SOUTHERN CALIFORNIA ACADEMY OF SCIENCES 175 
This method does not require the use of any instrument or 
mechanical device except a pair of compasses and a straight 
edge. 
THEOREM. 
The side of the square is the mean proportional between the 
‘are of the quadrant and the diameter. 
DEMONSTRATION. 
When the radius is R. The area of the circle—=R*°P and the 
side of the square—R/y P. 
2RP RP 
The are of the quadrant— —— — and 
Aa 2 
Are of 90° Side Side Diameter 
RP 
— Tits e ———e rea\y/alee 2k 
) 
Ie 
or — WIP VP 2 
2 
iedale Cc 
The Length of Jupiter’s Day. 
Jupiter’s axial rotation period according to Neweomb is 9 
hours 55 min. 37 see. (See Eneyclopedia American). 
Assuming that 92,996,000 miles is the true value of the 
astronomical unit. 
The product of Jupiter’s sideral period mass and mean dia- 
meter is the length of his day. viz: 
d. sideral 
d. mass. diam. 
1 86552 
4332.58 & —— « —— X 1.0027379 = .413623668 = 
1048 867450 
li m. see. 
9 By’) 37.08. sideral time. 
Observe that in the foregoing equation the mass and mean 
diameter of Jupiter are expressed in solar terms. 
Also that to reduce a time interval in mean solar time to 
sideral time it is necessary to multiply by 1.0027379. 
. R. H. CHITTENDEN. 
Urbita, Oct. 18, 1906. 
