176 Captain Owen on Circummeridian Altitudes. 
4" she would have fallen 24” and risen 36”, difference 12”; at 5™ 
she would have fallen 37”.5 and have risen 45”, difference 7.5; 
and at 6™ have fallen 54”, and have risen 54”, difference 0; there- 
fore at T on the meridian, and at six minutes after, the altitude 
was the meridian altitude, but at A, the apex, at 3™ after the me- 
ridian passage, the moon was 13”.5 higher than when on the merid- 
ian, and which was her greatest altitude. 
d. Let the rate of motion in declination be expressed by d, pre- 
fix to it the sign + if the object be advancing towards the observer, 
or rather towards his zenith, by its motion in declination ; but pre- 
fix to d the sign — if the object be, by its motion in declination, 
receding from the observer’s zenith, and note d in seconds of are, 
equal to the amount of motion in declination in (1™) one minute 
of time, at the observer’s place and time. 
a. Let a express the number of seconds of arc, of the proper 
motion of the object in altitude (supposing it to have no motion 
in declination) in (1™) one minute of time from the meridian, to 
which prefix the contrary sign to that of d© (¢™). Let ¢ express 
the minutes of time corresponding to any arc in question. 
Now at T and at ¢ in the diagram, the object observed is shown 
to have equal altitudes, equal to the meridian altitude. To find 
the value of the are T’™=d#, we have seen by analyzing the dia- 
gram, and by hypothesis, that the proper motion in altitude at f= 
—at’, and that the excess of motion in declination at the same 
point ¢= + dl; and dt—at’=o. 
“t=, and if d=9” as in the diagram, and a=1”.5, then (= 
are T¢™ —6" ; now, although by the foregoing analysis it may 
be seen that the apex A exactly bisects the are Té, this may be 
