748.

LET 6 be the capacity of the barrel, r that of the receiver; then since the defect from the standard altitude has become

h

a

and density ∞ defect (Vince, Prop. LXVII.)
we have
h: h

PNEUMATICS.

:: x2 + (l and x =

:. 2b + r : b + r :: hù : (h − a)= and b:r :: h2-—(h—a) : 2 (h — a) — z

h

a density at first

ditto after n turns :: (2b + r)" : (b + r)"

749.

Let x denote the space through which the elasticity of the air in the tube depresses the mercury; L the length of the tube, h the standard altitude, and a the depth of the air in the tube before inversion; then (Vince's Hyd. Prop. LXI.) we have

h: x :: x + d h: a

h) × x = ah

-(l - h) ± √ (1 − h)2 + 4ah)

2

one value of which indicates the space through which the air depresses the mercury, and the other the altitude of mercury it would support if placed above it, by its pressure against the top of the tube.

750.

Lets be the specific gravity of the air at the Earth's surface, A the area of the given plane, supposed to bear no proportion to the great circle of the earth, and r the radius of the earth; then ns is the weight of a cubic foot of the given uniform medium at the earth's surface, and by the law of gravitation, its weight at the distance

R+ x

from the earth's centre is

R2

(R + x)2

Hence the increment of the pressure or weight of the medium upon the plane is

Rins

(R + x)2

AR'ns x

x ns

× Adx

and that pressure or weight is

GR

Let xoo. Then

R

1

R+x

1

751.

and the force of gravity F c

=

and the pressure required is measured by

nsAR.

Hence it appears that the pressure of the infinitely high column of any substance whatever is equal to the weight of a column of the same base, and whose altitude is equal to the radius of the earth.

1

(dist.) q

Generally, let the compressing force ∞ (density) ∞ D,

0.

of D.

Let ♪ be the density at the surface of the earth, and R the ra

dius of the earth; then since

M =

and we get

D=

which gives

a = e

and its value is that with which this fraction vanishes when for p and D we put R and . Hence by the rule for estimating such

P

fractions

1

Rx

p-1 pla

9-1

R2 ♪~2 × (R'~^ — p'~') + ♪~ft.

SP-2

752.

Let yxa be the equation to the DensityCurve, (see Vince's Hyd. p. 82), y representing the density at the distance x from the surface of the earth, and ♪ the density at the surface. Then if h be the height of the homogeneous atmosphere, as determined by the barometer, we have (see Vince),

hydx

dy

la

:: y = axe

(1).

Let D be the density of mercury, determined at the time this formula is applied, and b the height at which it then stands in the tube of the barometer; then

(2).

Hence the distance from the surface of the earth of the centre

of gravity of the cylinder whose altitude is x, is

น

dP

dQ

when x = 0.

응.

=

d.e

1

To verify this result, we should have u 0, when ≈ = 0.

But we then have

ebD

If, however, we find u by the rule for Vanishing Fractions, we

have

+ C′

-8

d. ebD

+ C

753.

Let W be the weight of the balloon and all its appendages, the specific gravity or density of the atmosphere at the surface of the earth when the barometer stands at b feet, and

a

that of the gas. Also let y denote the density of that stratum

N

whereat the balloon will cease to ascend, and let c3 be the capacity of the balloon in cubic feet. Then since the weight of the air displaced by the gas is

C3 X Y

and that of the balloon, its appendages and

gas,

W + cx =

n

.. by the question (Vince. Prop. XIV.)

is

of 1000); and ♪ =

6

5

n

x l.

noc

3

nW + dea

x= 42057 × 1.

the height required in feet.

Ex. 1. Let the gas be hydrogen, or n = 13; b = 30 inches =

5

feet; D = 14019 (water being supposed to have the density

See Barlow's Tables.

x = 42057 x 1.

n

42057 x 1.

Then

78c3

65 W + 6c3

Let it further be supposed that W 20 stone = 4480 ounces, and c3

200000 cubic feet.

Then

156000

149 12

4875

466

42057 × 2.34768 feet nearly.

.. x = 98736 feet

-32912 yards

(1)

(2)

18 miles 1232 yards.

This prodigious altitude is owing to the disproportion of the balloon to the weight of the load and materials. The surface of such a machine, even under the form of a sphere, is about 24000 square feet, and the weight of that surface would consequently be considerably greater than the weight here attributed to it.