in the upper parts of it has been neglected, the whole height ascending 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 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 circumstance.

Now, the nature and measure of this decreasing density, of ascents in the atmosphere, has been explained and determined in prop. 76. Pneumatics. 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 = 63551X log. of in fect, when the temperature of the air is 55°;

D

d

D

and 60000 X log. for the temperature of freezing cold;

d

we may therefore assume for the medium x=62000 × log. for a mean degree between the two.

But to get an expression for the density d, in terms of x 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 ď do not greatly differ, invented by Dr. Halley, and explained in the Introduction to our Logarithms, p. 110, that"

D

X D

D

nearly, is the number answering to the log. 1 of the ratiowhere n denotes the modulus 434294 18 &c. of the common

D

logarithms. But, we before found that x=62000×log. of d'

x

D

or is the log. of which log. was denoted by in the

62000

expression just above, for the number whose lug. is lor

x

62000; substituting therefore,

x

62000

for l, in the expression n-il

124000

124000n124000n+x ting D= the density at the surface. Now put 124000n or

-=d, the density of the air at the altitude x, put

nearly 54000 c; then will be the density of the air c+x

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

C-x

ratio of cx to c-x; therefore av2 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

·av2 X- +w for the whole resistance, both from the air

c+x

and the ball's mass; conseq.

the accelerating force of the

small part -or 1, within the

X + will denote W c+x w

'c+x'

which will make

no sensible difference in the result, but be a great deal simpler

Now the fluent of the first side of this equation is evidently-x+2c ×h. I. (c+x); and the fluent of the latter

fore the general fluential equa. is −x + 2exh. I. (c+x)=

-W

64a

Xh. I. (+). But when x=0, and v the initial

(v2+).

velocity, this becomes 0+2c Xh. l. c= Xh. l. (v2+~~);

theref. by subtraction the correct fluents are -x+2cxh. 1.

c+x

C

64a

av2+w
av2+w'

when the first velocity v is dimi

nished to any less one v; and when it is quite extinct, the

state of the fluents becomes -x+2c × h. l.

C

h. l.

Here, in the quantity h. 1.

c+x

the term x is always small

C

in respect of the other term c; therefore, by the nature of

Now the latter side of this equation

is the same value for x as was found in the 5th problem, which therefore put = b; then the value of x will be easily found

2c-x

from the formula x=b, by a quadratic equation. Or,

2c+x

still easier, and sufficiently near the truth, by substituting b for x in the numerator and the denominator of

-x=b, and hence x =

2c+b 2c-b

2c-x

> 2c+x

then

2c-b b, or by proportion, as 2c+b 2c-b: 2c+bb: x: that is, only increase the value of x, found by prob. 5, in the ratio of 2c - 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, for the 2d example to the 5th prob. where x was = 6420; therefore as 2c-b: 2c+b or as 105045: 119055:: 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.

The fluxion of the time is t= But the general equa

tion of the fluxions of the space x and velocity v, in the last

x= X
c+x 32 av2+w

hence == ㄨ V 32

C-x av2+w'

C-X.

prob. was

W c+ x
; ther.x+: X

x

W c+x

But x, which is al

be substituted for may

with

ways small in respect of c, is nearly b as determined in the

last problem; theref.

c+b c-b out sensible error; and then becomes = X

X

c+b 32 c-b 6, in the constant ratio also in the same con

Now, this fluxion being to that in prob. of c b to c + b, their fluents will be stant ratio. But, by the last prob. c 54000, and b = 2955 for the first example in prob. 5; therefore c-b=51045, and c+b=56955, 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 obiect 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 increases, 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 diminu tion of motion by the air's resistance. It is also evident that 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

*This reasoning is n t conclusive. The velocity of the descending body ins creases continually, but never becomes equal to a certain determinate velocity.

therefore

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 first of the three tables, the weight of the ball being 1.05lb. we perceive that the resistance increases in the 2d column, till 0.69 opposite to 200 velocity, and 1.56 answering to 300 velocity, between which two the proposed resistance 1.05, 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 60870, theref. 1.05, we must say, as 0.69: 1·05:: 2002: v2 = 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 0000172572r, near the end of problem 2, where r is the resistance to the velocity v, by making 105

1·05=r; for then v2 =

for 2 as before.

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 :: d 2462 1.965

2462:

d=v2 and hence v=246/ 1.965

=

175 ✓ d. If we take here the 31b 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, And if we take the with which the 3lb ball will descend. 6lb ball, whose diameter is 3.53 inches, as in the 3d table of 330, resistances: then 3·53 = 1·88, and 1751 × 1·88 = being the greatest velocity that can be acquired by the 61b ball, and with which it will afterwards uniformly descend. For a 91b ball, whose diameter is 4.04, the velocity will be 175 × 2.01 = 353. And so on for any other size of iron Where the first column conball, as in the following table. tains the weight of the balls in lbs; the 2d their diame

ters