Archimedes Project

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of the point from the said line PL; that is, a body falling freely through PA, acquires the velocity in the curve at A, EF, " at F, KD, " at D, LH, " at H. The reason of which is, that the line PL is what is called the Directrix of the parabola, the property of which is, that the perpendicular to it, from every point of the curve, is equal to one-fourth of the parameter of the diameter at that point, viz, PA = 1/4 the parameter of the diameter at A, EF = " at F, KD = " at D, LH = " at H.

2. If a body, after falling through the height PA, which is equal to AB, and when it arrives at A if its course be changed, by reflection from a firm plane AI, or otherwise, into any direction AC, without altering the velocity; and if AC be taken equal to 2AP or 2AB, and the parallelogram be completed; the body will describe the parabola passing through the point D.

3. Because AC = 2AB or 2CD or 2AP, therefore AC2 = 2AP . 2CD or AP . 4CD; and because all the perpendiculars EF, CD, GH are as AE2, AC2, AG2; therefore also AP . 4EF = AE2, and AP . 4GH = AG2, &c; and because the rectangle of the extremes is equal to the rectangle of the means, of four proportionals, therefore it is always, , and , and , and so on.

IV. Having given the Direction of a Projectile, and the Impetus or Altitude due to the sirst velocity; to determine the Greatest Height to which it will rise, and the Random or Horizontal Range.

Let AP be the height due to the Projectile velocity at A, or the height which a body must fall to acquire the same velocity as the projectile has in the curve at A; also AG the direction, and AH the horizon. Upon AG let fall the perpendicular PQ, and on AP the perpendicular QR; so shall AR be equal to the greatest altitude CV, and 4RQ equal to the horizontal range AH. Or, having drawn PQ perpendicular to AG, take AG = 4AQ, and draw GH perpendicular to AH; then AH is the range.

For by the last cor. ... , and by sim.triangles, ... , or ; therefore AG = 4AQ; and, by similar triangles, AH = 4RQ.

Also, if V be the vertex of the parabola, then AB or 1/2AG = 2AQ, or AQ = QB; consequently AR = BV which is = CV by the nature of the parabola.

Hence, 1. Because the angle Q is a right angle, which is the angle in a semicircle, therefore if upon AP as a diameter a semicircle be described, it will pass through the point Q.

2. If the Horizontal Range and the Projectile Velocity be given, the Direction of the piece so as to hit the object H will be thus easily found: Take AD = 1/4AH, and draw DQ perpendicular to AH, meeting the semicircle described on the diameter AP in Q and q; then either AQ or Aq will be the direction of the piece. And hence it appears, that there are two directions AB and Ab which, with the same Projectile velocity, give the very same horizontal range AH; and these two directions make equal angles qAD and QAP with AH and AP, because the arc PQ is equal to the arc Aq.

3. Or if the Range AH and Direction AB be given; to find the Altitude and Velocity or Impetus: Take AD = 1/4AH, and erect the perpendicular DQ meeting AB in Q; so shall DQ be equal to the greatest altitude CV. Also erect AP perpendicular to AH, and QP to AQ; so shall AP be the height due to the velocity.

4. When the body is projected with the same velocity, but in different directions; the horizontal ranges AH will be as the sines of double the angles of elevation. Or, which is the same thing, as the rectangle of the sine and cosine of elevation. For AD or RQ, which is 1/4AH, is the sine of the arc AQ, which measures double the angle QAD of elevation.

And when the direction is the same, but the velocities different, the horizontal ranges are as the square of the velocities, or as the height AP which is as the square of the velocity; for the sine AD or RQ, or 1/4AH, is as the radius, or as the diameter AP

Therefore, when both are different, the ranges are in the compound ratio of the squares of the velocities, and the sines of double the angles of elevation.

5. The greatest range is when the angle of elevation is half a right angle, or 45°. For the double of 45 is 90°, which has the greatest sine. Or the radius OS, which is 1/4 of the range, is the greatest sine.

And hence the greatest range, or that at an elevation of 45°, is just double the altitude AP which is due to