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x rad.

COS.

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di?.

TIL fro

SP

COS. BDPS

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COS. BSP

SB

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COS. BSP

LMDB= DBS and the LNDP= DPS ; hence the three angles necessary for this reduction are BDP, DBS, and dps.

PB? + BD? - Dp? It is shewn (K. 39.) that cos. DBP=

2BD X PB

DBP Hence PB? + BDP-(2BD X PB X

rad.
(N) That is, The square of the side subtending any acute L
of a triangle, is less than the squares of the sides containing that
acute Li by double the rectangle of those sides, multiplied by the
cosine of the acute 2 dividend by radius.
Therefore, in the triangle BDP
PB?=BD? +PD'—(2BD XPD X

rad.
And, in the triangle SPB
PB =BS + PS (2BS X PS X

rad.
By making the values of PBR equal to each other, and reducing
the equation, we get (BD? - B52)+(pd? – Psa)=(2BD X PDX

-(2BS + P8 + rad.

rad.

:). But BD2-Bs 2 = PDSPS’ =SD? (Euclid 47 of I.), therefore sdo x rad=

(BD X PD X COS. BDP)-(BS X PS X cos. BSP), And hence cos. BSP

(BD X PD X COS. BDP) -(8D% x rad.)

BS X PS rad. : BD :: sine DBS : SD

rad. : PD :: sine DPS : SD. Hence, rad. : BD XPD:: şine DBS x sine DPS: sds. (KEITH'S Geom. 14 of VII.)

1 And, sp2 x rad.=BD XPD X sine DBS x sine DP8 X

rad. Again, rad. : BD :: cos. DBS : BS

rad. : PD: :: cos. DPS : PS Hence rad.? : BD X PD :: cos. DBS X COS. DPS : BS X PS,

1 And BS X PS = BD X PD X.cos. DBS X COS. DPS X

rad. Therefore by substitution the cosine of the BSP= cos.. BDP-(sine DBS x sine DPS X

rad,

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rad:)

2

rad.

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COS. DBS X COS. DPS X
(cos. BDP X rad: 2) (sine Dbs x sine dps x rad.)

COS. DPS X cos. DPS

(O) M. de Lambre gives the following formula * for finding the angle BSP; which he demonstrates by spherical trigo nometry, viz. Sine į Bsp = rad. V

sine { (BDP + DBS — DPs) sine { (BDP + DPS-DBS) The sides Bs and Ps of the triangle BSP are easily found, from the height is and the angles DBS and dps being given.

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EXAMPLE L.

Wanting to know the height of an object sD (standing on the horizontal plane PSB), and also the horizontal distances SP and so, I measured a base PB=300 yards, at p the Lof elevation des was 8o.15', and the LDPB, inclined to the horizon, was 57.15': at B the L of elevation DBS was 89.45.50", and the DBP, inclined to the horizon, was 630.15'. Required the horizontal Z. BSP, the horizontal distances sp, and SB, and the height of the object above the plane.

Answer. L BPS= 560.51',50"; _ PBS=620.54.30"; (L. 91.) BS=289-41 ; PS=307*69; and ds=44.614 yards.

The height ds, and the distances Bs and Ps, may be found without reducing the angles. For, in the triangle DBP all the angles are given, and the side BP, whence BD and PD may be found; then in the triangle dps, the LDPs and the side pd are given to find us and sp; also, in the triangle DBS, DB and the DBS are given, from which es may be found.

EXAMPLE LI. From the top of a tower 44.614 yards high, the angle BDP, subtended by two distant objects B and p, was 59o.30'; and the angles of depression muB = 8°.45'.50", and nDP= 8o.15. Hence it is required to find the horizontal distances PB, BS, and Ps.

Answer. The t MDB= L DBS =8°, 45'. 50"; and LNDP = DPS = 8o.15', and BDP = 59° .30. By the formula (N. 92 or 0.93.) cos. BSP=49655, hence L BSP=60°. 18. 40" • 4. and Bs=289.41, Ps=307•69, and PB = 300. The same answer may be obtained without reducing the angles, and without the formula.

(P) The formulæ at the conclusion of N. 92, and 0.93, may be applied to any two triangles, whether their bases be in the same horizontal plane and their vertices be elevated, as in Examples L. and Li, or their vertices be in the same horizontal plane, and the extremities of the base of the one triangle be elevated above the extremities of the base of the other.

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COS, PDS X cos, BDO

at d, takes the altitude of a celestial object/ AKIT (9)

In the annexed geiwolaitaal geving end se pob.MY figure, if the an- 2980233tob an doitw Post gle PDB be measured with a sex (2007 - ) tant, and the ver- and tical angles BDO, le ser.com

29 PDS be measured but with a theodolite, De cosine L SDO = (cos. PDB X rad.2) – (sine PDS X sine BDO x rad.)

COS, PDS X COS. BDO or by M. de Lambre's formula*, viz. sine | sdo = rad.

sine (PDB + PDS - BDO) ~ sine į (PDB + BDO - PDS

EXAMPLE LII. From a station at d in the horizontal plane pso, I took the angle PDB, subtended by the tops of two towers,=370.53'.20" ; and also the angles of elevation bpo=4°23.55", and pos= 4o. 17.21". The height of the tower bo is known to be 40 yards, and that of ps 30 yards; from which it is required to find the horizontal distance of my station from each of the towers, and their horizontal distance from each other.

Answer. By the formula (P. 94.) SDO=370.59'; DS=400 yards; Do=520; and so=320 yards.

(Q) The same answer may be found without the formula. For, with the BDO and height Bo, find DB=521.0536, and DO=520; with the Pds and height Ps, find DP=401•1234, and Ds=400. Then with dp, DB and the _ PDB, find PB

PB2-BOPS so, the horizontal distance=320 yards. OF THE DIP, OR DEPRESSION OF THE HORIZON AT SEA. (R) The dip or depression of the ho

E 210 rizon at sea, is the angle contained between the horizon of the observer, and the far

CBO thest visible point on the surface of the sea.

For, if an observer whose eye is situated by a sextant, or Hadley's quadrant, and brings that object to the surface of the water at B, instead of the horizon DF, he evidently makes the altitude too great by the 2 FDB.

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320-1562 ; lastly, Vi

se

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*Traité de Trigonométrie par M. Cagnoli, Appendice, page 467; or, Prop. 7. Chap. XI. Book III. folļowing

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AC

AC

Now the Lady is a right angle, and the L ABD is likewise a right angle, therefore the LPdB= DAB. Hence, in the rightangled triangle ABD, there is given AB the radius of the sphere, AD=AC+DC, the radius of the sphere increased by the height of the eye above the surface of the sea, to find the , DAB, or quantity to be subtracted from the observed altitude of any ce

rad. X AB lestial object. Hence cos. LDAB = cos. LFDB=

AC + CD Where AC=AB is the radius of the sphere, and do the height of the observer's eye above the surface of the sea.

OR, (EC + CD) X DC=DB2 (Euclid 36 of III.), that is (EC X DC)+ DC®=DB%; or (2AC X DC) + DC? =DB?, and rejecting DC? as being small w 2AC X DC = DB, but AB = Ac : rad. :: DB = 72AC X DC : tangt. DAB=tangt. Z FDB, therefore rad. x 72AC

12 x rad.2 tangent FDB=

XV/DC =

XVDC. Now the first of these terms is a constant quantity, and if the diameter of the earth be 7964 miles, the logarithm of this quantity in feet will be 6.48915, hence log, tangt. Z FDB = 6:48915+ log. dc in feet.

OF THE PARALLAX OF THI (S) That part of the heavens

visi

2 in which a planet would appear,

H if viewed from the surface of the earth is called its apparent place : and the point in which

-:K it would be seen at the same instant from the centre of the BA

N earth, is called its true place, the difference between the true and Dl.. apparent place is called the parallax in altitude.

Let c be the centre of the earth, A the place of an observer on its surface, whose visible horizon is AB, true horizon cd, and zenith z. Let zird be a portion of a great circle in the heavens, and a the absolute place of any object in the visible horizon; join ce and produce it to F; then F is the true place of the object, and B its apparent place in the heavens; and the angle BEF=AEC is the parallax.

The parallax is the greatest when the object is in the horizon; for the angle AEc is greater than agc, hence the more elevated an object is (its distance from the earth's centre continuing the

CELESTIAL BODIES.

FA

same) the less is the parallax. When the object is in the zenith the parallax vanishes, for then ac and az are in the same straight line cz.

Since the apparent place (H) of a planet is more distant from the zenith (z) then the true place (1) it therefore follows, that the parallax in altitude must be added to the observed altitude, in order to obtain the place of a planet as seen from the earth's centre. The stars on account of their immense distance from the earth have no sensible parallax, and the sun's mean parallax is only 8“.6. The moon's greatest horizontal parallax is 61'.32". least 54'.4".

(T) The horizontal parallax being given, to find the parallax at any given altitude. In the right-angled triangle EAC

Radius : Éc::sine AÉC : AC And in the triangle GAC

GC-EC : sine Gac :: AC : sine AGC And by comparing these two proportions

radius : sine GAC :: sine AEC : sine AgĆ. But the sine of an arc is equal to the sine of its supplement, therefore sine GAC=sine GAK, and GAK is the complement of GAE, therefore sine of GAK=cosine GAE, hence

radius : cosine GAE :: sine AEC : sine AGC. The last two terms being small, the arcs may be substituted for their sines without sensible error.

Hence the following rule:
Radius
: Cosine of the apparent altitude,
:: The horizontal parallax in seconds,
: The parallax in altitude in seconds.

EXAMPLE LIII. The apparent altitude of the moon's centre is 24° 29'.44". the horizontal parallax 55'.2". Required the parallax in altitude. Radius, sine of 90°

10•
: cosine moon's altitude 240.29'.44". 9.95904

: horizontal parallax 55.2"=3302"log.= 3.51878

: parallax in altitude 3005" log.= 3.47782 Hence the parallax in altitude is 3005"=50'.5". OF THE ADMEASUREMENT OF ALTITUDES BY THE BARO

METER AND THERMOMETER. (U) One of the most simple and easy practical rules for measuring the elevations and depressions of objects by means of

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