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. theref. 1.50958 × 4256 = 6424r the height, being a little more than a mile. We may now examine what will be the height ascended, considering the resistance always as the square of the velocity. PROBLEM V. To determine the Height ascended by a Ball projected as in the two foregoing problems; supposing the Resistance of the Air to be as the Square of the Velocity. Here it will be proper to commence with selecting some experimented resistance corresponding to a medium kind of velocity, between the first or greatest velocity and nothing, from which to compute the other general resistances, by considering them as the squares of the velocities. It is proper to assume a near medium velocity and its resistance, because, if we assume or commence with the greatest, or the velocity of projection, and compute from it downwards, the resistances will be everywhere too great, and the altitude ascended much less than just; and, on the other hand, if we assume or commence with a small resistance, and compute from it all the others upwards, they will be much too little, and the computed altitude far too great. But, commencing with a medium degree, as for instance that which has a resistance about the half of the first or greatest resistance, or rather a little more, and computing from that, then all those computed resistances above that, will be rather too little, but all those below it too great; by which it will happen, that the defect of the one side will be compensated by the excess on the other, and the final conclusion must be near the truth. Thus then, if we wish to determine, in this way, the alti tude ascended by the ball employed in the 1st table of resistances, when projected with 2000 feet velocity; we perceive by the table, that to the velocity 2000 corresponds the resistance 98 lb; the half of this is 49, to which resistance corresponds the velocity 1400 in the table, and the next greater velocity 1500, with its resistance 574, which will be properest to be employed here. Hence then, for any other velocity v, in general, it will be, according to the law of the 57v2 squares of the velocities, as 15002 2:: 57: 1500% *0000254v2 av2, putting a = '000025, which will denote the air's resistance for any velocity v, very nearly, counting from 2000. Now let a denote the altitude ascended when the velocity is v, and w the weight of the ball: then, as above, av2 is the resistance resistance from the air, hence av + w is the whole resisting w a the fluent of which, by form 8, ish. log. (v2 + =~~); which when x = 0, and v = v the first or projectile velocity, becomes 0= x h. 1. (v2+); theref. by subtract 4ga w a ing, the correct fluent is x = x h. l. 4ga a when the velocity is reduced to v; and when v = 0, or the w 4ga av2+ w x h. 1. velocity is quite exhausted, this becomes Ex. 1. The values of the letters being w=1051b, 4g=64, a=0000254, the last expression becomes 645 x hyp. log. v2+41266 vel. v being 2000, the same expression 1484 x log. 41266 becomes 1484 x log. of 97.93 = 2955 for the height ascended, on this hypothesis; which was 2922 by the former problem, being nearly the same. W Ex. 2. Supposing the same ball to be projected with the velocity of only 1500 feet. Then taking 1100 velocity, whose tabular resistance is 27.6, being next above the half of that for 1500. Hence, as 1100: 2 :: 27·6: 00002375v2=av2. This value of a substituted in the theorem also 1500 for v, and 105 for w, it brings out = 2728 for the height in this case, being but a little above the ratio of the square roots of the velocities 2000 and 1500, as that ratio would give only 2560. 4ga w Ex. 3. To find the height ascended by the first ball, projected with 860 feet velocity. Here taking 600, whose resistance 6.69 is a near medium; then as 6002: 6.69 :: 1 : 0000186a. Hence 2334 the height; which is less than half the range (5100) at 45° elevation, but more than half the range (4100) at 15° elevation, in pa. 161 vol. 2; being indeed nearly a medium between the two. |
