of the diameters of the balls, for any other balls whatever. This formula then, though serving quite well for some particular resistance, or even for constructing a complete series or table of resistances, is not proper for the use of problems in which fluxions and fluents are concerned, on account of the mixed number 2, in the index of the velocity v. We must therefore have recourse to an expression in two terms, or a formula containing two integral powers of the velocity, as v2 and v, the first and 2d powers, affected with general coefficients m and n, as mv2 + nv = r the resistance. Now, to determine the general numerical values of the coefficients m and n, we must adapt this general expression mo2 + nv = r, to two particular cases of velocity, at a convenient distance from each other, in one of the foregoing tables of resistances, as the first for instance. Now, after making several trials in this way, I have found that the two velocities of 500 and 1000 answer the general purpose better than any other that has been tried. Thus then, em→ ploying these two cases, we must first make v = 500, and r4lb, its correspondent resistance, and then again v = 1000, and r = 21.88 lb, the resistance belonging to it: this will give two equations, by which the general value of m and of n will be determined. Thus then the two equations being 500m+500n = 4.5, and 1000 m + 1000n 21.88; dividing the 1st by 500, and the S 2d by 1000, they are the dif. of these is 500m + n = '009, 1000m + n = .02188; 500m 01288, and therefore div. by 500, gives m=00002576; hence n ·009 – 500m009-01288 —— -'00388 = n. Hence then the general formula will be 00002576 v2 00388vr the resistance nearly in avoirdupois pounds, in all cases or all velocities whatever. Now, cess, and again a little in defect, 1700 67.85 74.13 1800 76.48 82.44 1900 85.62 90.44 2000 95.28 98.06 by very small differences; so that, on the whole, they will nearly balance one another, in any particular instance of the range or flight of a ball, in all degrees of its velocity, from the first or greatest, to the smallest or last. Except in the first two or three numbers, at the beginning of the table, for the velocities 100, 200, 300, for which cases another theorem may be employed. Now, in these three velocities, as well as in all that are smaller, down to nothing, the theorem '00001725v2 r the resistance, will very well serve, as it brings out for the first three resistances 17, and 69, and 1554, differing. in the last only by a very small fraction. Corol. 1. The foregoing rule 000025760-00388v=r, denotes the resistance for the ball in the first table, whose diameter is 1.965, the square of which is 3.86, or almost 4; hence to adapt it to a ball of any other diameter d, we have only to alter the former in proportion to the squares of the diameters, by which it becomes 3-86(00002576v — •00388v) = ('00000667v2 — ·001v)d2 = (·00000žv2 — ·001v)d2, which is the resistance for the ball whose diameter is d, with the velocity v. dd Corol. 2. And, in a similar manner, to adapt the theorem '00001725v2 = r, for the smaller velocities, to any other size of dd of ball, we must multiply it by the ratio of the surfaces, 3.86 by which it becomes '00000447d2v2 = r. We shall soon take occasion to make some applications in the use of the foregoing formulas, after considering the effects of such velocities in the cases of nonresistances. PROBLEM III. To determine the Height to which a Ball will rise, when fired from a cannon Perpendicularly Upwards with a Given Velocity, in a Nonresisting Medium, or supposing no Resistance in the Air. By art. 73 pa. 151 vol. 2, it appears that any body projected upwards, with a given velocity, will ascend to the height due to the velocity, or the height from which it must naturally fall to acquire that velocity; and the spaces fallen being as the square of the velocities; also 16 feet being the space due to the velocity 32; therefore the space due to any proposed velocity v, will be found thus, as 32: 16 :: v2 : s the space, or as 64 : 1 :: v2: 42s the space, or the height tow hich the velocity v will cause the body to rise, independent of the air's resistance. = Exam. For example, if the first or projectile velocity, be 2000 feet per second, being nearly the greatest experimented velocity, then the rules becomes x 2000=62500 feet 11 miles; that is, any body, projected with the ve locity 2000 feet, would ascend nearly 12 miles in height, without resistance. Corol. Because, by art. 88 Projectiles vol. 2, the greatest range is just double the height due to the projectile velocity, therefore the range, at an elevation of 45°, with the velocity in the last example, would be 233 miles, in a nonresisting medium. We shall now see what the effects will be with the resistance of the air. PROBLEM IV. To determine the Height to which a Ball projected Upwards, as in the last problem, will ascend, being Resisted by the Atmosphere. Putting to denote any variable and increasing height ascended by the ball; u its variable and decreasing velocity there; d the diameter of the ball, its weight being w; m='000003, and n001, the coefficients of the two ternis denoting the law of the air's resistance. Then (mo-nv)d2, by cor. 1 to prob. prob. 2, will be the resistance of the air against the ball in avoirdupois pounds; to which if the weight of the ball be added, then (mv2-nv)d2+w will be the whole resistance to the ball's motion; this divided by w, the weight of the ball in (mv2―nv)d2 + w mo? nud2 + 1 =ƒ the retardd2+1=f motion, gives 20 ing force. Hence the general formula vv = 2gfx (theor. 10 pa. 342 vol. 2, edit. 6) becomes vv=2gx × (mv2―nv)d2 + w making v negative because v is decreasing, where g = 16 ft.; and hence Now, for the easier finding the fluent of this, assume Then the general fluents, taken by the 8th and 11th forms vol. 2 pa. 307, give x = to rad. q and tan. z] = P -W 2gmd2 -to 2gmd2 × [ log. (v2 - v + d) + m md2 x arc to rad. q and tang. v-p]. But, at the beginning of the motion, when the first velocity is v for instance, and the space x is 0, this fluent becomes 0 = --W 2gmd2 tan. v × [ log. (v2 -v + ;) + × arc radius q P p]. Hence by subtraction, and taking v = 0 for the end of the motion, the correct fluent becomes x= × [ log. (vv) log + 1/2/2 X 2gmd2 (arc tan. v- parc tan. p to rad q)]. But as part of this fluent, two arcs to tans. y-p and denoted by the dif, of the p, is always very small in comparison parison with the other preceding terms, they may be omitted without material error in any practical instance; and then the most height to which the ball will ascend, when its motion ceases, and is stopped, partly by its own gravity, but chiefly by the resistance of the air. w But now, for the numerical value of the general coefficient 4gmda w md2 and the term ; because the mass of the ball to the diameter d, is 5236d3, if its specific gravity be s, its weight will be 5236 sd3 = w; therefore 78540sd, this divided by 4g or 64, it gives m = 1227.2sd w d2 ='5236sd, and w 4gmd2 w md2 - for the value of the general coefficient, to any diameter d and specific gravity s. And if we further suppose the ball to be cast iron, the specific gravity, or weight of one cubic inch of which is 26855, it becomes 330d, for that coefficient; also 78540sd = 21090d= = 150. And hence the foregoing fluent becomes 330d x hyp. log. 72 -150v+21090d w and V2 m 150v+21090d 21090d or 760d x com. log. 21090d changing the hyperbolic for the common logs. And this is a general expression for the altitude in feet, ascended by any iron ball, whose diameter is d inches, discharged with any velocity v feet. So that, substituting any values of d and v, the particular heights will be given to which the balls will ascend, which it is evident will be nearly in proportion to the diameter d. Exam. 1. Suppose the ball be that belonging to the first table of resistances, its weight being 16 oz. 13 dr. or 1'05 lb, and its diameter 1.965 inches, when discharged with the ve locity 2000 feet, being nearly the greatest charge for any iron ball. The calculation being made with these values of d and v, the height ascended is found to be 2920 feet, or little more than half a mile; though found to be almost 12 miles without the air's resistance. And thus the height may be found for any other diameter and velocity. Exam. 2. Again, for the 24 lb ball, with the same velocity 2000, its diameter being 5'6 = d. Here 760d = 4256, and 150v+21090d 21090d 58181 the log. of which is 1.50958; theref. |
