512 Lieut. Col. Lambton’s abstract of the results deduced 
arc on the meridian extending from the equator to the lati- 
tude of the point F ; where A denotes the arc of latitude in 
parts of the radius 1. 
Let F' be any other point on the meridian, whose arc of 
latitude is A'. Then AF = a A'— e ^ A'. Sin. 2 A) and 
therefore F F'=tf(A'— ■ A)-— + 4- Sin. 2 A' — Sin. 2 A. j 
Let F", F", be any other two points on the meridian whose 
respective arcs of latitude are A" and A'". Then from the 
same reasoning as above, we have F" a (A w — A")~ 
— e 4 - |. Sin. 2 A'" — f Sin. 2 A"} 
Now FF' and F" F" are here supposed to be measured arcs 
on the meridian, whose respective lengths in fathoms may be 
called L and L', corresponding with the celestial arcs A'— A, 
and A'" — A To shorten the operation, put A' — A = r; 
A'" - A" = r\ Also + 4 Sin. A'— £ . Sin. A = S, and 
4. A . Sin. 2 A'" — \ . Sin. 2 A" = S'. Then we have 
L = ar — es ; L f — ar 1 — • es\ And therefore a — ; — . 
rs‘ — r's ’ 
r' L — =. r L' ■> e r ' L — r L' , ,, 
— e = rs , - 5 an d - = - s , L __- s L i equal the compression 
expressed in fractional parts of the semi-equatorial diameter. 
To apply this to the case in question, 
Let A = the latitude of Punnae =8 9 38,4 
A = the lat. of Daumergidda= 18 3 23,6 
A' — • A = r - - g 53 45,2 =,1727158 
A f/ = lat. of Montjouy - 41 21 44,96 
A'''=dat. of Dunkirk - 51 02 09,2 
A"' —- A" = / 
g 40 24,24 =,1688327 
