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Er. 4. With the same ball, and 1640 velocity. Assume 1200, whose resistance 34·13 is nearly a medium. Then as 12002 : 34′13 :: 1:·0000237=a. Hence x h. 1. 64a =2854; again less than half the range (6000) by experiment in vol. 2, even with 15° elevation.
Ex. 5. For any other ball, whose diameter is d, and its =bd22 bd2v2 + w w
weight w, the resistance of the air being
putting b =
and x = X
the cor. flu. x =
x h. 1.
x h. 1.
4gbd2 for the whole height when v = 0. Now if the ball be a 24 pounder, whose diameter is 5.6, and its square 31-36; then bd2: =0002091, and = =1794; and bdy 836, and
24 64bd2 836+ 24 24
therefore 1794 x h. l. = 1794 x 3.57888= :6420, being more than double the height of that of the small ball, or a little more than a mile, and very nearly the same as in the 2d example to prob. 4.
To determine the Time of the Ball's ascending to the Height determined in the last prob. by the same Projectile Velocity as there given.
By that prob. x=- X
x (arc tang.
X arc tan.
N x arc to
- arc tang.
; or by cor
the time in general when the first velocity v is reduced to And when v = 0, or the velocity ceases, this becoTRES
t= ✔. x arc to tang.
for the time of the whole
THEORY AND PRACTICE
Now, as in the last prod v=2000, "=1-05, a='0000234
x 203:14 X
of 80* 25', whose length is 1.5606 ; then
2g 203 14 x 1.5606 15606 =
- 9":91, the whole time of ascent.
S? Remark. The time of freely ascending to the same height 2955 feet, that is, without the air's resistance, would be 2983
= 1,/2955 = 13":59; and the time of freely ascending, commencing with the same velocity 2000, would be
= 62'4 =1'2".
In the preceding problems, relating to the height and time
Now, the nature and measure of this decreasing density,
has been explained and deter-
com Com ting D= nearly 511 =1. PE at any genera za
But, in se si sistance to the retom 0:2 - Der of air the same as a tix suriat ratio of c + at C-2. Test resistance at the hero ti to veic *000025%. To this auding 3, that are av? x
+ 7 for the whole res:sm. Free and the ball's mass; conse
vil cemete the accelerating force of the ba. Or, if we include the small part or l, within the factors which sz. mit: no sensible difference in the result, bet bei padidapat in the process, then is **** = ; acesting
Now, as in the last prob. v=2000, w=1·05, a='0000234 = 41266, and √/= = 203·14, and
= 98-445 the tangent, to which corresponds the arc
of 89° 25′, whose length is 1-5606; then 2 × 203-14 ×
203.14 × 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 ** = 62′′4 = 1′ 2′′.
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
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 x; then is x = 63551 × log. of in feet, when the temperature of the air is 55°; and 60000 x log.for the temperature of freezing cold;
we may therefore assume for the medium x=62000x log. for a mean degree between the two.
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, 7, whose terms do not greatly differ, invented by Dr. Halley, and explained in the Introduction to our Logarithms, p. 110, that i 7 m
nt fi nearly, is the number answering to the log. I of the ratio 2 where n denotes the modulus •43429448 &c of the common logarithms. But, we before found that x=62000 x log. of as or 62600 is the log. of , which log. was denoted by l in the expression just above, for the number whose log. is l or 62000.; substituting therefore
for l, in the expression
* D, it gives the natural number
с + 3
=d, the density of the air at the altitude x, put. 124000n + x ting D= I 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.
But, in the 5th prob. it appears that av? 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
X; therefore auz x will be the resistance at the height X, to the velocity v, where a = *000025. To this adding w, the weight of the ball, gives arz x + w for the whole resistance, both from the air and the ball's mass; conseq.
+ W will denote the accelerating force of the ball. Or, if we include the small part oor 1, within the factors, which will make no sensible difference in the result, but be a great deal simpler in the process, then is
= f the accelerating
с + x
C + x