w

av24w

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.

พ

1

Ex. 5. For any other ball, whose diameter is d, and its =bd22 bd2v2 + w w

ad2va d2v2
3.86 150000

weight w, the resistance of the air being
the retarding force will be

putting b =

1 150000

-W

thence vv=2gxx

bd2v2+20
พ

and

and x = X
2g bd2v2 + w
bd2v2+w
= 4gbda
bd2v2 + w

το

พ

bday2+wo

the cor. flu. x =

x h. 1.

x h. 1.

W

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

w

3

4gbd2

8bd2

24 64bd2 836+ 24 24

62.72 300000

bd2v2+w

w
215
6.

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.

PROBLEM VI.

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.

-w

By that prob. x=- X
2ga

a

w

28

~

ther. i

w
a

x (arc tang.

V

X arc tan.

"W

2ga

860
24

บ

(

N x arc to

15

a

พ

215 ;

6

i

- arc tang.

; or by cor

a

a

the time in general when the first velocity v is reduced to And when v = 0, or the velocity ceases, this becoTRES

V.

1

W

V

t= ✔. x arc to tang.

for the time of the whole

28

THEORY AND PRACTICE

Now, as in the last prod v=2000, "=1-05, a='0000234
Hence = +1266, and v* = 203•14, and

=
100000
= 98 445 the tangent, to which corresponds the arc

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".

6

32

PROBLEM VII.
To determine the same as in prob. v, taking into the ac-
count the Decrease of Density in the Air as the Ball asćends
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 den-
sity 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 geo-
metrical progression, at altitudes which rise only in arithme-
tical progression ; by which it happens, that the altitudes
ascended are proportional only to the logarithms of the de-
crease 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 consi-
dered 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 deter-
mined in prop. 76, pa. 244, &c, vol. 2. It is there shown,
that if o denote the air's density at the earth's surface, and
d its density at any altitude a, or r; then is x = 68551 X
log. of in feet, when the temperature of the air is 55";
for the temperature of freezing cold;

We

D

d

N

19:01

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

229

10

Hence 20

9000000

= 98-445 the tangent, to which corresponds the arc

of 89° 25′, whose length is 1-5606; then 2 × 203-14 ×

X

1.5606=

203.14 × 1-5606
32

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′′.

2955
16

2g

2000

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 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

D

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

nal ,

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

D

62000

n

124000

1-11 2

* D, it gives the natural number

, n+1

*d=d, or

n +

124000

C+

с + 3

124000n=-*

=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

force.

CX

av

с + x

au2

Cr
Х

C + x

W