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Doc. 200.] 573
REPLY TO CELSIUS.
22. These 3° 7’ are to be as 36° 44’ the side now found
subtracted from the south is to 29° 17’ the complement
angle 153° 50’ ; hence 1530 of the apparent south angle,
50—3° 7’ = 150° 43’, which is 80 26° 37’ the apparent side
the apparent (south] angle of the calculus, to that which
situated at the west; of which is required.
the complement is 290 17’. Sine 26° 37’= 965129
23. For (n c) Sine 29° 17’= 968942
Sine 59° 50’=993679 1934071.
Sine 29° 17’=968942 Sine 36° 44 ’ = 977676
Sine 250 1’=962621. Sine 21° 30’= 956395.
24. For ( n h ) 27. 75° 35’ 24 ° 53’
Sine compl. 29° 17’= 994062 143° 14’ 21° 31’
Tang. compl. 59° 50’= 979435 218° 49’ 46 °23’.
Tangent 54 ° 29’=1014627. From this arises the pro
25. For the side (gc) sub- portion:
tract 218° 49’ : 46 ° 23’ :: 75 ° 35 ’: 16 ° 1’.
54 ° 29’—26° 37’=27° 52’. 28. If 16° 1’ be subtracted
Sine compl. 24° 58’= 995739 from the angle 24° 53’, there
Sine compl. 27° 52’= 994647 remains the declination west,
Sine compl. 36° 44’ = 1990386 . go 52’.
If its complement is taken, Between the declination ob
the required side (c g ) is ob- served by Professor Celsius
tained = 143 ° 14 ’. 8° 49’ and that obtained by
26. For the angle (g c h) computation 8° 52’, there is a
the proportion must be used : difference of 3 minutes.
>
It hence appears plainly that my computed declination
agrees so thoroughly with the Professor’s accurately made ob
servation which he has communicated to us, that scarcely one
minute, much less 8 degrees 13 minutes or 493, minutes,
separates us.
The Professor’s observation, as may be seen in the Trans
actions of the Academy, p. 387, is as follows:
In 1740, on July 28, I observed at Upsal the
declination of the needle towards the west from
its northern direction . 8° 53’.
After the needle had been drawn from its
place by a key, it stood at 8° 50’.
. . .
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