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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,

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

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
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

(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.
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
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
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.

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

(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

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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