SURVEY OF MASSACHUSETTS. 51 
(2.) A = 90° 25' 44.58 
d= 446 .54 
5=90 26 31 .12 
3 = 89 46 35 .03 
SF = 89 56 .09 
(4) ° = = 19" 58.05 tan = 7.7640366 
K= 94.402.'74 log. = 4.9749846 
H= 548.31 log. = 2.7390212 
A = 89° 45! 35.84 
5 = 89 46 35 .08 
§=90 26 31 .12 
S40 —180°= 18 06.16 
(aot 8. - 1607 a 967 89°06 
* 
= 7 46 11 
10 C 
= = .078 C= 1’ 18’.08 
me 
“The difference of level, therefore, between the tops of the signals at Blue Hill and 
Nahant stations being 548.31 ft., and the signal on Blue Hill being three feet higher 
above the bolt than that in Nahant, this quantity must be deducted from H, and the 
remainder, = 548.31 — 3 = 545.31 ft., will be the elevation of the bolt at Blue Hill 
above the bolt at Nahant. 
“The bolt at Nahant has been ascertained, by levelling, to be 89.83 ft. above the mean 
level of the sea; therefore 89.83 + 545.31 = 635.14 ft. will be the height of Blue Hill 
station above the sea by this observation. 
“Where but one observation has been made, the following formula was applied; namely, 
H= K cot (A +7 — 4 )i in which H = difference of level between the telescope at the 
place of observation, and the top of the signal observed. The difference of level of the 
bolts is obtained by applying a correction to H, equal to the difference between the 
* To determine the value of C, we divide K by the value of a second deduced from the mean radius of the earth as 
determined by Mr. Ivory, its log == 2.005465]; thus: 
K se log. 4.9749846 
1” = log. 2.0054651 
et 
C = 982,22 = 2,9695195 
