Captain Owen on Circummeridian Altitudes. 177 
demonstrated by the fluxional differentials of the original equation. 
Thus dt/—at’=0o, where their difference A x is a maximum, then 
d=2at, and i=i—are TA™=the equation of the apex =«. 
Now to find the excess of the greatest altitude above the me- 
ridian altitude = A zx in the diagram, let us substitute = for ¢ in the 
a? a = ds 
Sm . 2 
original equation, then 5,— },=Awv=r=5=—5—=Q2; fore=S. 
This expression reduced gives “=r=Ar=Q2 =f = *, We 
thus obtain the following rules, viz. 
1. To find the equation of the apex, or its distance in time from 
the meridian = ¢ = 4: — 
Divide the motion in declination in (1™) one minute of time by 
the proper motion in altitude for one minute from the meridian, both 
motions being of the same denomination; the quotient will be the 
equation of the apex in minutes of time = «='T™ A, being the 
difference of time between the passage of the object over the me- 
ridian, and of its attaining its greatest altitude. 
2. To find the excess (~r) of the greatest altitude above the 
meridian altitude (= Az) =Q 2=£=%:— 
Divide the square of (d’) the motion in declination in one min- 
ute of time by four times the proper motion in altitude (4a) in 
(1™) one minute from the meridian, the quotient will be (r) the 
apicial excess of altitude (= Az = dé) above the meridian altitude, 
of the same denomination as the motions (a and d). 
In the foregoing analysis, it was assumed that the object was 
approaching the zenith of the observer. Now let us examine and 
analyze the motions by reference to the same diagram, when the 
object is receding from the zenith, d=9”; Q=1”.5 as before. 
Now suppose the (moon) object to enter on her track at o in 
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