Now, as in the last prob. v=2000, w=1·05, á='0000234 Hence 41266, and 203·14, and = √ 229 9000000 a a 98445 the tangent, to which corresponds the arc 1 of 89° 25', whose length is 1-5606; then × 203.14 X 1.5606 2955 203.14 x 1.5606 2g 9"-91, the whole time of ascent. Remark. The time of freely ascending to the same height 2955 feet, that is, without the air's resistance, would be = √2955 = 13" 59; and the time of freely ascending, commencing with the same velocity 2000, would be 2000 =62′′ 1′2′′. 2g 16 32 PROBLEM VII. To determine the same as in prob. v, taking into the account the Decrease of Density in the Air as the Ball ascends in the Atmosphere. In the preceding problems, relating to the height and time of balls ascending in the atmosphere, the decrease of density in the upper parts of it has been neglected, the whole height ascended by the ball being supposed in air of the same density as at the earth's surface. But it is well known that the atmosphere must and does decrease in density upwards, in a very rapid degree; so much so indeed, as to decrease in geometrical progression, at altitudes which rise only in arithmetical progression; by which it happens, that the altitudes ascended are proportional only to the logarithms of the decrease of density there. Hence it results, that the balls must be always less and less resisted in their ascent, with the same velocity, and that they must consequently rise to greater heights before they stop. It is now therefore to be considered what may be the difference resulting from this cir cumstance. Now, the nature and measure of this decreasing density, of ascents in the atmosphere, has been explained and determined in prop. 76, pa. 244, &c, vol. 2. It is there shown, that if D denote the air's density at the earth's surface, and d its density at any altitude a, or r; then is r = 63551 × log. of in feet, when the temperature of the air is 55°; and 60000 × log.for the temperature of freezing cold; we we may therefore assume for the medium x=62000 × log. for a mean degree between the two. D But to get an expression for the density d, in terms of r out of logarithms, without which it could not be introduced into the measure of the ball's resistance, in a manageable form, we find in the first place, by a neat approximate expression for the natural number to the log. of a ratio,, whose terms do not greatly differ, invented by Dr. Halley, and explained. in the Introduction to our Logarithms, p. 110, that n-1 n + l X D nearly, is the number answering to the log. 1 of the ratio where n denotes the modulus 43429448 &c of the common logarithms. But, we before found that x=62000 × log. of, or 62000 is the log. of which log. was denoted by / in the expression just above, for the number whose log. is lor ; substituting therefore for l, in the expression 62000 62000 124000 124000n+x d, the density of the air at the altitude x, put, ting D = 1 the density at the surface. Now put 124000n or nearly 54000 = c; then will be the density of the air at any general height x. c+ x But, in the 5th prob. it appears that av2 denotes the resistance to the velocity v, or at the height x, for the density of air the same as at the surface, which is too great in the ratio of c + x to c 20 ; therefore a2 x will be the resistance at the height x, to the velocity v, where a = 0000254. To this adding w, the weight of the ball, gives C-X av2 + for the whole resistance, both from the air c+ x av2 and the ball's mass; conseq. - X W the accelerating force of the ball. Or, if we include the no sensible difference in the result, but be a great deal simpler a Now the fluent of the first side of this equation is evidently 2c × h. l. (c+x); and the fluent of the latter το side, the same as in prob. 5, is × h. I. (v2 + ); there --W 640 64a 64a fore the general fluential equa. is-x+2c x h. 1. (c+x)= × h. 1. (2). But, when a = 0, and v=v the initial velocity, this becomes 0+2cx h.l.cx h. 1. (v2+); theref. by subtraction, the correct fluents are x + 2c x h.l. when the first velocity v is diminished to any less one v; and when it is quite extinct, the c+x w = x h. 1. C 64a av2+w h. l. ave+, for the greatest height x ascended. Here, in the quantity h. 1. +, the term x is always small C in respect of the other term c; therefore, by the nature of logarithms, the h.l. of is nearly Now the latter side of this equation is the same value for r as was found in the 5th problem, which therefore put = b; then the value of x will be easily found from the formula xb, by a quadratic equation. Or, still easier, and sufficiently near the truth, by substituting b for r in the numerator and the denominator of then 2c+x 2c -x 2c+x 20-b b, and hence x = 2c+bb, or by proportion, as 2cb: 2c+b::b: x; that is, only increase the value of r, found by prob. 5, in the ratio of 20 b to 2c + b. Now, in the first example to that prob. the value of x or b was there found 2955; and 2c being = 108000, theref. 2c - b⇒ 105045, and 2c + b = 110955, then as 105045 : 110955: 2955: 3121 = the value of the height x in this case, being only 166 feet, orth part more than before. Also, Also, for the 2d example to the 5th prob. where x was = 6420; therefore as 2c-b : 2c + b or as 105045 :'110955 :: 6420: 6780 the height ascended in this example, being also the 18th part more than before. And so on, for any other examples; the value of 2c being the constant number 108000. PROBLEM VIII. To determine the Time of a Ball's Ascending, considering the Decreasing Density of the Air as in the last prob. บ The fluxion of the time is t =. But the general êquation of the fluxions of the space r and velocity v, in the last ways small in respect of c, is nearly =b as determined in the · + ს c-b c+x last problem; theref. may be substituted for with c-x Now, this fluxion being to that in prob. 6, in the constant ratio of cb to c + b, their fluents will be also in the same constant ratio. But, by the last prob. c=54000, and b=2955 for the first example in prob. 5; therefore c-b51045, and cb56955, also, the time in problem 6 was 9"-91; therefore as 51045: 56955 :: 9′′·91 : 11′′·04 for the time in this case, being 1"13 more than the former, or nearly the 9th part more; which is nearly the double, or as the square of the difference, in the last prob. in the height ascended. PROBLEM IX. To determine the circumstances of Space, Time, and Velocity, of a Ball Descending through the Atmosphere by its own Weight. It is here meant that the balls are at least as heavy as cast iron, and therefore their loss of weight in the air insensible; and that their motion commences by their own gravity from a state of rest. The first object of enquiry may be, the utmost degree of velocity any such ball acquires by thus descending. Now it is manifest that the ball's motion is commenced, and uniformly increased, by its own weight, which is its constant urging force, being always the same, and producing an equal increase of velocity in equal times, excepting for the diminution of motion by the air's resistance. It is also evident that VOL. III. this U this resistance, beginning from nothing, continually increases, in some ratio, with the increasing velocity of the ball. Now, as the urging force is constantly the same, and the resisting force always increasing, it must happen that the latter will at length become equal to the former: when this happens, there can afterwards be no further acceleration of the motion, the impelling force and the resistance being equal, and the ball must ever after descend with a uniform motion. It follows therefore that, to answer the first enquiry, we have only to determine when or what velocity of the ball will cause a resistance just equal to its own weight. Now, by inspecting the tables of resistances preceding prob. 1, particularly the 1st of the three tables, the weight of the ball being 105 lb, we perceive that the resistance increases in the 2d column, till 069 opposite to 200 velocity, and 1.56 answering to 300 velocity, between which two the proposed resistance 105, and the correspondent velocity, fall. But, in two velocities not greatly different, the resistances are very nearly proportional to the squares of the velocities. Therefore, having given the velocity 200 answering to the resistance 0·69, to find the velocity answering to the resistance 1.05, we must say, as 0.69: 1:05 :: 2002: v2 = 60870, theref. v=60870 246, is the greatest velocity this ball can acquire; after which it will descend with that velocity uniformly, or at least with a velocity nearly approaching to 246. The same greatest or uniform velocity will also be directly found from the rule 00001725v2r, near the end of problem 2, where r' is the resistance to the velocity v, by making 1.05= r; for then v2 = for vas before. 1.05 .00001725 -=60870, the same value But now, for any other weight of ball; as the weights of the balls increase as the cubes of their diameters, and their resistances, being as the surfaces, increase only as the squares of the same, which is one power less; and the resistances being also in this case, as the squares of the velocities, we must therefore increase the squares of the velocity in the ratio of the diameters of the balls; that is, as 1.965: d: 2462: dvand hence v = 246 √ = 175/d. 2462 1.965 d 1.965 If we take here the 3 lb ball, belonging to the 2d table of resistances, whose diameter d is 2.80; then √✓✅/2·80=1·673, and 175 x 1·67 = 294, is the greatest or uniform velocity, with which the 3 lb ball will descend. And if we take the 6lb ball, whose diameter is 3.53 inches, as in the 3d table of resistances: |