178 Captain Owen on Circummeridian Altitudes. 
the diagram, moving thence to T, she will, by the hypothesis, have 
an altitude equal to the meridian altitude, now at ¢, from which she 
is distant T™ ¢ (=6™) and is rising by her proper motion in alti- 
tude (~Q?#*); at 5™ she will have to rise 37”.5 by her proper 
motion in altitude, and from T she will have to rise 54” (= 6’ X 
1”.5) to her meridian altitude at ¢, the difference 54” —37”.5 = 
16”.5, but she has fallen by her motion 9”, and 16”.5—9= 
7”.5, or she will have risen 7”.5 in one minute above her me- 
ridian altitude at T or ¢ In like manner, at four minutes from 
¢ she will have risen from T 54” — 24” = 30” by her proper mo- 
tion in altitude, and in two minutes will have fallen by her motion 
in declination, now receding, 18”, the difference = 12” higher than 
the meridian altitude at T and ¢. 
Again, at 3" (the apex in this example), 54” — 13”.5= 40.5 
the quantity she has risen by her proper motion in altitude from 
T in 3 minutes, and she will in the same time have fallen by her 
motion in declination 3 X 9” = 27, the difference 40”.5 — 27” = 
13”.5, her excess of altitude above the meridian altitude at 3 minutes 
before her meridian passage (in this example the maximum excess). 
At 2™ from ¢, the point of meridian transit, she will have risen 
from T 54” — 6” = 48", and have fallen by her motion in declina- 
tion 4 X 9 = 36”, and 48” — 36” = 12”; so that at 2 minutes 
before her meridian passage, her altitude will be 12” higher than 
at the meridian. Again, at 1™ before meridian passage 54” — 1”.5 
= 52”.5 —45 =7".5, and at ¢ the meridian 54” — 0” — 54” =0, 
or the altitude at ¢ the meridian is equal to the altitude at T (6™) 
six minutes before she came to the meridian. In the first case 
of a motion in declination advancing to the zenith, she came to 
the meridian before she attained her greatest altitude (3") three 
