511 
is the arch of the nautical meridian. But fince the me¬ 
ridional parts in the tables are expreffed in minutes, there. 
fore 2= .2—1—--—x io g- cot - i co • tat - — %• rad ' 
3 '> 4>59 
cot. 4- co. lat. , 
— 79 1 S> 7°5 X >og. -- = 79 I 5 > 7°5 X log. 
radius 
FLUXIONS. 
rat x 1 + ft 
rt 
rcct — r*t 
•. Putrr- 
rr + att y/ i-{-tt \/ 1 -f tt rr+att f 1 ftt 
rt rfft 
ccz=.ff,a^L cofect lat.then Z=- 
V 1 -{-tt rr-^cctty/ 1 +tt 
tan. -J- co. lat. 
Cor. 1. Hence the meridional parts for the difference 
of latitude of two places is = 7915,7 X difference of the 
log. tangents of half their complements of latitude. 
Cor. 2; Since radius : meridional difference of lati¬ 
tude :: fo tangent of the courfe : to the difference of 
longitude: and--— = ,000126331, therefore, As 
■ 7915-705 
,000126331 : to tangent of the courfe :: fo the diff. of 
the log. tangents of half the comp, of the lat. : 10 the 
difference of longitude. 
dtherwife thus.—Let t— tang, of the latitude AB, 
DC=x ; then by the nature of the circle yy=.rr—xx and 
rff X 
—2 t —2 
cc-J- rt y/ 1 -ft 
-r-Ti—i.—rff x cc — rr+rrv* x 
-— — V — -;whofe flu.isZ—2_ t £ 1 l. 
— 2 2 
rr — cc — r) v 
2 ry/rrc—c 
rf-v-\-y/rr—cc 
X log. - - = z—• if L X log. 
ry/v —3/ rr—cc 
r<r+f 
7 rr , 
r V 1 4 ~ —~f 
= 2 ~if L X log. 
ra—f 
, becaufe by trigonometry 
yy~— xx, and y 
But by 
y 
trigonometry, y y/i+~ or =<r . I n minutes Z =2—7915.7 
.: rad. (1) : t—-, and y—tx. Likewife 
r ■; «"• v = —. And by the nature of Mercator’s 
J x 
-rrx _ , . 
■-. But (V 
• fry \ . rrj. 
projection, x : r :: v J : z — ~ 
rzzyy—rr—xx^ whence x~ 
.+1 _ 
, , r r-. f \I rT - CC 
Xlog. \J - • Or putting - or -2 - —d, then 
J <i r tr 
cr - 
r 
xy 
, and y — 
rt 
will Z=2- 
7915.7^ 
—rtt , rrt 
, and xy— 
1 +('\§ 
.*+«— Tt 
rrt 
1 /i+tt 
therefore z— 
and 
3/1 -{-tt 
— rr X —rtt 
X log. 
■ + d 
Where 
natural 
y/ 1 
I - j" tt 1 4 tt\st 
Therefore, z = 2.30258r x log. t + 
fi _j_i« : but if t is the tangent, f 1 fit is the fe- 
cant ; and by the Elements of Trigonometry, fecant + 
tangent = cotan. of half the complement of AB,—There¬ 
fore z = 2.30258c x log. cotangent of half the comple¬ 
ment of the latitude AB. 
133. To find the meridional parts for the oblate 
fpheroid.—Let the femitranfverfe CA —r, femiconjugate 
CG=c, arch AB-r, ordinate DB —y, CD=«, meridional 
parts of AB, put =Z ; and l— tan. of the lat. at B, as 
before. 
Then by the nature of the figure ccrr — ccxx=.rryy, and 
But by trigonometry, y : —x ^ ^ 
cofecant of the latitude to the radius 1, and 2= the me¬ 
ridional parts for the fphere. 
a + d 1 -\-dy 
a — d 1 —dy 
therefore Z=2—7915.7^ x log. y/l—^Z, Butbytrigo- 
Or 
thus.—Since _-= 
ST 
y* 
therefore 
metry, f 
1 4 - dy. 
1 —dy 
cotan. | arch whofe natural cofine is 
— ccxx = rryy. 
:: rad. (1) : /= 
rry 
ccx ' 
And, by Mercator’s Projection, 
» : r:: v : Z = —. And fince ccxt=rry, we (hall have 
x 
rr cct , . rrcct 
and y— ■ 
y/ rrfeett 
rreett 
rr-J-a/fll 
rrcct f 1 + tt 
; but v— y/ x 2 +y 2 
y/ rr-{-cctt y/ rr+cctt (2 
r A c 4 t 2 t 2 -f- rV/* 
rr+cc //) 3 
rr + cctt yf rr + cctt 
- ; therefore Z 
( 5 ) - 
r 3 cct yf 1 4. u X y/ rr 4- cctt rcct f i -{- tt 
rr 4* cctt y/ rr 4- ecti X rr 
rr 4- CC M 
or Z = 
dy. Therefore let y— nat. S. lat. A = arch whofe nat„ 
cofine is dy , to the rad. 1 ; and B= log. cotan. |A—-io, 
as had in the /ables ; then Z=2—7915.7dB. 
In the fpheroidal figure of the earth, we fhall have 
<£=.093, or thereabouts, according to fir I. Newton ; or 
d—. 148, according to Maupertuis. 
Ex. 12. To find the nature of the curve which a heavy 
flexible line will form itfelf into by its gravity. 
134. Let the line be fufpended on the two fixed points 
D, P, and difpofe itfelf into the curve DAP 5 A its ver¬ 
tex, AQ its axis, and BC an 
ordinate. 
1. The part of the curve 
ABD is kept in its pofition by 
a certain force at A aCting in 
direction AZ parallel to the 
horizon for if the line be 
cut through at A it will re¬ 
duce itfelf to a perpendicular 
pofition. And t^tis force act¬ 
ing at A is always the fame whatever length the curve 
be of; for if the line be cut through at B, and then the 
point B faftened to the plane ; it is evident the force at 
A is neither greater nor lefs; for the relifiance of the 
point B does the fame as the tenfion of the line in B did 
before 5 and the force in A or the tenfion of fhe line in 
A muff remain the fame. 
2. Let a— the given part of the line whofe weight is 
equal to the tenfion of the line in A, and-AC=*, BCry, 
2 AB=2, 
Si s* 
