the diameter of the shell being d; that is, the terminal velocities will be all less in the ratio of 147.3 to 175.5. Now, computing these several velocities by this rule, to all the different diameters, they are found as placed in the 6th col.; and in the 7th or last column are set the altitudes which would produce these velocities in vacuo, as computed from this theorem UU 64 Having now obtained these terminal velocities, and their producing altitudes, for the shells, we can, from them and the former table of ranges and elevations, easily compute the greatest range, and the corresponding angle of elevation, for any mortar and shell, in the same way as was done for the balls in this problem. Thus, for example, to find the greatest range and elevation, for the 13 inch shell, when projected with the velocity of 2000 feet per second, being nearly the greatest velocity that balls can be discharged with. Now, by the method before used, = 3.796; opposite to this, found in the first column of the table of ranges, corresponds 34° 49′ for the elevation in the 2d column, and the number 2.764 in the 3d column; this multiplied by the altitude 4340, gives 11995 feet, or more than 24 miles, for the greatest range. 2000 This however is much short of the distance which it is said the French have lately thrown some shells at the siege of Cadiz, viz, 3 miles, which it seems has been effected by means of a peculiar piece of ordnance, and by loading or filling the cavity of the shell with lead, to render it heavier, and thus make it fitter to overcome the resistance of the air. Let us then examine what will be the greatest range of our 13 inch shell, if its usual cavity be quite filled with lead when discharged, with the projectile velocity of 2000 feet. Now the diameter of the cavity, being about of that of the mortar 13, will be nearly 9 inches. And the weight of a globe of lead of this diameter is 139.31b; which added to 187.8, the weight of the shell empty, gives 3271b, the whole weight of the shell when the cavity is filled with lead, which was found 286 when supposed all of solid iron, their ratio or quotient is 8783. Then, as before, the theorem will be .8783 187.3d for the terminal velocity; which, 128, becomes 670 for the terminal velocity; therefore its producing altitude is. 2000 the same method as before, 670 2985; which number found found in the first column of the table of ranges, the opposite number in the 2d col. is 37° 15′ for the elevation of the piece, and in the 3d column 2·14, multiplied by 7014, gives 15010 feet, or nearly 3 miles. So that our 13 inch shells, discharged at an elevation of about 374 degrees, would range nearly the distance mentioned by the French, when filled with lead, if they can be projected with so much as 2000 feet velocity, or upwards. This however it is thought cannot possibly be effected by our mortars; and that it is therefore probable the French, to give such a velocity to those shells, must have contrived some new kind of large cannon on the occasion. 4. Having shown in the preceding articles and problems, how, from our theory of the air's resistance, can be found, first the initial or projectile velocity of shot and shells; 2dly, the terminal velocity, or the greatest velocity a ball can acquire by descending by its own weight in the air; 3dly, the height a ball will ascend to in the air, being projected vertically with a given velocity, also the time of that ascent; 4thly, the greatest horizontal ranges of given shot, projected with a given velocity; as also the particular angle of elevation of the piece, to produce that greatest range. It remains then now to enquire, what laws and regulations can be given respecting the ranges, and times of flight, of projects made at other angles of elevation. Relating to this enquiry, the Encyclopædia Britannica mentions the two following rules: 1st." Balls of equal density, projected with the same elevation, and with velocities which are as the square roots of their diameters, will describe similar curves. This is evident, because, in this case, the resistance will be in the ratio of their quantities of motion; therefore all the homologous lines of the motion will be in the proportion of the diameters." But though this may be nearly correct, yet it can hardly ever be of any use in practice, since it is usual and proper to project small balls, not with a less, but with a greater velocity, than the larger ones. 2dly, the other rule is, "If the initial velocities of balls, projected with the same elevation, be in the inverse subduplicate ratio of the whole resistances, the ranges, and all the homologous lines in their track, will be inversely as those resistances.' This rule will come to the same thing, as having the initial velocities in the inverse ratio of the diameters, as distant perhaps from fitness as the former. Two tables are next given in the same place, for the comparison of rangesand projectile velocities, the numbers in which appear to be much wide of the truth, as depending on very erroneous effects of the resistance. Most of the accompanying remarks, however, however, are very ingenious, judicious, and philosophical, and very justly recommending the making and recording of good experiments on the ranges and times of flight of projects, of various sizes, made with different velocities, and at various angles of elevation. Besides the above, we find rules laid down by Mr. Robins and Mr. Simpson, for computing the circumstances relating to projectiles as affected by the resistance of the air. Those of the former respectable author, in his ingenious Tracts on Gunnery, being founded on a quantity which he calls F, (answering to our letter a in the foregoing pages), I find to be almost uniformly double of what it ought to be, owing to his improper measures of the air's resistance; and therefore the conclusions derived by means of those rules must needs be very erroneous. Those of the very ingenious Mr. Simpson, contained in his Select Exercises, being partly founded on experiment, may bring out conclusions in some of the cases not very incorrect; while some of them, particularly those relating to the impetus and the time of flight, must be very wide of the truth. We must therefore refer the student, for more satisfaction, to our rules and examples before given. in vol. 2 pa. 162 &c, especially for the circumstances of different ranges and elevations, &c, after having determined, as above, those for the greatest ranges, founded on the real measure of the resistances. CHAPTER CHAPTER XIV. PROMISCUOUS PROBLEMS, AS EXERCISES IN MECHANICS, STATICS, DYNAMICS, HYDROSTATICS, HYDRAULICS, FROJECTILES, &c. &c. PROBLEM I. Let AB and AC be two inclined planes, whose common altitude AD is given = 64 feet; and their lengths such, that e heavy body is 2 seconds of time longer in descending through AB than through AC, by the force of gravity; and if two balls, the one weighing 3 and the other 2lb, be connected by a thread and laid on the planes, the thread sliding freely over the vertex A, they will mutually sustain each other. Quere the lengths of the two planes? THE lengths of planes of the same height being as the times of descent down them (art. 133 vol. 2), and also as the weights of bodies mutually sustaining each other on them (art. 122), therefore the times must be as the weights; hence as 1, the difference of the weights, is to 2 sec. the diff. of 13: 6 sec. the times of descending down the two times, :: 12: 4 sec.) planes. And as 16:64: 1 sec. : 2 sec. the time of descent down the perpendicular height (art. 70). Then, by the laws of descents (art. 132), as 2 sec.: 64 feet feet, the lengths of the planes. { 6 sec.: 192 4 sec. : 128 Note. In this solution we have considered 16 feet as the space freely descended by bodies in the 1st second of time, and 32 feet as the velocity acquired in that time, omitting the fractions and, to render the numeral calculations simpler, as was done in the preceding chapter on projectiles, and as we shall do also in solving the following questions, wherever such numbers occur. Another Solution by means of Algebra. Put the time of descent down the less plane; then will +2 be that of the greater, by the question. Now the weights being as the lengths of the planes, and these again as the times, therefore as 2 : 3 :: x : x + 2; hence 2x + 4 = 3x, and 4 sec. Then the lengths of the planes are found as in the last proportion of the former solution. PROBLEM 2. If an elastic ball fall from the height of 50 feet above the plane of the horizon, and impinge on the hard surface of a plane inclined to it in an angle of 15 degrees; it is required to find what part of the plane it must strike, so that after reflection, it may fall on the horizontal plane, at the greatest distance possible beyond the bottom of the inclined plane? Here it is manifest that the ball must strike the oblique plane continued on a point somewhere below the horizontal plane; for otherwise there could be no maximum. Therefore let BC be B A M the inclined plane, CDG the horizontal one, p the point on which the ball impinges after falling from the point A, BEGI the parabolic path, E its vertex, BH a tangent at B, being the direction in which the ball is reflected; and the other lines as are evident in the figure. Now, by the laws of reflection, the angle of incidence ABC, is equal to the angle of reflection HBM, and therefore this latter, as well as the former, is equal to the complement of the C the inclination of the two planes; but the part IBM is = c, therefore the angle of projection HBI is the comp. of double the 4c, and being the comp. of HBK, theref. HBK = 24C. Now, put a = 50 = Ap the height above the horizontal line, tang. DBC or 75° the complement of the plane's inclination, r = tang. HBI or H-60° the comp. of 24c, s = sine of 2 4HBI = 120° the double elevation, or sine of 4 c; also x= AB the impetus or height fallen through. Then, and = BI = 4KH2Sr, by the projectiles prop. 21, BK7X KH = STX CD = t x BD = t(xa)} by trigonometry; also, KD = BK BD == STX -a); x+a, and KE = 4BI = S≈ ; then, by the parabola, √ BK : ✔DK :: KE : FG = KE X KD KB 5 •~?~22] = 2b (ax-b2x2), putting b = sine of 24c sine of 30°. Hence CGCD+DFBG= txta + sx ±2b√ (ax—b2x2) ♬ maximum, the fluxion of which made, tion, reduced, gives x = where p = s VOL. III. n +t, |