Ex. 2. Given W 20 stone, and the other elements as before, to determine the magnitude of the balloon necessary just to lift that weight from the ground.

Here x 0. Consequently

78c3

65 × 4480+6c3

65 × 4480

72

= 1

Hence if r be the radius of a spherical balloon of this capacity, we have

-4044

4044 cubic feet.

4πr3

3

and r3 = 1000 nearly.
10 feet nearly.

..

Even such a vehicle, it appears, would be bulky and expensive. Its exterior would require upwards of 1200 square feet, or 133 square yards of silk.

755.

Professor Playfair, (p. 254, vol. 1, of Outlines of Natural Philosophy,) has erroneously treated this subject. The weight of the gas is there wrongly computed.

754.

Since the altitudes of fluids in a tube are inversely as their specific gravities, if a be altitude of the water sustained by the air, we have

1

: 30 ::

13568

1

1000

.. a 413.824 inches.

and the required length is ..

a

1=

sin. 30°

: a

=2a827.648 inches.

0

where a is the depth of air in the tube before inversion, h the standard altitude, and a the defect from that altitude, caused by the elasticity of the air.

But by the question,

x = 33

( 4

133 inches, and a =

Therefore

W-E

Then

F= 1

PNEUMATICS.

756. Let E denote the elasticity of the fluid when occupying a given space rh, r being the radius of the base of the tube; then its elasticity, when occupying the space

Tr2x

is Ex

h

Hence if W denote the weight of the piston, the moving force upon

it is

Hence (vdv=

v2

v2 = 2

and the force which accelerates its descent is

E

h

X
W 20

Fdx)

Eh

{
{

Let xh-S

=2

+28) = 4

W

Eh
W

Eh
W

× lx

3

4

inch.

inches

--X + C

x }

{

which gives the velocity required.

x l. 22+h

x l.

h-S

x}

S
+ s}

If the fluid be atmospheric air, its elasticity in a natural state, or when the barometer stands at 30 inches, is about 151b. upon the square inch, and we have

E2 x 15 lbs. ;

and if in addition to the piston there is pressing upon it the whole atmospherical column of air, for W we must put W + r2 x 15, and we then get

v2 = 2.

If the time of acquiring this velocity be required, we have

dx dt =

1

dx

√2 Eh
(h-

W

Let b=

757. ceiver after t turns is (Vince, p. 112,)

(r+tb).

d = D x

r

= D' x

h-S

157r2h
15%12+ W h

=

In the condenser, the density of the air in the re

(1)

D being the density at first, and r and b the capacities of the receiver and barrel.

Again, in the air-pump, after t turns the density, is (Vince, 107,)

(2)

(b+r)
(2b+r)*

t = 3; then we have

""

10 + 3

10

or the density has increased

Also D' x

D' - D' x

7. + S}

353

1654

113

123

or the density has decreased

1.h +

ths.

353

Eh

W

ths.

Also

758. Let R be the radius of the body of the pump, r that of the suction-tube. Also let H be the height of a column of water whose weight = weight of a column, having the same base, of the atmosphere, s the length above the water of the suction-pipe, p the play of the piston, u the height of a column of water equivalent to the pressure of the air below the piston, and r the elevation due to the first stroke. Then

x+uH ..

(1)

Hence

= {A

2

by supposition.

Hence

R$

r2 − (H + 22 p + s) x = −

TXS

TRp+πr2. (s-x)

72sH
Rp+r2 (s-x)

A ± √Ãa—B}

and u, u x

× H

น.

u = — { 2H — A = √ A2 — B2

X2 =

Ug =

R2

7 PH
7.2

7. (s-x) R3p + r2 (s—x1)

Again, let x, be the elevation of the water in the tube at the end of the second, and u, the elastic force of the air within the pump at that period; then as before

X2+Ug = H

·P+s)2 - 4

R2

— { A' ± √/A'—B−4≈ (H+s—x)}

1⁄ { 2H − A ‡ √ A2 — B — 4x (H+s−x)}

2

H

and so on to x

and Un

Since x and u must be both less than H, it follows that the

lower sign of the root is alone applicable.

759.

after m turns of the air-pump, we have (Vince, p. 107,)

♪ D x

1

If D be the density of the air at first, and that

- D X

Hence

But y = y

Hence

(1)

Again, let a be the required depth of air in the tube at first, and y the standard altitude; then (Vince, p. 92,)

y y - al-ɑ : x

:. xy = (l − a) (y — a)

7 being the length of the tube.

which gives

x=

xy =

Also if y denote the standard altitude when under the receiver, we have, by the same rule

y' : y'

bl-b: x

Therefore

y =

=

(b+r)TM

(2b+r)

d

D

n+1

n+2

718

.

D (1-6) (y /1/ - b)

음

bD. (1 − b) + ad (1 − a)

(a - b)

(la) (ya)

y

= (l—a)(a−b)(n+1)" ×

-

b. (n + 2)" (1 − b) + a (n + 1) = (1 − a),
(n + 1)" (a - b)

b(l—b)(n+2)TM+a(l+b)(n+1)* b(l—b)(n+2)"+a(l−a)(n+1)"

760. Let D, d, be the densities at first of the two airs, and ♪ that of the atmosphere when its elasticity is E. Also let D', d', denote the densities at any other time t.