u74Take AC (in these three figures) proportional to the gravity, and AK to the resistance; but take them on the same side of the point A, if the body is descending, otherwise on the contrary. Erect Ab, which make to DB as DB2 to 4BAC: and to the rectangular asymptotes CK, CH, describe the hyperbola bN; and, erecting KN perpendicular to CK, the area AbNK will be augmented or diminished in an arithmetical progression, while the forces CK are taken in a geometrical progression. I say, therefore, that the distance of the body from its greatest altitude is as the excess of the area AbNK above the area DET.

For since AK is as the resistance, that is, as AP2 × 2BAP; assume any given quantity Z, and put AK equal to ; then (by Lem.[Pg 284] II of this Book) the moment KL of AK will be equal to or , and the moment KLON of the area AbNK will be equal to or .

CASE 1. Now if the body ascends, and the gravity be as AB2 + BD2 BET being a circle, the line AC, which is proportional to the gravity, will be , and DP2 or AP2 + 2BAP + AB2 + BD2 will be AK × Z + AC × Z or CK × Z; and therefore the area DTV will be to the area DPQ as DT2 or DB2 to CK × Z.

CASE 2. If the body ascends, and the gravity be as AB2 - BD2, the line AC will be , and DT2 will be to DP2 as DF2 or DB2 to BP2 - BD2 or AP2 + 2BAP + AB2 - BD2, that is, to AK × Z + AC × Z or CK × Z. And therefore the area DTV will be to the area DPQ as DB2 to CK × Z.

CASE 3. And by the same reasoning, if the body descends, and therefore the gravity is as BD2 - AB2, and the line AC becomes equal to ; the area DTV will be to the area DPQ as DB2 to CK × Z: as above.

Since, therefore, these areas are always in this ratio, if for the area[Pg 285] DTV, by which the moment of the time, always equal to itself, is expressed, there be put any determinate rectangle, as BD × m, the area DPQ, that is, , will be to BD × m as CK × Z to BD2. And thence PQ × BD3 becomes equal to 2BD × m × CK × Z, and the moment KLON of the area AbNK, found before, becomes . From the area DET subduct its moment DTV or BD × m, and there will remain . Therefore the difference of the moments, that is, the moment of the difference of the areas, is equal to ; and therefore as the velocity AP; that is, as the moment of the space which the body describes in its ascent or descent. And therefore the difference of the areas, and that space, increasing or decreasing by proportional moments, and beginning together or vanishing together, are proportional. Q.E.D.

COR. If the length, which arises by applying the area DET to the line BD, be called M; and another length V be taken in that ratio to the length M, which the line DA has to the line DE; the space which a body, in a resisting medium, describes in its whole ascent or descent, will be to the space which a body, in a non-resisting medium, falling from rest, can describe in the same time, as the difference of the aforesaid areas to ; and therefore is given from the time given. For the space in a non-resisting medium is in a duplicate ratio of the time, or as V2; and, because BD and AB are given, as . This area is equal to the area and the moment of M is m; and therefore the moment of this area is . But this moment is to the moment of the difference of the aforesaid areas DET and AbNK, viz., to , as to , or as into DET to DAP; and, therefore, when the areas DET and DAP are least, in the ratio of equality. Therefore the area and the difference of the areas DET and AbNK, when all these areas are least, have equal moments; and are therefore equal. Therefore since the velocities, and therefore also the spaces in both mediums described together, in the beginning of the descent, or the end of the ascent, approach to equality, and [Pg 286]therefore are then one to another as the area , and the difference of the areas DET and AbNK; and moreover since the space, in a non-resisting medium, is perpetually as , and the space, in a resisting medium, is perpetually as the difference of the areas DET and AbNK; it necessarily follows, that the spaces, in both mediums, described in any equal times, are one to another as that area , and the difference of the areas DET and AbNK. Q.E.D.

SCHOLIUM.

[Pg 287]

The resistance of spherical bodies in fluids arises partly from the tenacity, partly from the attrition, and partly from the density of the medium. And that part of the resistance which arises from the density of the fluid is, as I said, in a duplicate ratio of the velocity; the other part, which arises from the tenacity of the fluid, is uniform, or as the moment of the time; and, therefore, we might now proceed to the motion of bodies, which are resisted partly by an uniform force, or in the ratio of the moments of the time, and partly in the duplicate ratio of the velocity. But it is sufficient to have cleared the way to this speculation in Prop. VIII and IX foregoing, and their Corollaries. For in those Propositions, instead of the uniform resistance made to an ascending body arising from its gravity, one may substitute the uniform resistance which arises from the tenacity of the medium, when the body moves by its vis insita alone; and when the body ascends in a right line, add this uniform resistance to the force of gravity, and subduct it when the body descends in a right line. One might also go on to the motion of bodies which are resisted in part uniformly, in part in the ratio of the velocity, and in part in the duplicate ratio of the same velocity. And I have opened a way to this in Prop. XIII and XIV foregoing, in which the uniform resistance arising from the tenacity of the medium may be substituted for the force of gravity, or be compounded with it as before. But I hasten to other things.

Of the circular motion of bodies in resisting mediums.

LEMMA III.

Let PQR be a spiral cutting all the radii SP, SQ, SR, &c., in equal angles. Draw the right line PT touching the spiral in any point P, and cutting the radius SQ in T; draw PO, QO perpendicular to the spiral, and meeting in O, and join SO. I say, that if the points P and Q approach and coincide, the angle PSO will become a right angle, and the ultimate ratio of the rectangle TQ × 2PS to PQ2 will be the ratio of equality.

For from the right angles OPQ, OQR, subduct the equal angles SPQ, SQR, and there will remain the equal angles OPS, OQS. Therefore a circle which passes through the points OSP will pass also through the point Q. Let the points P and Q coincide, and this circle will touch the spiral in the place of coincidence PQ, and will therefore cut the right line OP perpendicularly. Therefore OP will become a diameter of this circle, and the angle OSP, being in a semi-circle, becomes a right one. Q.E.D.

Draw QD, SE perpendicular to OP, and the ultimate ratios of the lines will be as follows: TQ to PD as TS or PS to PE, or 2PO to 2PS; and PD to PQ as PQ to 2PO; and, ex æquo perturbatè, to TQ to PQ as PQ to 2PS. Whence PQ2 becomes equal to TQ × 2PS. Q.E.D.

PROPOSITION XV. THEOREM XII.

If the density of a medium in each place thereof be reciprocally as the distance of the places from an immovable centre, and the centripetal force be in the duplicate ratio of the density; I say, that a body may revolve in a spiral which cuts all the radii drawn from that centre in a given angle.

Suppose every thing to be as in the foregoing Lemma, and produce SQ to V so that SV may be equal to SP. In any time let a body, in a resisting medium, describe the least arc PQ, and in double the time the least arc PR; and the decrements of those arcs arising from the resistance, or their differences from the arcs which would be described in a non-resisting medium in the same times, will be to each other as the squares of the times in which they are generated; therefore the decrement of the[Pg 288] arc PQ, is the fourth part of the decrement of the arc PR. Whence also if the area QSr be taken equal to the area PSQ, the decrement of the arc PQ will be equal to half the lineola Rr; and therefore the force of resistance and the centripetal force are to each other as the lineola and TQ which they generate in the same time. Because the centripetal force with which the body is urged in P is reciprocally as SP2, and (by Lem. X, Book I) the lineola TQ, which is generated by that force, is in a ratio compounded of the ratio of this force and the duplicate ratio of the time in which the arc PQ is described (for in this case I neglect the resistance, as being infinitely less than the centripetal force), it follows that TQ × SP2, that is (by the last Lemma), , will be in a duplicate ratio of the time, and therefore the time is as ; and the velocity of the body, with which the arc PQ is described in that time, as or , that is, in the subduplicate ratio of SP reciprocally. And, by a like reasoning, the velocity with which the arc QR is described, is in the subduplicate ratio of SQ reciprocally. Now those arcs PQ and QR are as the describing velocities to each other; that is, in the subduplicate ratio of SQ to SP, or as SQ to ; and, because of the equal angles SPQ, SQr, and the equal areas PSQ, QSr, the arc PQ is to the arc Qr as SQ to SP. Take the differences of the proportional consequents, and the arc PQ will be to the arc Rr as SQ to , or . For the points P and Q coinciding, the ultimate ratio of to is the ratio of equality. Because the decrement of the arc PQ arising from the resistance, or its double Rr, is as the resistance and the square of the time conjunctly, the resistance will be as . But PQ was to Rr as SQ to , and thence becomes as , or as . For the points P and Q coinciding, SP and SQ coincide also, and the angle PVQ becomes a right one; and, because of the similar triangles PVQ, PSO, PQ becomes to as OP to . Therefore is as the resistance, that is, in the ratio of the density of the medium in P and the duplicate ratio of the velocity conjunctly. Subduct the duplicate ratio of the velocity, namely, the ratio , and there will remain the density of the medium in P, as . Let the spiral be given, and, because of the given ratio of OS to OP, the density of the medium in P will be as . Therefore in a medium whose[Pg 289] density is reciprocally as SP the distance from the centre, a body will revolve in this spiral. Q.E.D.