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by the diurnal motion to E, where she would riseabout 50^ minutes later than on the former day (andabout 13° in advance of her former place). But heractual motion is nearly on the ecliptic; and when shewas at E on the first day the ecliptic must have beenJ n one of the two positions e E or e' E. (In otherwords, E must be the point where the ecliptic crossesthe equator, either descendingly or ascendingly.)*N.ow in the former case, the moon on the secondnight will be as at M, and will be carried by thediurnal motion to the point h on the horizon; in thelatter she will be as at M', and will be carried tothe point h'; and obviously M h is a much longer arcthan M' hf. In fact, if K E K' be part of the equinoctialcolure (or circle square to the equator through theequinoctial point E), the two arcs MK and M'K' areobviously equal,f and we see that M h exceeds, whileM' li falls short of the common length of those equalarcs by the very appreciable equal arcs K h and K' h’.Thus the hour of rising in the former case will belater than in the latter, by the time corresponding totwice the diurnal arc K h or K' li, as well as by a notinconsiderable increment of time due to the fact thatthe moon is all the while moving on her orbit, andmoves farther, of course, the longer she is delayed.The hour of rising will in both cases be later than the
* The direction in which we follow the ecliptic is contrary tothat of the diurnal motion, because the sun’s annual motion in theecliptic is from west to east.
f They are also each very nearly equal to E m.