Captain Owen on Circummeridian Altitudes. 179 
minutes ; and the excess of her greatest altitude above her me- 
ridian altitude was 13”.5, and she rose through the meridian. 
In this case of a receding motion in declination, she falls through 
the meridian, and comes to it after she has attained her greatest 
altitude ; in this case, also, the apicial excess of altitude (~r) is 13”.5. 
So that in all cases where the object has any motion in dec- 
lination, its greatest altitude is always higher than the meridian 
altitude, and the excess (~r) and the equation (= e) of the apex, 
are the same in quantity in both cases, and found by like process. 
And r is always — to the greatest altitude. But « is + to the 
time of meridian passage, when the motion in declination is to- 
wards the zenith, and — when receding. 
The foregoing development unfolds to us another very pretty 
and equally important practical principle. Whether the object by 
its change in declination move to or from the zenith, or the ze- 
nith move to or from the celestial object, the relative effect is the 
same, and the greatest altitude near the meridian would be in 
excess, or greater than the meridian altitude. 
Exampie.— A steamer going 13 knots, steering due south, or 
directly towards the moon, whose motion towards the zenith is 
17” per minute (a case that may well happen), the sum of these 
two motions would be 30”, and suppose her proper motion in al- 
titude, when motionless in declination, was 1”.5 in one minute from 
the meridian, then her equation of the apex would be 10” (=e 
= sa7;) and she would get her greatest altitude ten minutes after 
the calculated time of the moon’s transit, and the excess (ae= 
150° = 5 = F= 150" = 4 = MH 150”) of her greatest above 
her true meridian altitude would be (2.5) two minutes and a half. 
Had the steamer been steering north, and the moon’s motion 
in declination receding from the zenith, the effect would have been 
