DEPRESSION. 
The following method is i by Delambre for finding 
the quantity of terreftrial r ion. 
LetCbe thec eat of tie cath (Pl IX. oe fs. )s 
A and B an ee f from the t A, we obferve 
the point B, it will aces « "Bi by ile effet of the res 
fraction ; the point - A will, when feen from B, for the fame 
reafon appear at 
et the apparent “zenith diftances 
=dandVBA'=7 
and the angles of vcfrabiion, 
BAB = 
ABAT=r 
The true zenith diftances will be 
cf AB=S34+r=D 
VBAS=e4r=D\. 
eka ais +VBA=34%'%+r+4+r. (1) 
And becavfe the fam of the exterior angles of a triangle 
is equal to the two interior and oppofite 
ZZAB=CH 
ABC 
VAB=C4+BAC, 
and ZAB+VBA = 180°+C=D+D’. (2) 
andd + +r+7r = 180°+C 
fince r is nearly equal +’ 
> | — §(d+3'— 180). 
os 
Co 
(3) 
=n. (4) 
pone r= C. 
ntity 7 varies extremely, according to the ftate 
of the atmofphere. In the trigonometrical furvey it was 
found to ve omitog,. .- 
The n may be taken J. or #= 0.08, 
above eauscas we obtain 
LZAB=3+r=90°+ EC + 40-98). 
Fa ee eee 1C-c4 £(3— 8). 
Example. 
Let § be equal to go° 15! 30". 
ld 
From the 
9 57 5% 
K the diftance between the two fignals = 93.522 
feet. _ 
Firft find 2 C - - Log. K = 4.97091 
—_ ; _ R= 5.31442 
Co-log. ¢ the mean radius of the earth ¢ = 2.680 
923.8 = 2.96533 
Angle C = 15’ 23".8 
I= go? 1s 30" 
. d’= 89 57 50 
3+2— 180= 0 13 20 
5 a 180 sea 
2 
Cc _ us 
rn 74 
g=a2C= 1,2 
Log. r (62”) = 1.79239 
Co-log. C = 7.03466 
= 0.067 = 882705 
To find the difference in the heights of bale — above 
fer 
the level of the fea, by obferving their dep 
C be the centre fy, the earth nei eer a a {phere, 
» B, two points unequally diftant from the centre. If AB’ 
a true level line or terreftrial are, B B/ = H will be ae 
diffrence of the level of the two peints A and B. If, m 
r, ZAB= e the true zenith diftance of he 
=90°—FC, 
+r 
poet B, then, fines! BAC= 
If w 
have very nea 
yaaa ae 
+ £C = go° 
AB y= 
$C=D— 
In the triangle BA B’ making A B’= 
a firs A Kiin.(90°+3C—D) _ Root Ge. D) 
BPAC = 18)? — D — go° 
oe. ore go° ~ $C —g90e° + D— 
Cc. 
os"is. B fin (D—C) in, (D—=C)  “™) 
ee a the angle B’ a right angle, then we fhall 
ny 
H=Keor (8+r—ZC) (2.) 
Relatively likewife : Aa level of the fo A. 
H cot. (3) +r —1C). 
Tf we employ the ee. ee 
it will be relatively to the point B. 
It has been demonftrated above, that 
ons (0 +r—iC) 
ZAB= 90° + 3043 0-2) 
VBA = 90° Be yore, 
therefore, B AC = 180 ~Z A B=90° — aan 
BPA C = go? 
BAB BAC. BAC=1(3 = 
BBA = 189°—VB = 90° — 4 dl ee vy 
aa & fa. BAB _ _K fin. 3(Y — 9) 
nee = Ge ABB ook i@—spcy (+) 
This le is exact, but we may often take 2C = 0, then, 
fince — = = tang. 
H = K tang. (cm 8). (5. 
Hen may obtain ang height of a ftation above the 
we 
level Of the fea, by obferving, when it is vifible, the hori- 
zon of the fea. For let A Bbe a tangent from the ob- 
ferver at Bto the feaat A. Let. CA = N. 
en, in the triangle C A B, 
és 
C8 =r = ae 
rw. fim ool. =); 
BB = N= ¢( cof. C 
_ but 1 — cof. C = fin. C tang. 2C, 
__- therefore, N = ¢ wig. C tang. 1C: 
Allo, C = a == go? eee eae D— 0°, and 
; +r ‘Therefore, N= = e tang. (2 + r — go") tang 4 
3 + e— 90°). When « is aicnaee, it may be deduc 
ut in practice it will be more convenient to- transform 
N into fome funétion of 2, which may be effe&ted thus. 
—-go e€ting the refra€tion, which may be 
done wou fenfible error, then 7 = nC = °); 
fubftituting this laf thee in equation, (6.) and recolleing 
that tang. mr = m tan when x is So and m 
does not much exceed unity, we fhall hav 
N = $e tang? (9 — 90° + 2)(3 — go°) = Le tang. (+2) 
(t — 90°)) and very nearly 
N= $¢ (1-2) tang.* (3 — 90°). 
-Example I. 
At the ftation A, ‘the zenith diftance of the point B was 
obferved 90° 15’ 30”; and at the ftation B, the zenith di- 
ftance of A was Coo ed 89° 57! 50”, the diftance from 
A to B 93,522 feet ; 3 required the height of A above the 
level of B. 
cof, Ta —3+ oF 
(7-) 
xf. By the exa& formula H = 
3 
