Now this Problem is not of that difficulty as appears at first sight, for in _Tab. 4. Fig. 3._ let the Cylinder ABCD be cut obliquely with the Ellipse BKDI, and by the Center thereof H, describe the Circle IKLM; I say, the Curve Surface IKLB is equal to the Rectangle of IK and BL, or of HK and 2 BL or BC: And if there be supposed another Circle, as NQPO, cutting the said Ellipse in the Points P, Q; draw PS, QR, parallel to the Cylinders Axe, till they meet with the aforesaid Circle IKLM in the Points R, S, and draw the Lines RTS, QVP bisected in T and V. I say again, that the Curve Surface RMSQDP is equal to the Rectangle of BL or MD and RS, or of 2 BL or AD and ST or VP; and the Curve Surface QNPD is equal to RS × MD----the Arch RMS × SP, or the Arch MS × 2 SP: Or it is equal to the Surface RMSQDP, substracting the Surface RMSQNP. So likewise the Curve Surface QBPO is equal to the Sum of the Surface RMSQDP, or RS × MD, and of the Surface RLSQOP, or the Arch LS × 2 SP. This is the most easily demonstrated from the Consideration, That the Cylindrick Surface IKLB is to the inscrib'd Spherical Surface IKLE, either in the whole, or in its Analogous Parts, as the tangent BL is to the Arch EL, and from the Demonstrations of _Archimedes de Sphæra & Cylindro, Lib. I. Prop._ XXX, and XXXVII, XXXIIX. which I shall not repeat here, but leave the Reader the pleasure of examining it himself; nor will it be amiss to consult Dr. _Barrow_'s Learned Lectures on that Book, Publish'd at _London_, _Anno 1684_, _viz._ _Probl._ IX. and the Corollaries thereof. Now to reduce our Case of the Sum of all the _Sines_ of the Sun's Altitude in a given Declination and Latitude to the aforesaid Problem, let us consider (_Tab. 4. Fig. 4._) which is the _Analemma_ projected on the Plain of the _Meridian_, Z the Zenith, P the Pole, HH the Horizon, ææ the Æquinoctial, ♋♋, ♑♑ the two Tropicks, ♋1 the _Sine_ of the Meridian Altitude in ♋; and equal thereto, but perpendicular to the Tropick, erect ♋I, and draw the Line TI intersecting the Horizon in T, and the Hour Circle of 6, in the Point 4, and 64 shall be equal to 6R, or to the Sine of the Altitude at 6: And the like for any other Point in the Tropick, erecting a Perpendicular thereat, terminated by the Line T I: Through the Point 4 draw the Line 4, 5, 7 parallel to the Tropick, and representing a Circle equal thereto; then shall the Tropick ♋♋ in _Fig. 4._ answer to the Circle NOPQ, in _Fig. 3._ the Circle 457 shall answer the Circle IKLM, T4I shall answer to the Elliptick Segment QIBKP, 6R or 64 shall answer to SP, and 5I to BL, and the Arch ♋T, to the Arch LS, being the semidiurnal Arch in that Latitude and Declination; the _Sine_ whereof, tho' not expressible in _Fig. 4._ must be conceived as Analogous to the Line TS or UP in _Fig. 3._