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19ʻ.20"1-141=5.20") is the error of the instrument, or the quantity by which the Lof elevation was too small, or the cof depression too large.

THE NATURE OF TERRESTRIAL REFRACTION AND ITS EFFECTS

ON ANGLES OF ELEVATION.* (S) As terrestrial refraction arises from the gross vapours, and exhalations of various kinds, which are suspended in the air near the surface of the earth, and which are perpetually changing, it is very difficult to ascertain the exact quantity of it at any particular time.

(T) The course of a ray of light in its passage through the atmosphere is, in general, that of a curve which is concave towards the earth, and the observer views the object in the direction of a tangent to this curve; hence the apparent, or observed angle of elevation is always greater than the true angle.

(U) The altitudes of the heavenly bodies when within 50 or 6° of the horizon, should never be used where a very accurate result is required. The figures of the sun and moon, when near the horizon, are sometimes elliptical, having the minor axis perpendicular to the horizon, and the major axis parallel to the horizon. This change of figure arises from the refraction of the under limb being greater than that of the upper. But a perpendicular object, situated on the surface of the earth, will not have its length altered by refraction, the refraction of the bottom being the same as that of the top.

(W) The allowances usually made for refraction are too uncertain for any reliance to be placed on them, as scarcely two writers agree on this subject., Dr. Maskelyne makes it to of the intermediate arc Ps between the observer and the object; Bouguer }; Legendre t'; General Roy from į to it; and in the second volume of the Trigonometrical Survey, the variation is found to be from } to ste of the intermediate arc.t This difference does not arise from inaccuracy of observation, but from circumstances which cannot be avoided, as the evaporation of rains, dews, &c. which produce variable and partial refractions.

(X) The following method is used in the Trigonometrical Survey & for ascertaining the quantity of refraction.

Let c be the centre of the earth, P and s two stations on its

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* See a paper by Mr. Huddart, in the Philosopbical Transactions for 1797, page 29 ; and another by the Rev. S. Vince, 1799, page 13. . Also the Trigonometrical Survey of England and Wales, Vol. I. page 175. † Page 177 and 178, Part I.

Volume I. page

175.

surface; PB, SA the horizontal lines at right angles to ca and CB; also suppose A and B to be the true places of the objects observed, and a and 6 their apparent places. Then the brs will be the refraction at P, and the Lasp that at s.*

In the quadrilateral figure csop, the angles at Pand s are right angles, therefore the sop+ Zc=two right angles, but the three angles of the triangle sop=two right angles, hence _ OSP + SOP+ LOPSS 2 SOP+ LC, consequently Żosp+ LOPS= LC, which is measured by the intermediate arc ps.

Now (L ASP + LBPS)-(_ Asa + 2 Bpb)=(LOSP + LOPS) -( asa + LBPb) = 2C-(Asa + LBPb)= LasP+ 2bPS the sum of both the refractions. Hence the following

(Y) RULE. Subtract the sum of the two depressions from the contained arc, and half the remainder is the mean refraction.

(Z) If one of the objects (a) instead of being depressed, be elevated, suppose to the point e, the Lof elevation being e sa; then the sum of the angles msP, MPs will be greater than ZOP8+ LOSP (the 2 c, or contained arc Ps) by the Lof elevation e sa. Hence Zc+ Lesa= L MSP + LMPs, from each of these equals take the BPb, then Lc+ LesA-LBPb= L.msp+ LPs the sum of the two refractions; that is, subtract the depression from the sum of the contained arc and elevation, and half the remainder is the mean refraction. Perhaps it may be necessary to remark, that previous to the observations the error of the instrument must be accounted for. (Q. 85.)

EXAMPLE XLVIII. (A) The refraction between Dover castle and Calais church was thus determined.+

Let c be the centre of the earth, Ps the surface; p the station on Dover castle; A the top of the great balustrade of Calais steeple; EDF the horizontal line; also let PG=SD; then the 4 FDG=} Lc, or half the arc PS (Q. 86.). The distance from Dover to Calais is 137455 feet, hence 364950 feet : 60':: 137455 feet: 22.35" the Lc, hence Z FDG=11.17").

The height of D above low-water spring tides=469 feet. The height of A (communicated from France)=140$ feet.

AG=3281

In observing these angles two instruments are used, one at pand another at s; and the reciprocal observations are made at the same instant of time by means of signals, or by watches previously regulated for that purpose. The observer at p takes the depression of s, at the same moment which the observer at s takes the depression of P.

+ Trigonometrical Survey, Vol. I. page 178.

The triangle DGA may be considered as isosceles, and DG or Da=137455 feet, the distance between Dover and Calais. Hence $ AG : rad. :: DA : secant 2 DAG = 899.55'.52".2, the double of which, deducted from 180°, leaves 8'.15" for the LGDA, to which add the FD=11'.171", and the whole angle FDA=19ʻ.321" supposing there was no refraction; but the

LFDA was determined from observation to be 17.59", hence the refraction was (19ʻ.32}" – 17'.59"=) 1'33}", being about Is of the contained arc.

(B) Mr. Huddart is of opinion, that a true correction for the effect of terrestrial refraction cannot be obtained by taking any part of the contained arc *; for different points, though nearly at the same distance from the observer, will have various refractions.

OF THE REDUCTION OF ANGLES TO THE CENTRE OF THE

STATION. (C) In surveys of kingdoms and counties, where signals on the steeples of churches, vanes of spires, &c. are used for points of observation, the instrument cannot be placed exactly at the centre of the signal, and consequently the angle observed will be different from that which would have been found at the centre. The correction is generally very small, and is only necessary where great accuracy is required.

The observer may be considered in three different positions with respect to the centre, viz, he is either in a line with the centre and one of the objects, or a line drawn from the centre through his situation would, if produced, pass between the objects; or a line drawn from the centre to the place of the observer, when produced, would pass without the objects.

(D) First, let the observer be at D, in a line between the objects B and c, viz. on one side of the triangle ABC; B being the proper centre of the station and . the ABC that required. It is plain that the 2 cda, being the

D exterior 2 of the triangle ADB, is too large by the interior DAB. Therefore, ABC=LCDA - LDAB.

(E) SECONDLY. Let the observer be at o, within the triangle ABC, and let B be the centre of the station, and the L ABC that required. Now Laoc+ LOAC+ LOCA= L ABC+

0

А A

OAB+ 20AC+ LOCA + LOCB, each of the sums being equal to two right angles; therefore the LAOC ZABC + LOAB+ 4 OCB, that is, the Laoc is greater than the 2 ABC by the sum of the angles OAB and OCB. Therefore 2 ABC* = LAOC-(LOAB + LOCB).

(F) THIRDLY. Let the observer be at E, without the triangle ABC, and let a be the centre of the station, and the CAB that required.

Now , CEB + LECA + LACO + LOCB + LEBO + LOBC=

CAB+ LACO + LOCB + LABE + LEBO + LOBC, each of the sums being equal to two right angles; therefore

L CAB+ Z ABES LCEB+ L ECA, and consequently
LCABL CEB + LECA - LABE.

EXAMPLE XLIX. Let A and B represent the vanes on two steeples; E the situation of the theodolite upon the steeple A, and o its situation upon the steeple B. Then, suppose AE=12 feet

BÓ=10 5 feet
LCEB=740.32

Laoc=49o.27
LCEA=1390.39

LCOB=1379.55'.
It is required to find the angles CAB and ABC, the distance
AB being 5000 feet.

SOLUTION. The CEB= L CAB nearly, and ZAOC=LABC nearly, with these angles and AB, find ac and BC (as in Example I. Chap, III.) = 4581.8 and 5811.6. Then,

AC : sin. LCEA:: AE : sin. Z ECA=5.50"

AB : sin. <BEA:: AE : sin. LABE=7'.29". Hence 4 CAB=74°.30'21".(F. 90.)

BC • sin. 2 COB::BO: sin. LOCB=4.10".

AB : sin. LAOB::BO: sin. 20AB=0?.56". Hence L ABC=490.21'.54" (E. 89.). With the corrected angles CAB, ABC, and the distance AB, the sides ac and Bc may be determined.

OF THE REDUCTION OF ANGLES FROM ONE PLANE TO

ANOTHER.

(G) Angles which are inclined to the horizon, may be reduced to the corresponding horizontal angles, in cases where very great accuracy is required. Let the lines Ps, PB, Bs be tbree chords of terrestrial arcs, that is, let the points P, B, and s, be all equally distant from the centre of the earth, and

* If the proper centre of the station were at 0, and the observer at B, it is plain. that the angles OAB and oce must be added to the ABC to obtain the LAQC.

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let the point d be elevated so as to be farther from the centre of the earth than any of the points P, B, s, by the quantity sd, it is required to reduce the triangle BDP to the triangle BSP.

(H) The line so may be supposed to be perpendicular to each of the chords so and sp without sensible error, though strictly speaking the angles DSB and dsp are each equal to 90° + the arc which the chords so and sp subtend (Q. 86. and Note.)

Likewise the chords SB and sp may be used instead of their corresponding arcs. (G. 83.) By inspection of the figure it is plain that BD is greater than es, and PD greater than Ps; but the base PB is common to the two triangles

Б. BDP and BSP, therefore the / BSP is greater than the / BDP. (Eu

S CLID 21 of I.)

PE (1) Now if a perpendicular be drawn from d upon the

BD x cos. DBP, base PB, the segment GB=

rad.

(K. 39.); and if a perpendicular be drawn from the point s upon the base PB, the segment intercepted between us and the perpen dicular will be =

(K. 39.):

rad. (K) Hencé cos. DBP X BD = cos. SBÚ X BS, and cos. sbr=

y also BD : rad. :: BS ; cos. DBŞ, for the tri, angles Bsb and dsp are right angled at s.

COS. DBP x rad. Conseq. cos. SBPS

s or cos. DBS : COS. DBP:: rad. : cos. SBP.

(L) That is, The cosine of the 2 of elevation : the cosine of the L inclined to the horizon :: radius : cosine of the hori, zontal L.

Exactly in the same manner the ZSPB may be found.
Then 180°-(SBP+ SPB)= L BSP.

(M) But the Bpr taken on the elevated situation d, may be reduced to its corresponding 2 BSP, by using only the ob

BS X COS. SBP

BD cos. DBP X

BS

COS. DBS.

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