Captain Owen on Circummeridian Altitudes. 169 
Rue. — Convert the interval of time into degrees and decimals, 
and multiply it by the (natural) cosine of the declination; call the 
product 2°; and take the difference of the altitudes in degrees and 
decimals; call this Q°; add the log. of z‘-+Q to the log. of 2\—Q 
and halve the sum, to which add the log. secant of the middle 
altitude; this last sum is the log. of the corresponding arc in azi- 
muth in degrees and decimals. 
Cael: h, m. 8 
Exampce. First altitude 49.42.4 Time 1.20.00 P. M. 
Second altitude 49. 7.5 1.40.46 
34’.9 20.46 
= 0°.58 = Q. = 5°.2 
o1 
The declination at the Middle Time 11.23 
or Polar Distance 78.37 
° 
Now, the interval 5.2 x 0.98 Nat. cosine of declination 
89.0 
468 
42 
5.10 = 2\ 
° fe} 
2+ Q =5.1+4 0.58 = 5.68 log. 0.754 
2\— Q=5.1— 0.58 = 452 log. 0.655 
sum 1.409 
half 0.7045 
Middle altitude 49°.25/ secant 0.1867 
Are of azimuth 7°.784 log. 0.8912 
Corresponding to 20™- 46°. 
Interval of time = 7° 47/ 
If 2 be put for an are in azimuth and ¢ for its corresponding arc 
in time, and a= difference of altitude, 
t X cos. decl.” = z X cos. alt.°+ @ 
for small arcs with a mean altitude, which is Mr. Raper’s rule. 
32 
