-known formula

.'.

= c,

of

Spherical Trigonometry,

= cosec PZ cosecPXx/hav (ZX+PZ-PX) hav (ZX-PZ^PX). .

NOTE.

It is better for beginners to use this formula, as the quantities used represent the sides of the triangle PZX, and the difference of and

PX

PZ

ahcays to be found. The work is, however, usually shortened by the use of the formula is

hav h = sec

I

since

PZ PX= l^d when

and

+ d when

l

*>

.

sec

ha v (2 + ^+0?) hav (zand declination are

<

latitude

of the

same name,

they are of different names.

If the table of haversines is not at hand, the formula can be adapted to the ordinary table of log sines, etc., by writing /3

sin2 - = hav

Ji

.

.

.

sin 2 ~

D_ = cosec PZ

DT PA

,

.

cosec

.

sin

(9,

ZX + PZ^PX =

.

sin

ZX-PZ-PX

If the haversine table is used, the hour angle is taken from the top of the page when the heavenly body is west of the meridian, and from the bottom when the heavenly body is east of the meridian.

The hour angle having been found, which is the apparent time in the case of the sun, the application of the equation of time will produce the mean time required.

A

different

method

is

heavenly body. Let QAQ' represent the

m the A

mean

QQ' the meridian,

celestial equator,

AR

the heavenly body, point of Aries. or the arc QR is the hour angle in

being the

QPR

sun, X

necessary in the case of any other

its

right ascension,

first

the hour angle in

(1),

and

is

24

h.

(2).

Am =RQ+AR-Am

mean time = mQ = ^Q

= heavenly

in (1),

body's hour angle + its right ascenR.A. mean sun.

PKOBLEMS ON TIME.

135

FIG. 59.

24

mean time = mQ

h.

= QR-Rm in (2), = QR-(AR-Am) = 24 h. body's hour

+ R.A.

mean

mean time = hour angle + right

or

angle

right ascension

its

sun,

ascension

figures for other cases

will be

R.A.

mean

sun.

found that the

By drawing above expression (24 hours being added or subtracted if necessary) holds good generally. It is really this expression that is used in the case of the sun, as the equation of time is the difference between the it

mean and apparent suns. found in the next chapter, on finding the

right ascensions of the

Examples

will be

longitude.

To find 96.

Let

I

the length of the

day

at

any

place.

be the latitude, d the sun's declination, h the Then 2h = length of day.

sun's hour angle at rising or setting.

In the quadrantal triangle tan I tan d, if cos h = .

cos h = tan If .-.

1 = 0, = 90

7*,

I

.

tan

d, if

I

PXZ (fig.

cos/?, = 0,

or 6 hours;

at the equator, therefore, the If .-.

60),

and d are of the same names, and d are of different names. I

d = Q,

cosh = 0,

h = 90

or 6 hours;

day

is

always 12 hours long.

at the time of the equinox, therefore, the sun rises at 6 o'clock all

over the earth, and day and night are each 12 hours long.

NAVIGATION AND NAUTICAL ASTRONOMY.

136 If

l

.-.

ft

= 90-d, cos/i=-l, = 180 or 12 hours;

the sun, therefore, does not set, but just reaches the horizon below the elevated pole, at midnight.

x'

>90

which

impossible; the sun, therefore, neither rises nor sets, but continues above the horizon, and there is continuous daylight. If

(Z,cosA>>

1,

is

,

= 90 d, when I and d h = 0, and the sun comes

If .-.

meridian and does not

>90

If

when

names, cosh = l, horizon when on the

different

the

to

rise at all.

and d have

I

different names,

cos/t>l,

impossible; the sun, therefore, neither rises nor sets, continues below the horizon, and there is continuous

which but

d,

have

is

darkness.

The formulae i.e.

(when and

I

= tan I tan d, - h) = tan I. tan d

cos

cos (180

and d are

of the

JL

.

same name),

cos h' = tan

(when I and d are Show that

cos

(

1

SO

when

tan

d

-

//

)

= c< >s

// ',

12 hours-/i = / .......................... (1) cos p = cos

Also .'.

cos

x cos .

XM = cosp

.

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

MZX

In right-angled triangle

cos z = cos

y cos .

= cos y .'.

sec

X M,

.

XM

cos p sec x .

;

from

(2),

= cos2.sec_p.cos# ............ cos2/

Equations (1) and (3) determine x and

.

........ (3)

In order to avoid

y.

the ambiguity in the signs of tan^, tana, etc., let 9Q^x = k, and 90

m.

h.

s.

-

6 14 32

r

6 25

m.

s.

E.A. a Cygni, 20 37 53 10

2

1 '6

30'1

Elapsed

20 48 24-7 18 33 25'5 2 14 59

N.

in sidereal time.

10 30

R.A. Vega,

S.

\elapsed time expressed J

=CPV

178

NAVIGATION AND NAUTICAL ASTRONOMY. To find ZC.

To find ZV.

53 45' 40"

68

2 30

26' 30"

2 30

53 43 10 4 31

53 38 39

68 19 29

43

23

53 37 56

68 19

90

90

36 22

4

21 40 (1)

PC 45 PF51

6'

54=ZF

To find CV.

sin 9'850242

vers 6 00930781

90/

19 sin 9-892435 versditf.

0005880

6 13

cliff.

CP F=2 h.

6

14 m. 59

s.

hav 8'925544

0099048

Z=34

8'

37"

90 Latitude,

55 51 23

N

EXAMPLES.

179

Examples. (I) April 30th, in latitude by account 45 N, longitude 140 30' W, the following double altitude of the sun was observed: Chronometer time. h. m. s.

App. time, nearly. h. m. 2 12 P.M.

257

4 49 P.M.

4 41 25

The rim Index

True bearing. S 55

sun's L.L.

alt.

48

50'

30"

23

2

40

W

West

of the ship in the interval was South (true), 6 miles per hour. 20"+; height of eye, 20 feet.

error, 3'

June

(2)

Obs.

by account 50

latitude

in

3rd,

30'

Chronometer time. h. m. s.

App. time, nearly. h. m.

9 55A.M. 11 15 A.M.

The run

of

the ship

Index

hour.

error,

7

43 27

9

2 26

in the interval

10"-; height

1'

Obs.

longitude 72 E, the

N,

following double altitude of the sun was observed

:

sun's L.L.

alt.

51

18'

59

31

was

N

10"

True bearing. S 50 E

E

S 20

40

W (true),

79

10 miles per

of eye, 18 feet.

(3) April llth, in latitude by account 57 N, longitude following double altitude of the sun was observed

10

30' E,

the

:

App. time, nearly. h. m.

OA.M. OP.M.

11

2

(4)

Obs.

sun's L.L.

alt.

1021 28

40

15'

20"

12231

35

22

10

Tru

bearing.

E

S 20 S 38

W

E

The run Index

Chronometer time. h. m. s.

of the ship in the interval was |N (true), 11 miles per hour. height of eye, 21 feet.

error, 2' 50"

June

;

29th, in latitude

by account 33 S, longitude was observed

ing double altitude of the sun App. time, nearly. L. m.

17 E, the follow-

:

Chronometer time. h. m. s.

8 10 A.M.

11 14 46

11

38'

50"

True bearing. 52 E

1135A.M.

23936

33

26

30

12

N 57

W,

Obs.

The run

of the ship in the interval was error, 1'50"-; height of eye, 22 feet.

sun's L.L.

alt.

N N

E Index

10 miles per hour.

(5) December 27th, in latitude by account 36 30' N, the following simultaneous double altitude was observed :

Obs.

Star.

/3Orionis,

-

a Hydrge,

Index

alt.

-

31

25' 30"

-

40

16

10

Height of

error.

2'

10"-

2

40

+

True bearing.

S 50 S 40

W

E

eye, 21 feet.

(6) June 15th, in latitude by account taneous double altitude was observed

4150'N,

the following simul-

:

Obs.

Star.

aTauri, a Pegasi,

Index

alt.

error.

-

-

19

31' 20"

2'

50"-

-

-

62

43 40

3

10

Height

of eye, 24 feet.

+

True bearing.

N 85 E S 15

E

NAVIGATION AND NAUTICAL ASTRONOMY.

180

(7) June 5th, was observed

by account 37 N, the following double

in latitude

altitude

:

Chronometer time. Obs.

Star.

a Aquilse, a Scorpii,

Index

EbN

24

53

(true), 12 miles

December 17th, was observed

(8)

48'

True bearing,

s.

S 86 S 17

7 14 26

10"

7 29

48

10"-; -height of eye, 24 feet; run

2'

error,

interval,

19

m.

h.

alt.

altitude

of the

E E

ship in

the

per hour.

in latitude

by account 34

5' S,

the following double

:

Chronometer time. Obs.

Star. a Eridani, a Piscis Australis, -

Index

error,

interval,

NE

20"+;

2'

65

24'

40"

59

36

50

height

(true), 12 miles

To find

the latitude

shown

122. It is

the pole star

is 7

l

h.

alt.

of

True bearing, S 13 E

s.

S 87

7 18 27

run of the ship

21 feet;

eye,

W

in

the

per hour.

by observation of the Pole Star.

183) that the latitude by altitude of to be obtained from the expression (

civ, 2 ^2 "HI ~

= a p cos h'Lit-' +*-

a being the corrected altitude, hour angle.

To the

m.

6 57 54

p

/. '

.

tan a sin

i

1

ft ,

the polar distance, and h the

therefore, must be applied the from the above expression, and inserted in the Nautical Almanac, depending on the sidereal time of observation and the polar distance, 1' being always subtracted from the corrected altitude, so as to allow the third correction

corrected

altitude,

corrections calculated

to be always additive.

The work should be done

in accordance with the example Nautical Almanac, remembering, however, that on board ship the time kept is apparent time, which must be reduced to mean time by applying the equation of time to the nearest minute be/we the sidereal time is found. The first correction being calculated for each 10 m., if the

supplied in the

sidereal time

found

falls

between two of the tabulated times

the correction must be corrected for the interval. of this

is

(The neglect a frequent source of error in the working of examples.)

Example. March 6th, apparent time at ship,

462.V20". latitude.

Index

longitude 37

in

the

error, 2'

observed

W,

at about 7 h. 35 m.

altitude

30"-; height

of

the

of eye, 20 feet.

P.M.

was Required the

pole

star

EXAMPLES. h.

ship apparent time,

m.

h. m. s. 22 55 52

s.

R.A. mean SUD,

7 35

+

12

equation of time,

181

correction for 10 h.,

38'6

1

15 m., S.M.T.,

7 47

Long,

2 28

Gr. date,

22 57 33 7 47

S.M.T.,

Mar. 6th.

10 15

sidereal time, obs. alt.,

2-4

6 44 33

-

46 25' 20"

-

index error,

corrected altitude,

2 30

1st correction,

-

46

16' 31"

34-

11

-

46 22 50 4 57

46

4 24

-

dip,

50 +

2nd 46 18 26

3rd

1

14 +

55

refraction,

46 17 31 Latitude,

1

Corrected

alt.,

46 16 31

-

Some tables have a correction for the pole star inserted for ten year intervals, which give fairly correct results if care is taken to obtain a properly proportioned correction but to obtain an accurate correction often involves more work than ;

expended in taking the Nautical Almanac.

is

separate

corrections

out of the

Examples. (1) April 5th, about 2 h. 15 m. A.M. apparent time, in longitude 67 21' W, the obs. alt. Polaris was 47 45' 30". Index error, 2' 20"+ ;

height of eye, 21

feet.

April 27th, about 11

(2)

30' E, the obs. alt. Polaris

h.

24

in.

was 40

P.M.

13' 30".

apparent time, in longitude 141 Index error, 2' 30"- height of ;

eye, 24 feet,

June

(3)

W,

10'

7th,

the obs.

about 9

alt.

h.

27 m. P.M. apparent time, in longitude 124 Index error, 1' 40" + height

Polaris was 46 33' 10".

;

of eye, 22 feet. (4)

June

the obs.

25

alt.

25th, about 1 h. 2 m. A.M. apparent time, in longitude 13 53' E, Polaris was 40 27' 40". Index error, 2' 20" + height of eye, ;

feet.

December

(5)

W,

19'

the obs.

6th, about 11 h.

alt.

Polaris

19m.

was 32

1'

P.M.

50".

apparent time, in longitude 63 Index error, 3' 10" + height of ;

eye, 18 feet. (6)

December

10th, about

1 h.

23 m. A.M. apparent time, in longitude

E, the obs. alt. Polaris was 21 of eye, 19 feet. height

114

11'

35'

30".

Index

error,

2'

10"-

;

NAVIGATION AND NAUTICAL ASTRONOMY.

182

Fwmulce (1)

"

in

Latitude Problems" collected for reference.

Circumpolar body.

= sum

lat.

above and below the pole (measured from

of nier. alts,

the same point of the horizon). (2)

Body below lat.

(3)

Body above lat.

the pole.

= 90 + mer.

- dec.

alt.

the pole.

= mer.

zen. dist.

+ dec.

(4) Ex-ineridian altitude (using estimated latitude). vers mer. zen. dist. =vers. obs. zen. dist. -vers

where hav 0=cos then use

(5)

cos dec.

;

hav hour angle

;

(3).

If the body

is

and 12

h.

then use

est. lat.

below the pole, substitute 90 for hour angle,

+ mer.

alt.

(2).

Ex-meridian altitude (not using estimated latitude). cos y = sin alt. cosec cot & = cos hour angle, cot dec. then lat. = + y. ;

(6)

Reduction

meridian.

to

x

.

.

-2 cos

.

(in seconds of

arc)=

=

or

Double

I

cos

.

cos

I

cos

.

:

.

-

dec. sin

k

;

d hav h .

:

:

=r,

smz.sml

sin z

(7)

for zen. dist.,

-hour angle

d-

j

sin 2 h -

.

-

.

2

sin 1

altitude.

ZForarc(l). vers

XY= vers (PX - PY) + vers $ = sin PX. sin PT hav XPY. ;

where hav

.

PZ or arc (5). Similar PX For arc (2).

formula.

hav PXY=cosecPX. cosec XY\\w(PY+PY^XY)\w(PY-PX * XY).

ZX Y or arc (3). PXZor

Similar formula.

arc (4) = arc (2) + arc (3) according as Jfl'does not or does

pass between

P and

Z.

If tab'es containing haversinesand versines are not available,

arc (1)

may

VA PX+Py

sin

Similarly for

.

PX

.

PZ or arc

And PA" F from

sin

.A

PY. cos 2

(6>

the formula,

cosec

.

.

'Z

And

.

(PX + PY 2 /')'

Z

II

DV sin PY+(PX~XY)'sin PY-(PX~A PX. cosec PY. .

* .-in-

A'Yor

be found from

similarly for

ZXY or arc (3).

)

>

'

CHAPTER

TRUE BEARING OF TERRESTRIAL

COMPASS ERRORS. OBJECT. To find

the deviation

XIII.

MAXIMUM AZIMUTH.

when

123. Bearings should

the true

bearing has been found.

always be measured from the north,

as the compass error is defined to be the angle between the directions of the true and compass north, or, more exactly,,

between the planes in which these directions lie. If, therefore, a true bearing is found from the south, take and measure it from the north. The compass it from 180 error can, of course, be found equally well with reference to the south point, but it is better, for a beginner especially, to adhere to one method.

Draw

a line to represent the true meridian.

From

the point

which represents the zenith lay off the true bearing. This will give the direction of the sun or other heavenly body. Measure back from this direction an angle equal to the body's compass bearing, and so obtain the direction of the compass in

it

north.

The compass error will be the difference between the true and compass bearings, and will be east or west, according as the compass north falls to the right or the left of ^the true north.

N

Examples. 75 W.

(1)

N 60 W,

The sun bore by compass

the true bearing being

make the angle NZS = 75. S is in the direction of the in SZ make the angle SZC=60. the point ZC is clearly the direction of the north point of the compass, which, falling to the left of the true north, shows that the compass error is west. Fig. 86 (1).

At

sun.

Z

in

At

NZ

Z

75

60

Compass

error,

15

W

NAVIGATION AND NAUTICAL ASTRONOMY.

184 (2)

N 80

The sun bore by compass

W,

the true bearing being S 120

W.

180 120

N

60 "W true bearing from north.

N

N

(2)

\

(I)

The figure being drawn clear that the compass error

similarly to the one is

20 E.

Fig. 86

in

the last case,

it

is

(2).

a figure is always drawn in this manner no mistake can occur, whereas mistakes do constantly occur if verbal rules are given, which are usually derived from figures with two directions of the sun (which cannot be the case), and only one If

direction of the north point (of which there are two at least). When the variation is given or, in practice, taken from the

and it is required to find the deviation, the figure be modified as follows

chart,

is

to

:

Draw

ZM

to represent the direction of the magnetic north. will be east or west according as C falls to the

The deviation

amount will be the algebraical and the compass error. variation is 17 W. Fig. 87 (1). that the (1) suppose

right or the left of difference

Thus Here

between

in case

M, and

its

-the variation

NZM=lT,

deviation = 2, and is east because the compass north falls to the right of the magnetic north. .'.

COMPASS ERRORS. In case (2) suppose that the variation

22

E.

falls

to

is

Fig.

8.7 (2).

NZM=22,

Here

.'.

and

185

deviation = 2,

west because the compass north

is

the left of

the magnetic north.

C

N

NM

(3)

FIG. 87.

As a

third case suppose that the compass error

In this case the deviation

is

2

W, and

87 (3)) between 2 to the MZN+NZC=4< (algebraical difference right and 2 to the left),

the variation

and

is

2

is

E.

C

west as

falls to

is (fig.

the left of M.

To find the compass error^ or the deviation, by amplitude, or by the true bearing of the sun at rising or setting. 124. Let

horizon, different

PZX

X

when name

or

or

X

f

be the position of the sun on the is of the same name as, or

the declination to,

the latitude.

Fig. 88.

and PZX' are quadrantal .

and

and

.

cos

PX = sin PZ

.

triangles.

cos

PZX

cosPX = cosPX.cosecP^ cos PX' = sin PZ cos PZX, cosP^' = cosPZ'.cosecP

(1)

.

(2)

186

NAVIGATION AND NAUTICAL ASTRONOMY.

But

= sin amplitude, cosPZX'=-siuEZX' = sin amplitude, cos

PX = sin dec.

cos

PX' =

cosec

sin dec.

PZ= sec lat.

Therefore (1) and (2) each become sin amplitude

= sin dec. sec lat.,

FIG. 88.

W

the amplitude being marked E or according as the sun is or S according to the name of the rising or setting, and

N

declination..

The true bearing is then 90 amplitude, or 90 + amplitude, according as the latitude and declination are of the same or different name, and the compass error or the deviation may be deduced, in accordance with the explanation just given. Examples. (I) April 13th, about 5 h. 15 m. A.M. apparent time, in latitude 52 20' N, longitude 11 15' W, the sun rose by compass E f S. The variation being 24 ; required the deviation.

W

NOTE. As courses and bearing are now usually measured in degrees from the north or south point of the compass, it would appear to be an advantage to do away with the term "amplitude," in which the bearing is measured from the east or west point (confusion thus being frequently <

caused),

and write the formula, cos true bearing at rising or setting = sin dec. sec lat,

remembering that the bearing must be taken from 180 when the declination is south, so as to obtain the true bearing from the north point.

EXAMPLES. m.

h.

Long.,

Declination.

17 15

April 12th,

45

-

187

9

W

Greenwich app. time, 18 24

Change.

3'25"N

54-5"

5 27

6

8 57 58

327-0

6

sin dec.,

-

9-192734

sec

-

10-213911

lat.,

sin amp., or cos true bearing,

Amplitude,

-

E

14 46' 45"

N

90

True bearing, Compass

Compass

error.

Variation,

Deviation,

N

75 13 15

N 98 -

26 15

E E

23 13 24

OW W

47

E

(2) April 30th, about 5 h. 5m. P.M. apparent time, in latitude 41 35' S r The variation being longitude 85 15' E, the sun set by compass 18 "W" required the deviation.

NWbW.

;

NAVIGATION AND NAUTICAL ASTRONOMY.

188

April 30th, -

Long.,

h.

ni.

-

5

5

-

5 41

Declination.

14 46' 55"

E

Greenwich app. time, 23 24 April

29.

Change.

N

6

14 46 27

27-6

sin dec., sec

46"

28

-

-

lat.,

sin amp., or cos true bearing,

\ /

True bearing = N 70

FIG. 90.

Amplitude,

-

True bearing,

or

Compass bearing, Compass

error,

Variation,

Deviation,

-

N N

4W

70 56 15 13

W 49 W

18

4 11

E

FIG. 91.

9-406581

10-126104 .

g 9-532685 4'

W.

EXAMPLES. Examples. 41

41'

189

April 4th, about 6 h. 20 m. P.M. apparent time, in latitude 20' W, the sun set by compass 60 20' W. The

(I)

N

N, longitude 35

variation being 24

W

;

required the deviation.

(2) April 22nd, about 6 h. 30 m. A.M. apparent time, in latitude 27 47' S, longitude 93 25' E, the sun rose by compass east. The variation being 10 15' required the deviation.

W

(3)

;

June

being 17 (4)

47'

E

20'

June

about 7

5th,

longitude 129

;

h.

15 m. P.M. apparent time, in latitude 37 51' N, 76 20' W. The variation

the sun set by compass required the deviation.

W,

26th, about 6 h. 15 m. P.M. apparent time, in latitude 10 15' N, 22' E, the sun set by compass 65 30' W. The variation

N

longitude 110

being 2 (5)

20'

E

required the deviation.

;

December 2nd, about 6h. 20m. A.M. apparent time, in latitude 12 5'N, 50' E, the sun rose by compass S 63 30' E. The variation

longitude 62

being

N

1

To find

10'

W

the

compass error or

125. Let

;

required the deviation.

the deviation

by altitude azimuth.

X

be the position of the sun or other heavenly on the not horizon, the altitude and compass bearing body at observed the same time. In the triangle the being

PZX

three sides are known.

Hence the angle

PZX may

be deter-

mined by the usual formula,

PZX = cosec PZ cosec ZX x/hav (PX +PZ* ZX) hav (PX -PZ*> ZX).

haversine

.

NOTE.

As

in the case of longitude

by chronometer, the formula

PZX = sec lat. sec alt./v/hav(P.D. + lat.

haversine

may

be made use

of.

^ alt.) hav (P. D.

- lat.

*

alt.)

NAVIGATION AND NAUTICAL ASTRONOMY.

190

The

true bearing being thus found and

marked

W

N

or

S accord-

ing to the latitude, and E or according as the body or of the meridian, the compass error or the deviation

W

E may

is

deduced as before. altitude azimuth problem is a convenient one, since the best time for longitude observations is also suitable for observa-

TDC

The

tions for azimuth.

The compass bearing being taken

at the

same time as, or immediately after, the observation for longitude, the same altitude will determine the longitude and the compass has also the advantage of being independent of the chronometer, the dead reckoning longitude being sufficiently accurate for the determination of the declination. error.

It

error of

however, seldom used in actual practice, the true bearing from the " Azimuth Tables," which are calculated taken being for every four minutes of ship apparent time, so long as the It

is,

than 60, for limits of latitude 60 N to to 23, the ship apparent time being S, found in working the observations for longitude or else deduced from the time by chronometer. These tables have been calculated by "Time Azimuth."

.sun's altitude is less

and declination

60

June

Example. 27 22

5

1st,

45' S, longitude 18',

W

;

about 3

104

13' E,

h.

10 m. P.M. apparent time, in latitude

the true altitude of the sun's centre was

when the sun bore by compass

N

40

20'

W.

The

variation being

required the deviation. h.

June

1st,

-

-

m.

Declination.

22

3 10

1

18

3'8

22 90

2

37

77'9

112

2

37 = polar

Greenwich app. time,) 20 13

May

31st,

/ 24 3 47

co-lat.,

62

15'

zen. dist.,

67

42

6

27

112

2

polar dist.,-

-

20 5

54'8"

6 57

Long.,

Change.

N

3'

0"

dist.

cosec 10'053063

cosec 10-033760

37

117

29 37

106

35 37

hav 4-931 902

^hav

4-904030 9-D22755

EXAMPLES.

191

CMN Sun

S 13223'W

True bearing,

180

N 47 N 40

Compass bearing, Compass

7

error,

W W 17 W

37

20

OW

Variation,

5

Deviation,

2 17

W

FIG. 93.

Examples. (1) April 9th, about 8 h. 30 m. A.M. apparent time, in latitude 89 40' E, when 47 27' N, longitude 129 31' W, the sun bore by compass the obs. alt. sun's L.L. was 30 13' 10". Index error, ]' 50"+ height of

N

;

The

eye, 21 feet. (2)

June

variation being 22

12th, about 3 h. 20

m.

30'

P.M.

E

;

required the deviation.

apparent time, in latitude 20

15' S,

^/*

N

36 20' W, when the obs. longitude 37 29' W, the sun bore by compass alt. sun's L.L. was 24 34' 50". Index error, 1' 30"height of eye, 18 feet. The variation being 12 ; required the deviation. ;

W

(3)

December

3rd,

about 7

h.

50 m. A.M. apparent time, in latitude

longitude 109 50' E, the sun bore by compass S 79 20' E, when the obs. alt. sun's L.L. was 34 30' 50". Index error, 30"+ height of eye, 19 feet. The variation being 14 30' W; required the deviation.

42

15' S,

;

s/

December

9th, about 8 h. 50 m. A.M. apparent time, in latitude N, longitude 123 20' E, the sun bore by compass S 45 30' E, when the obs. alt. sun's L.L. was 24 33' 20". Index error, 2' 40"height of (4)

23

25'

eye, 26 feet.

To find

the

The

variation being 1

compass error or

50'

W

;

;

required the deviation.

the deviation

by time azimuth.

126. When the heavenly body's bearing by compass is observed, note the time by chronometer, and from it obtain the body's hour angle from the meridian. We then have in

the triangle

which

ZPX, PZ, PX, and

to find the true bearing

1st method,

by the use

the included angle

PZX

ZPX, with

(fig. 92).

of Napier's Analogies.

tan \ (Z+ X) = cos J (PX *

PZ) sec J (PX + PZ) eot

ZPX

'

_

NAVIGATION AND NAUTICAL ASTRONOMY.

192

X) = sin J (PX

tan i (Z *

7

-

-

-

and

!2

i

=

being found,

PX

PZ) cosec J (PZ + PZ) eot

PZX

is

sum

their

7P Y

or their differ-

is greater or less than PZ. ence according as It is to be noticed that when %(PX + PZ) is greater than 90, %(Z+X) will also be greater than 90. In this case we shall

have 180-i(^+^) instead of $(Z+X). 2nd method, by first finding ZX and then calculating PZX from the three sides of the triangle PZX. vers ZX = vers (PZ o PZ} + 2 sin PZ sin PZ hav ZPX,

havPZZ = cosecPZ cosecZZx/hav(PZ + PZ ^ ZX) hav(PZ - PZ

*> ZX). For the formulae without using versines and haversines, see

note, p. 175.

This method

is

sometimes preferred, as

formulae and has no distinction of cases. is,

however,

much

uses well-known

it

The former method

shorter.

The true bearing being found by error or the deviation

may

either method, the compass be deduced as usual.

Examples. (1) April 8th, about 7 h. 50 m. A.M. apparent time, in latitude 30 50' N, longitude 16 0' E, the sun bore by compass S 66 30' E, when a chronometer, whose error on G.M.T. was 1 h. 2 m. 19 s. fast, showed 7

49' 54".

The variation being 19

April 7th, Long.,

-

-

-

-

-

Greenwich app. time,

-

W

required the deviation. h. m. s.

;

Chronometer, 7 49 54

19 50 4

1

-

15'

m.

h.

E

-

error fast,

1

2 19

6 47 35

18 46

12 Declination.

7

12'

4

57"

N

Change. 56-1

52

-

G.M.T.,

Long.

18 47 35

140

E.,

5'2 -

S.M.T.,

19 51 35

eq. of time,

7

1

57-

90 19 49 38

82

1

55 = PAT

24

29172 4'

52"

4 10 22 sun's hr. angle. Equation of time,

m. 1

1

2 Change.

a.

53'6

7

3-6

5-2

67

-M.T.

Latitude,

-

5

U

EXAMPLES. 1st

Method.

PX= PZ= sum, diff.,

^ sum, i

diff.,

193

NAVIGATION AND NAUTICAL ASTRONOMY.

194

To find PZX. 59

10' cosec 10-0661 78

-

62

53 cosec 10*050571

3

43

-

82

52

86

35

79

9

co-lat.,-

zen. disk,

polar

dist,,

hav 4'836142 hav 4-804199 9-757090

PXZ=98 (2)

In which

PX+PZ is

14' as before.

greater than 90.

July 23rd, about 4h. 45m. P.M. apparent time, in latitude 19 30' S, longitude 31 35' W, the sun bore when a chronometer, whose error was 1 h. 49m. 16s. fast on G.M.T., showed 8h. 44m. 21s. The variation

NWfW,

being 14

W

;

required the deviation.

h.

m.

July 23rd,

4 45

Long.,-

2

Gr.app.time,

6 51

6W

h.

m.

s.

Declination.

chron.,

8 44 21

error,-

1

G.M.T.,

6 55

5

20

Long.,

2

6 20

90

S.M.T.,

4 48 45

PX=llO

eq. of time,

5'

49 16

Change.

32"N

30-7

3 32

6-9

2

27(53

1842

21T83

2

6 15 4 43 30

ZPX

20

3'

Equation of time. m. B.

6 14-8 2 21

5

15 I!

I.'.

-M.T.

32"

PX=

70

NAVIGATION AND NAUTICAL ASTRONOMY.

196

(2)' April 19th, about 4 h. 15m. P.M. apparent time, in latitude 27 25' S, 43 20' W, when a chronolongitude 39 14' E, the sun bore by compass meter, whose error on G.M.T. was 1 h. 54 m. 37 s. slow, showed 11 h. 42 m.

N

19

The

s.

June

W

variation being 22

about 8

;

required the deviation.

20 m. A.M. apparent time, in latitude 15 15' S, 46 40' E, when a chronolongitude 80 10' W, the sun bore by compass meter, whose error on G.M.T. was 2 h. 3 m. 38 s. fast, showed 3 h. 43 m. 16 s. The variation being 12 E required the deviation. (3)

8th,

h.

N

;

June

26th, about 5 h. P.M. apparent time, in latitude 13 42' N, 111 12' E, the sun bore by compass 69 W, when a chronometer, longitude whose error on G.M.T was 1 h. 55 m. 13 s. slow, showed 7 h. 42 m. 18s. (4)

N

The

variation being 2

20'

E

required the deviation.

;

December

9th, about 8 h. A.M. apparent time, in latitude 11 25' N, (5) longitude 58 20' E, the sun bore by compass S 61 20' E, when a chronometer, whose error on G.M.T was 1 h. 1 m. 57 s. fast, showed 5 h. 1 m. 8 s.

The

variation being

December

1

15'

W

;

required the deviation.

50 m. P.M. apparent time, in latitude (6) 17 21' S, longitude 81 15' W, the sun bore by compass S 64 W, when a chronometer, whose error on G.M.T. was 1 h. 26 m. 49 s. slow, showed 7 h. 49 m. 28

s.

The

27th, about 3

variation being 12

To find

X

h.

the true

E

;

required the deviation.

bearing of a terrestrial

object.

be the object projected on the celestial concave, 127. Let the true place of the sun's centre. To find PZO, the true bearing of 0, we must measure the

angular distance

OX

and the

altitude of 0.

Hence the angle

FIG. 97.

at the zenith

OZX

can be found.

The

sun's true bearing

PZX

360- (PZX + OZX).

having been calculated, PZO If had been E of the meridian, as at 0', the true bearing of 0' would be PZX + O'ZX, but the construction of a ii-un will equal

TRUE BEARING OF A TERRESTRIAL OBJECT. when

will in each case determine the true bearing of

and OZX have been found. The angular distance OX

is,

of course,

apparent place of the sun, but in practice being measured from the true place. If

is

on the horizon,

197

PZX

measured from the it

OZ=9Q, and OZX

is

considered as

becomes a quad-

rantal triangle.

True bearing and Mercatorial bearing. 128. The azimuth or true bearing of a heavenly body is the angle at the zenith 'between the celestial meridian and the circle of altitude through the body and similarly as to the ;

true bearing of a terrestrial object. The Mercatorial bearing is the bearing as calculated by the method of Mercator's sailing, and is the angle between the meridian and the line on the chart from the position of the

observer to that of the object. On account of the distortion of the Mercator's chart, these two bearings are not the same, but the difference is not appreciable unless the distance is more than a few miles. Thus, three distant peaks which appear to an observer to lie in the same line from him (i.e. on the same great circle) will

not appear to on the chart.

lie

on the same straight

line

from his position

true bearing of a mountain peak be If, therefore, the calculated so as to obtain the compass error, the peak must

not be at a great distance it must be far enough to render inappreciable the effects of parallax as a ship swings, but not far enough to render appreciable the difference between the ;

true and Mercatorial bearing.

An

expression for this difference

may

be

found as follows

:

The true bearing of a distant object differs from the angle made by the rhumb line with the meridians by one-half the convergency of the meridians. Let PAC, PBD be the meridians

passing through two the of the arc places, being parallel of middle latitude between them, CD the arc of the equator, EA, tangents to PA, PB, falling on the diameter produced.

AB

EB

E

The convergency

AEB.

of the meridians

is

measured by the angle

NAVIGATION AND NAUTICAL ASTRONOMY.

198 Since

AB

is

very small,

may

it

Be considered as a straight

line.

since

.'.

BE are equal,

AE,

2r cot mid.

Again, since the triangle

.

\ cot mid.

AOB

lat.

sin J

AEB.

is isosceles,

= 2r sin J departure, lat. sin J AEB = sin \ departure,

sin \ convergency

= sin A departure X sin mid. .

= sin J difF. Of. Raper's Practice

,

cos mid. long,

,

x sin mid.

of Navigation, 10th Edition,

lat.

,

lat. lat.

395.

Example. In latitude 28 N, at 4 h. 25 m. P.M. ship apparent time, an was observed to be 81 40' distant object 0, whose true altitude was 5, from the nearer limb of the sun, whose declination is 2 S and semi-diameter is to the left of the sun, and E of the meridian ; required the true 16'. bearing of 0.

To find co-lat.,

polar

-

-

dist.,

h. 25

sin 9-945935

92

sin !)-99!)7.V>

30

diff.

hour angle, 4

Z.\.

62

m.

EXAMPLES. '0

= 61

46' 15"

30

199

obs. dist.,

vers0526937) 64 f vers 01 33975

-

81 40'

-

semi-diameter,

16

-

OX=8l

56

0660976

ZX=10ll'

To find PZX. co-lat.,

62

-

0'

zen.dist.,

70 11

pol. dist.,

92

To find OZX.

cosec 10-054065

OZ,

85

cosec 10'0265 11

xz,

70 11

ox,

81 56

cosec 10-001656

0'

cosec 10-026511

14 49

8 11

^hav 4-884836 ^hav 4-824738

100 11

83 49

96 45

^ hav 4-873616

67

|

7

9-790150

9-644340

OZX = 83

10330'30"

OZX=

hav 4742557

12' 45"

83 12 45 186 33 15

360 173 26 45 true bearing of is S 6 33' 45" E. had been on the horizon we should have had /.

If

cos

and

OZX

OZX = cos OX. cosec ZX,

would have been 81

19'.

From

the following data find the error of the compass Examples. (1) Altitude sun's centre, 45 20'

Altitude moon's centre, Sun's declination,

-

-

8

79 50

distance,

Moon's compass bearing, Time,

8

23

Latitude,

Lunar

30 50

-

-

-

S 40 30

N N

W

(about) 9 A.M.

NAVIGATION AND NAUTICAL ASTRONOMY.

200 (2)

In latitude 50

angular distance of the horizon was 79 meridian.

20' N, at 8 h. 45 m. 58 s. A.M. ship apparent time, the the nearer limb of the sun from an object on

being to the right of the sun and west of the

47',

Find the true bearing of Sun's semi-diameter, Sun's declination,

0,

haying given

-

16'

S

8

-

At

a place in latitude 30 25' N, when the sun's declination was N, the angular distance of its centre from an object to the left of it, elevated 20 above the horizon, was found to be 85 at 3 h. 15m. P.M. apparent time. Find the true bearing of the object. (3)

10

29'

Maximum 129.

When

Azimuth.

the declination of a heavenly body is greater it will cross the meridian between

than the latitude of the place,

the zenith and the pole. If be the position of such a body at rising,

X

XSX'

of

its

it

is declination, parallel clear that the azimuth will in-

x

crease

the

until

the

body reaches which the

position 8, at circle of altitude is to

the

of

parallel

after

which

until

the

meridian, the body

will

it

body the is

a tangent declination,

decrease

reaches

azimuth,

the

when

at T, being equal

to that at rising.

PSZ is

a

maximum

is

being a right angle, the

hour angle when the azimuth

FIG. 99.

found from the expression cos

ZPS = cot PZ tan PS = tan lat. cot dec.,

and the maximum azimuth from the expression

PS = sin PZ sin PZS sin PZS = cos dec. sec lat. sin

or If

to

a stick were placed at Z, its shadow at first would appear the west until the sun reached S, when it

move towards

would return and finally point to the south, a similar occurring in a reverse order in the afternoon.

result

MAXIMUM AZIMUTH.

201

is

The time during which the shadow would thus retrograde and ZPS. the difference between the hour angles At the time of the maximum azimuth, the heavenly body

is

said to be

is

ZPX

moving

vertically in azimuth.

Examples. (1) At a place in latitude 15 N, a star, whose right ascension 23 h. 56 m. and declination 20 N, has its maximum azimuth at the same

N is upon the prime Find the right ascension

time that another star of declination 5 stars being east of the meridian.

vertical, both of the second

star.

(2)

Find the time during which the shadow of a stick when the sun's declination is 15 N.

will retrograde in

latitude 5 30' N, (3)

sun's

N, when the sun's declination is 19 50' N, find the it will have this azimuth.

In latitude 5

maximum

azimuth, and at what time

Formulce in " Compass Error Problems" collected for (1)

reference.

Amplitude.

Sin amplitude or cos true bearing = sin dec. sec

lat.

(2) Altitude azimuth.

hav azimuth = cosec c cosec s\/hav ( p + c ~ .

.

or

(3)

sm

azimuth 2

= cosec c

cosec z

.

hav (p - c

**

22

p+c sin r .

.

z)

2

10 h. 18 m. 25 s. and 1 h. 3m. 18 s. ; chronometer slow on G.M.T.,

30 m. 42 s. altitudes of sun's centre, 9 52' 40" and 25 23' 50" declina22 4' S and 22 5' S; equation of time, 10m. 50s. and 10m. 47s. subtractive from apparent time run of the ship in the interval, south 30' ; bearings, S 48 30' E and S 1 1 30' E. 1 h.

;

;

tion,

;

To

find the longitude with latitude 41

48 112

10'

0"

4

5tf N.

10-127792 10-033039

63 54 80 144

7 20

20

4-978233

16 13 20

4-149506

1

9-288570 S.

N.

210

NAVIGATION AND NAUTICAL ASTRONOMY. G.M.T. h.

in.

s.

10 18 25 1

30 42

EXAMPLES.

78 30'

41 42', longitude 52 18', draw the position line dco the line ab S 30' to a'6', and dco cuts a'b' at the position Project

c latitude

Through \l8 41' 40" sun un, / 1 20

'

'

>

W

2 50

58 11 50

18 43

4 37

4 37

58

18 38 23

alt.,

7 13

15 11

15 58 app.

14' 40"

moon, /

18 54 21

app. alt,

58 22 24

28 11

2 43

Auxiliary angle

60 18 51 38

58 50 35

90

90

zen. disk, 71

8 22

zen. disk,

To calculate

Moon's

zen. disk,

Sun's

31

A

25' 27"

27 2

60 25 56

9 25

the true distance.

-

31

9' 25'

-

71

8 22 Versines.

1212746

223

vers.

1739631

134

vers.

0042849

vers.

1940684 0226160 662

sum, 102 17 47 vers.

Moon's app.

-

58 22' 24"

-

18 54 21

alt.,

Sun's

Auxiliary angle

.4,

sum,

77 16 45

-

60 25 56

sum, 137 42 41 16 50 49 diff., distance,

Angle J,

-

99 44' 49"

-

60 25 56

sum, 160 10 45 39 18 53

diff.,

vers.

75 162

662

5162732

4000000 1162732

465 267 true distance,

distance at

XV.

h.,

99 21' 56" 99 22 43 47

Prop. log. of

33410 P.L., 2-36133

prop. log. of elapsed time, 2 '02723

diff.

EXAMPLES.

235 h.

elapsed time,

-

-

-

m.

s.

1

41

1

41

15

G.M.T.,

In practice

(N.B.

it

is

-

-

-

15

-

the

sufficient to take the last five figures of

and of their sum. The required distance will then be found in the column headed by the degrees in the observed distance, or in one of the columns on either side of it. The above versines will then be versines

arranged thus

:

12746

223

89631

134

42849 40684 26160 662

68 75 162

662

.

62732

The nearest corresponding five figures are 62465, and the distance is 99 21' 56" as before.) To find S.M.T. and

in the

column headed 99,

the longitude. h.

57 37'

0"

100

cosec 10'073409

Sun's hour angle,

-

cosec 10*006649

eqn. of time,

-

-

42 23 71

8 22

113 31 22 28 45 22

| hav A hav

-

s.

8 8

..-'

S.M.T.,

-

-

-

-

20

G.M.T.,

-

-

-

-

15

Long, in time,

-

4*922416

m.

20

1

41

4 58 19

4*395043

Longitude, 74

34' 45" E.

9*397517

154. If the lunar distance of a star or planet is observed, It is, however, necessary is similar.

the method of calculation

whether the distance of the star Is observed from the moon the moon's semi-diameter being added to the observed distance in the former case and subtracted from it in the latter, in order to obtain the distance

to notice

nearer or farther limb of the

;

of the centres.

There is no distinction of cases with regard to the sun, the bright limb of the moon being always turned towards it, and the observed lunar distance being always that of the nearer limbs.

NAVIGATION AND NAUTICAL ASTRONOMY.

236

Examples. 47

(1)

April

1st,

about 3

observed

Obs.

32

26'

40"

2

20-

error,

June

3rd, about 7 h.

Obs.

(W

30m.

10'

22'

Obs.

54

30"

Index

December

Obs. dist. nearer limbs.

93

30 +

2

40"

moon's L.L.

Obs. dist. nearer limbs.

26

112

alt.

9'

50"

30"

51'

130-

130-

of eye, 22 feet.

24th, about 4 h. P.M. apparent time, in latitude 37 E, the following lunar was observed

longitude by account 93 Obs.

37 error,

42' S,

:

Obs.

alt.

alt.

moon's L.L.

sun's L.L.

Index

50"

33'

20-

:

Obs.

20-

error,

20"

N,

h.

Height (4)

35'

10' :

of eye, 18 feet.

alt.

58'

alt.

1

sun's L.L.

23

40 +

2

30 m. A.M. apparent time, in latitude 18 45' N, E, the following lunar was observed

April 15th, about 7 by account 136

longitude

Obs.

30"

56'

30+

eye, 20 feet.

50+

error,

20"

moon's L.L.

Height (3)

75

14'

P.M. apparent time, in latitude 17

of meridian).

26

58

E, the following lunar was observed

Pollux

alt.

Obs. dist. nearer limbs.

alt.

1

longitude by account 70

Index

following lunar was

moon's L.L.

Obs.

alt.

Height of (2)

the

:

sun's L.L.

Index

W,

30'

latitude

apparent time, in

P.M.

h.

N, longitude by account 67

15'

36

9'

10"

1

30+

54'

91

40"

59'

10-

2

Height

30"

Obs. dist. nearer limbs.

40 +

1

of eye, 23 feet.

(5) December 2nd, about 9 h. 30m. P.M. apparent time, in latitude 13 longitude by account 60 W, the following lunar was observed

5'

N

}

:

Obs.

a Pegasi

alt.

Obs.

45

Index

error,

42

1'

30"

3

10Height

(6)

December 3rd, about

longitude by account 15 obn.

51

1 h.

50'

25'

0"

2

20+

Obs. dist. F.L.

87

35' 1

20"

50 +

of eye, 25 feet.

30m.

W,

A.M. apparent time, in latitude 32" 47' N, the following lunar was observed:

alt.

Jupiter's centre (E of meridian).

Index error,

alt.

moon's L.L.

(VV of meridian).

49' 1

Obs.

78

40"

20 -

alt.

Obs.

1'

50"

1

50+

Height of eye, 18

feet.

dist.

N.L.

moon's L.L.

48

13'

2

10"

10-

EXAMPLES. (7)

December

24th, about

237

30 m. P.M. apparent time, in latitude

1 h.

35 12' S, longitude by account 105 E, the following lunar was observed Obs.

Obs.

moon's L.L. (E of meridian).

alt.

sun's L.L.

66

Index

16

58'

0"

1

30 -

error,

Height 155. If

altitudes

of

:

alt.

4'

10"

1

50 +

Obs. dist.

N.L.

90

48' 1

20"

40 -

of eye, 21 feet.

from any cause, such as an indistinct horizon, the the moon and other heavenly body cannot be

conveniently observed, they may be calculated, provided that the error of the chronometer on ship mean time is known

with some accuracy. This error

may

be

obtained

by an observation

of

some

convenient heavenly body taken shortly before the lunar observation, any change of longitude in the interval being allowed for.

The estimated longitude

will be sufficiently accurate for

obtaining a Greenwich date with which to correct the necessary astronomical elements.

The hour angles of the moon and other heavenly body being obtained from the expression hour angle of body

= ship mean

time + R.A.

mean sun

R.A. of body,

or (in the case of the sun)

= ship the zenith distances

mean time + equation

may

of time then be calculated by means of the

ordinary rule for finding the third side the included angle are known.

Thus the true

altitudes are

;

when two

known, and the apparent

sides

and

altitudes

are obtained by adding the correction in altitude in the case of the sun or a star, and subtracting it in the case of the moon. It

must be here noticed that the correction for the moon's taken out for the true altitude, is only an approximation,

altitude,

as the correction

true

correction

is

calculated

for the apparent altitude, the out for the approximate

being that taken

apparent altitude.

The same

principle

holds in the case of

the sun or a star, but the corrections are so small in all probable cases that a second approximation is usually unnecessary.

The apparent

mean time

is

altitudes having been obtained, the Greenwich found as before, and, the ship mean time being

known, the longitude can be obtained.

NAVIGATION AND NAUTICAL ASTRONOMY.

238

156. In

the example which follows,

the same as that worked out for the

which

first

is

practically

method, the S.M.T.

is

taken 28 seconds different from that which is there obtained, in order to show that a somewhat considerable error in the S.M.T. does not cause a large error in the resulting longitude, provided that the lunar distance

correctly observed.

is

April 16th, about 8 h. A.M. apparent time, in latitude 32 23' S, longitude by account 74 45' E, the obs. dist. of the nearer limbs of the sun and moon was 99 11' 30", when a chronometer, whose error on

Example.

S.M.T. was 2

was

2'

10"

+

h.

15 m. 17

s.

fast,

showed 10

15 m. 45

h.

s.

The index

error

.

Equation of time, h.

m.

s.

Sun's declination.

chron.,

-

10 15 45

error fast,

-

2 15 17

7 59

28

9 59 58

8

10

7'

s.

57"

N

53'2

13'1

'61

9

5'5

9

7-6

5-49

478-8

+

12

Sun's S.D.,

20

S.M.T.,

15' 58"

28 Moon's S.D.

4 59

Long., E,

14' 59"

Gr. date,

-

15

1

28 15th.

12 15 11

Moon's R.A. h. m. s.

to M.T.

Moon's hor. par. 54' 54"

EXAMPLES. h.

20

m.

239 h.

s.

36 hav 9'395968

hav 6

0=54

9-315910 7'

m.

s.

2 22 29

hav B 8-847073

0=30

15"

hav 8'971069

45' 15"

37 vers vers 42

23'

vers 5 15' 43" 0004 195

0261348

18

0675270 57

zen. dist.,

true

alt.,

a pp.

alt.,

vers 6 0140594

0413863 59

NAVIGATION AND NAUTICAL ASTRONOMY.

240

h. .-.

G.M.T. = 15

and

S.M.T. = 20

m. 1

s.

59

28

4 58 29

Longitude = 74

/.

which

differs

by

2'

37' 15" E,

30" from the previous result, although the error on

S.M.T. was 28s. in error.

157. An error as large as this is not likely to occur in actual practice, so that this method would appear to be preferable to the former, unless the altitudes can be observed with

great accuracy, which is seldom the case with a star except in the dusk. If the lunar distance is accurately observed with a well-adjusted instrument, other small errors of observation or time do not appear to affect the result to any great degree, but an error in the distance (which includes incorrect index error) appears to produce an error about thirty times as great in the resulting longitude. The index error should therefore

be carefully ascertained before a lunar distance Examples.

(1)

June

is

observed.

3rd, about 10 h. 2 m. P.M apparent time, in latitude

longitude by account 76 15' E, the observed distance of Antares (E of meridian) from the further limb of the moon was 42 44' 30", index error 2' 50" + when a chronometer, whose error on S.M.T. was 3 h. 18 m. 47 s.

35

15' S,

,

showed 6

slow, (2)

June

h.

42 m. 3

s.

3rd, about 7 h.

30m.

P.M.

apparent time,

in latitude 17

10'

N,

longitude by account 70 10' E, the observed distance of Pollux (W of meridian) from the nearer limb of the moon was 93 33' 50", index error 2'

30" f,

fast,

when a chronometer, whose h. 15m. 45 s.

error on S.M.T. was

1 h.

48m. 29s.

showed 9

(3) December 3rd, about 1 h. 30 m. A.M. apparent time, in latitude 32 47' N, longitude by account 15 50' W, the observed distance of Jupiter's centre (E of meridian) from the nearer limb of the moon was 48 13' 10", index error 2' 10"-, when a chronometer, whose error on S.M.T. was 3 h. 29 m. 16 s. slow, showed 9 h. 51 m. 23 s.

December 2nd, about 9 h. 30 m. P.M. apparent time, in latitude N, longitude by account 60 W, the observed distance of a Pegasi (W of meridian) from the further limb of the moon was 87 35' 20", index error 1'50" + when a chronometer, whose error on S.M.T. was 2 h. 38 m. -2\ M. fast, showed 11 h. 58 m. 33 s. (4)

13

5'

,

CHAPTER

XVII, >

INVESTIGATION OF CORRECTIONS (DIP, PARALLAX, REFRACTION, ETC.). INVESTIGATION OF ERRORS IN LATITUDE, HOUR ANGLE, COMPASS, ETC., ARISING FROM SMALL ERRORS IN LATITUDE, ALTITUDE, ETC. To

calculate the correction for

above the level of the 158.

"

"

dip

for a height of h feet

sea.

Let the heavenly body be in the direction of S, then feet above the sea, will measure the angle

an observer at P, h

SPD when

he brings the body down to his horizon.

FIG. 112.

The body's altitude, if observed at A, at sea level, vertically below P, would be measured by S'AH, where is in the

AH

s.

N.

Q

NAVIGATION AND NAUTICAL ASTRONOMY.

242

plane of the sensible horizon. parallel (as the distance of the

SPK=S'AH,

with PA),

KPD

Hence

is

PS

since

and AS' are

infinitely great is parallel to AH.

PK

where

compared

from the would have been if

the correction to be subtracted

observed altitude to reduce observed at A. This correction

And, body is

is

named

what

to

it

it

the correction for "the dip of the

sea horizon," or briefly, the correction for "dip." It may be calculated thus :

Let the earth's radius = a, then

cos

6= a

fi

0= 1 --r-r=

h T-T

a+h a+h

= h-

a very nearly

;

h o 2sm22#s =-. 2 a

.'.

as

suppose.

+ T, a

cos

1

But

OP = a + h.

KPD = W-OPD = POD =

The angle

is

very small, we .-.

write 6 sin

may

02 sin*

1'

for sin

;

1'=-; a

in minutes of arc

=

^/

sin 1

p=

a

Vj a = 3960 miles nearly

;

= 1980. 1980 5280

3-296665

3722634 2)7-019299

sin

V

3-509650 4-463726 1-973376

1-063

The dip raises

from

is

-026624

diminished by the terrestrial refraction

(\vliicli

the horizon), the amount being variously estimated at

^

Taking

to -^ of itself. -fa as an average value

INVESTIGATION OF CORRECTIONS. from sin^ = A/~-, Starting * Za /\ s\ L of ^ for sin -

we

by writing the

243

circular

measure

obtain

i

7T

180x60

2

180 x

,

2/i

e==

3960 x 5280

which produces the same result as before,

To find if

the distance at

which an

height above sea level

its

159.

is

h

But the arc

= circ.

where d

is

also be obtained

2

2

If refraction is

is

is

it

may

be shown that

^J

meas. of

*

practically

x a

from the fact that

DP

2

=(d+h)h,

the earth's diameter.

DP = 2ah -f h = 2ah

distance

be seen at sea

a and the distance PD, which are

AD

equal,

may

can

feet.

As in the calculation of the dip

the circular measure of the angle

This

object

= l'

viz.

since h2 is so small

compared with a

;

taken into account, it will be found that this by about the same amount that the dip

increased

decreased:

In the formation of the table in Inman's Tables, the distance given = ^f*/2aA. Refraction.

When

a ray of light passes through a transparent medium of uniform density it moves in a straight line. But 160.

if

passes obliquely from one such

it

path

medium

to

another

will be bent at the point of incidence, in accordance

the two conditions (1)

The

its

with

:

directions before

and

after incidence

and the persame

pendicular to the surface at that point will lie in the plane.

(2) The angles which these directions make with the perpendicular will have their sines in a constant ratio, so long as the media remain the same. When the ray passes from a

244

NAVIGATION AND NAUTICAL ASTRONOMY.

vacuum

into a

medium

coefficient of refraction

passes from one

"

the ratio of the sines for that

medium

medium

to another

;

is

called

"the

and when the ray

the sines of the angles

will be inversely as the coefficients of refraction. The earth is surrounded by an atmosphere of variable density, which may, however, be considered as consisting of an infinite

number

of concentric layers, each of uniform density throughout the layer, while the densities decrease rapidly the further we

recede from the earth.

Hence a ray of light from a heavenly body S, which is not in the zenith, will not travel directly to the eye of an observer at E, but will travel along the path S...

ABODE.

FIG. 113.

The body ES'

will therefore appear in the direction is the tangent to the path at E.

Let the radius OE be produced to meet the Produce 8A to meet OZ at F. at Z.

Then ZFS or ZE8

(as

ES

is

parallel to

ES'

celestial

FS owing

.

t

where

concave to the

the true zenith great distance of the star), will be very nearly distance, while ZES' is very nearly the apparent zenith distance. Hence refraction causes heavenly bodies to appear above their true places, and the correction

observed altitude.

must be subtracted from the

INVESTIGATION OF CORRECTIONS. 161.

follows

The

correction

may

be calculated

245

approximately as

:

Let r be the mean refraction in seconds, and c the constant between the sines of incidence and refraction for the

ratio first

and

and

last layers,

let

ZES'

sin (z

sin z cos r .

But since r SO',

we may

is

+

Then

z.

+ T) = c sin z

cos z sin r=.c sin .

always small,

its

z.

greatest value being

over

little

write

cosr=l, sin r = r sin 1"

r=

c

sin

1 ^77 /f

& - U. tan ^CVA-l z VUlU. a tan

.

.

0, ^,

1

Hence the correction

for

refraction

it -10-

= a-

c

Ul

i

;

1 -TTf*

sin 1

decreases as the zenith

distance decreases, as might have been expected, the rays of light from a body with a low altitude having to pass through

a larger amount of atmosphere of greater density than those from a body of greater altitude. A more exact formula is found to be

By

repeated observations of circumpolar stars, Bradley found

57-54" as the value of

4

sin 1

Hence the formula

for the

mean

refraction from

which the

tables are calculated becomes

r = 57-54" tan (0-3r). This

is

F.

Hence refraction corrected for a barometric height h = ^r Taking into account the increase of temperature, a final formula

is

of

volume of

arrived at

x ^~-> x 57'54" tan (2! *\j'\j

where h

and

calculated for a barometric height of 29'6 inches

a temperature of 50

r.

air for increase

:

3r),

the height of the barometer in inches and temperature Fahrenheit. is

^

t

is

the

NAVIGATION AND NAUTICAL ASTRONOMY.

246

To show how

= atanz

to determine the value of a in the expression by observations of circumptolar stars.

and

162. Let z

z

be the apparent zenith distances of a

circumpolar star at inferior and superior transits.

Then

PZ = ZX-PX = + a tan 2

p.

Similarly

PZ= z' + a tan z +p In the same manner,

;

z

if

and

are the apparent zenith distances of another circumpolar z

star,

FIG. 114.

2PZ = z -f z' + a (tan z + tan z') + z + a(tan0 + tan0') = 2 + s' + ri(tanz + tan0'); ;

tan z + tan z & (tan z + tan z)

To show

the

effect

distance at which 163. If

A

and

an

B

of the terrestrial refraction (1) on the object can be seen at sea, (2) on the dip.

be two places on the earth's surface,

ADB

the the angle between the perpendiculars at A and B, curved path of the rays by which A and B are visible to

each other, it r = aC, where circular

is

a

found that, if r be the has a mean value =

measure of the angle

AOB.

terrestrial refraction, '078,

and

C

is

the

INVESTIGATION OF CORRECTIONS. (1)

PA=h, and

Let

is visible

Then

from

B PT ;

'P = 90 + r

247

PDB

be the curved ray by which the tangent to the ray at P. Join PB. let

(since

PDB

touches the earth at

P

B\

B

FIG. 116.

a+h

COST

sin(90+r)

h _ cos r cos (G r) __ a~ r) cos((7 2/i,

^

_ " C(C

2r)

(7

1-2 sin

2

and r being expressed

in circular

measure.

((7_ r )2'

.

But

C

f

if

there

is

no refraction,

/y

is

for this value,

.

p=

Then

z'.

z-,

z = z'p.

.'.

Substituting this value of z

we

obtain

R'-R = R may

from which R'

be found.

173. Considering that, as an average value, the horizontal = 3*6 69 7 times the horizontal semi-diameter, and that

parallax

p = hor.

par.

x sin z' the above expression may be put

form without

t

If these substitutions are

made and small terms

takes the form, augmentation in seconds

it

= ARZ cos z where

into a

p.

;l

= 3'6697

H

x sin

^ 2

h

2

A.M.,

PM

JT.Hl.j

))

consider the declination as increasing, and the hour angles

S'PZ being

equal,

SPZ,

make

PX = PS, ZX = ZS. centre Z and

then

distance ZS\ then S'DX a is the practically plane right-angled triangle, and = = = in correction altitude while S'X dz, required dp change of

Describe an arc S'D with

XD

is

declination in the interval, which interval from the meridian.

dz = dpcosPXZ

Now An

PSZ

approximate value of

of course, twice the A.M.

is,

174, 176)

(cf.

can easily be found, and thence

ZS'.

The

altitude thus obtained being placed

the time taken

when

on the sextant, and

the sun has this altitude, the

mean

of

the chronometer times will be the time of apparent noon. Example. April 16th, about 8 h. 12 m. A.M. mean time, the true zenith distance of the sun was 56 25' 6", when a chronometer showed 8 h. 46 m. sec. (approximate error, 4 h. 25 m. slow on G.M.T.), latitude 30 N, longitude 75 W. Required the correction in altitude necessary to obtain the same P.M. hour angle. The A.M. hour angle from noon will be found to be 3 h. 47 m. 46 s.,

57

declination 10 app. time.

9'

N, equation of time at app. noon 16

The elapsed time being twice the hour

and the change

of declination per

hour 52 '96

angle,

"=6'

-006649

56

'081426

PZ

60

7 h. 35 m. 22

s.,

42'5".

To find

80

subtractive from

i.e.

s.,

To find PSZ.

PS ZS

s.

402 '5"

60

dz.

2 '604766

26' cos 9'693231

2-297997

24

60

84

4-825511

36

4-489982

)

198-6

tfc=3' 19"

9-403568 26'

As

the declination .-.

and the

is

increasing, dz

P.M. zen. dist.

P.M. declination is 10

subtractive

15' 43".

/

;

19" = 56 21'47",

Calculation produces the same hour of apparent noon will therefore be

The chronometer time

angle as before.

20

is

= 5625'6"-3

h.

46 m. 57

s.

+ 3 h.

47 m. 46

s.

=0 h.

34 m. 43

s.

MISCELLANEOUS PEOBLEMS, To find h.

m.

the error

ETC.

275

of chronometer,

s.

000 apparent time

of

apparent noon.

16 equation of time.

23 59 44 mean time of apparent noon.

500

4 59 44 G.M.T. of apparent noon. 34 43 chronometer time of apparent noon. 4 25

as an equal altitude, the chronometer times are 8 h. 46 m. 57 of equal altitudes is found to be 8s. s., and the equation

Worked 4

chronometer slow on G.M.T.

1

s.,

22 m. 45

h.

h.

m.

s.

8 46 57 A.M. time.

3 47 54 | elapsed time.

34 51 8 - equation of equal altitudes.

time of apparent noon o/i /IQ /chronometer 34 43 as before.

1

*To

find

the

Moon's

Hour Angle when

at its

maximum

altitude.

188. If the [declination be constant, a heavenly body will attain its maximum altitude when on the meridian; but when is increasing rapidly, the decrease of altitude meridian passage, due to the earth's revoluafter immediately tion on its axis, will be more than counterbalanced by the

the declination

due to the increase of declination; and a greater altitude than that on the meridian will be attained. Similarly, if the

rise

is decreasing rapidly, the maximum altitude will the meridian passage. precede Hence the moon is at its maximum altitude when its rate

declination

of

change of altitude arising from change in declination

is

equal and opposite to its rate of change of altitude arising from the diurnal motion.

Now, change

of altitude

= 15 cos lat. sin azimuth x interval

in

time;

change in

1 sec.

= 15 cos lat. sin azimuth

in seconds.

d" = moon's change of declination in 10 minutes,

If 7//

= change

in 1.

NAVIGATION AND NAUTICAL ASTRONOMY.

276

As the body

is

so near the meridian

and

in declination

change

we may

consider the

in altitude as practically equal;

sin h sin p --^= -sm z .

d"

;

d"

600

;

sin z

9000 cos d"

I.

d

cos

sin (ltd)

9000*008 J.COBci" In the same manner the hour angle of any other heavenly body, when at its maximum altitude, may be obtained.

To find

ihe

time when the sun will dip, when the change of

altitude is considerable, 189. Consider,

course

is

directly

to the

owing

firstly,

for

rapid motion of

simplicity,

the

case

or almost directly towards or

the ship.

when the away from

the sun.

As the

change of declination is practically in the may be considered to have the same effect as a and it is evident that the sun will reach of altitude, change its greatest altitude when the change of altitude due to the combined effect of the ship's speed and the sun's change of sun's

it

meridian,

declination

counterbalanced by the change of altitude in

is

the opposite direction, due to the revolution of the earth. Let c = the algebraic sum of the ship's speed in miles, and the sun's hourly change of declination.

60 x 60

^ 60

corresponding change in seconds of arc per second of time.

If

dz = change of altitude owing to the revolution of the earth in time dh,

dz

dh cos lat. sin azimuth

= 15 cos lat. sin azimuth

in seconds, per second of timt

The sun

will

have

,

15 cos

sin h

maximum

its

=

I

sin h cos .

-

:

K

\