After the same method, from five observed latitudes, we may find the latitude to a given time.
If the differences of the observed longitudes are small, suppose of 4 or 5 degrees, three or four observations will be sufficient to find a new longitude and latitude; but if the differences are greater, as of 10 or 20 degrees, five observations ought to be used.
LEMMA VII.
- Through a given point P to draw a right line BC, whose parts PB, PC, cut off by two right lines AB, AC, given in position, may be one to the other in a given proportion.
From the given point P suppose any right line PD to be drawn to either of the right lines given, as AB; and produce the same towards AC, the other given right line, as far as E, so as PE may be to PD in the given proportion. Let EC be parallel to AD. Draw CPB, and PC will be to PB as PE to PD. Q.E.F.
LEMMA VIII.
- Let ABC be a parabola, having its focus in S. By the chord AC bisected in I cut off the segment ABCI, whose diameter is Iμ and vertex μ. In Iμ produced take μO equal to one half of Iμ. Join OS, and produce it to ξ, so as Sξ may be equal to 2SO. Now, supposing a comet to revolve in the arc CBA, draw ξB, cutting AC in E; I say, the point E will cut off from the chord AC the segment AE, nearly proportional to the time.
For if we join EO, cutting the parabolic arc ABC in Y, and draw μX touching the same arc in the vertex μ, and meeting EO in X, the curvilinear area AEXμA will be to the curvilinear area ACYμA as AE to AC; and, therefore, since the triangle ASE is to the triangle ASC in the same proportion, the whole area ASEXμA will be to the whole area ASCYμA as
