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NOTES ON PRACTICAL ASTRONOMY 
AND GEODESY 



By L. B. STEWART 



SECOND EDITION 
Revised and Enlarged 



UNIVERSITY OF TORONTO PRESS 

1921 




Si 

tap 3, 



COPYRIGHT, CANADA, 1921 
BY L. B. STEWART 



PREFACE 

These Notes form the substance of a course of lectures in 
Practical Astronomy and Geodesy given to students of the 
Third Year in Civil Engineering of the Faculty of Applied 
Science . and Engineering of the University of Toronto. 
These lectures are only designed to meet the requirements of 
the engineer or land surveyor in his ordinary practice, or 
those of the explorer wishing to determine his position in an 
unsettled country, and therefore barely touch upon the precise 
methods of the geodetic surveyor. The subjects treated, 
however, and the methods of treatment, are fundamental; 
so that the course serves as a useful introduction for those 
who desire to extend their studies to the higher parts of the 
subject. 

LOUIS B. STEWART. 



NOTES ON PRACTICAL ASTRONOMY 
AND GEODESY. 

PRACTICAL ASTRONOMY. 

In these notes it is proposed to set forth in outline the most 
useful methods for determining positions and directions on the 
surface of the earth. It is assumed that the observer is provided 
with an engineer's transit, or a nautical sextant, so that the 
methods described are only such as are adapted to the use of 
those instruments. More precise methods, necessitating instru- 
ments of the highest class, are therefore entirely omitted, or but 
briefly referred to. 

1. Spherical Co-ordinates. Solution of the Astro- 
nomical Triangle. 

Determination of the position of a point. 



Fig. J 




In Fig. 1 CAO and ABO are fixed planes of reference; 
is the point of observation. The direction of the line OS 
is determined when the angles AOB and BOS are known; 
also when the spherical angle ACS and the arc CS are known. 

Planes of reference 

The planes of reference used in astronomy are those of the 
equator, the ecliptic, the meridian, and the horizon. 

The plane of the equator is that of the earth's equator. 
As the direction of the earth's axis is nearly fixed in space, 
being subject only to slow changes of direction due to pre- 
cession and nutation, therefore the plane of the equator is, 
nearly a fixed plane. 

The plane of the ecliptic is the plane of the earth's orbit. 



The plane of the observer's meridian is a plane determined 
by the earth's axis and the point of observation. 

The plane of the horizon is a tangent plane to the earth's 
surface i.e., to the surface of standing water at the point 
of observation. It is therefore perpendicular to the observer's 
plumb line. 

The celestial sphere 

This is an imaginary sphere of infinite extent, whose centre 
is coincident with the centre of the earth. Upon its surface 
the heavenly bodies may be assumed to be, as they apparently 
are, set like brilliants. 

The reference planes above denned are assumed to be 
produced to intersect this sphere in great circles. The plane 
of the horizon, as above denned, may be assumed to intersect 
the sphere in the same circle as that determined by a parallel 
plane through the earth's centre, owing to the infinite extent 
of the celestial sphere. 




Fie. 2 



Fig. 2 shews a projection of the celestial sphere on the 
plane of the meridian, the reference circles being represented. 
Thus 

PZHR is the meridian, 

EFR the equator, or equinoctial, 

HDN the horizon. 



The ecliptic is not shewn, but V is a point in which it 
intersects the equator. 

If S now be the position of a star (by that term denoting 
any heavenly body), and secondaries ZSD and PSF to the 
horizon and equator respectively be drawn through it, these 
arcs, with the meridian PZ, form a spherical triangle PZS, 
which, from its frequent use in the solution of astronomical 
problems, is termed the astronomical triangle. 

Definitions 

The circle ZSD is a vertical circle; PSF a declination or 
hour circle. P is the celestial pole; Z the zenith. SD is the 
altitude of S; ZS its zenith distance; SF its declination; PS its 
polar distance; the angle PZS its azimuth; and ZPS its hour 
angle. PSZ is generally called the parallactic angle. 

As the observer's latitude is the angle between the direction 
of the plumb line at the place of observation and the plane 
of the equator, it follows that the latitude is the angle ZOE 
or the arc ZE. This is also equal to the arc PN. 
The following notation will be used: 

h denotes the altitude SD of 5. 

" denotes the zeiith distance ZS. 

8 demotes the declination SF. 

p denotes the polar distance PS. 

t denotes the hour angle ZPS. 

A denotes the azimuth PZS. 

C denotes the parallactic angle. 

<t> denotes the observer's latitude EZ or PN. 

a denotes the right ascension VEF. 

Systems of Spherical Co-ordinates. 

1st system Altitude and azimuth. 

The arcs SD and DN serve to determine the position of 5 
with reference to the horizon and the meridian. 

A small circle parallel to the horizon is termed an almu- 
cantar. 

A vertical circle is a great circle perpendicular to the 
horizon. 

The prime vertical is that vertical circle which passes 
through the east and west points of the horizon. 

2nd system Declination and hour angle. 

The arcs SF and FE determine the position of S with 
reference to the equator and the meridian. 

A parallel of declination is a small circle parallel to the 
equator. 

3rd system Declination and right ascension. 



The planes of the equator and the ecliptic intersect in a 
right line called the line of the equinoxes. This line inter- 
sects the sphere in the vernal and autumnal equinoxes. 
The vernal equinox is the point through which the sun 
passes in going from the south to the north side of the equator; 
it is shewn at V, Fig. 2. 

The equinoctial colure is the declination circle passing 
through the equinoxes. The solstitial colure is the declination 
circle passing through the solstices the points of greatest 
north and south declination on the ecliptic. It is therefore 
at right angles to the equinoctial colure. 

The co-ordinates in this system are the arcs SF and FV. 

4th system Celestial latitude and longitude. 

V 




Fig. 3 



In Fig. 3 VEAR is the equator, VHAK the ecliptic, VA 
the line of the equinoxes, VPA the equinoctial colure, and 
EPR the solstitial colure. 

The co-ordinates in this system are SG, the latitude of S, 
and GV the longitude. These are denoted by /3 and X respec- 
tively. 

In the first system the co-ordinates change continually 
and irregularly on account of the diurnal rotation of the earth. 
In the second system the declination is unchanged by that 
rotation, and the hour angle changes uniformly with the time. 
In the third and fourth systems the co-ordinates are unchanged 
by the diurnal rotation. 



The third system of co-ordinates is for this reason used in 
the construction of ephemerides. 

Although unchanged by the diurnal rotation, the co-ordi- 
nates of the third and fourth systems are changing continually 
though slowly on account of precession and nutation. 

Solution of the Astronomical Triangle. 

(1) Given the altitude and azimuth of a star, and the latitude 
of the place, to find the star's declination and hour angle. 




Fie. 4 

If we denote the angular points of the astronomical triangle 
ZP and Shy AB and C, respectively, then in Fig. 4 we have 
given 

A=A, 6 = 90-/z, c = 9O-0; 
and it is required to find 

a = 90 -5, and B = t. 
These are given by the first of (1) and (5), Sph. Trig., p. 69 
which become 

sin 5 = sin h sin 0+cos h cos cos A. 
sin A cot t = cos tan h sin cos A. 
The first of these may be written 

sin 5 = sin h (sin 0+cot h cos cos A) 
Then introducing the auxiliary such that 

tan = cot h cos A (1) 

it becomes 

sin 5 = sin h (sin^-fcos < tan 6) 

_ sin h sin (0+0) (2) 

COS0 

The second equation may be written 

sin A 



tan t = 



cos <f> tan h sin cf> cos A 

sin A 
tan h(cos <j> sin </> cot h cos A) 

sin A 

tan h(cos sin tan 0) 

sin A cos 
tan h cos (0+0) 



Eliminating tan h by (1) this becomes 

tan A sin (3) 

tan T = (j. i a\ 

COs(0 + 0) 

Equations (1), (2) and (3) give the solution. 

(2) Given the declination and hour angle of a star, and 
the latitude of the place, to find the altitude and azimuth of 
the star. 

In the spherical triangle, Fig. 4, we have given 
a = 90-8, c = 90-<t>, and = r 
and b = 90-handA 

are required. These are given by the second equations of (1) 
and (5), Sph. Trig., which become 

sin h = sm 8 sin 0+cos 5 cos < cos t 
sin t cot A = cos tan 8 sin <f> cos t 
These may be written 

sin h = sin 6 (sin < + cos 4> cot 8 cos t) 
. sin t 

tan A = r? 1 = i Ti ^ 

tan 5(cos sin <f> cot 5 cos t) 

Then substituting cot Q\ = cot 8 cos t (4) 

they become 

. , sin 5 cos(0i </>) (5) 

sin h = r 

sin 0i 

sin t sin B\ 



tan .4 = 



tan 8 sin (0i 0) 

Then eliminating tan 5 from this last by (4) it becomes 

. tan t cos 0i 

tan A = sin^-0) (6) 

0i being given by the equation 

tan 8 (7) 

tan 0i = 

COS T 

These two problems serve for the transformation from the 
first system of co-ordinates to the second; and conversely. 

(3) Given the altitude and declination of a star, and the 
latitude of the place, to find the azimuth and hour angle. 
In this case we have given 

a = 90-5, b = 90-h, and c = 90-<f> 
and are required to find 

A =A, and B = t 
These are given by the first and second of either set of 
equations (6), (7) or (8), Sph. Trig. In these equations we 
have 

s =Ka+&+c) =9O-K0+5-r) 
s-a =$(-c+6+c)= Kr+5-0) 



s-b = \(a-b+c) =90-i(r+<+5) 
sc = (a+b-c) = Kf+0-8) 

so that on substituting 5' = |(f+0+5) they become 



sin 2 \A = 


cos s' sin (5' 5) 


(8) 


cos <f> sin f 




cos 2 \A = 


cos(s' f)sin(s' </>) 
cos sin f 


(9) 


tan 2 \A = 


cos 5' sin(/ 5) 
cos(5' f ) sin (V 0) 


(10) 


^in 2 ir 


sin (^' <)sin(s' 5) 


(11) 




cos < cos 5 




cos 2 \r = 


cos (V f) cos s' 
cos < cos 5 


(12) 


+ o-i2 i^ 


sin(s' <f>) sin(s' 8) 


(13) 



cos s' cos(.s' f) 

(4) Given the altitude, declination, and hour angle of a 
star, to find its azimuth, and the latitude of the place. 

The data here are 

o = 90 -8, b = 90-h, and B = r\ 

and the required quantities 

A=A, and c = 9O-0 

These may be found by (3) and the second of (1), Sph. 

Trig., which become 

. sin r cos 5 (14) 

sin A = - 7 v 

cos h 

sin & = sin 8 sin 0+cos 5 cos <f> cos t 

This last becomes (see eq. 5) 

. T sin 5 cos (01 0) 

sin h = : zr 

sin di 

Then transposing, we have 

fa x\ sin \ sin 6l ( 15 ) 

COS(0i <$>)= : z 

sin 5 
0i being given by the eq. 

tan 5 

tan 0i = 

cos r 

There may be two solutions of this problem; but the 

ambiguity may be removed by first determining <j> and then 

A by either of the equations (8), (9) or (10). 

(5) Given the declination and azimuth of a star, and the 
latitude of the place, to find the hour angle and altitude. 

Thus we have 

a = 90-5, A =A, and c = 9O-0; 



and are required to find 

B = t, and b = 90-h. 

The first of these is given by the second of (5), Sph. Trig., 

which becomes 

sin t cot A = cos 4> tan 5 sin <j> cos t 

or sin r cot A-\~ sin (f> cos r = cos <f> tan 5 

which may be thus transformed: 

cot A (sin r + tan A sin < cos t) =cos tan 5 

or, substituting tan 2 = tanVl sin <f> (16) 

this becomes 

cot A sin(r+0 2 ) . s 

; = cos <b tan 5 

cos 02 

or, transposing 

sin(T+0 2 ) =cos 4> tan 8 cos 2 tan A 

Then eliminating tan A by (16) we have 

sin(r-|-0 2 ) =cot <f> tan 5 sin 0j (17) 

Equations (16) and (17) determine r. 

We may now find h by applying one of equations (3), 

Sph. Trig., to the astronomical triangle, which gives 

7 sin t cos 5 (18) 

cos h = ; - A 

sin A 

We may also find h directly from the data by means of 

the first of (1), Sph. Trig., which gives 

sin 8 = sin h sin 0+cos h cos <j> cos A ; 

which may be written 

sin 8 = sin </>(sin h-\-cos h cot <j> cos A); 

in which substituting 

cot 3 = cot (f> cos A 

we have 

sin 8= sin <(sin & + cos h cot 03) 

sin <f> cos(h 63) 

(19) 

Also . tan <j> (20) 

tan 3 = -. 

cos A 

(6) To find the altitude, hour angle, and azimuth of a 
circumpolar star when at elongation, or maximum azimuth. 

It is assumed that the latitude of the place is known. 

When a star is at elongation the angle C, Fig. 4, is a right 

angle, and the solution is given by equations (26), (28) and 

(27), Sph. Trig., which become 

. 7 sin<*> tan* . A cos 8 (21), (22), (23) 

sin h= . r, cos t= r, sin A = . 

sin 8 tan 8 cos * 





sin 


03 






cos(h- 


-w- Sln 


8 sin 
sin 


03 


tan 3 


tan 4> 









(7) To find the altitude and hour angle of a star when on 
the prime vertical. 

Here the azimuth A is equal to 90, and it is assumed 
that 4> and 8 are given. Then applying equations (26) and 
(28), Sph. Trig., we find 

tan 5 . , sin 5 (24), (25 j 



COS T = 



tan $ 



sin h = 



sin <f> 



(8) Given the right ascension and declination of a star, 
and the obliquity of the ecliptic, to find the latitude and 
longitude of the star. 

In the triangle PP'S, Fig. 3, 

PS = 90 -5 P'S = 90-P 

SPP' = 90 + a SP'P = 90-X 

PP' = t 
and we have by equations (1), (4)and (3), Sph. Trig., 
sin /3 = sin 8 cos e cos 5 sin e sin a 
cos /3 sin X = sin 8 sin e+cos 8 cos e sin a 
cos 13 cos X = cos 5 cos a 
Then substituting 

m sin M = sin 5 

m cos M = cos 5 sin a 



they become 

sin /3 = m sin (Jlf e) 
cos /3 sin X = m cos(Af c) 
These may be written 



} 



(26) 



(27) 



(28) 
(29; 



tan M = 



tan 5 
sin a 



. sin 5 sin(M e) 

sin /3 = = ,j 

sin M 

tan a cos(Af e) 
tan X = 



(30) 



cos M 

The quadrant in which M is situated is determined by 
equations (27), m being assumed always positive. 

(9) Given the latitude and longitude of a star, and the 
obliquity of the ecliptic, to find the right ascension and declin- 
ation of the star. 

As in the last case we have 

sin 8 = sin /5 cos e+cos /3 sin e sin X _ "j 
cos 5 sin a = sin (3 sin e cos /3 cos e sin X J- (31) 

cos 5 cos a = cos /3 cos X ; J 

in which substituting 

n sin iV = sin /3 ) (32) 

w cos N = cos sin Xf 



9 



they become 

sin 8 = n sin(7V+e) 
cos 5 sin a = n cos(iV+e) 
cos 5 cos a = cos jS cos X 
From these we derive 

tan 



(33) 
(34) 
(35) 



tan N = 



sin 5 = 



tan a = 



sin X 

sin j3 sin(7V+e) 

sin N 
tan X cos(7V+6) 
cos N 



(36) 



It 



2. Time. 

The sidereal day. 

The earth's motion of rotation, as far as can at present be 
ascertained, is uniform; though theoretical considerations 
point to a possible retardation of its, velocity. If such retarda- 
tion exists, its amount must be extremely minute, as up to 
the present time none has been detected. The time of ap- 
parent rotation of the starry sphere is therefore sensibly 
constant, and may consequently be adopted as a unit of time 
and be denoted the sidereal day. Owing to the proper 
motions of the fixed stars the practical sidereal day is the 
time of rotation of the vernal equinox. 

Sidereal time. 

The sidereal day is assumed to begin at the instant of 
upper meridian transit of the vernal equinox, which point 
will in future be denoted by and referred to as the point V; 
and the sidereal time at any instant is the hour angle of V 
at that instant. It is thus equal to the right ascension of 
any star which is on the meridian of the observer at that 
instant. 

The solar day. 

A unit of time dependent on the sun is necessary for the 
purposes of daily life. 




F/g.5 



On account of the earth's orbital motion about the sun 
the latter body has an apparent motion among the stars, 

11 



so that it returns to the meridian of a place nearly four 
minutes later on any given day than on the previous day, as 
shewn by a clock regulated to sidereal time. 

This apparent motion of the sun, however, is not uniform. 
The earth moves in an ellipse, of which the sun occupies 
one of the foci, and its angular velocity about the sun varies 
inversely as the square of its radius vector; the angular 
velocity of the sun on the ecliptic therefore varies in the 
same manner. An inequality in the lengths of the solar days 
results from this; but a further irregularity is due to the 
obliquity of the ecliptic; for, even if the sun's motion on the 
ecliptic were uniform, its motion in right ascension would 
not be so. 




F/g. 6 



This is illustrated in Fig. 6, which is a projection on a 
plane perpendicular to the earth's axis. Pi and P% are two 
consecutive positions of the earth in which the sun is on a 
given meridian. The earth in the interval has performed a 
complete rotation on its axis plus the angle M'PiM" ' , which 
equals PiSP 2 , which is the angle through which the pro- 
jection of the radius vector has revolved during the interval. 
This angle varies from day to day, owing to the causes above 
mentioned, viz., the eccentricity of the earth's orbit and the 
obliquity of the ecliptic. The solar day, being equal in length 
to the time of an absolute rotation of the earth on its axis 
plus the variable angle P1SP2, is therefore variable in length. 
The angle P1SP2 is clearly the motion of the sun in right 
ascension in the solar day. 

To obtain an invariable unit of time dependent upon the 
sun astronomers invented a fictitious sun, called the mean 
sun, and denoted by S in Fig. 7, which is assumed to move 
at a uniform rate on the equator and to return to the vernal 
equinox at the same instant as another fictitious sun Si, 

12 



assumed to move at a uniform rate on the ecliptic. Sy i* 
also assumed to pass through perigee, and therefore apogee, 
at the same instant as the true sun. 




Fig. 7 



The relative positions of the three suns at different times 
of the year are shewn in Fig. 7. There, the points VBXA 
shew the positions of the sun when the earth is at correspond- 
ing points in Fig. 5. 

Solar time. 

Apparent solar time at any instant is the hour angle of 
the true sun at that instant. 

Mean solar time is the hour angle of the mean sun. 

Apparent noon is the instant when the sun is on the meridian 
of a place. Mean noon is the instant when the mean, sun is 
on the meridian. 

The equation of time is the difference between apparent 

13 



and mean solar time ; or, it is the difference of right ascen.si.yn 
of the true and mean suns. 

Tracing out the relative positions of the three suns in 
Fig. 7 shews that the equation of time changes its algebraic 
sign four times in the year, about April loth, June 14th, 
Aug. 31st, and Dec. 24th. It therefore has four maximum 
values. 

Civil and astronomical time. 6-t*f+-9*<~ /?&& . 

The civil day begins at the instant of lower meridian 
transit of the mean sun, or at midnight; while the astro- 
nomical day of the same date begins at upper meridian 
transit 12 h. later. 

Time at different meridians. 




F/g.8 



At any instant at two places in different longitudes, the 
hour angles of the sun, or of V, differ by an amount equal 
to their difference of longitude; consequently the difference! 
between the local times of the two places, either solar cr 
sidereal, is equal to their difference of longitude. 

This as shewn by Fig. 8. Thus if PA and PB are the 
meridians of two places, S and V the mean sun and the 
vernal equinox, respectively; then the M.T, at A is the 
angle APS , and the sidereal time the angle APV. The 
corresponding times at B exceed these by the angle APB 
(denoted by L). 

Standard time. 

For convenience, since 1883 the time used ac any place in 
N. America, instead of being the local time of the place, is 
theoretically the time which differs by the nearest who!e 
number of hours from Greenwich time. This is called st.n- 



14 



The change in the astronomical day. 

Since the beginning of the year 1925 the astronomical day 
has been discontinued, or made coincident with the civil day. 
This change necessitates certain modifications in the method 
of converting sidereal to mean time. 

Equations (37) and (38) remain unchanged in form, thus: 

Q = (T+L)(l+k') + V -L (37) 

T=(0+L)(l-k)+M-L (38) 

In the former of these equations T-\-L denotes the G.M.T. 
corresponding to the given time, and reckoned from midnight, 
and V the G.S.T. at the previous Greenwich O h . 

In the latter equation, if 0+L > 24* it must be diminished 
by 24 /f before being reduced to the equivalent mean time inter- 
val. The value of the quantity M must be that given in the 
ephemeris for the immediately preceding Greenwich sidereal 
noon. Thus, if, when M is taken for the given date, the 
quantity 

(e+L)(i-&)+M 

is less than L, then M must be taken for the next following 
date ; but, if that quantity is greater than 24 h -\-L, then M must 
be taken for the previous date. For intermediate values M 
must be taken for the given date. 



dard erne. Thus, the time which differs by 5 h from Gr. 
time 2s used at all points whose longitudes lie between i* 
30 m and 5 h 30 m W. The following standard times ?ve 
used in N America: 

Atlantic, differing by 4 h from Gr time; 

Eastern, differing by 5 h from Gr. time; 

Central, differing by 6 h from Gr. time; 

Mountain, differing by 7 h frorc Gr. time; 

Pacific, differing by 8 h from Gr. time; 

Yukon, differing by 9 h from Gr. time. 

Relation between the lengths of the solar and sidereal units of time. 

The tropical year is the interval of time between two cdb 
secutive passages of the mean sun thiough the niran vern^ 
equinox 




Fig. 9 



In Fig. 9 let 5 and S' be the positions of the mean sui 
relatively to 7 at the instants of two consecutive 1 ransiti 
over the meridian of some place. Then it is evident that the 
mean solar day is equal to the sidereal day plus the motion 
of the mean sun in right ascension in one mean solar day. 
(See also Fig. 6.) Therefore if 

D the length of the solar day, and 

D' = the length of the sidereal day, and 

n =the number of mean solar days in 1 tropical year; 
then 

1 tropical year = rZ> 

rv{D'+D') 
n 

= (-r l)D'. 



15 



2 



But 1 tropical year = 365.24222 mean solar days 
.'. 1 tropical year = 366.24222 sidereal days, 
Tf then 

M any interval of time expressed in mean solar days, 
eS = the same interval expressed in sidereal days; 
M 365.24222 , , 
TS^ 366^4222 ==1 -^ aSSUme; 
. S 366.24222 , . ,, 

*** M '365.24222 = 1+ * 

in which = 0.00273043 

'=0.00273791 

Also 

24 h M.S.T. = 24 h 03 m 56 s .555 Sid. T. 

24 Sid. T. = 23 56 04 .091 M.S.T. 
The cor. version of a given interval of M.S.T. into 
the corresponding interval of Sid. T., or conversely, is 
best effected by means of tables given in the Nautical 
Almanac. 

To convert the mean time at a given meridian into the corre- 
sponding sidereal time. 

Let T the given local M.T.; 

G -the corresponding Sid. T ; 
L =the longitude of the place; 
V, =the Gr Sid. T. at the previous Gr. mean noon. 
Then 

e = (r4-L) (l+k') + V -L (37) 

V is taken from the ephemeris. Instead of using the 
factor k' the reduction of T-k-L to the equivalent sidereal 
interval is made by means of tables given in the N. A 

To convert the sidereal firz> at a given meridian into the 
corresponding mean time. 

Using e same notation, and in addition denoting by 
M the mean time at Gr. of the previous Gr sidereal noon, 
we have 

r-(e-fL) {l-k)+M-L (38) 

Here again the tables of the N A. are used instead of the 
factor k. 

The value of M, to be taken from the N. A., is to be that 
for the date of the transit of V immediately preceding the 
given time. Thus if 

(9-j-L) (l-k)+M>2& 
then the. value of M must be taken for the previous date. 

16 



To convert the apparent solar time at a given meridian into 
the corresponding sidereal time. 

This may be done by first reducing to M.T. by applying 
the equation of time to be taken from the N. A. and then 
reducing to sidereal time by the method given above. 

A more convenient method, however, is to interpolate 
from the N. A. the value of the sun's right ascension at^ the 
Gr. instant corresponding to the given time. Then if in Fig. 8 
So represents the true sun it is clear that if / = the hour angle 
of the sun, or the apparent time, then 

= ;+a (39) 

To determine the hour angle of a heavenly body at a given 
time. 

If in (39) it is assumed that the hour angle / may have 
any value up to 24 h , then that equation is general and 
applies to every case and any heavenly body. It may be 
necessary in some cases to deduct 24 h from t -\- a. Transpos- 
ing we have 

t = Q-a (40) 

Here it may be necessary in some cases to increase by 
24 h to render subtraction possible. 

The hour angle denoted by r, found by solving the astro- 
nomical triangle the parts of that triangle being limited 
to values less than 180 , being given, we have 

/ = r if west 
= 24 h -r if east. 

The hour angle of the sun may be found by equation (40) 
if the sidereal time is given. If the mean time is given, it 
may be reduced to apparent time by applying the equation 
of time, thus finding the required hour angle. 

Reduction of time to arc ; and conversely. 
These reductions may be made by means of the following 
numerical relations: 

l h =15 l = 4 m 

l m = 15' l' = 4 s 

I s =15" 



17 



3. Determination of Time by Observation. 

Correction and rate of a chronometer. 

As the term implies the correction of a chronometer is 
the amount that must be added to the chronometer time to 
give the true time. 

The rate of a chronometer is the amount it loses in a unit 
of time. 
Thus, if 

7\ and !T 2 = the true times at given instants; 
T\ and r' 2 = the chronometer time at those instants; 
A7"i and A2" 2 = the chronometer corrections; 
57"= the chronometer rate. 
Then LT X =T X -TA (41) 

AT 2 = T 2 -T 2 'f 

Ar 2 -Ar x (42) 

7Y-7Y 

These equations give the corrections and rate with their 
proper algebraic signs. The rate is thus given in terms of the 
chronometer interval. 

1st method By transits. 

(a) Meridian transits. 

A transit instrument having been adjusted in the meridian, 
the time of transit of any heavenly body across the vert, 
wire may be observed by a chronometer whose correction is 
to be found. 

If the chronometer is regulated to sidereal time the true 
sidereal time of transit is at once given by the right ascension 
of the body, whence the chronometer correction at once 
follows by (41). If regulated to mean time, the sidereal 
time of transit of the body must be reduced to mean time. 

If the sun is observed, the time of transit of each limb 
should be noted and the mean taken; thus finding the time 
of transit of the centre. If only one limb can be observed, 
then the observed time must be corrected by the "time of 
semi-diameter passing the meridian", which may be taken 
from the N. A., or computed by the equation 

5 (43) 

/=- =- sec 5 v ' 

15 

in which 5 is the angular semi-diameter of the sun. 

If the correction of a M.T. chronometer is to be found 
by a transit of the sun, the true M.T. of transit may at once 
be found by applying the equation of time to the apparent 
time of transit (fi 1 . 

18 



(b) Transits across any vert, circle of known azimuth. 

In this case the latitude of the place and the declination 
of the heavenly body must be known; then the hour angle 
may be computed by means of (16) and (17), which may be 
written 

tan = tan A sin </> 
sin(r+0) =cot </> tan 5 sin 6 
The sidereal time then follows by (39), or the M.T. by apply- 
ing the equation of time as already shewn. 

The rate of a chronometer may be found by observing two 
consecutive transits of a star across the same vert, circle. 
The true interval between the transits is 

24 h Sid. T. or 23 h 56 m 04 s .09 M.T. 

(c) Transits across the vertical circle of Polaris. 
This method will be described under Azimuth. 

2nd method By a single altitude. 

The method of observing an altitude of a heavenly body is 
described below, p. 65 et seq. 

Corrections to be applied to an Observed Altitude. 

(a) Refraction. 

The ray of light that reaches an observer from a star, in 
traversing the earth's atmosphere is continually bent down- 
wards from a rectilinear path by the increasing refractive 
power of the air with density as the surface of the earth is 
approached. In consequence, the apparent direction of a 




F/g. /O 



star is that of a tangent to the curved path of the ray at the 
point where it reaches the observer. This is illustrated in 
Fig. 10. 

19 



An observed altitude must then be diminished by an 
amount equal to the angle between the final direction of the 
ray and the straight line drawn to the star, as appears in the 
figure. The magnitude of r decreases as the altitude increases, 
and its value is best found from tables. These contain 
corrections depending upon the readings of the barometer 
and thermometer. An approximate value of r may be found 
by the equation 

r = bl".l tan f 
or a closer approximation by the formula 

9836 

r= 460+7 tanr 

in which 

b =the barometer reading in inches; and 
/ = the temperature of the air in degrees F. 

(See Field Astronomy for Engineers, by Prof. G. C. Comstock). 

(b) Semi-diameter. 

In observing the sun or moon the altitude of its upper or 
lower limb is observed. To find the altitude of its centre a 
correction for semi-diameter must be applied. This may be 
found in the N. A. 

(c) Parallax. 

As the centre of the celestial sphere is coincident with that 
of the earth, if the directions of a heavenly body from that 
point and from a point on the earth's surface differ sensibly, 




F/G // 



then a correction must be applied to any observed co-ordinate 
to reduce it to the centre of the earth. This is only necessary 
with members of the solar system. 

20 



In Fig. 11 .S is the centre of the heavenly body observed, 
the centre of the earth, A the point of observation. The 
triangle A SO gives 

sin p = sin C 

p being the parallax in altitude. If f' = 90 the resulting 
value of p is the horizontal parallax. Denoting it by ir we 
have 



sin 7r = 



a 



sin p = sin w sin $"' ; (44) 

or very nearly 

P = t sin f ' = it cos &' (45) 

This gives the correction for parallax with sufficient accuracy 
for any body except the moon. 

(d) Dip of the horizon. 

At sea the altitude of a heavenly body is measured with a 
sextant from the sea horizon, the observer standing on the 
deck of a ship. A correction must therefore be applied to 
the observed angle on account of the dip of the visible below 
the true horizon. 




Fte. /2 




F/G./3 



In Fig. 12 we have from PI. Geom. 

AB J(2a + h)h V2ah 



tan D' = 



a 



a 



a 



nearly 



or 



D' = 



V 



2)i 



a 



This gives the dip uncorrected for refraction; but, as shewn 
in Fig. 13,. the ray of light which reaches the observer from 
the horizon follows a curved path, so that the apparent dip 



21 



D is less than D' . The mean value of the ratio of D to D' 
is .9216 : 1, so that 



D = .9216 -i/ 



2h_ 
a 



or in seconds of arc 

.9216 /IT 



D = 



\" 



sinl' v a 

Substituting a mean value of a in feet, this becomes 

D = 58". 82 JT (46) 

h being in feet. 

The rule known to navigators: "Take the square root of 
the height of the eye above sea level in feet and call the result 
minutes", is thus very approximately correct. 

Having applied the necessary corrections to the observed 
altitude, the reduction may be made by either of the equa- 
tions (11), (12) or (13). If a number of observations are to 
be reduced an equation derived as follows is more convenient: 
Taking the equation 

cos f = sin 5 sin 0+cos g cos cos t 
it may be written 

1 versin f = sin 5 sin 0+cos 5 cos $(1 versin r) 
= cos(0 5) cos cos 5 versin t 
= 1 versin (0 5) cos cos 5 versin t 

versin f versin (0 5) (47) 

versin r = r 

cos cos 5 

This requires the use of tables of natural and logarithmic 

versins. In the absence of such a table the following form of 

the equation may be used 

. , cos(0-5)-cos f (48) 

Sin^ f r = r 

2 cos cos 5 
Example. The following observations were taken with a 
sextant and artificial horizon on Aug. 1, 1892, at a place 
in latitude 52 31' 04", and approximate longitude 7 h 50 m W.; 
to find the watch correction. 

2 -alt. O Watch 

52 11' 30" 7 h 21 m 29 s A.M. 

10 22 54 

30 24 27 

30 25 28 

00 28 18 

10 29 52 

10 30 54 

Index error =+20". 



52 


38 


53 


05 


53 


24 


54 


16 


54 


44 


55 


03 



First find the approximate Gr. M.T., thus: 

Mean of extreme times = 7 h 26 m ll 8 
Ast. time, July 31 =19 26 11 

Long = 7 50 

Gr. M.T., Aug. 1 = 3 16 11 



For this time we take from the N. A. 

8 =+17 48' 56" 
S= 15 48 

E= 6 ra 03 s .6 

Reduction of first observation: 

Obs'd. 2-alt. =52 

Index error 



h' 
r 



P 

h 

f 

Eq. (13) s' 

s'<f> 

s'-S 



log sin (s' 4>) 

log sin (s' 8) 

log cos s' 

log cos 0' f) 



11' 30" 

+20 



2)52 
= 26 


11 50 

05 55 

1 52 


26 


04 03 
15 48 


25 


48 15 
8 


= 25 
= 64 


48 23 
11 37 



67 


15' 48" 


.5 


14 


44 44 


.5 


49 


26 52 


.5 


3 


04 11 


.5 



9.405738 
: 9.880708 

9.587143 
; 9.999377 



9.286446 
9.586520 



log tan 2 \t 


= 9.699926 


log tan \t 


= 9.849963 


h 


= 35 17' 38".9 


T 


= 70 35 17 .8 




= 4 h 42 m 21 s .2 



23 



'. App. Time 


= 7 17 38 


.8 


E 


6 03 


.6 


Mean Time 


= 7 23 42 


.4 


Watch 


= 7 21 29 




AT 


= + 2 13 


.4 



Having reduced the remaining observations the complete 
results are as follows: 

AT 



+2 m 13 s 


.4 


16 


.0 


13 


.2 


15 


.0 


15 


.3 


14 


.3 


15 


.2 


Mean =+2 14 


.6 



Another example will be found on p. 43. 

To find the effect of errors in the data on the time computed 
from an observed altitude. 

Taking the equation (see Fig. 4) : 

cos b cos a cos c sin a sin c cos B = 
and differentiating by means of the expression 

%dB= M- da+-^-db+ # dc 
dB da db dc 

we find 

sin a sin c sin BdB = 

(sin a cos c cos a sin c cos B)da 
sin bdb-\-{cos a sin c-f-sin a cos c cos B)dc 
= sin b cos Cda sin M&+sin & cos ylrfc 
by applying equations (4), Sph. Trig. Then substituting in 
the left-hand number 

sin a sin B = sin b sin ^4 
we have 

JT> cos CV/a d& </c 

dB = ; t ; j + 



sin c sin ^4 sin c sin ^4 sin c tan A 

Then introducing the astronomical co-ordinates, and re- 
membering that 

da=d8 db=dh dc=d(j) 
we have finally 

, cos Cd8 dh d(j> (49) 

cos <t> sin A cos < sin A cos < tan .4 

24 



The errors being small may be regarded as differentials, 
so that (49) gives the effect of errors in 8, h, and <f> upon the 
resulting hour angle r. It shews moreover that the effect of 
those errors is least when A and C are both large, or when the 
star observed is near the prime vertical. 

3rd method By equal altitudes of a heavenly body. 

Method of observation with a transit or sextant. 

Equal altitudes of a heavenly body on opposite sides of 
the meridian correspond, generally speaking, to equal hour 
angles. This is the case of a fixed star, whose change of 
declination between the two positions may be neglected. 
The mean of the times of equal altitudes is then the time of 
meridian transit. The method is therefore an indirect one 
for observing the time of meridian transit. 

In the case of the sun, however, allowance must be made 
for the change of declination in the interval between the two 
observations. An expression for the correction to be applied 
to the mean of the observed times is derived as follows: 




F/e. 14 



Fig. 14 shews a projection of the celestial sphere on the 
plane of the horizon. S\ and 52 are the two positions of the 
sun's centre at the instants of the two observations; S\ the 
position it would have occupied if there had been no change 
of declination. The two triangles PZSi and PZS't are then 
equal in all respects. It is therefore required to find the 
change of hour angle resulting from a small change of declin- 
ation. Taking the equation 

cos f sin 5 sin cos 5 cos <t> cos r 

25 



we find by differentiation 

dr cos 5 sin sin 5 cos # cos t 
d8 cos 5 cos <j> sin t 

tan <f> tan 8 



If we now write 
this becomes 



sin r tan t 
dr=-2AT d8 = 2A8 



-2Ar =(^-^Y2A5 
\ sin t tan r / 



or in seconds of time 



. AS /tan </> tan 5\ 

15 \sin r tan t/ 



(50) 



This is the "equation of equal altitudes." 
In this equation 

AT = the correction to be applied to the mean of the 

observed times to find the time of meridian transit; 

A5 =half the change in the sun's declination in the 

interval between the observations, positive if the 

sun is moving north. 

t may be assumed equal to half the elapsed interval between 

the observations. Attention must be paid to the algebraic 

sign of 8; it is positive if north. 

The advantages of this method are that the absolute 
altitudes need not be known; and small errors in 4> and 8 
have no appreciable effect. 

To find the time of rising or setting of a heavenly 
body. 

Take the equation 

sin A = sin 8 sin #+cos 5 cos <f> cos t; 
which may be written 

cos r = sin h sec <f> sec 5 tan <j> tan 5 (51) 

In the case of the sun, when its upper limb is just visible 
in the horizon it is in reality below the horizon by the amount 
of the refraction, 34' approximately; and its centre is below 
the limb by the amount of the semi-diameter, which may 
be taken as 16'; parallax may be neglected. Therefore 
}i = 50', and sin 50' = 0.0145 ; .". the above equation becomes 
cos t= 0.0145 sec <t> sec 5 tan <f> tan 8 (52) 

The time of rising of the moon's centre is usually computed. 
In this case the effect of parallax is important. Assuming its 
amount as 57', the altitude of the moon's centre when it is 

26 



apparently in the horizon = 57' -34' = 23'. Also sin 23' = 
0.0067; so that (51) becomes 

cos t = 0.0067 sec <f> sec 8 tan <f> tan 8 (53) 

Having computed the hour angle, the time readily follows. 

Construction of sun dials. 

The horizontal dial and the prime vertical dial only will 
be considered. 

In any form of dial the edge of the gnomon which casts the 
shadow must be parallel to the earth's axis, as the position 
of the shadow cast upon any plane is then independent of 
the sun's declination 




Fig. 15 shews the construction of the horizontal dial. 
The edge of the gnomon if produced will intersect the celestial 




F/0 /6 



sphere in the pole P, Fig. 16. PON is the meridian plane, 
NOL a horizontal plane, and POL a plane through the 
sun's centre. LON (denoted by a) is the angle which an 
hour line, corresponding to a given hour angle t, makes with 
the noon line. The triangle PLN then gives 

tan a = sin cf> tan t (54) 

27 



The construction for a prime vertical dial is shewn in 
Fig. 17. OPZ' is the meridian plane; OMZ' that of the 




Fig. 17 



prime vertical; and OP'M a plane through the sun's centre. 
is the required angle corresponding to the hour angle t. 
The triangle P'MZ' gives 

tan /3 = cos tan t (55) 

A sun dial gives apparent solar time. 



fi 



4. Determination of Latitude by Observation. 

As shewn on p. 3, the latitude of a place is equal to the 
altitude of the pole, or the declination of the zenith, i.e., to 
either arc PN or EZ, Fig. 2. 

1st method By meridian altitudes or zenith distances. 




Fte/8 



If the altitude or zenith distance of a heavenly body be 
observed when crossing the meridian, and the necessary 
corrections be applied, the latitude at once follows by one 
of the following equations, depending upon the position of 
the body. For the star 

So. . . .0 = f + 5 (5 being negative) (56) 

S 3 ....<j>=5-{ = h-p 
S A . . . .0 = 18O-5-f = fc+p 
If S 3 and S 4 are the positions of the same star observed at 
both culminations, then by taking the mean 

h+h' p-p' (57) 

* = ~2 2T 

the accented letters belonging to lower culmination. 

If S\ and 53 are two stars observed at nearly equal zenith 
distances, we have by taking the mean of the first and third 
of (56) 

5 + 5' r-f (58) 

</,= ~2~ + ^r 

the accented letters belonging to the north star. This formula 
is the basis of Talcott's method of determining latitude, the 
observed quantity being the difference of zenith distance of 
the two stars, which are selected so that that difference is 
small enough to be measured by a filar micrometer placed in 
the focus of a telescope. Details of method outlined. 



29 



If the direction of the meridian is not known the maximum 
altitude of the heavenly body may be observed. If that 
body is the sun the maximum altitude differs slightly from 
the meridian altitude, owing to its rapidly changing declin- 
ation. The resulting error is entirely negligible, especially 
if instruments of only moderate precision are used; its 
value is given by the expression 

()' ^^ or [5.54861, (^(tan,-tan 8 ) 

in which A5 is the hourly change in the declination expressed 
in seconds. The correction is always subtractive. 

Example. On July 10, 1914, the meridian altitude of 
the sun's upper limb was observed (Cir. L) to be: 

68 11' 30". 
To find the index error of the transit used the following 
V.C.R's were taken on a terrestrial point: 

Cir. L 034'30" 

Cir. R 31 

Diff. 
I.E. 

Obs'd alt. 
I. E. 

h' 

r 



3 30 
1 45 


= 68 


11' 30" 
1 45 


= 68 


09 45 
23 


68 


09 22 
15 46 


67 


53 36 
3 


= 67 
= 22 
= 22 


53 39 
06 21 

17 48 


= 44 


24 09 



h 

f 
5 



2nd method By an altitude observed out of the meridian, 
the time being known. 



30 



To the observed altitude the necessary corrections must 
be applied, and the hour angle derived from the observed 
time. The latitude then follows by means of (15) 

, i. sin h sin 

COS(d> d)= : 

v J sin 8 

6 being found by the equation 

a tan 5 

tan = 

cos r 

To find the effect of errors in the data we have by trans- 
posing (49) 

d<t>= cos C sec Add + sec Adh-\- cos "<j> tan Adr (59) 

This equation shews that the effect of errors in the data is 
least when A is small and C large, though the second con- 
dition is unimportant, as the error in the declination is always 
small in comparison with the other errors. These conditions 
are fulfilled, however, by observing a close circumpolar star 
near elongation. 

Hence the method by means of the pole star. 
As the altitude of this star never differs much from the 
latitude, the method consists in computing a correction to 
apply to the former to give the latter. An expression for this 
correction is derived as follows: 
Taking the equation 

sin h = sm < sin 5+cos 4> cos 5 cos t 
and substituting in it 

(j) = h-}-x 
8 = 90-p 
we have 

sin /* = sin(/z+x)cos p-\-cos(h-\-x)s'm p cos t 
Then expanding the sin and cos of h+x, and again expanding 
the sin and cos of x and p and neglecting the powers of their 
circular measures above the second, we have 

sin h = \ sin h ( 1 -~- ) + x cos h > ( 1 ~ ) 

+ ) cos fe I 1 J x sin h Kp cos r 

x 2 . p 2 

= sin h sin h-\-x cos h ~ sin h + p cos f cos h 

z 

px cos t sin //. 
Whence 

x cos h= p cos h cos T-\-\(x 2J rp"--\-2px cos t) sin h 
or x= p cos T-\-\{x 2 -\p 2 -\-2px cos t) tan h. 

31 

3 



Assuming now as a first approximation 

x= p COS T, 
and substituting in the right-hand member, we have 
x= -p cos r + i( 2 cos 2 r+p 2 -2p 2 cos 2 t) tan h 
= p cos r + 2 sin' t tan h 
or in seconds of arc 

x= p cos T-\-hp 2 sin 1" sin 2 t tan /z 
We have then finally 

<f> = h-p cos t + | 2 sin 1" sin 2 r tan h (60) 

The effect of the omission of the smaller terms in the 
above expansions can never amount to 0".5. 

Example. The following observations of Polaris were 
taken on June 14, 1904, with a small transit: 
Cir. V. C. R. Watch 

R. 45 44' 14 h 50 04 s 

L. 45 43 53 46 

R. 45 45 57 10 

L. 45 44 59 44 

The watch was regulated to sid. time, and its correction was 
20 s . The star's co-ordinates were: 

a = ih 24 m 26 s 

8 =88 47' 27" (."./) = 4353"). 
The mean of the first and second observations being taken, 
and that of the third and fourth, the reduction is made as 
follows: 

Eq. (60) r = 14 h 51 m 55 s = 14 h 58 m 27 s 

AT = -20 = -20 



6 
a 

t 

T 

V 
r 

h 



log p 

log COS T 



= 14 51 35 

= 1 24 26 

= 13 27 09 
= 10 32 51 
= 158 12' 45" 

= 45 43 30 
56 

= 45 42 34 

= 3.638789 
= 9.967813 



log 1st term = 3.606602w 



= 


14 58 07 
1 24 26 


= 


13 33 41 

10 26 19 

156 34' 45" 


= 


45 44 30 
56 


= 


45 43 34 


= 


3.638789 
9.962659w 


= 


3.601448w 



32 



log 0.5 

log 2 
log sin 1" 
log sin 2 r 
log tan h 

log 2nd term = 

h 

1st term 
2nd term 



<t> 



1.698970 

7.277578 

6.685575 

9.139134 

10.010756 



= 


0.812013 





45 
1 


42' 
07 


34" 

22 

6 


= 


46 


50 


02 



= 1.698970 

= 7.277578 

= 6.685575 
= 9.198634 
= 10.011009 

= 0.871766 

= 45 43' 34" 
= 1 06 34 

7 

46 50 15 



Mean = 46 50' 08" 

Circum-meridian Altitudes. 
If a number of altitudes of a star be observed in quick 
succession when near the meridian, each will differ by but a 
small amount from the meridian altitude. A correction may 
then be computed for each altitude which, when applied to 
it will give a value of the meridian altitude. The mean of 
these resulting values having been taken the latitude then 
follows by means of one of the equations (56). 

To find an expression for this correction we return to the 
equation 

sin h = s\n <$> sin 5 + cos cos 5 cos t, 
which is transformed as follows : 

sin A = sin (/> sin 5+cos <f> cos 5(12 sin-^r) 
= cos(</> 5) cos <f> cos 8 . 2 sin 2 f t 
= cos f cos <j> cos 8 . 2 sin 2 ^ t 
= sin h cos 4> cos 8 . 2 sin 2 | r 
by (56), f being the meridian zenith distance and h the 
meridian altitude. If we now write 

h = h y 
we have sin h = s\n{h ^y)=sin h y cos h 

by expanding and discarding powers of y above the first. 
Substituting in the above expression for sin h, it becomes 

sin h y cos h = sin h cos < cos 8 . 2 sin 2 \ r 
or cos d> cos 8 

y = 



2 sin 2 h t 



cos h 
or in seconds of arc 

cos </> cos 5 2 sin 2 \ t 

cos h sin 1" 

This gives the required correction. 



(61) 



33 



If higher powers of small quantities be retained in the 
above expansions the expression for y becomes 

y = y'~ ^^tan * y* + 25"J*(i+3 tan \) y' - (61') 



in which y' = 



6 

cos 4> cos 5 2 sin 2 \ t 



cos h ' sin 1 



// 



and log^- =6.3845449, log S -^- = 12.5929985. 

Then h = h-\-y 

In applying this method stars giving values of h approach- 
ing 90 5 must be avoided. If = 45 a and h = 60 the second 
term of (61') will not exceed 1" for values of t less than 13 m , 
and the third term for values less than 33 m . 

The value of log m is given in Table III. 

Example. The following observations were taken with a 
sextant and artificial horizon on Sept. 2, 1893: 
2 -alt. _ Watch 

89 59' 15" ll h 53 m 36 9 

90 00 15 56 37 

90 00 45 59 28 

89 59 15 12 03 57 

89 58 30 05 46 

89 57 30 07 11 

89 55 15 09 13 

Index error = 0; watch correction = 8 s . 
An approximate value of the latitude is found by regarding 
the maximum observed altitude as the meridian altitude, as 
follows : 

Max. 2 -alt. =90 00' 45" 

Eccentric error +2 00 



Obs'd alt. 

r 



90 
= 45 


02 45 
01 22 

58 


45 


00 24 
15 54 


45 


16 18 
6 



p 

h =45 16 24 

34 



f =44 43 36 

5 = 7 37 54 



(approx.) = 52 21 30 



The hour angles corresponding to the observed times may 
be found by first finding the watch time of culmination, thus 
App. time of culm'n = 12 h 00 m 00 s 
E = -21 



M.T. =11 59 39 

AT = 08 



Watch time of culm'n =11 59 47 



From this follow the hour angles tabulated below. The 
corrected zenith distances are also found as above. We then 
proceed as follows: 

log cos 4> =9.785843 

log cos 8 =9.996136 



log cos h =9.847403 



9.781979 
9.934576 
logra =1.87545 



log (h -h) =1.81003 

h -h =64".57 



The remaining corrections are computed in a similar 
manner, and are tabulated below. 

f t h h f 

44 44' 21" 6 m 11 s 1' 05" 44 43' 16" 

43 51 3 10 17 34 

43 36 19 00 36 

44 21 4 10 29 52 

44 43 5 59 1 00 43 

45 13 7 24 1 32 41 

46 21 9 26 2 30 51 



Mean =44 43 39 
6 = 7 37 54 



<f> =52 21 33 



35 



3rd method By two observed altitudes of a star, or the 
altitudes of two stars, and the elapsed time between the 
observations. 

In addition to the latitude this method also serves to 
determine the time and azimuth. 




F/g/9 



Let S\ and S 2 be the positions of the star or stars at tht 
instants of observation. The first step in the reduction is 
to determine the difference of hour angle S1PS2. If the sun 
is observed twice, this angle is equal to the elapsed interval 
of apparent time between the observations, though usually 
the effect of the change in the equation of time may be 
neglected. If one fixed star has been observed the angle 
S1PS2 is equal to the elapsed sidereal interval between the 
observations. If two stars are observed at the times T\ and 
r 2 , the right ascensions being ai and a 2 , then 

S.PS, = ( ai - a,) - (TV- r 2 ) (62) 

Si being the more easterly star. The interval 7\ T 2 must be 
in sidereal time. 

Then, PS\ and PS? being known, the triangle S1PS2 may 
be solved, finding S1S2 and PS1S2. The three sides of the 
triangle ZS1S2 are now known, so that it mav be solved, 
finding the angle ZSiS 2 . Then PSiZ = ZS 1 S 2 -PS 1 S 2 . The 
triangle PZS is finally solved, finding PZ the co-latitude. 
Completing the solution gives also the hour angle ZPS and 
the azimuth PZS. 

This method is further developed in works on navigation, 
in which graphical solutions are given. 

4th method By transits of stars across the prime vertical. 

A star whose declination lies between the limits and < 
will cross the prime vertical above the horizon twice in its 
diurnal course. 

The times of transit across the p. v. may be observed by 
means of a transit adjusted in the p. v. If S\ and S 2 are the 
two positions of a star at the instants of observation, then 



the elapsed sidereal interval between the observations is 
equal to the angle SiPS 2 , and half that interval is the hour 




F/G.20 



angle of the star at either observation. Transposing eq. (24) 
we have 



tan 4> 



tan 8 



(63) 



COS r 

by which the latitude may be found. 

This method is little used with small instruments, but 
when applied to the astronomical transit instrument it is 
one of the most precise methods known for determining 
latitude. 

5th method -By observations of stars at elongation. 

If two circumpolar stars be selected, whose times of elonga- 
tion, one east and the other west of the meridian, are not 
widely different, we have for the two stars, applying eq. (23) 

cos 5i A cos <5 2 (64) 



whence 



sin A\= sin A 2 = 

cos 4> 

sin A 1 cos 5i 
sin A2 cos 8 2 



cos 



From this by composition and division 

sin ^4i+sin A2 cos 81 -f- cos 5 2 
sin A\ sin A 2 cos 81 cos 82 



or 



tan %(A!+A 2 ) 



= - cot \ (5i + 5 2 ) cot \ (5i-5 2 ); 



tan %(Ai-A 2 ) 
from which finally 

tan %(Ai Ai) = -tan ^(^4 1 +i4 2 )tan |(5i + 5 2 )tan J(*i k) (65) 
From this may be found the difference of the azimuths of 
the two stars when their sum is known. The sum of the 
azimuths may be observed by poirting the telescope of a 
transit to each star in turn, when at elongation, noting the 



37 



readings of the horizontal circle and taking their difference. 
From the sum and difference of A\ and A* their separate 
values may be found. The latitude then follows by either 
equation 

cos 5i cos 5 2 (66) 



cos 4> = 



sin^4i sin^lj 



This method was due to Prof. J. S. Corti. 

The best stars for observation are those having large 
azimuths when at elongation, or whose declinations do not 
greatly exceed the latitude. Their elongations then occur 
at high altitudes, and therefore this principle must not be 
pushed to an extreme, as the effect of an unknown inclina- 
tion error of the horizontal axis of the transit increases 
rapidly with the altitude. 



5. Determination of Azimuth by Observation. 

1st method By meridian transits. 

The time of meridian transit of any star may be computed 
as shewn on pp. 11 and 16. If the correction of a chrono- 
meter be known, the chronometer time of transit may be 
found. By directing the sight line of a well adjusted transit 
to the star at that instant, it will thus be placed in the meri- 
dian plane, and a meridian line may then be established on 
the ground; or by horizontal circle readings when pointing 
to the star and a mark, the azimuth of the latter may be 
determined. 

It is clear that a slow-moving circumpolar star is best ior 
this observation, as then the effect of an error in the computed 
time of transit is a minimum. The rate of change of azimuth 
of a star when crossing the meridian is given by the relation 

A A 1 - A cos 5 (67.) 

AA = lo.Ar -r 7- - rr 

sin \4> 8) 

(see eq. 75) AA being expressed in arc and At in time. In 
the case of the pole star over 2 m are required for a change 
of azimuth of I', when crossing the meridian. 

2nd method By transits across any vertical c'~Je, the 
latitude being known. 

Having computed the hour angle from the observed time, 
the data of the problem are r, 8, and <i>, and the azimuth of the 
star may be computed by means of (6) and (7), or 

tan 5 . tan t cos 6 

tan 9 = , tan A - n r- 

cos t sin 'yd <p) 

The same considerations as in the last method lead to the 
choice of a close circumpolar star for this observation. The 
equation from which (6) was derived may be placed i \ more 
convenient forms. Thus it may be written, 

tan A = 

cos tan 5 sin <f> cos 

then multiplying the right-hand member through by sec j> 
cot 5, this becomes 

. sec <j> cot 8 sin r (68) 

tan A ~z - . 

1 tan <f> cot 5 ccs r 

This form is convenient when subtraction log's are available. 
(See Manual of Survey of Dominion Land*.) 



39 



Again, the above equation (68) may be written 

. tan A ' 

tan A = 

1 m 

in which 

tan A ' = sec </> cot 5 sin r (a) 

and m=ta.n </> cot 5 cos r 

= tan A' sin 4> cot r (&) 

Taking logarithms and expanding, we have 
log tan A =log tan A' log (1 m) 

= log tan yl'+/x (m-\-\ m 2 -\-\ m 3 -\-) 
= log tan A' +ii m-\-\ /j, m--\-l /x w 3 + (69) 

H being the modulus of the common system of logarithms- 
We also have 

log M =1.6377843 

log |/i = 1.3367543 
log \n =1.1606631 

The algebraic sign of m is the same as that of cot t. The 
third term of (69) is seldom required. 

In taking the observation the procedure is as follows: 

Point to the reference point and note H.C.R. 

Then point to the star, note time and H.C.R. 

Then reverse instrument and again point to the star and 
note time and H.C.R. 

Then point to the reference point and note H.C.R. 

The means of the H.C.R's on the star and reference point 
are then taken, increasing or diminishing one in each case by 
180; and also the mean of the times of pointing to the star, 
from which the hour angle is derived. 

Having computed the azimuth by (68) or (69), let: 

A s denote the azimuth of the star reckoned from the north 
in the direction ESW; 

A p that of the reference point. 

R s the H.C.R. on pointing to the star 

Rp that on pointing to the reference point. 
Then A p -A, =R P -R s 

or A p =A S +R P -R s (70) 



40 



Example. The following observations were taken in 
Aug., 1904, at a place in latitude 46 54': 

PL obs'd. Cir. H.C.R. Watch 

R-P. R. 178 14'.5 

Polaris R. 57.5 15 h 55 m 08 s 

Polaris L. 181 02 .5 16 01 05 

R.P. L. 358 14 .5 

The watch correction was found by observing the meridian 
transit of a Scorpii, as follows: 

Watch time of transit = 16 h 23 m 00 s 

R't ascension of star =16 23 34 



Watch corr'n = +34 

From the N.A. 

a (of Polaris) = l h 25 m 03 s 

8 =88 47' 28" 

.'. p =4352" 
The computation then proceeds as follows: 

Mean of obs'd times = 15 h 58 m 06 s 

Watch corr'n = -f 34 



Eq. (40) Sid. time =15 58 40 

a = 1 25 03 



1 =14 33 37 

T = 9 26 23 

= 141 35' 45" 



Eq. (68) log sec = 10.165405 log tan <j> = 10.028825 

log cot 8 = 8.324328 log cot 8 = 8.324328 

log sin r= 9.793235 log cos r= 9.894122w 

8.282968 8.247275w 
Subt. log= 0.007610 

log tan 4= 8.275358 
A= 1 04' 48" 



Eq. (70) A s = 1 04' 48" 
R P =178 14 30 



179 19 18 
Rs = 1 00 00 



A p -178 19' 18" 

41 



The computation by (69) is as follows: 
Eq. (a) 



Eq. (6) 



iog sec 
log cot 8 
log sin t 


= 10.165405 

= 8.324328 
= 9.793235 




log tan A' 
log sin 
log cot r 


= 8.282968 
= 9.863419 
= 10.100887 


8.282968 


log m 


= 2.247274 




log n 


= 1.637784 




log ^ w 


= 3.885058 


- 0.007674,6 


log m 2 


= 4.494548 




log | m 


= 1.336754 


a 


log ^ m w 2 


= 5". 83 1302.. 


..0.000067,8 


log tan A 


= 1 04' 48". 


8.283035,8 
8.275361,2 


A 


07 



This method may be used to advantage in finding the 
variation of a compass. An explorer's instrumental equipment 
may consist of a sextant and a compass. With the former 
instrument an observation of the sun for time may be taken. 
If the compass bearing of the sun's limb then be taken, the 
true azimuth of that body may be computed in terms of 
t 8 and 0, which, compared with the magnetic azimuth, will 
give the variation. The quantity 6 1 sec h must be added to 
or subtracted from the azimuth of the sun's centre to obtain 
the azimuth of the limb, h is given by (5) and need only be 
known approximately. 
The equations then are: 

tan 5 . 7 sin 6 cos (0 0) 

tan = sin h = -.^ '- 

cos r sin 

. tan r cos A . , 
tan A = . ,_ ^ AA=S sec h 

sin (0 0) 

The best time for this observation is when the sun is near 
the prime vertical. 

3rd method By an observed altitude. 

The method of observation is described on p. 65 et seq. 

The data are h 8 and 0, and the reduction is made by one 
of the equations (8), (9) or (10). 

Example. The following observations of the sun for 
azimuth and time were taken on July 30, 1914, at a place 
in latitude 44 24' 09", and approximate longitude 5 h 18 m 15 s W.: 

42 



Pt. obs'd. Cir. H.C.R. V.C.R. Watch 

R.P. R. 23 26'.5 

_| R. 219 22 2826'.5 4 h 56 m 39 s p.m. 

|o~ L. 40 34 27 23 .5 59 47.5 

R.P. L. 203 26 



The reduction is as follows: 
To find the azimuth: 
Mean of V.C.R's. =27 55' 00" 
r 1 49 



P 

h 

r 

4> 
8 

s' 

s'-4> 
s'-8 
s'-Z 

log cos s' 
log sin (V 5) 

log cos(>'-f) 
log sin (s' <t>) 



27 53 11 

8 

= 27 53 19 

= 62 06 41 
= 44 24 09 
= 18 34 07 

= 62 32 28 

= 18 08 19 

= 43 58 21 

= 25 47 

= 9.663807 
9.841555 

9.999988 
9.493203 

9.505362 
9.493191 



To find the time: 
log sin (>'-</>) =9.493203 
logsin(5'-5) =9.841555 



log cos s' =9.663807 

logcosO'-f) =9.999988 



log tan \t 



h 



T 

A.T. 
E 



M.T. 



Stand. T 
Watch 



AT 



9.334758 
9.633795 



logtan^r =9.670963 
= 9.835481 
= 34 23' 54" 
= 68 47 48 
= 4 h 35 m ll s .2 
= +6 16 .0 



= 4 41 

18 


27 .2 
15 


= 4 59 
= 4 58 


42 .2 
13 .2 


= +1 


29 .0 



log tan 2 \A 


= 10.012171 


log tan \A 


= 10.006085 


\A 


= 45 24' 05" 


A 


= 90 48 10 


A s 


= 269 11 50 


Rp 


= 23 26 15 




292 38 05 


Rs 


= 219 58 


A, 


= 72 40 05 



43 



Example. The following observations were taken with a 
small transit in Sept., 1899, to determine azimuth, time and 
latitude. 

PL obs'd. Cir. H.C.R. V.C.R. Watch 

R 157 10' 27 21' 7 h 43 m ll s p.m. 
L 337 58 27 06 46 23 

R 158 42 26 46 .5 49 35 
L 339 37 26 19 53 24 

R 160 22 25 59 .5 56 43 
L 341 16 25 33 8 00 29 
R 225 45 
L 



Arcturus 

Arcturus 

Arcturus 

Arcturus 

Arcturus 

Arcturus 

R.P. 

Altair 

Altair 

Altair 

Altair 

Altair 

Altair 



R 
L 
R 
L 
R 



34 23 

34 25 

34 27 .5 

34 29 

34 30 

34 32 



16 55 

19 11 

21 18 

23 00 

24 45 
27 19 



A mean time watch was used. Arcturus was to the west of 
the meridian, and Altair near the meridian and east of it. 

The approximate meridian altitude of Altair was observed 
to be 

34 39' 30" 
whence a value of the latitude for reducing the azimuth 
observations was found as follows: 

h' = 34 39' 30" 
r = 1 23 



h 


= 34 


38 


07 


.<- 


= 55 


21 


53 


5 


= +8 


36 


23 


$ 


= 63 


58 


16 



The apparent places of the two stars were: 

a 5 

Arcturus 14 h ll m 05 s +19 42' 24" 

Altair 19 45 55 +8 36 23 

The reduction of the first azimuth observation is as follows: 
V.C.R.,Cir.i? =27 27' 

V.C.R., Cir. L =27 06 



Mean 
r 



h 



= 27 


16 


30" 


= 


1 


51 


= 27 


14 


39 


= 62 


45 


21 



44 



Eq. 


(10) 

(70) 


s' 

s'<f> 

s'-h 

*'-r 

log cos s' 

log si n(Y 5) 

log cos^ f) 
log sin(V <j>) 

log tan 2 \A 
log tan \A 

u 

A 

A s 

Rp 

Rs 
Ap 


= 73 13' 00".5 
= 9 14 44 .5 
= 53 30 36 .5 
= 10 27 39 .5 




= 9.460524 
= 9.905235 




= 9.992721 
= 9.205930 




9.365759 
9.198651 


Eq. 


= 10.167108 
= 10.083554 
= 50 28' 40".5 
= 100 57 21 .0 
= 259 02 39 
= 225 45 




484 47 39 
= 157 34 




= 327 13 39 



Reducing the remaining azimuth observations in a similar 
manner, and taking the mean, the result is 

A p =327 13' 31" 

In order to reduce the latitude observations it is necessary 
to find the hour angle of Altair corresponding to each of the 
observed times. This may be done by computing the hour 
angle of Arcturus from the observations of that star, and 
combining it with the difference of right ascension of the two 
stars. Thus: 

Eq. (13) logsin(>'-0) =9.205930 

logsin(s'-5) =9.905235 



log cos 5' =9.460524 

logcos(s'-r) =9.992721 



9.111165 
9.453245 



log tan 2 A' =9.657920 



45 



log tan \r =9.828960 

\t =33 59' 54" 

t =67 59 48 

= 4 h 31 m 59 s .2 

Reducing the remaining observations in the same way, the 
hour angles are: 

4 h 31 m 59 s .2 
38 45 .1 
45 54 .4 



Mean = 4 38 52 .9 



The difference of r.a. of the two stars is 

5 h 34 m 50 s ; 
therefore the hour angle of Altair 

= 4 h 38 m 53 s 
-5 34 50 



= -55 57 
(the star being east of the meridian) at an instant equal to 
the mean of the observed times, or 

7 h 51 m 37 s .5 
Then as the change of hour angle of a star is equal to the 
change in the sidereal time, the hour angle of Altair at the 
time of the first latitude observation is found as follows: 
Observed time, 1st obs'n = 8 h 16 m 55 s 

Mean of times of az. obs'ns = 7 51 37 .5 



Diff. = 25 17 .5 

Equivalent sid. interval = 25 21 .7 

Hour angle at mean of times = 55 57 



Hour angle at 1st lat. obs'n. = 30 35 



The hour angles of Altair are thus found to be 

-30 m 35 s 

28 19 

26 12 

24 29 

22 44 

20 10 

The latitude observations are now reduced as follows: 

Eq. (61) h' =34 23' 00' 

r 1 24 



\" 



= 34 21 36 
46 



f 


resu 
= 63 c 


Its ; 

1 58 


ire: 
'40" 

59 

28 

28 

53 

46 


= 55 38 24 




8 

ho 


= 63 58 16 
= 8 36 23 
= 34 38 07 


log COS 
log cos 5 


= 9.642291 
= 9.995082 


log cos h 


= 9.915287 


log m 


9.637373 

9.722086 

= 3.263353 


logy 

y=mvt 

r 


= 2.985439 
= 16' 07" 
= 55 38 24 


5 


= 55 22 17 
= 8 36 23 





= 63 58 40 


The complete latitude i 

= 




Mean = 


= 63 


58 


42 





The inclusion of the small term in the expression for 
increases this result by less than 1". 

The effect of the error of 26" in the value of the latitude 
used in the computation of A is found by the formula 

d A = - d(j) 

cos tan r 

to be about 21". 5. 

4th method By an observation of a circumpolar star at 
elongation. 

The azimuth and hour angle of the star may be found by 
(22) and (23). From the former the time of elongation may 
be computed. 



47 



4 



Description of method of taking the observation. 

In the case of the pole star, assuming a = l h 26 m , 5 = 88 50', 
we find t = 5 h 58 m 20 s , and .'. G = 7 h 24 m 20 s , the sidereal time 
of western elongation. This may be used to compute ap- 
proximately the time of either elongation at any time of the 
year. 

5th method By transits of stars across the vertical circle 
of Polaris. 

From the observed times of transit of two stars across the 
same vertical circle, the azimuth of that circle may be com- 
puted. 




F/g.21 



To find the azimuth: In Fig. 21, S\ is the position of Polaris 
at the time of transit and 6 1 that of an equatorial star. SZS\ 
is then the vertical circle of the instrument, and PZ the 
meridian. The angle SPS\ (denoted by A) differs from the 
difference of r.a. of the two stars by the sidereal interval 
between their transits, or 

A = (a 1 -a)-(r 1 -D (71) 

T\ and T being the observed times of transit of Polaris and 
the other star, respectively, ai and a their right ascensions. 
In computing A the subtractions should be algebraic; A will 
then be affected by the + sign if the star S is west of the 
meridian, and by the sign if east. 
We next take the equations: 

sin A cot C = cos 8 tan Si sin 5 cos A 
sin r cot C = cos 5 tan <f> sin 8 cos r 

. cos 5 sin C 

sin A = 

cos (f> 

which are obtained from (5) and (3), Sph. Trig. From the 
first of these we have 



tan C = 



sin A 



cos 5 cot p sin 8 cos A 

sin A 
cos 8 cot p(l tan p tan 8 cos A) ' 



4S 



= tan (l + tan p tan 5 cos A+), 

cos 5 

_ p sin (14-^ tan g cos a) (72 ) 

cos 5 v ' ; 

neglecting p 3 . Again, from the second equation we have 

cot C . tan 

sin t . - + cos t = - , 

sin 5 tan 5 

cot C .. t 2 tan ^ 
r T ^inT + 2 "tan 5 ' 

again neglecting the cube and higher powers of small quanti- 
ties; /. 

cot C f_ tan _ _ sin (0 8) 
sin 5 2 tan 5 cos sin 5 ' 

Then assuming as a first approximation 
cot C __ sin (0 5) 
sin 5 cos sin 5 ' 

sin (0 5) _ 

or r = , tan C, 

cos 

we have by substitution for t 2 in the above equation 

sin (0 5) 

t = , tan v ,- 

cos 

sin (0 5)sin A /1 , s A \ 

s= : - (1 i p tan 5 cus A) 

cos cos 5 

by (72) ; or in seconds of arc 

p sin(0 5)sin A /i , , . ir ,. f ... (73) 

t=- - (l-*-/>sinl "tan 5 cos A) 

cos cos 5 

If the time star be observed below the pole, then 5 changes 
its sign, and r becomes the hour angle reckoned from lower 
culmination. 

To find the azimuth we have from the third of the above 
equations 

COS 

or by (72) , p sin A ,., . s . .,, . s (74.) 

y v J A = -(l + sm 1'' tan 5 cos A) 

COS 

A and p being in seconds of arc. 

Comparing equations (73) and (74) we see iu^t 

, sin (0-5) (75) 

t = A 

cos 5 

49 



Example. The following observations were taken at 
Toronto, Mar. 29, 1899: 

PL obs'd. H.C.R. Watch 

R.P. 45 18' 

Polaris 73 33 .5 8 h 30 ra 51 s 

r Hydrae 73 33 .5 8 34 43 

The apparent places of the stars were: 

a 8 

Polaris l h 21 m 21 s +88 46' 23" 

f Hydrae 8 50 06 + 6 19 35 

We have then the following data: 

A =111 13' (Eq. 71.) 
4> = 43 39 36" 
8 = 6 19 35 
p =4417"; 
so that the computation proceeds as follows: 
Eq. (74) log sin A =9.96952 

logp =3.64513 



Eq. (70) 



Eq. (75) 



log COS <j> 


= 9.85941 




3.61465 


log 5692 
log tan 5 
log cos A 
logp 

log sin 1" 


= 3.75524 
= 9.04480 
= 9.55858 
= 3.64513 
= 6.68557 


log -5 


= 0.68932 


.'. .4=5687'^ 
As 
R P 


= 1 34' 47" 
= 358 25' 13" 
= 45 18 


Rs 


403 43 13 
= 73 33 30 


A P 


= 330 09' 43" 


log A 

log sm(4> S) 


= 3.75488 
= 9.78280 


log cos 5 


= 9.99735 


log 3470 

50 


3.53768 
= 3.54033 



r 3470" -231 s = 3 m 51 s 

o(f Hydra) = 8 50 06 



G r 8 46 15 

L = 5 17 35 



9(atGr.) =14 03 50 

Equiv. M.T. int'l - 14 01 32 
M.T. of sid. noon =23 33 23 



37 34 55 

13 34 55 

Standard Time = 8 34 55 

Watch = 8 34 43 



Watch corr'n = +12 



6th method By the observed angular distance of the sun 
from a terrestrial point. 

This method is useful when the sextant is the only instru- 
ment available. 




F/g.22 



In Fig. 22 5 is the centre of the sun, and the terrestrial 
point. The observation comprises: 

Measuring the angular distance SO, 
Noting the time of observation, and 
Measuring the altitude of 0. 
The latitude being known, the altitude and azimuth of 
the sun's centre are computed by (4), (5) and (6). The 
apparent altitude is then found by subtracting the parallax 
and adding the refraction The measured angular distance 
is corrected for semi-diameter. We have then 

01 sin (s ZS) sin (s ZO) 

tan^ 4 a = - 



in which s = 

51 



sin 5 sin (s SO) 
ZS+ZO+ SO 



If then, h' =the apparent altitude of the sun 
H = the altitude of 
D = the angular distance SO 
we find on substituting 

,_ h'+H+D 
S ~" 2 
! sin(s'-#)sin(/- /Q (76) 

cos s cos(s D) 
If H is so small that it may be neglected, as is often the 
case in hydrographic surveys, then (76) becomes 

tan 2 |a = tan %(D+h') tan h(D-h') (77) 

The azimuth of then is 

A * a 

If the correction of the watch is not known the observer 
may proceed as follows: 

Measure the altitude of the sun, then the angular distance 
SO, then again the altitude of the sun, noting the watch 
time of each of the three measurements. The altitude of 
the sun at the instant of measuring SO may then be inter- 
polated. The altitude of is measured as before. A may 
then be computed from the data h 8 and <f> by either (8), (9) 
or (10). The remainder of the reduction is as before. 



82 



6 Determination of Longitude by Observation. 

The engineer is seldom called upon to determine longi- 
tude, so that only some methods useful to the explorer will 
be here :' escribed, and also in outline the most precise method 
known, \ z., that by the electric telegraph. 

The difference of longitude between two places may be 
defined as the angle between the planes of their meridians. 

It was seen p. 14 that the local times of two places 
differ by an amount equal to their difference of longitude, 
expressed in time. Any method, therefore, that serves to 
compare the local times of the two places, at the same absolute 
instant of time, will determine their difference of longitude. 

1st method By portable chronometers. 
If the correction of a chronometer on the local time of a 
place A is found by observation, and also its rate, and the 
chronometer is then transported to another place B, and its 
correction on the local time of that place found, the local 
times of the two places may be thus compared : Let 

AT, 57" = the correction and rate found at A at the time T; 
AT' = the correction found at B at the time T ( = 7"+/) 
Then at the instant T the true time 

itA = T+t+AT+t . 8T, 
atB = T+t+AT; 
the difference of which is 

AL = AT+t . ST -AT, 
or the difference of the corrections of the chronometer on 
the times of the two places at an assumed instant of time. 

2nd method By signals. 

Any signal that may be seen at the two places may be used 
to compare their local times. A chain of observing stations 
may be established between the extreme stations, with inter- 

* (VfV > x* f 

F/o.23 

mediate signal stations, so that the method may be used 
between points at a considerable distance apart. The signal 
used may be the disappearance of a light, a flash of gun- 
powder, etc. 

Let A ard B be the terminal stations, C and D intermediate 
stations, and S\ S* and S3 signal stations (Fig. 23). Then if 
a signal be made at Si which is perceived at A at the titr:? 7\ 

53 



and at C at the time T%', and if then a signal be made at 5 2 
which is perceived at C at the time T 3 and at D at the time 
D i} etc.; then, A being the more easterly station, we have 
AL = (T l -T i ) + (T 3 -T i ) + (T b -T 6 ) 
= T l -(T 2 -T 3 )-(T i -T 5 )-T 6 ; 
which shews that it is not necessary to know the corrections 
of the chronometers at the intermediate stations, but only 
their rates. The times T\ and Tq, are the true local times at 
A and B, respectively. 

Eclipses of Jupiter's satellites are also used in longitude 
determinations. As the satellite appears to fade out gradu- 
ally the observed time of an eclipse will depend upon the 
power of the telescope used. But for this objection this 
method would be a useful one for finding longitude. 

Reference may be made to the ephemeris. 

3rd method By the electric telegraph. 

The observer at each station must be provided with a 
transit instrument, chronometer, and electro-chronograph, 
for determining time with precision, and also a portable 
switchboard by waich connections can be made with the 
main telegraph line for sending signals to the other station. 




F/e.24 

The connections for observing the transits of stars in 
determining time are shewn in diagram in Fig. 24, and for 
sending arbitrary signals in Fig. 25. 

The procedure at each station is to observe a set of stars 
for determining time and the instrumental constants. Then 
a series of signals is sent to the distant station, which are 
also recorded on the local chronograph. A second set of 
stars is then observed. By means then of the two time sets 
the correction of the chronometer on local time at the epoch 
of the signals can be interpolated. 

These operations may be repeated on as many mutually 
clear nights at the two stations as may be considered neces- 
sary, say five nights. 



54 



In Figs. 24 and 25 

C is the chronometer, 
By the chronometer battery, 
Rx the chronometer relay, 
B t the chronograph battery, 
M the chronograph magnet, 
K the transit key. 




F/g.25 



Also in Fig. 25 

LL is the main line, 
R 2 the sounder relay, 
S the sounder, 
Rz the signal relay, 
Rh a rheostat, 
G a galvanometer, 
K' the telegraph and signal key. 
A signal is made by breaking the main line circuit by means 
of the signal key, which may be a special break-circuit key. 

If now at a time T\ at. station A a signal is made which 
is recorded at B at the time T\ ; and if ATi A7Y are the chrono- 
meter corrections on local time at the two stations, and x the 
time of transmission of the signal; then the difference -of 
longitude is: 

AL=--(7\4-r-(7Y4-A7Y-*) 
= AZ,i+x 
in which AL 1 = (r 1 +A7 , 1 )-(2V'f A7Y) 



55 



If a signal now be made at B at the time 7Y, and recorded 
at A at tne time T 2 ; then 

AL = (r 2 +Ar 2 -.r) - (7Y+A7Y) 
= AL 2 x 
in which AL 2 = (r 2 +Ar 2 ) -(7Y+A7Y) 

Taking the mean of these values of AL x is eliminated, and 
we have 

. T AL t +AL 2 
- 2 

4th method By moon culminations. 

An examination of the moon's hourly ephemeris contained 
in the N. A. will shew that the motion of that body in right 
ascension is very rapid. If then a value of that co-ordinate 
be found by observation, and the corresponding Gr. time be 
interpolated from the ephemeris, the error in the time due 
to the error in the observed quantity will not be excessive. 
The Gr. time being thus found at the instant of the observa- 
tion, which also serves to determine the local time, the longi- 
tude follows by taking the difference of the two times. 

To determine the moon's r.a. the meridian transit of the 
moon's limb and that of some neighbouring star are observed. 
Then let 

and 6' = the sidereal times of transit of the moon's centre 
and a star. 

a and a' = their right ascensions 
and we have 

a-a' = 9-e' 

or a = a'+e-e' 

which gives the moon's right ascension. 

To find the sidereal time of the semi-diameter passing the 
meridian in order to correct the observed time of transit of 
the limb, let 

a = the sid. time of the S.D. passing the meridian 

.S = the moon's angular S.D. 

Aa = the increase of the moon's r.a. in l m of M.T. 

then r =the increase of the moon's r.a. in 1 sid. second; 
oU.lol 

and 

Act . . . , . 

o" hn ifu ~ lts increas e in the interval a; 

j . Act 5 sec S 

and -' '-'60T64 = "IS" 

as each side of the equation expresses the time of S.D. passing 
the meridian if there were no change of r.a.; .". 

56 



5 

a 



15 cos5 O-ecok) 



60.164 S 



15 cos 5 (60. 164 -Aa) 

This quantity is given in the N.A. 

To interpolate the Gr. M.T. corresponding to an observed 
value of the moon's r.a., let 

a =the ephemeris value nearest to a, 

To =the corresponding Gr. M.T., 
T = the Gr. M.T. corresponding to a, 
x = T T (in seconds) , 

Aa =the increase of a in 1 minute of M.T. at the time T , 
ha = the increase of Aa in 1 hour. 
Then the increase of Aa in the interval x is 

x * 
. da; 



3600 
.*. the value of Aa at the middle instant of the interval x is 

Aa + ^oiT 5a 

and .*. the increase of a in the interval x is 

~6cr( Aa + w 5a )' 

X / X "\ 

and.'. a = a + -^r-f Aa + ^qq 3a )' 

Then x= 60 ( a ~ a ) _ 60(a-a o ) 



60 (a a ) /, X 6a \ , 

C 1 - -7200- Aa"/ nearly 



Aa 

, x' 2 8a . 

= X ~ "720T -Aa~ nearly 

... , 60 (a a ) 

in which x = 

Aa 

Then T=T + x 

If then 6 is the Gr. sid. time corresponding to T we have 

L = e-a 
A more accurate method than the foregoing is to take 
observations for determining a on the same night at the 

57 



station whose longitude is required and also at another 
station whose longitude has been well determined. Thus 
the increase in a while the moon is passing over the interval 
between the two meridians is determined. This increase, 
divided by the increase in 1 hour of longitude, gives the 
difference of longitude in hours. Thus if 

en and a.i =the values of a found at the two stations, 

H = the increase of a in 1 hour of longitude while 
passing over the interval between the two 
meridians; 

then ^^P" 

H may be taken troin the ephemeris. 



6S 



7. The Theodolite and the Sextant. 

The Theodolite. 

For a knowledge of the construction and method of ad- 
justment of the engineer's transit theodolite reference may 
be made to any standard work on surveying. 

A well constructed and adjusted transit should fulfil the 
following conditions: 

(1) The vertical and horizontal axes should pass through 
the centres of the horizontal and vertical circles, respectively, 
and should be perpendicular to their planes. 

(2) The axis of the alidade of the horizontal circle should 
coincide with the axis of the circle. 

(3) The line joining the zeros of the verniers of either 
circle (assuming that each is read by two verniers) should 
pass through the centre of the circle. 

(4) The extreme divisions of each vernier should coincide 
at the same time with divisions of its circle. 

(5) The horizontal axis should be perpendicular to and 
intersect the vertical axis. 

(6) The sight line of the telescope should be perpendicular 
to and intersect the horizontal axis, and in all positions of 
the focusing slide. It should also intersect the vertical axis. 

(7) The two threads in the telescope, whose intersection 
determines a point on the sight line, should be truly hori- 
zontal and vertical, respectively, when the instrument is 
adjusted for observation. 

(8) The levels attached to the horizontal plate should read 
zero when the vertical axis is plumb. 

(9) When either vernier of the vertical circle reads zero, 
and also the level attached to the alidade of that circle, the 
sight line should be horizontal. 

conditions 1, 2, 3, 4 and the second part of 6 are fulfilled 
by the maker in the construction of the instrument; the 
others, and sometimes 3, can be attended to by the observer. 
With regard to 9, the alidade of the vertical circle of a transit 
intended for astronomical observation should be provided 
with a level capable of detecting a change of inclination 
considerably smaller than the least count of the vernier. 
The position of the alidade should be adjustable by means 
of a slow-motion screw, so that the bubble of its level may 
readily be brought to the centre, after plumbing the vertical 
axis of the instrument. 

It is proposed to examine the effects of these errors of 
C(ns ruction and adjustment, shewing how in most cases 
they may be eliminated. 

59 



(1) The effect of an inclination of the horizontal axis. 

In Fig. 26, which is a projection of the celestial sphere on 
the plane of the horizon, the horizontal rotation axis of the 
transit is assumed to be inclined at a small angle to the 
horizon, so that the collimation axis defined as a right line 
through the optical centre of the objective perpendicular to 
the horizontal axis traces on the celestial sphere the great 
circle A'PZ'. P being any point and APZ a vertical circle, 




F/g.26 



the true altitude of P is the arc AP; and the apparent altitude, 
affected by the inclination of the axis, the arc A' P. Z' is 
the zenith of the instrument, and ZZ' is equal to the inclin- 
ation b. It is clear that the effect of b on the H.C.R. is 
shewn by the spherical angle AZA' . To find an expression 
for this angle we have in the triangle PZZ' 

+ p77 , tan PZ' 
tan PZZ = 



or 



cot AA X 



sin ZZ' . ' 
cot ti 



sin b ' 

or tan Ayli = tan h' sin b; 

or, as AAi and b are small, we may write this 

AA^btanh', (79) 

or the effect of an inclination of the horizontal axis on the 
H.C.R. varies as the tangent of the altitude of the point 
sighted. 

In measuring the horizontal angle between two points it 
is evident that the effect of b is nil if the altitudes of the two 



60 



points are equal, and that it increases with the difference of 
the altitudes. A reversal of the instrument reverses the 
algebraic sign of A^li, so that its effect on a horizontal angle 
is eliminated by the reversal. 

To find the effect of b on the measurement of a vertical 
angle we again refer to the triangle PZZ', from which we 
have 

cos PZ = cos PZ' cos ZZ' 
or sin h =sin h' cos b 

Then denoting h' h by Ah and expanding cos b we have 

sin(h'-Ah)=sm h' (l - -y) 



or 



sin h' Ah cos h' = s'm h' ^-sin h' 



by expanding the sin and cos of Ah and neglecting its square 
and higher powers; .*. 

b 2 
Ah = tan h' 

It appears then that the effect of & on a vertical angle 
varies as the square of b. Introducing the values of Ah and 
b in seconds we have 



., sin 1'' 

Ah = - b- tan h 



(80) 



This is a very small quantity; for, assuming 6 = 1' and h' = 
45, we find A/z = 0".0087; it may therefore be safely neg- 
lected. It is not eliminated by reversal. 




fie. 27 



61 



(2) The effect of a collimation error; i.e., an error arising 
from non-coincidence of the sight line and the collimation 
axis as above defined. 

Assuming that there is no inclination error the sight line 
in this case will trace on the celestial sphere a small circle 
parallel to the great circle traced out by the collination 
axis. In Fig. 27 PZ' is the small circle, and A'BZ the great 
circle traced out by the collimation axis. H and H' are the 
poles of those circles, Z' being the zenith of the instrument; 
ZZ' or PB is the collimation error, denoted by c. 

To find the effect of c on a H.C.R., denoted by AAt, we 

have in the triangle BZP 

_ tan BP 
tan BZP =~.^jr=- 
sin BZ 



or 



tan AAz = 



tan c 



cos h' 

or very nearly AA 2 = c sec h' (81) 

or the effect of a collimation error on a H.C.R. varies as the 
secant of the altitude of the point sighted. 




F/&.88 



The effect of this error on the measurement of a horizontal 
angle evidently also increases with the difference of the 
altitudes of the two points sighted, and is eliminated by a 
reversal of the instrument. 

To find the effect of c on the measurement of a vertical 
angle we have in the triangle BZP 

cos PZ. = cos BZ cos BP 
or sin & = sin h' cos c 

As this is the same equation as was derived in the discussion 

62 



of the last error, it follows that equation (80) also epressexs 
the error in this case. 

(3) To find the effect of a non-fulfilment of condition 1, 2 
or 3, so that the line joining the zeros of the two verniers 
does not pass through the centre of the circle. 

The circle in Fig. 28 represents the graduated circle, of 
which is the centre. 0' is the centre of the alidade. Also 
the line joining the zeros of the two verniers does not pass 
through the point 0'. It is clear from the figure that if in any 
position of the alidade the reading of the vernier Vi is less 
than what it would be if the line ViV 2 occupied a parallel 
position passing through 0, then the reading of F 2 will be 
in excess by the same amount. By taking the mean of the 
two values of an angle, found by taking readings of both 
verniers, the effect of eccentricity is therefore eliminated. 

By a different process it may be shewn that the effect of 
eccentricity may be eliminated by any number of equi- 
distant verniers. 

With regard to condition 8, it is convenient that the plate 
levels should be in good adjustment, but in any case it is 
advisable to use the more precise level attached to the alidade 
of the vertical circle, or the telescope level, in plumbing the 
vertical axis. The effect of the error arising from imperfect 
leveling may be shewn as follows: 




fiG.29 



In Fig. 29 Z is the zenith, Z' the point to which the vertical 
axis is directed. P is any point. The triangle PZZ' gives 
the equation 

sin 6' cot = sin d tan h' cos d cos d' 

63 
5 



Then expanding sin d and cos d and neglecting all but the 
first power of d we have 

sin 6' cotd = d tan A'+cos 0' 
or sin 0' cos cos 0' sin 6 = d tan h' sin 

or sin (9' d)=d tan A' sin 

or as 0' is small 

d'e = d tan A' sin 
Now if there are two points sighted in turn, and 0/ and 0/ 
are the values which 0' takes, respectively, we have 

6i6i = d tan A/ sin di 
e-i'-d 2 = d tan A 2 ' sin 2 
so that, taking the difference 

(0 2 / -0i / ) - (0 2 -0i) =d(tan A 2 ' sin 2 -tan /*/ sin 00 (82) 

This expresses the error in the horizontal angle between 

the two points. It appears to be a maximum when 2 = 27O 

and 01 = 90, and for high altitudes its value may exceed d. 

It is not eliminated by reversal. 

To find the effect on a vertical angle, we have in the triangle 
APA' 

, tan h 

cos /= rr, 

J tan h' 

, , tan k tan h i /, , f*\ 

or tan V = 7- = ^ = tan h I 1 + ~ I, 

cos/ 1 _ /_. V 2 / 

2 

nearly. Then writing h' = h-\-Ah we have 

tan /*' = tan(A+Aft) 

= tan h-\-Ah sec 2 & 
by Taylor's theorem. 

f 2 
AA sec 2 h = ^r- tan A 

ft 

or Ah= tan A cos 2 A 

Zi 

Again, in the triangle PZZ' 

sin 0' sin J 



cos A 
d sin 0' 



sin/ = 

or / = 

cos h 

Substituting in the above expression for Ah we have 

. , d 2 sin 2 0' 7,7 

A/* = - T~r- tan h cos- # 
2 cos 2 h 



64 



(83) 



= sin 2 6' tan h 

Zi 

sin 1" 
or in seconds Ah = = d 2 sin 2 8' tan h 

Zi 

This is never appreciable. 

(5) It is convenient that adjustment 9 be nearly perfect, 
the ugh not essential, as the effect of imperfect adjustment is 
elin inated by reversal. 




F/g30 



In Fig. 30 the circle represents the vertical circle of the 
transit; OP is the sight line, directed to some point P. The 
error of VA , the reading of the vernier V, is evidently = 

e+e'. 
If the telescope now be transited, turned in azimuth, and 
again directed to the point P, it amounts to the same thing 
as transiting and directing to a second point P' which has the 
same absolute zenith distance as P. The reverse reading is 
then VA' whose error is = 

~(e+e') 
The mean of VA and VA', the two readings of vernier V, 
is therefore the altitude of P freed from the effect of index 
error. 

To observe an altitude of a heavenly body with a transit. 
It has been shewn that errors of adjustment have no 
ippreciable effect upon a vertical angle, except the index 

65 



error, whose effect may be eliminated by reversal. In ob- 
serving the altitude of a star, therefore, the method is to make 
two pointings to the star, reversing the instrument between 
the pointings. The telescope is first directed so that the 
star is very near and approaching the horizontal thread at 
a point a little to the right or left of the centre. The time of 
crossing the thread is then noted, and also the V.C.R. The 
instrument is then reversed and directed as before, with the 
star at about the same distance on the opposite side of the 
centre, thus eliminating the effect of any inclination of the 
thread. The time of passage across the thread is again 
noted, and the V.C.R. If azimuth is required as well as 
time, the star must be observed on the intersection of the 
horizontal and vertical threads. The mean of the two V.C.R's. 
is then the observed altitude freed from the effect of index 
error corresponding to the mean of the observed times. 

It is thus assumed that the change of altitude of a star, 
during short intervals of time, is proportional to the time. 
This assumption will seldom lead to an error exceeding 3 .1 
for an interval of 3 m between the observations. 

In observing the sun the same general method is followed 
as in observing a star, but as there is no definite point at 
the sun's centre that can be observed, the procedure is as 
illustrated in Fig. 31. The sun's image is first brought to the 




F/g3/ 



oosition shewn by the broken circle Si, so as to be in contact 
.vith the horizontal thread and slightly overlapping the 
vertical thread. It may then be kept in contact with the 
horizontal thread by turning the altitude tangent screw; 
its own motion will then bring it into contact with the vertical 
thread, as shewn by the full circle Si. After noting and 



66 



recording the time and the readings of the circles the instru- 
ment is reversed and the observation repeated, bringing the 
sun into the position S 2 . The figure represents an afternoon 
observation for time and azimuth, taken with an inverting 
telescope. If time alone is required the contact of the sun's 
image with the vertical thread is not important. The means 
of the readings of the two circles may now be regarded as 
corresponding to a pointing to the sun's centre at an instant 
equal to the mean of the times. 
A form of record is shewn on p. 43. 

The Sextant. 

The principle and construction of the instrument. 

In Fig. 32 AB is the graduated arc, Mi the index mirror, 
M2 the horizon mirror, M{V the index arm to which the 
mirror M\ is attached, and carrying the vernier V at its 
extremity. The instrument embodies the principle that if 
a ray of light SMi be incident upon the mirror Mi, then 




Fig. 32 



reflected from it to the mirror M 2 , from which it is again 
reflected, then the angle c between the first and last direc- 
tions of the ray is equal to double the angle d between the 
mirrors. This is readily proved, for in the triangles MiM 2 C 
and MiM 2 D we have, respectively 

2b = 2a+c 
and b = a+d 

or 2b = 2a+2d 

c = 2d 
The mirror M 2 is attached permanently to the frame of 
the instrument, and half of its surface is unsilvered, while 

67 



Mi is attached to the index arm and turns with it. The 
sighting telescope is directed along the line CM 2 . The mirrors 
are so placed that when their planes are parallel the index 
V is at the zero A of the graduated arc AB. The arc is divided 
into twice the number of degrees that it subtends at its centre 
Mi. 

To measure the angle between two points the instrument 
is held so that its plane passes through the two points, and 
the left-hand point is seen in the field of the observing tele- 
scope through the unsilvered half of the mirror M-i. The 
index arm is then turned until the other point, seen by double 
reflection from the two mirrors, appears to coincide with 
the first. The reading of the arc is the angle subtended by 
the two points at the point C. It is to be remarked that C is 
not a fixed point for all angles. 

Adjustment of the sextant. 

To observe an altitude of the sun with a s?xtant and 
artificial horizon. 

The artificial horizon is a horizontal reflecting surface, 
usually the surface of mercury contained in an iron trough. 
In observing the altitude of a heavenly body the angle is 
measured between its image, seen by reflection in the arti- 
ficial horizon, and that seen by reflection from the mirrors of 
the instrument. Fig. 32 shews that this angle is equal to 
double the apparent altitude of the body. In observing the 
sun, instead of superposing the two images seen in the field 
of the telescope, it is best to bring them into external contact, 
thus observing either the upper or the lower limb. As the 
horizon image appears erect in the field of an inverting 
telescope, and the other image inverted, the identification 
of either image shews which limb has been observed. 

To determine the index error of the instrument after 
observing the sun, set the vernier nearly at zero and then 
direct the sight line to the sun; the two images will now be 
seen nearly in coincidence. Then turn the tangent screw 
until the images are in external contact, and read the arc. 
Then reverse the motion of the screw, causing the images 
to pass one over the other until they are again in contact, 
and again read the arc. One of the readings will be on the 
extra arc. Half the difference of the two readings is the 
index error, positive if the reading on the extra arc is the 
greater. The sum of the readings is twice the sun's angular 
diameter. 



6S 



8. Formula of Spherical Trigonometry. 
cos a =cos b cos c+sin b sin c cos A 
cos b =cos a cos c+sin a sin c cos B 
cos c =cos a cos 6+sin a sin b cos C 
cos .4 = cos B cos C+sin B sin C cos a 
cos B = cos ^4 cos C+sin A sin C cos b 
cos C = cos A cos 2?+sin ^4 sin B cos c 
sin A sin 5 sin C 

sin a sin 6 sin c 

sin a cos 5 =sin c cos & cos c sin b cos ^4 
sin a cos C =sin 6 cos c cos 6 sin c cos A 
sin 6 cos A =sin c cos a cos c sin a cos 5 
sin b cos C = sin a cos c cos a sin c cos B 
sin c cos ^4 =sin b cos a cos Z> sin a cos C 
sin c cos 5 =sin a cos b cos a sin b cos C 



where 



sin A cot 5 = sin c cot 6 cos c cos ^4 
sin 5 cot A = sin c cot a cos c cos B 
sin 5 cot C = sin a cot c cos a cos B 
sin C cot jB =sin a cot 6 cos a cos C 
sin .4 cot C =sin 6 cot c cos b cos ^4 
sin C cot .4 =sin 6 cot a cos b cos C 



\ 



sin 



sin- 



sin* 



L . sin (s b) sin (? c) 
sin b sin c 

1R sin (s a) s'm(s-c) 
sin a sin c 

kr _ sin (s a) sin (s b) 
sin a sin 6 



5 = 



cos 2 \A 



cos- 2 



cos' 



iB = 



^C 



a+b-\-c 

2 
sin 5 sin(.y a) 

sin 6 sin c 
sin 5 sin(s b) 

sin a sin c 
sin s sin(s c) 



sin a sin b 
= sin (5 6) sin (5 c) 

sin 5 sin(s a) 
1 r sin (s a ) sin (5 c) 

sin 5 sin(s b) 
lr sin (s a) sin (5 6) 

2^- ; - 

sin s sin (5 c) 
69 



tan 2 \A = 



tan 2 



tan 2 



(1) 

(2) 
(3) 

(4) 



(5) 



(6) 



(7) 



(8) 



. 9 i COS 

sin 2 a = 



Scos(S-A) ' 
sin B sin C 



sin 2 2 



i, cos 6 1 cos (S 

sin A sin C 



(S-B) 



sin 2 \c = 



cos 



where 



5 cos (S 
sin yl sin B 



C) 



cos 2 a = 
cos 2 \b = 



cos 2 \c = 



^ 



c ,4+5 + C 
= 2 

cos (5-5) cos (S -C) 

sin 5 sin C 
cos (S- A) cos (S-C) 

sin .4 sin C 
cos (S- A) cos (S-B) 

sin yl sin B 

S cos(S-A) 

(S-C) 



i _ cos 5 cos(S-A) 

' cos (S-B) cos (S-C, 

tan 2 \h = cos 5 cos (<>-) 

cos (5 -^) cos (5- C) 

tan 2 A c = - _ cos ^ cos (^-Q 
2 cos (S-A) cos (5-5) 

s analogies 



cos (S-A) cos (5- 

Delambre's analogies 

sin h(A+B) cos i( g _ft) 

cos \C cos c 

sin \(A -B) sin Ha~ft) 

o ir 1 = - 



cos \C 



sin ^c 



"-"j 2*-' 0111 2' 

cos Q4+) cos | (a +6) 

sin \C cos fc 



cos \ 



HA-B) _ 
sin |C 
Napier's analogies- 



sin 



i _ 



sin \c 



HA+B) 



tan |(. 

tan h(A-B) 



cos ^ 

1 



(a-b) 
cos ^(a+6) 
sin_(a &) 



cot \C 

cot C 






(9) 



(10) 



(11) 



(12) 
(13) 
(14) 
(15) 

(16) 
(17) 
(18) 
(19) 



70 



os a cos b 


(C = 90) 


(20) 


sin a 


. sin b 

sin jD = : 

sin c 


(21) 


sin c 




tan & 
tan c 


tana 
cos B = 

tan c 


(22) 


tan a 
sin b 


D tan b 

tan 5 = . 

sin a 


(23) 


cot ^4 cot 5 


(24) 


cos B s 


. cos A 
sin 2> = 


(25) 



Formulae for right-angled triangles 
cos c = 

sin A = 

cos A = 

tan ^4 = 

cos c = 

sin ^4 = 

cos b cos a 

Solution of oblique-angled triangles. 

Case 1. Given a, b and c, the three sides 
Solution by means of equations (6), (7) or (8). 

Case 2. Given A, B and C, the three angles. 
Solution by means of equations (9), (10) or (11). 

Case 3. Given two sides and the included angle, as a, b 
and C. 

1st solution By means of equations (16), (17) and (3). 
2nd solution By means of equations (1) and (5), or 
cos c = cos a cos 6+sin a sin b cos C 
sin C cot A = sin b cot a cos b cos C 
sin C cot .B = sin a cot b cos a cos C 
These equations become, when adapted for logarithmic 
computation 

tan = tan a cos C tan Q x tan b cos C 

cos a cos (b Q) tan C sin 0i 

cos c = * tan B - 



tan yl = 



cos 6 s'm(a 6i) 

tan C sin 6 



sin (6-0) 

Case 4. Given two angles and the included side, as A, B 

and c. 

1st solution By means of equations (18), (19) and (3). 

2nd solution By means of equations (2) and (5), or 

cos C cos A cos .B+sin A sin B cos c 

sin A cot B = sin c cot b cos c cos ^4 

sin B cot ^4 = sin c cot a cos c cos B 

These equations become 

tan 62 = tan A cos c tan 3 = tan B cos c 

^ cos A cos 0B-|- 2 ) , tan c sin 3 

cos C = tan b = . . . , * 

ccs 2 sin(/l + 3 ) 

tan c sin 2 

sin( + 2 ) 

71 



tan a = 



Case 5. Given two sides and an angle opposite one of 

them, as a, b and A. 

1st solution By means of equations (3), (16) and (18), or 

. _ sin b sin A 

sin B = - : 

sin a 

, cos| (a b) . , . , _ N 

tan^C= 1) , , ; cot$(A+B) 

2 cos (a +6) v ' 

, cos h(A+B) .. , ,. 

tan|c = cosK^-^) tanKa+&) ' 
2nd solution By means of equations (3), (5) and (1), or 

sin b sin A 

sin B = : 

sin a 

sin C cot A =sin b cot a cos fr cos C 

cos a = cos 6 cos c + sin b sin c cos A 

These equations may be thus adapted for log's. 

tan 4 = tan A cos b 

sin(C+0 4 ) =tan b cot a sin 4 

tan 5 = tan b cos A 

. n . cos a cos 5 

cos(c 5 )= 7 

cos b 

Case 6. Given two angles and a side opposite one of 
them, as A B and a. 

1st solution By means of equations (3), (16) and (18), as 
in the last case, (3) being written 

. . sin B sin a 

sin b = -. -. 

sin A 

2nd solution By means of equations (3), (2) and (5), or 

. , sin B sin a 

sin b= : -. 

sin A 

cos A = cos B cos C+sin B sin C cos a 

sin B cot A = sin c cot a cos c cos B 

Adapting for log's, we have 

tan 6 = tan B cos a tan 7 = tan a cos B 

//- i \ cos yl cos 06 / n . . 

cos (C+0 6 ) = 5 sin (c e-i) = tan B cot ,4 sin 0, 

cos B 



72 



GEODESY. 

1. Figure of the Earth. 

In any survey the extent of which is such that the curva- 
ture of the earth's surface must be taken into consideration, 
the figure of the earth may be regarded as that of an oblate 
spheroid, the elements of a meridian section of which are, 
as determined by Col. A. R. Clarke, 1866: 

Major semi-axis, a = 20926062 ft. 
Minor semi-axis, 6 = 20855121 ft. 
Denoting the eccentricity by e we have 

. _ a?-b> (1) 



e- 



a 2 



The following log's are of frequent use: 
log a =7.3206875 

log e =2.9152513 

loge 2 =3.8305026 

log(l-e 2 ) =1.9970504 

log ^ =3.8334522 
le i 

lo S l/i 2 =2.9167261 



V 1= 



Radii of curvature Any section of the spheroid by a 
plane is an ellipse. If the plane contains the normal, or 
plumb line, at a point, the resulting section is a normal 
section. Any straight line so called traced on the earth's 
surface is therefore a portion of an elliptic arc; for practical 
purposes, however, if its length does not exceed 100 miles, 
it may be regarded as a circular arc whose radius is the 
radius of curvature of the normal section, of which it is a 
portion, at its middle point. If the normal section coincides 
with the meridian an expression for its radius of curvature is 

0(1 -e 2 ) (2) 

pm ~' (l-e 2 sin 2 </>)f 

If the normal section is perpendicular to the meridian its 
radius of curvature is 

a (3) 

Pn (l-e 2 sin 2 </>) 

This is also the length of the normal AN or BN', Fig. 39 
terminated in the minor axis of the spheroid. These are 
termed the "principal radii of curvature" at a point whose 
latitude is 4>. The radius of curvature of a normal section 
whose azimuth is a may be expressed in terms of these; thus 

73 



1 cos 2 a sin 2 a (4) 

Pa P P n 

or = (1+ q ; cos 2 cos 2 a ) 

Pa Pn \ l-e- J 

By substituting in (2) and (3) 

sin 6 = e sin 
they become p m =a{l-e 2 ) sec 3 (6) 

p n =a sec (7) 

Eq. (4) may also be placed in a convenient form for com- 
putation. Thus writing it 

Pm Pn 

p n cos 2 a + pmsin 2 a 
it may be thus transformed 

Pn Pn 



Pa = 



sin 2 a + -^- cos 2 a sin 2 a (l + ^- cot 2 a^) 

Pm \ Pm / 



Then writing cot 2 a = cot 2 x 

Pm* 

it becomes 

Pn Pn 



Pa " sin 2 a(l+cot 2 x) sin 2 a cosec 2 * 



sin 2 x 

Pn ~ n 

sin 2 a 



p a is then given by the equations 

/ p^~.._ sin 2 * (8), (9) 



tan x = -i/ cHL tan a P a = Pn 



sin 2 a 



Pn 

By expansion in series the log's of these radii of curvature 
may be thus expressed: 

log Pm =7.3199482 -[3.3448221] cos 20 

+ [6.27371. .] cos 40- (10) 

log p = 7.3214243 -[4.8677005] cos 20 

+[7.79659..] cos 40- (11) 

log p a =log p n [3.4712365] cos 2 cos 2 a 

+ [5.00366. .] cos 4 cos 4 a- (12) 
The numbers in brackets are the log's of constant numerical 
coefficients. 

For tables giving the values of p m p , etc., see the Supple- 
ment to the Manual of Dom. Land Surveys, also Table IV. 

74 



A Trigonometric Survey. 

Objects of such a survey. 

Choice of stations. Well-conditioned triangles. The base 
net. 

Height of stations in order to overcome the earth's curva- 
ture: 




f/e.33 



Let A and B be two stations whose heights above sea 
level are H x and H 2 , and distance apart 5. is the centre of 
curvature of the arc s. The curved line AB is the path of the 
ray of light between the two stations, z is the zenith distance 
of B observed at A. We have then in the triangle ABO: 

B0_ _ sm BAO 
AO '' = sin ABO 
P+Hj sin(z+r) sin(z-fr) 

p-\-Hi sin(z+r <r) sin(z-}-r)cos a cos(z+r) sin a' 

Ml 
P 1 



or 



or 



or 



1 + 



1+^ 

P 



1 - 



o- cot(z-{-r) 



Q + v) 0~ 7 1 ) =l+*cot( 2 + r) + 



75 



Hi-Hi. , . , , <r 2 

or =o-(cot s r cosec- 2) + 



P v ' ' 2 ' 

expanding by Taylor's theorem. Then as 

per s, r ma, 
m denoting the coefficient of refraction, and 2 is nearly 90, 
we have 

c-2 



H 2 -H x = s (cot 2 j + 



2 P 



s 2 



eq. 


and 


(13), 


we 


find 












Hi-Hi 


H' 


-Hi 










5 




s' 


Then so' 


Iving 


for 


Hi we 


have 










Hi 


H's 
s 


-H*s' 

-s' 


+ ks 



= 5 cots + -jr- (l-2m) (13) 

If i7' now be the height of the ray AB at a distance 5' 
from A, we have 

H'-Hi = s' cotz+4?- (1-2/n) 

2p 

1 2m 

Writing & for and eliminating cot 2 between this 

Zp 



= k(s-s') 



(14) 

This gives the height necessary for a station at A in order 
that a distant station B, of known height H 2 , may be visible 
over an intervening elevation H'. 

If we solve for H' we have 

H' = H '~ Hx s'+Ht-ks'is-s'); (15) 

which gives the height of the ray of light at a given distance 
from A . 

Clarke gives the following values for m: 

For rays crossing the sea, m .0809 

For rays not crossing the sea, m = .0750 

Measurement of a base line Geodetic base lines are now 
measured with tapes or wires of invar, an alloy composed 
of iron and nickel in the proportion of 64 to 36. This material 
has an extremely small coefficient of expansion, so that the 
difficulty experienced in determining the temperature correc- 
tion, when other materials are used, is thus obviated. Good 



7B 



results may also be obtained with a well standardized steel 
tape by working in cloudy weather or at night so as to avoid 
sudden changes of temperature. 

In making a measurement the tape is stretched clear of 
the ground by applying a considerable tension, and rests at 
its zero points on supports in the form of tripods or stakes 
driven firmly into the ground. The rear zero division of the 
tape having been placed in coincidence with a fine mark on 
the head of its support, the relative positions of the forward 
zero division of the tape and the mark on its support may 
then be measured with a scale. The distance between the 
marks on the two supports may be found by applying certain 
corrections to the tape length. These corrections are: 

For temperature, 

For tension, 

For sag, and 

For grade. 
Correction for temperature: 

cx = aL(t to) (16) 

in which 

L =the standard length of tape; 
t = the temperature at which it is standard ; 
/ = temperature at time of measurement; 
a = coefficient of expansion. 
Correction for tension: 

c 2 = eTL (17) 

in which 

e = extension of unit length due to unit tension. 
T= tension in lbs. 
Correction for sag: 

i _L_ / W\* (18) 

24 T 2 24 V T ) 



cz = 
in which 



w =wt. of unit of length of tape 
W = wt. of tape. 
Correction for grade: Denoting the difference of elevation 
of the end supports, determined by levelling, by h, we have 

a = L-(L 2 -h 2 )i 



~ L L y 2 U 8 U 16 u " ) 



1 W 1 h 4 

2 L + 8 U + 



t + t(t)'+ 



h h , h / h\* 
2~ 

This first term in this expression is nearly always sufficient. 

The following may be used as the coefficients of expansion 
for steel and invar tapes: 

Steel, 0.0000114 
Invar, 0.00000041 
In the absence of experimental data the extension of a 
steel tape may be computed from its modulus of elasticity, 
28000000 lbs. The extension of invar may be taken to be 

0.00000004394 ft. 

per lb., per foot, per sq. in. of cross section. 

The distance between the supports, reduced to the horizon- 
tal, then is 

L =L+Ci-\-ci CzCi (20) 

Reduction of a base measurement to sea level 




F/g.34 



o 



Let, B = measured length of base, h being its height above 
sea level ; 
b =its length reduced to sea level. 
Then we have 

or b=B p 



B p+h P +h 

B-b =B (\ ^ =B 



V 1 P+h) B P+h 



=B - 
p 



p \ p p- p J 



78 



-tf-J-O- 



(21) 



The first term here is usually sufficient. 

A broken base It is sometimes necessary to measure a 
base line in two parts, deflecting through a small angle at 
their point of junction. 




Let a and b, Fig. 35, be the two parts, making the small 
angle C with one another. It is required to find the length c 
We have 

C 2 =a 2 +b 2 +2 ab cos C, 

= a?-+b 2 + 2ab(l \ nearly, 

= (a+byabC\ 



-^ -() 

= (a+b) i - | -^i), nearly, 



= a + b- 
or, if C is in seconds 



x abC 2 
2 a+6 ' 

sin 2 1" aC 2 



(22) 



* a+b 



sin 2 1" _ 
log -s- - =11.0701198 



7*0 interpolate a portion of a base Sometimes a portion 
of a base cannot be directly measured. In Fig. 36, a and b 
and the angles P Q and i? are measured ; it is required to find 
the length x. We have 

BE sin A CE sin A 

sin Q 



a 


sin P a-\-x 


BE 


a sin Q 


CE 


(a+jc) sinP 




79 



6 




F/g.36 



Again BE - s[n ( A + R ) E sin (A + 22) 
g ' 6+x sin (22-P) 6 " sin (22-0 

BE = (b+x) sin (22 -Q) 
CE bsin(R-P) 

.*. equating, we have 

ab sin Q sin(P-P) 
sinPsin(P-(2) -^+*M&+*) 

= aZ>-f-(a+&)#4-# 2 
Then write 

2 -, 4a&sin(gsin(P-P) (23) 

(o-&) 2 sinPsin(P-<2) 
and we have 

x 2 +(a+b)x+ab-Ua-b) 2 tan 2 K = 

x= ~%(a+b) y/\(a+b) 2 -ab+\{a-by tan 2 K 
= -\{a+b) \/|(a-&) 2 +(a-&) 2 tan 2 K 
= -\{a+b){a-b) secK 
If a = b this solution fails. In that case write 

2 , _ a& sin Q sin (22-P) 
" sinPsin(i2-0 
then we have 

x 2 + (a +b)x+ab -tan 2 K' = 
and x= -\ {a + b)V\(a+b) 2 -ab+tan 2 K' 

= - \{a +b) Vi(a-6) 2 +tan 2 K' 

= -|(a+6)tan K' (26) 

Measurement of angles The angles of a triangulation 
may be measured either with a direction theodolite, or one 
of the repetition pattern. The circle of the former instrument 
is usually read by three equidistant verniers or microscopes. 



(24) 
(25) 



80 



In measuring the angles at a station each of the distant 
stations is sighted in order, from left to right, and the micro- 
scopes read. The telescope is then transited, or reversed in 
the standards, and each station is again sighted, in the order 
from right to left, and the microscopes again read. A value 
of each angle is thus obtained from each microscope, and in 
each position of the instrument, direct and reversed. The 
mean value of the angle thus obtained is free from the effect 
of eccentricity and errors of adjustment of the instrument. 
With three microscopes the effect of reversal is to give, for 
each station sighted, six readings distributed at equal inter- 
vals round the circle, thus minimizing the effect of division 
errors of the circle. If the construction of the stand permits 
the circle may now be turned to a new position and the angle 
measurements repeated, etc., thus further diminishing the 
effect of division errors. 

A repetition theodolite is usually read by verniers, and 
with this pattern of instrument the repetition principle may 
be used to advantage. It may be thus described: 

Let A (the left-hand station) and B be two stations, the 
angle between which is to be measured. 

Point to A and read verniers. Loosen upper clamp and 
point to B and read verniers. Then loosen lower clamp and 
again point to A. Then loosen upper clamp and again point 
to B, thus obtaining a reading equal to double the angle. 
This process may be repeated until a final reading is obtained 
equal to, say, six times the angle between the two stations. 

Next loosen the lower clamp, transit the telescope, and 
point to B. Then loosen upper clamp, turn vernier plate 
in a clockwise direction, and point to A, thus diminishing 
the final reading of the first set of repetitions by the amount 
of the angle between the two stations. Repeat this opera- 
tion as often as in the first set, thus obtaining a final reading 
approximating closely to the initial reading. 

It is to be noted that in both sets of repetitions the vernier 
plate is always turned in a clockwise direction; that in the 
first set the instrument is turned from A to B with the upper 
clamp loose and the lower clamp tight; and that in the second 
set these conditions are reversed. 

The required angle is now found by taking the mean of the 
differences between the initial and final readings in the two 
sets, and dividing by the number of repetitions. This result 
is largely free from the effect of a drag of the circle by the 
vernier plate. 

Reduction of an observed angle to centre of station This 
reduction is necessary when for some reason the centre of a 
station cannot be occupied by the observer. 

81 



In Fig. 37 A is the centre of the station, the point occu- 
pied. The angles /3 and y are measured, and the distance 
m. The angle A is required. We have 




f/G.37 



A = BDC-x = 0-x+y; 

, . m sin /3 . m sin y 

and sin x = sin y = - 

c J b 

Then x and y being small we may substitute their circular 
measures for their sines, and write them in the form x sin 1" 
and y sin 1", x and y being expressed in seconds, so that we 
have 



,4=0- 



m sin /3 m sin y 



csinl" 



+ 



(27) 



b sin 1" 



Distant stations are rendered visible by means of acetylene 
lamps for night work, and heliotropes for day work. De- 
scription of some forms of heliotrope. 



62 



3. Computation of the Triangulation. 

The portion of the surface of the spheroid contained within 
a triangle is assumed to be a portion of a spherical surface 
whose radius is the geometric mean of the principal radii of 
curvature at the central point of the triangle. 

Spherical excess of a triangleIt is shewn in spherical 
geometry that the sum of the angles of a spherical triangle 
exceeds two right angles by an amount termed the "spherical 
excess" of the triangle. 

To find the spherical excess of a given triangle : 




f/G.38 



Let ABC be a spherical triangle, and A'B' and C points 
diametrically opposite A B and C. The surface of the hemi- 
sphere is made up of the three lunes ABA'C, BCB'A, and 
CAC'B this last being equal to the sum of the two triangles 
CAB and CA'B' less twice the area of the triangle ABC. 
Denoting these by Lune A, etc., and the area of the triangle 
by A, we have 

Lune A = 2wR 2 = 2AR 2 

K 

LuneJS =2BR* 

Lune C = 2CR 2 

2AR 2 +2BR 2 + 2CR 2 -2A = 2TrR t 

A 
or A+B + C tt= -= 

R* 



83 



or, denoting the spherical excess by e we have in seconds 

A (28) 



= 



R 2 sin 1" 
For a triangle on the earth's surface this may be written 

e = A (29) 

p m Pn Sin 1" 

The area of the triangle, in all but extreme cases, may be 
computed as if the triangle were plane, so that we may write 

ab sin C (30) 

2p m p n sin 1" 

a 2 sin B sin C (31) 

Or = ; 

2p m pn sin 1" sin (B-\-C) 
The value of 1/2 p m p sin 1" which we may denote by m 
may be computed by the expression 

log 2 Pm Pn sin 1" = 10^372023+ [3.469754] cos 20 (32) 

the number in brackets being the log. of a constant coefficient. 
The following table was computed by (32) : 

<f> log m 

50 10.37151 

51 141 

52 131 

53 121 

54 111 

55 101 

56 092 

57 082 

58 073 

59 064 

60 055 
Legendres theorem This theorem may be thus stated: 

If the sides of a spherical triangle are small in comparson 
with the radius of the sphere, it may be solved as a plane 
triangle by first diminishing each angle by one-third of the 
spherical excess of the triangle. 
To prove this, let 

A B and C be the angles of the triangle, 
a b and c the sides, expressed in radians, 
A'B' and C the angles of a plane triangle, whose sides 
a /3 and y have the same lengths expressed in feet as 
those of the spherical triangle. 



S4 






log m 


40 


10.37253 


41 


243 


42 


233 


43 


223 


44 


213 


45 


202 


46 


192 


47 


182 


48 


171 


49 


161 



Then we have 

cos a cos h cos c 



cos A = 



sin b sin c 



2r 2 _ 24r* V 2r * 24r V \ 2 ' 2 24 >V 

""* (A. _ ."\ /x _ t\ 

V r or % ) \ r Qr 3 J 

i__ , *L_/i_ , Jl t_ , V , y^\ 

2r 2 ^24r 4 \ 2r 2 ~ r 24r 4 2r 2 ^ 4r 4 "^ 24^/ 

<3t 187 3 ~^T 
r 2 " 6^ "" 6r* 
/3 2 + 7 2 -a 2 a 4-^4_ 7 4_ 6/3 2 7 2 

2r 2 + 24r* 

py A /3 2 +7 2 



r 2 



(1 - ^r\ 

\ 6r 2 J 



/ ^ + 7 2 -a 2 a 4-^4_ 7 4_ 6 ^2 7 2 X x ^+7 2 \ 

V 2/3 T 24/3 T r 2 / V 6r 2 ) 



^2 + 7 2_ a 2 B j8 4_ 7 _ 6/3 7 1 



207 ' 24/3 7 r 2 

l 8 4 4- / 3V-a 2 /3 2 + / 3 2 7 2 + 7 4 -a 2 7 2 
12/3 7 r 2 

= /3 2 + 7 2 -a 2 a 4 +/3 4 + 7 4 -2a 2 <3 2 -2a 2 7 2 -2 i 8 2 7 2 (c) 

2/37 24/37Z- 2 

Now in the triangle A'B'C we have 

., )8 2 +7 2 -a 2 (b) 

cosA = Wy 

sin 2 ,4' = l- 



/ j3 2 + 7 2 -a 2 Y 
V 2 /57 / 



a 4 +/3 4 + 7 4 ~ 2a 2 ff 2 ~ 2 "V ~ 2/3 2 7 2 (c) 

4/3 2 7 2 
.*. by (a) (6) and (c) we have 

cos A = cos A sm~ A -^-r 

Then assume ^4 = ^4'+0 

and we have cos A =cos A' B sin yl' 

by Taylor's theorem. Therefore comparing with (d) we have 

sin ^' = sin 2 ,4' -J^- 
or 1 

85 



By sin A' 1 . . , . 

or d= P7 6r2 = 3^. iPy sin A' 



3r 2 3 

This proves the theorem. 

If the three angles of a triangle are measured, the spherical 
excess may be computed by (30) or (31) using the values of 
the angles given by measurement. The closing error then is 

180 + - (A +B + C) 
which may be divided among the angles, giving to each a 
correction which is inversely proportional to its weight. One 
*hird of the spherical excess is then deducted from each angle, 
o.nd the triangle solved as a plane triangle. If the three 
jigles have equal weights the closing error may therefore be 
iound as if the triangle were plane and divided equally among 
them. 

For triangles the lengths of whose sides do not greatly 
exceed 6 miles the error due to the neglect of spherical excess 
is not likely to amount to 0.01 ft. 

In the case of a triangulation consisting of an intricate 
chain or network of triangles, the angles must be subjected 
to a rigid process of adjustment before the triangles are 
solved. The adjustment of a triangulation constitutes a 
subject in itself, which is beyond the scope of these notes. 
T eading principles outlined). 



88 



4. Geodetic Positions. 

The latitude and longitude of one of the stations, and the 
azimuth of a triangle side extending from that station, having 
been determined astronomically, the geographical co-ordinates 
of all the stations of the triangulation may now be computed. 
The problem thus presented for solution is: 

Given the latitude and longitude of a point on the earth's 
surface, and the length and initial azimuth of the line drawn 
from it to a second point, to determine the latitude and 
longitude of this point, and the azimuth of the first point as 
seen from the second. 




In Fig. 39 A is the first point and B the second ; C is the 
oole. AC and BC are the meridians of A and B. is the 
centre of the spheroid. AN and BN' are normals to the 



87 



spheroid at the points A and B. A'B'C is a spherical tri- 
angle, the centre of the sphere being at N. We have given 
then 

0i ai and s 
and are required to find 

</> 2 AL and a 2 
To find A0( = 2 0i) 

In the triangle A'B'C we have given b c and A'( = ai), and 
must find a( = 9O-0 2 '), C( = AL), and 5. 
We have 

cos a = cos 6 cos c-f sin b sin c cos A' 
or sin 02' = sin 0i cos c+cos 0i sin c cos ai 

= sin fa ( 1 s" j +c cos 0i cos ai 

c 2 
or sin 2 ' sin 0i = c cos fa cos ai ^-sin fa 

m 

But sin 2 ' sin 0i = sin(0i+A0') sin 0! 

= sin 0i / 1 j+A0' cos 0i sin fa 

, , A0' 2 . 
= A0 cos 0i jr- sin 0i 

A0' 2 . c 2 . 

. . A0 cos 0i sin 0i = c cos 0i cos en -~- sin 0i 

, A0 /2 c 2 

or A0 ~- tan fa= c cos ai tan fa. 

2 

Assuming as a first approximation 

A0' = c cos ai 

and substituting in the term in A0' 2 , we have 

c 2 c 2 

A0' = c cos ai s~ tan 0i+ tan 0i cos 2 a\ 

c 2 
= ccosai- -x- tan 0i sin 2 ai (33) 

Then substituting c= N 

we have (A0' being in seconds) 

, ,, 5 cos ai 1 /^cosaiN 2 , . .. /<3/n 

A * - AT sin-1" - 2 (iViETF') ta " * tan " "' Sm J (34) 

This gives the difference of latitude on an imaginary sphere 

whose radius is N( = p), whereas the radius should be 

88 



assumed equal to the value of p m for the mean of the latitudes 
of A and B, or, with sufficient precision, for the latitude 
0!-HA<'. We have then 

. t N (35) 

Pm 

Also <j> 2 = (t> 1 +A<t> (36) 

To find AL 

Again, in the triangle A'B'C, we have 

. _, sin c sin A' 

sin 6 = ; 

sin a 

. T sin c sin <n 
or sin AL = , 

COS fa 

or, substituting arcs for sines 

c sin a\ 



AL = 



cos fa' 



, AT 5 sin ai (37) 

or in seconds AL = -=r=~. 777 r 

N sin 1" cos 4> 2 

To find Aa( = a' ai) 
We have 

tan \{A'+B') - ^f^TS cot ff 
cos(a+o) 

But 4'+3' = ai + 180-a', 

= 180-(a'-a 1 ), 

= 180-Aa; 

a -& = 90-4>2-90 o +tf>i, 

= -(<fe-fo) = -A0; 

a+6 = 9O-0 2 +9O-0i 

= l8O-(0i+fc); 

cot|Aa= CO f ^ A0 cot \ AL; " 
sin m 

or tan ^Aa = sm 1 A w tan \ AL; 

cos f A0 

or, substituting arcs for tangents 

Aa = AL*^-. < 38 > 

cos %A<t> 

This is termed the convergence of the meridians of A and B. 
Then finally 

a 2 = 180-fa' 
= 180 o + a!+Aa (39) 

89 



An expression giving Aa directly in terms of the data is 
sometimes useful. It may be derived as follows: Taking the 
equation 

sin A' cot J3' = sin c cot b cos c cos A', 

it may be thus transformed 

sin ai 
tan B 



sin c cot b 


COS C COS di 






sin 


<n 




COS a\ ( 


cos c- 


sin c 


cot & 
cos aj 


tan a\ 







c 



2 



1 --^--e 



cot b 



2 cos ai 

c cot 6 c 2 c 2 cot 2 b ' 



(. , c cot b . c 2 c 2 cot 2 o \ 
cos o.\ 2 cos- 2 ai / 

. , / c cot b , c 2 , c 2 cot 2 6 \ 

.*. tan 5' + tan ai= -tan ail h H ; -J 

\ cos ai 2 cos^ ai / 

But 5 = 180 -a', .*. 

, / c cot 6 . c 2 . , c 2 cot 2 6 \ 

tan a tan a x = tan ai f -\ H 5 1 

V^ cos ai 2 cos 2 ai / 

Also a' = ai+Aa, .'. by Taylor's theorem 

tan a' = tan(ai+Aa) 

= tan ai+Aa sec 2 ai+Aa 2 tan ai sec 2 ai 

.'. , substituting, we have 

/ ' c cot b c 2 c 2 cot 2 b*\ 

Aa sec 2 ai+Aa 2 tan ai sec 2 ai = tan ai( h~sr + - * I 

\ COS ai 2 COS' 1 ai / 

c 2 
or Aa+Aa 2 tan ai = c cot b sin aiH =- sin ai cos ax 

-\-c 2 cot 2 6 tan oi 

Assuming as a first approximation 

Aa = c cot 6 sin oi 
and substituting in the term containing Aa 2 we find after 
reduction 

c 2 
Aa = c cot b sin a x -\ ~-sin ai cos ai(l+2 cot 2 b) 



or in seconds 
^ ta 
~N~ sin 1" ' 2\N J sinl" 



^ tan 0i sin ai .1/5 \ 2 sinai cos ai /t . , , J N 
Aa=- h 2(lvy sinl" ( 1 + 2tan "^ 

(40) 



90 



By writing 

x = s sin oi y = s cos a\ 

equations (34), (35), (37) and (40) become 

y x 2 tan </>i (41) 



A0 = 
AL = 



p w sinl" 2p m psinl" 

x (42) 

p cos fa' sin 1" 



A " = OT+wlrr (1+2tan! *' )(43 

These equations should not be used for distances exceeding 
20 miles. (38) should be used in preference to (40) or (43) 
when all the unknown quantities are required. 

For longer distances approaching 100 miles the following 
equations may be used : 

^ sin on 5 cos ai 

x = y = 

Pn Pn 

A , _ y y s tan 2 en x 2 tan <j>' (44) 

* = sin 1" + 3 sin 1" 2 sin 1" 

</>' = 0i + lst two terms 



Atf> = A0' -^ 

Pm 

...2 1// 

AL = 



(45) 



_____ sin 2 1" / _ cos 2 2 ' \ 

cos <h' sin 1" + 6 lAjL J V 1 sin 2 ai / 
AZ/ = lstterm 2 ' = </>i+A0' 

Afl -- AL sin *" _ sin2 *" rA ai /i _ cosHA0\ (46) 
cosiA0 12 K - a) \ l sin 2 <j> m J 

</>m=0i + |A0 Aa' = lstterm. 
The following log's are here useful : 

log 1/ sin 1" = 5.31442513 log sin 2 l"/6 = 1^.59300 
log 1/3 sin 1" =4.83730 log sin 2 1"/12 = 12.29197 

log 1/2 sin 1" = 5.0133951 

Example. Let s = 20 miles, 0i=44 30', ai = 48 20'. 
To find A0', eq. (34) 

log 5 (in ft.) = 5.0236639 

log cos ai = 9.8226883 



logp = 7.3214108 

log sin 1" = 6.6855749 



91 



4.8463522 
2.0069857 

log 690.8225 = 2.8393665 

5.67873 

log 0.5 = 1.69897 

log tan 0! = 9.99242 

log tan 2 ai =10.10129 

log sin 1" = 6.68557 



4>x 



log 1.4355 = 0.15698 



Ad/ = 689".387 

= 11' 29".387 



To find Ac/), eq. (35) 

log Ad/ - 2.8384631 

log Pn ~ 7.3214108 

\ ogPm = 7.3199151 

10.1598739 
logAd, = 2.8399588 

Ad, = 691".765 

= 11'31".765 



= 44 30' 



to -44 41' 31".765 

To find AL, eq. (37) 

log 5 =5.0236639 

log sin ai =9.8733352 

log Pn =7.3214108 

log sin 1" =6.6855749 

log cos to 1 =9.8518109 

4.8969991 
1.8587966 



log 1091.952 =3.0382035 

AL = 1091".952 

= 18' 11".952 



The second term in eq. (45) in this example =0".0005. 

92 



To find Aa, eq. (38) 
log AL 
log sin <t> m 


= 3.0382075 
= 9.8464016 


log cos |A< 


= 9.9999993 


log 766.672 
Aa 


2.8846091 
= 2.8846098 
= 12'46".672 



The second term in eq. (46) here amounts to 0".001. 

To find a 2 , eq. (39) 

ai = 48 20' 00" 

Aa = 12' 46".672 

180 00' 00" 



o 2 =228 32' 46". 672 



The above equations (41), (42), (43) and (38) may readily 
be adapted for the solution of a variety of problems. Thus 

given </>i </>2 and AL 
to find ai a 2 and s. 
We have x = AL . p n cos </> 2 sin 1" (47) 

A , 1 // I 1 X * tai1 <^1 1 // 

;y = A0.p m sin 1 + - -77 p w sin 1 

p m pn sin 1 

AJL . " /y x 2 tan^, (48) 

= A4> . Pm sin l"-i 

2p 

x (49) 



Then tan a x = 

Aa= AL 



sin 4> m 



cos | A0 
a 2 = 180 + ai+Aa 

x 3> (50) 



5 = 



sin ai cos ai 



Again, given 0i < 2 and on, 
to find .s AL and a 2 . 
We have from (48) and (49) 



y 2 tan 2 a\ tan 0! 



y= A<f> . p m sin 1" + 

2p 

* , i// i /-a j i//\9 tan- oi tan <j5>i 
= A$ . p m sin 1" + (A< . p w sin l") 2 - 

*Pn 

x y tan ai (51) 

93 



X 

s = 



AL = 



sin ai cos ai 

x 
p n cos 02 sin 1" 



Any other problem in which three of these six quantities 
are given may be solved in a similar manner. 

The foregoing equations may be used in reducing to differ- 
ences of latitude and longitude the courses of a traverse line. 
Only the first terms are here necessary, so that we may write 

x = s sin a y = s cos a 
A0 = 



AL = 



Pm sin 1 
x 



p n cos 4> sin 1" 



. x tan 4> 

Aa= : 777 = AL sin <j> (52) 

p n sin 1 

In latitude 45 the maximum values of the second terms of 
the above expressions, for a length of 1 mile, are, respectively 

0".0066 
.0093 
.0098 
The use to be made of Aa is to correct the azimuth of a 
course referred to the meridian of the initial station of the 
traverse, to refer it to the meridian of the initial point of the 
course. As a correction it is additive. The algebraic s gns of 
x and y must be carefully observed. 



94 



5. Certain Problems which occur in the Dominion 
Lands System of Survey. 

A general description of that system of survey. . 
(1) To find the amplitude of a meridian arc having a given 
length; and conversely. 
We have 

A0 = 



(53) 



Pm sin 1" 
A4> being in seconds; and conversely 

s = A(f) . p, sin 1" (54) 

If the arc is at a height H above sea level, then 

A0= (p M +tf)sinl" 



Pm (l+ V inl " 

V p>/ 

VO-S (55) 



Pm sin 



nearly. Conversely , #. (56) 



s = A<t> . p m sin 1 






Example. Find the amplitude of an arc whose length is 
24 miles, middle latitude 52, and height above sea level 
1200 feet. 

Eq. (55) 



log 24 
log 5280 


= 1.3802112 
= 3.7226339 


log 5 (in ft.) 


= 5.1028451 


log Pm 

log sin 1" 


= 7.3204817 
= 6.6855749 




2.0060566 


log 1249.650 
logH 


= 3.0967885 
= 3.07918 


log p m 


= 7.32048 


log 0.0717 


6.17597 
= 2.85549 


&4> 


= 1249.578 
= 20' 49".578 



95 

7 



For rinding the length of a meridian arc exceeding about a 
degree the following expression may be used: 
s = [5.56182842]A</> (in" degrees) 
-[5.0269884] cos 2< sin A< 
+[2.0527848] cos 4<t> sin 2A0 
-[1.17356. .] cos 6</> sin 3A0 + 
in which 

A0 = the difference of latitude of its extremities, 
4> = the mean of the extreme latitudes. 
The numbers in brackets are logarithms. 
This expression is sufficient for finding the length of a whole 
quadrant. 

(2) Given two points on the same parallel of latitude, at 
a given distance apart, to find their difference of longitude, 
and the convergence of their meridians. 




F/g.40 



A and B are the two points; ADB a normal section, and 
AEB a parallel of latitude. PD is drawn at right angles to 
ADB. The triangle PDB gives 

sin BD 



sin BPD = 



sin PB 
s 



or 



sin 



AL 



sin 



2N 



2 cos 4> 

or, as AL is assumed to be small, this rray be written 

s * 



AL = 



lV cos 



A AT S {L ' b} 

\>r in seconds aL = -r= - : 777 

N cos 4> sin 1 

If the higher powers of AL and s/27V are retained in the 
expansions, this becomes 

AL = Tt s . + ^^ (AL') 3 sin 2 <f> (59) 

N cos sin 1 24 

in which AL' is the first term. As 

TV cos cf> = P, 

the radius of the parallel of latitude, this may be written 

^ = ^7, + ^f-c^-h^)' * (eo) 

P sin 1 24 \P sin 1 / 

For a chord 6 miles in length, in latitude 52, the second 
term of (60) amounts to only 0". 00008, a quantity quite 
inappreciable, so that the first term may be considered exact. 
Again, in the triangle PDB we have 

__ tan BD (61) 









LUS J. J 


tan PB 


or 






Aa 
sin T 


tan 27V 

cot 


or 






A, 


5 tan 4> 
x= - N 


Aa 


being 


small- 


; or in 


seconds 

5 tan <t> 



(62) 
"" iVsinl" 

The higher terms are here also inappreciable. From (58) 
c n 1 (62) we have 

Aa = AL sin 
(See eq. 52). 

The deflection angle between two consecutive chords of 
the same length is clearly 

. s tan 4> 

Aa = , T . -j, 
TV sin 1" 

and the azimuth of a chord at either extremity 

90-^- 

To find the difference in length of s and the arc of the 
parallel p we have 

AL = -^ + ~ (?- Y sin 2 <f> t 

24 \N cos <j>J 

and AL = P 



N cos <f> 

_P . 

~N cos <f> ' 



97 



Equating these we have 

p-s = ( ) sin 2 0iVcos0 

24 y N cos </> / 



24 v^vy 



tan* <{> ^ 



To find the length of an offset from the chord to the parallel 
of latitude. 

Applying eq. (33) to the arc DE, Fig. 40, we have, denoting 
AD and DE by x and y, respectively, 

x 
N N' 2 V N_ 



y x l / x v 

-^ cos a- -,- 1 tan 4> sin- a 



and by*(61) cos a = tan <f> 

.'. writing sin 2 a = l we have 



i s 



'N'T ^N tan0 ~ 2^ tan< * 

x(5 x) 

= "2i^" tan<A 






6. Trigonometric Levelling. 

A and B are two stations whose difference of elevation is 
to be determined; A' and B' are the apparent positions of 
A and B, affected by refraction. The altitude h of B, observed 
at A , and the distance s, are assumed to be known. 




F/G.^-I 



Denoting the height BC of B above A by H, we have 

sin BA C 



H = AC 



sin ABC 



a 



BAC' = h-r + CAC' = h-r+- , 
= h ma + -tt , 

ABC = 90 -h + r- a 
= 90-h+tn<r-o 

= 90-{/; + (l-m)<x}. 



99 



sin {h + {\-m)c } (65) 

cos! A+(l m)a\ 

See Supp. to Manual of Dominion Land Surveys. 

For the numerical value of m see p. 76. 

In eq. (65) it is assumed that the distance 5 is equal to the 
chord AC. If A and B are stations of a trigonometric survey , 
and 5 is obtained by the solution of a triangle, then it is the 
distance AB reduced to sea level. The correction to 5 for 
elevation is 

H, 

s , 

P 

Hi being the height of A above sea level. Also the correction 
to reduce from th ; arc to the chord is 

24V, p) 

so that the length of the chord A C is 

<<+f)i>-i(i)T 

the second correction only becoming appreciable for con- 
siderable distances. 

Reciprocal zenith distances 

If the zenith distances z and z' be observed simultaneously 
at the two stations the effect of refraction is eliminated, if 
it can be assumed to affect the two zenith distances equally. 
Thus, returning to the above equation for H, we have 

BAG' = 90 -z-r+-^- 

ABC = 180 -z'-r 
But we have also 

A'AB = z + r = l?0-(z' + r)+a 

so that r = 

which therefore becomes known. Substituting this we have 
BA C = Z ~^ 

ABC = 90- -' 



2 

.". substituting in the first above expression for H gives 

H = s sin \{z'-z) (66) 

cos Kz'-s + o-) 
* having been corrected for elevation, and if necessary for 
curvature. 

100 



TABLE I. Mean Refractions. (Bar. 29.6 ins., Ext. therm. 48.) 



App. 


Refr. 


App. 


Refr. 


App. 


Refr. 


h 


ro 


h 


ro 


h 


ro 


/ 


/ it 


o / 


/ // 


/ 


i ir 


00 


34 10.5 


14 10 


,3 44.2 


27 20 


1 50.9 


30 


28 26.3 


20 


3 41.6 


40 


1 49.4 


1 00 


24 05.2 


30 


3 39.0 


28 00 


1 47.9 


30 


20 43.3 


40 


3 36.5 


20 


1 46.4 


2 00 


18 04.4 


50 


3 34.0 


40 


1 44.9 


30 


15 56.9 


15 00 


3 31.6 


29 00 


1 43.5 


3 00 


14 13.4 


10 


3 29.2 


20 


1 42.1 


30 


12 48.0 


20 


3 26.9 


40 


1 40.7 


4 00 


11 36.7 


30 


3 24.6 


30 00 


1 39.4 


30 


10 36.4 


40 


3 22.4 


31 00 


1 35. 5 


5 00 


9 45.0 


50 


3 20.2 


32 00 


1 31.9 


30 


9 00.7 


16 00 


3 18.1 


33 00 


1 28.4 


6 00 


8 22.2 


10 


3 16.0 


34 00 


1 25.2 


30 


7 48.6 


20 


3 13.9 


35 00 


1 22.0 


7 00 


7 18.8 


30 


3 11.9 


36 00 


1 19.1 


10 


7 09.7 


40 


3 09.9 


37 00 


1 16.3 


20 


7 00.9 


50 


3 08.0 


38 00 


1 13.6 


30 


6 52.4 


17 00 


3 06. 1 


39 00 


1 11.0 


40 


6 44. 3 


10 


3 04.2 


40 00 


1 08.5 


50 


6 36.4 


20 


3 02.4 


41 00 


1 06.2 


8 00 


6 28.9 


30 


3 00.6 


42 00 


1 03.9 


10 


6 21.6 


40 


2 58.8 


43 00 


1 01.7 


20 


6 14.5 


50 


2 57.0 


44 00 


59.6 


30 


6 07.7 


18 00 


2 55.3 


45 00 


57.5 


40 


6 01. 1 


10 


2 53. G 


46 00 


55.6 


50 


5 54.8 


20 


2 52.0 


47 00 


53.7 


9 00 


5 48.6 


30 


2 50.4 


48 00 


51.8 


10 


5 42.6 


40 


2 48.8 


49 00 


50.0 


20 


5 36.9 


50 


2 47.2 


50 00 


48.3 


30 


5 31.3 


19 00 


2 45.6 


51 00 


46.6 


40 


5 25.9 


10 


2 44. 1 


52 00 


45.0 


50 


5 20.6 


20 


2 42.6 


53 00 


43.4 


10 00 


5 15. 5 


30 


2 41. 1 


54 00 


41.8 


10 


5 10.6 


40 


2 39.7 


55 00 


40.3 


20 


5 05.8 


50 


2 38.3 


56 00 


38.8 


30 


5 01.1 


20 00 


2 36.9 


57 00 


37.4 


40 


4 56.6 


20 


2 34.2 


58 00 


36.0 


50 


4 52.1 


40 


2 31.5 


59 00 


34.6 


11 00 


4 47.9 


21 00 


2 28.9 


60 00 


33.2 


10 


4 43.7 


20 


2 26.4 


61 00 


31.9 


20 


4 39.6 


40 


2 23.9 


62 00 


30.6 


30 


4 35.7 


22 00 


2 21.5 


63 00 


29.3 


40 


4 31.8 


20 


2 19.2 


64 00 


28.0 


50 


4 28.1 


40 


2 17.0 


65 00 


26.8 


12 00 


4 24.4 


23 00 


2 14.8 


66 00 


25.6 


10 


4 20.9 


20 


2 12.7 


67 00 


24.4 


20 


4 17.4 


40 


2 10.6 


68 00 


23.2 


30 


4 14.0 


24 00 


2 08.6 


69 00 


22.1 


40 


4 10.7 


20 


2 06.6 


70 00 


21.0 


50 


4 07.5 


40 


2 04.7 


72 00 


18.8 


13 CO 


4 04.3 


25 00 


2 02.8 


74 00 


16.6 


10 


4 01.2 


20 


2 01.0 


76 00 


14.4 


20 


3 58.2 


40 


1 59.2 


78 00 


12.3 


30 


3 55.3 


26 00 


1 57.5 


80 00 


10.2 


40 


3 52.4 


20 


1 55.8 


85 00 


05.0 


50 


3 49.6 


40 


1 54. 1 


90 00 


0*00.0 


14 00 


3 46.9 


27 00 


1 52. 5 







TABLE II. Corrections to Mean Refraction. 



F* 


ctor B depending on 


the 


Factor t depending on the External 




Barometer. 






Thermometer. 




Ins. 


B 




B 


Fahr. 


t 


Fahr. 


t 




0. 845 


30.9 


1.044 


C 




O 




. 1 


0. 84* 


31.0 


17 


- 22 


1.163 


36 


1.026 


.2 


0. Sol 






- 21 


1.160 


37 


1.024 
1.022 


.3 


0. S55 






20 


i. : 


.4 


0. 85S 






- 19 


1. 1 


39 


1.020 


. 5 


0.81 - 






- 18 


1.152 


40 


1.018 


.6 


0. - 






- 17 


1.150 


41 


1.016 


7 


0. - 






- 16 


1.147 


42 


1.014 


- 


0.872 






- 15 


1 . 145 


43 


1.012 


.9 

- 


0. 87a 
0. 878 






- 14 

- 13 


1. 142 
1.139 


44 


1 010 


Fact r T 


depending 


45 


1 '. 007 


. 1 


0.882 


on the 


Attached 


- 12 


1.137 


46 


1 . 005 


2 


0.88a 


rmometer 


- 11 


1.134 


47 


1 . 003 


4 


0. 8* 
0.- _ 






- 10 

- 9 


l l! 


Jv 


i roi 


Fahr. 


T 


. 





0.999 


. o 


895 







- 8 


1.1.7 


50 


0. 


.6 





- 20 


1.' 


- 7 


1.124 


51 


0.996 


7 


0.1' - 


- 1C 


1.004 


- 6 


1.122 




0.994 


- 


0.91 





1.' 


5 


1. 119 


53 


0.992 


.9 


0.909 


- 10 


1.002 


- 4 


1.117 


.54 


0.990 


27 


0.9 - 


- 


-il 


- 3 


1.114 




0:98* 


. 1 


0.916 


30 


l.l 1 


_ 2 


1.112 


: 


0. 9 


2 


0. 919 


40 


0.999 


- 1 


1. 1 


57 


984 


3 


0.922 


50 


0.9 





1. 107 


58 


0. 9 - - 


.4 


0.92 


60 


0.' 


- 1 


l. : 





0.9* 


. 5 


0.929 


7 


0.997 


2 


l. li . 


60 


0.97* 




0.933 





0.9 


3 


1.100 


61 


0.97 


. 


0." 




0.9 " 


4 


1.0 " 


62 


0. 974 


8 


0. 939 


-100 


0.994 


5 


1.095 


63 


0.973 


.9 


0.943 
0. 9 






g 


1 093 


64 


0. 971 






7 


1.090 


: 


0. 


1 


0.949 






- 


1.088 


66 


0. '.' 7 




0. 9 ' 






9 


1.086 


" 


0. 965 


_ 


0.956 






10 


1.' 


68 


0. ' 


.4 


0.960 






11 


1.081 




0.961 


5 


0.963 






12 


1 . 079 


70 


0. 960 


.6 


0.966 






13 


1.070 


71 


0. 9 3 


7 


0. 97 






14 


1.074 


72 


C. 


- 


973 






15 


1.072 


73 


0. 954 


.9 


0. 97 






16 


1.069 


74 


0. 952 


2 


0.980 






17 


1.(07 


75 


0. 951 


. 1 


0.9* 









1.065 


76 


0.94 




0.9^7 




~ 


19 


1.063 


77 


0. 


_ 
3 


0.990 




20 


1.060 


7-- 


0. 945 


.4 


0.993 






21 


1. 05* 


79 


0. 9 


5 


- 




22 


1.056 





0.942 


.6 


1.000 




~ 


- 


1 . 0.54 


81 


0. 940 


7 


1.003 






1.052 


S2 


0.9 


- 


1 007 







_- 


1.049 





0. 936 




1.010 






- 


1.047 




0.9 " 


30.0 


1 . 014 






27 


1.045 


ss 


0. 933 


1 


1.017 






2* 


1.043 


86 


0. 931 




1.020 






29 


1.041 


-7 


0. 930 


_ 


1 . 024 






30 


1.039 


38 


0.92* 


i 


1.027 






31 


1.' 





0.926 


.5 


1.031 






_ 


1. ,34 


90 


0.924 




1 . 034 






33 


1.032 


91 


0. 92 


7 


l.< " 






34 


1 . 030 


92 


0.921 


- 


1.041 






" 


1.02^ 







TABLE III. m = 



2 sin 



2 I 



sin 1 



T 


m 


l m 


om 


gm 


4 m 


5 ra 


gm 


7 m 


8 m 


s 


// 


" 


" 


" 


it 


/ 


// 


1' 


// 





0.00 


1.96 


7 85 


17.67 


31.42 


49.09 


70.68 


96.20 


125.65 


i 


0.00 


2.03 


7.98 ' 


17. s7 


31.68 


49.41 


71.07 


96. 66 


126.17 


2 


0.00 


2.10 


S. 12 


18.07 


31.94 


49.74 


71.47 


97. 12 


126.70 


3 


0.00 


2.16 




18.27 


32. 20 


50.07 


71. 86 


97.58 


127 22 


4 


0.01 


2.23 


S. 39 


18.47 


32.47 


50.40 


72.20 


98.04 


127! 75 


5 


0.01 


2.31 


8.52 


18! 67 


32. 74 


50.73 


72. 66 


98. 50 


128. 2S 


6 


0.02 


2.38 


S. 66 


18.87 


33.01 


51.07 


73.06 


98.97 


128.81 


7 


0.(12 


2.45 


8.80 


19.07 


33.2 7 


51.40 


73.46 


99.43 


129.34 


8 


0.03 


2.52 


8.94 


10 2s 


33. 54 


51.74 


73.86 


99.90 


129. S7 


9 


0.04 


2.60 


9.08 


19.48 


33.81 


52.07 


74.20 


100. 37 


130.40 


10 


0.05 


2.67 


9.22 


19.69 


34.09 


52.41 


74.66 


100.84 


130. 94 


1 


0.06 


2. 75 


9.36 


19.90 


34.36 


52. 75 


75.06 


101.31 


131.47 


2 


(Mis 


2.83 


9.50 


20.11 


34. 64 


53.09 


75.47 


101.78 


132. CI 


3 


0.09 


2.91 


9.64 


20.32 


34.01 


53.43 


75.88 


102. 25 


132.55 


4 


0.11 


2.99 


9.79 


2C.53 


35. 19 


53 77 


70 29 


102.72 


133.09 


5 


0.12 


3.07 


9.94 


20. 74 


35.46 


54. 11 


76. 69 


103. 20 


133. 63 


6 


0. 14 


3.15 


10. 09 


20.95 


35. 74 


54.46 


77. 10 


103 67 


134.17 


7 


0. 16 


3.23 


10.24 


21.16 


36.02 


54. SO 


77.51 


104. 15 


134.71 


8 


0. IS 


3. 32 


10.39 


21.38 


36. 30 


55. 15 


77.93 


104.63 


135.25 


9 


0.20 


3.40 


10. 54 


21.60 


36.58 


55. 50 


78.34 


105. 10 


135. SO 


20 


0.22 


3.49 


10.69 


21. S2 


36. S7 


55. S4 


7S. 75 


105. 5S 


136.34 


1 


0.24 


3.58 


10. S4 


22.03 


37. 15 


50. 10 


79.16 


106.06 


136.88 


2 


0.26 


3.67 


11.00 


22.25 


37.44 


56. 55 


79.58 


106. 55 


137.43 


3 


0.28 


3. 76 


11. 15 


22.47 


37. 72 


56. 90 


SO. 00 


107.03 


137.98 


4 


0.31 


3.85 


11.31 


22 70 


38.01 


57.25 


so. 42 


107.51 


13S.53 


5 


0.34 


3.94 


11.47 


22.92 


38.30 


57.60 


SO. 84 


107.99 


139.08 


6 


0. 37 


4.03 


11.63 


23.14 


38. 50 


57. 96 


81.26 


108.48 


139. 63 


7 


0.40 


4.12 


11.79 


23. 37 


38. ss 


58. 32 


81.68 


108.97 


140.18 


8 


0.43 


4.22 


11.95 


23.60 


39. 17 


5S.68 


s2. 10 


109.46 


140.74 


9 


0.46 


4.32 


12. 11 


23.82 


39.46 


59.03 


S2.52 


109.95 


141.29 


30 


0.49 


4.42 


12.27 


24.05 


39.76 


59.39 


S2.95 


110.44 


141.85 


1 


0. 52 


4.52 


12.43 


24.28 


40.05 


59. 75 


S3. 38 


110.93 


142.40 


2 


0.56 


4.62 


12.60 


24.51 


40.35 


60.11 


S3. SI 


111.43 


14 2.00 


3 


0.59 


4.72 


12.70 


24.74 


40.65 


60.47 


S4.23 


111.92 


143.52 


4 


0.63 


4.82 


12.93 


24.98 


40.95 


60.84 


s4.66 


112.41 


144. OS 


5 


0. ('.7 


4.92 


13. 10 


25. 21 


41.25 


61.20 


S5.09 


112.90 


144.64 


6 


0. 71 


5. 03 


13.27 


25. 45 


41. 55 


61.57 


85. 52 


113.40 


145. 20 


7 


0. 75 


5.13 


13.44 


25.68 


41. S5 


61.94 


85 


113.90 


145.70 


8 


0.79 


5. 24 


13.62 


25 02 


42.15 


62.31 


86.39 


114.40 


140. 33 


9 


0.83 


5.34 


13.79 


26.16 


42. 45 


62.68 


86. 82 


114.90 


146. S9 


40 


0.87 


5. 45 


13. 96 


26. 40 


42.76 


63.05 


S7. 26 


115.40 


147.46 


1 


0.91 


5, 56 


14.13 


26. 64 


43.06 


63.42 


87. 70 


115.00 


14S. 03 


2 


(t. 96 


5. 67 


14.31 


26. ^ 


43. 37 


63. 79 


88. 14 


116.40 


14s. 00 


3 


1.01 


5. 78 


14.49 


27. 12 


43. 68 


64. 16 


ss.,-,7 


116.90 


149. 17 


4 


1 . 06 


5.90 


14.67 


27. 37 


43.99 


64.54 


89.01 


117.41 


149.74 


5 


1.10 


6.01 


14. So 


27.61 


44. 30 


64.91 


89.45 


117.92 


150.31 


6 


1.15 


6.13 


15.03 


27.86 


44.61 


65. 29 


89. so 


lis. 43 


150.88 


7 


1.20 


6 24 


15.21 


28. 10 


44.92 


65. 67 


90. 33 


IIS. 94 


151.45 


S 


1.26 


ti. 36 


15.39 


28.35 


45. 24 


66. 05 


90. 7s 


119.45 


152.03 


9 


1.31 


6. 18 


15. 57 


2S. 60 


45. 55 


66.43 


91.23 


119.96 


152.61 


50 


1 . 36 


6 60 


15. 76 


28.85 


45. 87 


66. 81 


91. 6S 


120.47 


153.19 


1 


1.42 


6 7.' 


15.95 


29.10 


46. is 


07. 19 


92. 12 


L20 us 


153. 77 


Q 


1.48 


6 84 


16. 14 


29.36 


46. .".< 


07 5s 


92.57 


121.49 


154.35 


3 


1 . 53 


6.96 


16. 32 


29.61 


46. s- 


67.96 


93. 02 


122.01 


154.93 


4 


1.59 


7.09 


16.51 


29.86 


47.14 


68.35 


93.47 


122 53 


155.51 


5 


1.65 


7.21 


16. 70 


30. 12 


47. 46 


68. 73 


93.92 


123.05 


156.09 


6 


1.71 


7 34 


16.89 


30. 38 


47. 70 


00. 12 


94 :.s 


123 57 


156.67 


7 


1.77 


7 47 


17 08 


30. 64 


is. 11 


69.51 


94. 83 


124.09 


157.25 


s 


1 . 83 


7.60 


17 28 


30.00 


4S.4:-! 


09 00 


95.20 


124 61 


157 84 


9 


1.S9 


7.72 


17 17 


31. 16 


4s. 76 


70. L!) 


95. 74 


125.13 


158. 43 



TABLE III. Cont. 



m : 



2 sin 2 |t 
sin 1 



T 


9 m 


10 m 


ll m 


r_ m 


lS m 


14 m 


15 m 


( 

16 m 


s 


// 


It 


// 


ft 


ft 


ft 


// 


tf 





159.02 


196.32 


237. 54 


282.68 


331.74 


384. 74 


441.63 


502. 46 


i 


159.61 


196.97 


238.26 


283.47 


332.59 


385. 65 


442.62 


503. 50 


2 


160.20 


197.63 


23S. 9S 


284.26 


333. 44 


386. 56 


443.60 


504. 55 


3 


160.80 


198. 28 


239.70 


285. 04 


334. 29 


387. 48 


444. 58 


505.60 


4 


161.39 


198. 94 


240.42 


285.83 


335. 15 


388. 40 


445. 56 


506.65 


r> 


161.98 


199. 60 


241.14 


286. 62 


336. 00 


389. 32 


446. 55 


507. 70 


6 


162.58 


200.26 


241.87 


287.41 


336.86 


390. 24 


447.54 


508. 76 


7 


163. 17 


200.92 


242.60 


288.20 


337. 72 


391.16 


448. 53 


509. 81 


8 


163.77 


201.59 


243. 33 


2S9.00 


338. 58 


392.09 


449.51 


510.86 


9 


164.37 


202.25 


244.06 


289. 79 


339. 44 


393.01 


450. 50 


511.92 


10 


164.97 


202.92 


244. 79 


290. 58 


340. 30 


393:94 


451.50 


512.98 


1 


165. 57 


203. 58 


245.52 


291.38 


341.16 


394.86 


452.49 


514.03 


2 


166. 17 


204. 25 


246.25 


292. 18 


342.02 


395. 79 


453.48 


515.09 


3 


166.77 


204.92 


246. 98 


292.98 


342.88 


396. 72 


454. 48 


516. 15 


4 


167.37 


205. 59 


247.72 


293.78 


343. 75 


397. 65 


455. 47 


517.21 


5 


167. 97 


206.26 


248. 45 


294.58 


344. 62 


3E8.58 


456.47 


518.27 


6 


168. 58 


206.93 


249. 19 


295. 38 


345. 49 


399. 52 


457.47 


519.34 


7 


169. 19 


207.60 


249.93 


296. 18 


346. 36 


400.45 


458. 47 


520.40 


8 


169.80 


208.27 


250. 67 


296.99 


347. 23 


401.38 


459.47 


521.47 


9 


170.41 


208. 94 


251.41 


297. 79 


34S. 10 


402.32 


460. 47 


522. 53 


20 


171.02 


209. 62 


252. 15 


298. 60 


348. 97 


403. 26 


461.47 


523. 60 


1 


171.63 


210.30 


252. 89 


299. 40 


349.84 


404. 20 


462. 48 


524. 67 


2 


172.24 


210.98 


253. 63 


300.21 


350.71 


405. 14 


463. 48 


525 74 


3 


172.85 


211.66 


254.37 


301.02 


351.58 


406.08 


464. 48 


526.81 


4 


173.47 


212.34 


255. 12 


301.83 


352.46 


407. 02 


465. 49 


527. S9 


5 


174.08 


213.02 


25.5. 87 


302. 64 


353. 34 


407.96 


466. 50 


528.96 


6 


174.70 


213.70 


256. 62 


303.46 


354.22 


408. 90 


467. 51 


530.03 


7 


175.32 


214.38 


257. 37 


304. 27 


355. 10 


409.84 


468. 52 


531. 11 


8 


175.94 


215.07 


258. 12 


305. 09 


355. 98 


410.79 


469. 53 


532. IS 


9 


176. 56 


215.75 


258. 87 


305. 90 


356.86 


411.73 


470. 54 


533. 26 


30 


177. 18 


216.44 


259. 62 


306. 72 


357.74 


412.68 


471.55 


534.33 


1 


177. 80 


217. 12 


260. 37 


307.54 


358. 62 


413.63 


472. 57 


535 41 


2 


178.43 


217.81 


261. 12 


308. 36 


359. 51 


414.59 


473. 58 


536. 50 


3 


179.05 


218.50' 


261.88 


309. 18 


360. 39 


415. 54 


474. 60 


537.58 


4 


179. 68 


219. 19 


262. 64 


310. 00 


361.28 


416.49 


475.62 


538.67 


5 


180. 30 


219.88 


263. 39 


310.82 


362. 17 


417.44 


476. 64 


539.75 


6 


180. 93 


220. 58 


264. 15 


311.65 


363. 07 


41S. 40 


477. 65 


540. 83 


7 


181.56 


221.27 


264.91 


312.47 


363. 96 


419.35 


478. 67 


541.91 


8 


182. 19 


221.97 


265. 68 


313. 30 


364. So 


420.31 


479. 70 


543.00 


9 


182.82 


222.66 


266. 44 


314. 12 


365. 75 


421.27 


480. 72 


544.09 


40 


183.46 


223.36 


267. 20 


314. 95 


366. 64 


422.23 


481.74 


545. 18 


1 


184.09 


224.06 


267. 96 


315.78 


367. 53 


423.19 


482.77 


546. 27 


2 


184. 72 


224.76 


268.73 


316.61 


368.42 


424. 15 


483. 79 


547. 36 


3 


185. 35 


225.46 


269.49 


317.44 


369.31 


425. 11 


484.82 


548.45 


4 


185. 99 


226. 16 


270.26 


318.27 


370.21 


426.07 


485.85 


549. 55 


5 


1S6. 63 


226.86 


271. C2 


319.10 


371. 11 


427.04 


486. 88 


550.64 


6 


1S7.27 


227.57 


271.79 


319.94 


372.01 


428.01 


487.91 


551.73 


7 


187.91 


228.27 


272.56 


320.78 


372.91 


428.97 


488.94 


552.83 


8 


188. 55 


228.98 


273. 34 


321.62 


373.82 


429.93 


489.97 


553. 93 


9 


189. 19 


229.68 


274. 11 


322.45 


374.72 


430.90 


491.01 


555.03 


.30 


189.83 


230. 39 


274.88 


323. 29 


375. 62 


431.87 


492.05 


556. 13 


1 


190.47 


231. 10 


275. 65 


324. 13 


376. 52 


432.84 


493. 08 


557. 24 


2 


191. 12 


231.81 


276. 43 


324.97 


377. 43 


433. 82 


494. 12 


558. 34 


3 


191.76 


232.52 


277.20 


325. SI 


378. 34 


434.79 


495. 15 


559.44 


4 


192.41 


233. 24 


277. 98 


326. 06 


379. 26 


435. 76 


496. 19 


560. 55 


") 


193. 06 


233.95 


278. 76 


327. 50 


3S0. 17 


436. 73 


497.23 


561.65 


fi 


193. 71 


234.67 


279.55 


328. 35 


381.08 


437. 71 


498.28 


562.76 


7 


194. 36 


235.38 


280.33 


329.19 


381.99 


438. 69 


499.32 


563. 87 


8 


105. 01 


236. 10 


281.12 


330. 04 


382.90 


439. 67 


500. 37 


564.98 


9 


195. 66 


236.82 


281.90 


330. 89 


383.82 


440. 65 


501.41 


566.08 



TABLE IV. Values of Log p n & Log p m (in feet). 





o / 


Log p 


Log p m 


4> 

/ 


Log p n 


Log p m 


40 00 


7.3212921 


7. 3195588 


50 00 


7.3215482 


7.3203271 


10 


2963 


5715 


10 


5524 


3398 


20 


3006 


5842 


20 


5567 


3524 


30 


3048 


5969 


30 


5609 


3651 


40 


3090 


6096 


40 


5651 


3777 


50 


3133 


6223 


50 


5693 


3903 


41 00 


3175 


6350 


51 00 


5735 


4029 


10 


3218 


6478 


10 


5777 


4155 


20 


3260 


6605 


20 


5819 


4281 


30 


3303 


6733 


30 


5861 


4407 


40 


3345 


6861 


40 


5903 


4532 


50 


3388 


6988 


50 


5944 


4657 


42 00 


3431 


7116 


52 00 


5986 


4782 


io- 


3473 


7244 


10 


6028 


4907 


20 


3516 


7372 


20 


6069 


5032 


30 


3559 


7501 


30 


6111 


5156 


40 


3601 


7629 


40 


6152 


5281 


50 


3644 


7757 


50 


6193 


5405 


43 00 


3687 


7885 


53 00 


6235 


5529 


10 


3730 


SOU 


10 


6276 


5652 


20 


3773 


8142 


20 


6317 


5776 


30 


381.5 


8271 


30 


6358 


5899 


40 


3858 


8399 


40 


6399 


6022 


50 


3901 


8528 


50 


6440 


6145 


44 00 


3944 


8656 


54 00 


6481 


6268 


10 


3987 


8785 


10 


6522 


6390 


20 


40.30 


8914 


20 


6563 


6513 


30 


4073 


9042 


30 


6603 


6634 


40 


4115 


9171 


40 


6644 


6756 


50 


4158 


9300 


50 


6684 


6878 


45 00 


4201 


9428 


55 00 


6725 


6999 


10 


4244 


9557 


10 


6765 


7120 


20 


4287 


9686 


20 


6805 


7240 


30 


4330 


9814 


30 


6845 


7361 


40 


4373 


7.3199943 


40 


6886 


7481 


50 


4416 


7. 3200072 


50 


6925 


7601 


46 00 


4459 


0200 


56 00 


6965 


7721 


10 


4502 


0329 


10 


7005 


7840 


20 


4544 


0458 


20 


7045 


7959 


30 


4587 


0586 


30 


7084 


S07S 


40 


4630 


0715 


40 


7124 


8196 


50 


4673 


0843 


50 


7163 


8314 


47 00 


4716 


0972 


57 00 


7203 


S432 


10 


4759 


1100 


10 


7242 


S550 


20 


4801 


1228 


20 


7281 


8667 


30 


4844 


1357 


30 


7320 


8784 


40 


4887 


1485 


40 


7359 


8901 


50 


4930 


1613 


50 


7398 


9017 


48 00 


4972 


1741 


58 00 


7436 


9133 


10 


5015 


1869 


10 


7475 


9249 


20 


5058 


1997 


20 


7513 


9364 


30 


5100 


2125 


30 


7552 


9479 


40 


5143 


2253 


40 


7590 


9594 


50 


5185 


2380 


50 


7628 


9708 


49 00 


5228 


2508 


59 00 


7666 


9S22 


10 


5270 


2635 


10 


7704 


7. 3209936 


20 


5313 


2763 


20 


7742 


7.3210049 


30 


5355 


2890 


30 


7779 


0162 


40 


5398 


3017 


40 


7817 


0275 


50 


7.3215440 


7.3203144 


50 


7854 


0387 








60 00 


7.3217891 


7.3210499 







f 1 



J II 









REFERENCE COP> 



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29 



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1 



2 
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1 





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