u71Now compare the series &c., with the series P - Qo - Roo - So3 - &c., and for P, Q, R and S, put e, , and , and for put or ; and the density of the medium will come out as ; that is (because n is given), as or , that is, as that length of the tangent HT, which is terminated at the semi-diameter AF standing perpendicularly on PQ: and the resistance will be to the gravity as 3a to 2n, that is, as 3AC to the diameter PQ of the circle; and the velocity will be as . Therefore if the body goes from the place F, with a due velocity, in the direction of a line parallel to PQ, and the density of the medium in each of the places H is as the length of the tangent HT, and the resistance also in any place H is to the force of gravity as 3AC to PQ, that body will describe the quadrant FHQ of a circle. Q.E.I.

But if the same body should go from the place P, in the direction of a line perpendicular to PQ, and should begin to move in an arc of the semi-circle PFQ, we must take AC or a on the contrary side of the centre A; and therefore its sign must be changed, and we must put -a for +a. Then the density of the medium would come out as . But nature does not admit of a negative density, that is, a density which accelerates the motion of bodies; and therefore it cannot naturally come to pass that a body by ascending from P should describe the quadrant PF of a circle. To produce such an effect, a body ought to be accelerated by an impelling medium, and not impeded by a resisting one.

EXAMPLE 2. Let the line PFQ be a parabola, having its axis AF perpendicular[Pg 272] to the horizon PQ, to find the density of the medium, which will make a projectile move in that line.

From the nature of the parabola, the rectangle PDQ is equal to the rectangle under the ordinate DI and some given right line; that is, if that right line be called b; PC, a; PQ, c; CH, e; and CD, o; the rectangle a + o into c - a - o or ac - aa - 2ao + co - oo, is equal to the rectangle b into DI, and therefore DI is equal to . Now the second term of this series is to be put for Qo, and the third term for Roo. But since there are no more terms, the co-efficient S of the fourth term will vanish; and therefore the quantity , to which the density of the medium is proportional, will be nothing. Therefore, where the medium is of no density, the projectile will move in a parabola; as Galileo hath heretofore demonstrated. Q.E.I.

EXAMPLE 3. Let the line AGK be an hyperbola, having its asymptote NX perpendicular to the horizontal plane AK, to find the density of the medium that will make a projectile move in that line.

Let MX be the other asymptote, meeting the ordinate DG produced in V; and from the nature of the hyperbola, the rectangle of XV into VG will be given. There is also given the ratio of DN to VX, and therefore the rectangle of DN into VG is given. Let that be bb: and, completing the parallelogram DNXZ, let BN be called a; BD, o; NX, c; and let the given ratio of VZ to ZX or DN be . Then DN will be equal to a - o, VG equal to , VZ equal to , and GD or NX - VZ - VG equal to . Let the term be resolved into the converging series , &c., and GD will become equal to &c. The second term [Pg 273] of this series is to be used for Qo; the third , with its sign changed for Ro2; and the fourth , with its sign changed also for So3, and their coefficients , and are to be put for Q, R, and S in the former rule. Which being done, the density of the medium will come out as , or , that is, if in VZ you take VY equal to VG, as . For aa and are the squares of XZand ZY. But the ratio of the resistance to gravity is found to be that of 3XY to 2YG; and the velocity is that with which the body would describe a parabola, whose vertex is G, diameter DG, latus rectum . Suppose, therefore, that the densities of the medium in each of the places G are reciprocally as the distances XY, and that the resistance in any place G is to the gravity as 3XY to 2YG; and a body let go from the place A, with a due velocity, will describe that hyperbola AGK. Q.E.I.

EXAMPLE 4. Suppose, indefinitely, the line AGK to be an hyperbola described with the centre X, and the asymptotes MX, NX, so that, having constructed the rectangle XZDN, whose side ZD cuts the hyperbola in G and its asymptote in V, VG may be reciprocally as any power DNn of the line ZX or DN, whose index is the number n: to find the density of the medium in which a projected body will describe this curve.

For BN, BD, NX, put A, O, C, respectively, and let VZ be to XZ or DN as d to e, and VG be equal to ; then DN will be equal to A - O, , , and GD or NX - VZ - VG equal to . [Pg 274]Let the term be resolved into an infinite series , &c., and GD will be equal to , &c. The second term of this series is to be used for Qo, the third for Roo, the fourth for . And thence the density of the medium , in any place G, will be and therefore if in VZ you take VY equal to n × VG, that density is reciprocally as XY. For A2 and are the squares of XZ and ZY. But the resistance in the same place G is to the force of gravity as to 4RR, that is, as XY to . And the velocity there is the same wherewith the projected body would move in a parabola, whose vertex is G, diameter GD, and latus rectum or . Q.E.I.

SCHOLIUM.

In the same manner that the density of the medium comes out to be as , in Cor. 1, if the resistance is put as any power Vn of the velocity V, the density of the medium will come out to be as . And therefore if a curve can be found, such that the ratio [Pg 275]of to or of to may be given; the body, in an uniform medium, whose resistance is as the power Vn of the velocity V, will move in this curve. But let us return to more simple curves.

Because there can be no motion in a parabola except in a non-resisting medium, but in the hyperbolas here described it is produced by a perpetual resistance; it is evident that the line which a projectile describes in an uniformly resisting medium approaches nearer to these hyperbolas than to a parabola. That line is certainly of the hyperbolic kind, but about the vertex it is more distant from the asymptotes, and in the parts remote from the vertex draws nearer to them than these hyperbolas here described. The difference, however, is not so great between the one and the other but that these latter may be commodiously enough used in practice instead of the former. And perhaps these may prove more useful than an hyperbola that is more accurate, and at the same time more compounded. They may be made use of, then, in this manner.

Complete the parallelogram XYGT, and the right line GT will touch the hyperbola in G, and therefore the density of the medium in G is reciprocally as the tangent GT, and the velocity there as ; and the resistance is to the force of gravity as GT to .

Therefore if a body projected from the place A, in the direction of the right line AH, describes the hyperbola AGK and AH produced meets the asymptote NX in H, and AI drawn parallel to it meets the other asymptote MX in I; the density of the medium in A will be reciprocally as AH, and the velocity of the body as , and the resistance there to the force of gravity as AH to . Hence the following rules are deduced.

RULE 1. If the density of the medium at A, and the velocity with which the body is projected remain the same, and the angle NAH be changed, the lengths AH, AI, HX will remain. Therefore if those lengths, in any[Pg 276] one case, are found, the hyperbola may afterwards be easily determined from any given angle NAH.