**Theory of Equilibrium of the Balloon**

The principle in virtue of which a balloon ascends is exactly the same as that which causes a piece of wood or other material to float partially immersed in water, and may be stated as follows, viz., that if any body float in equilibrium in a fluid, the weight of the body is equal to the weight of the fluid displaced. By the Îfluid displaced" is meant the fluid which would occupy the space actually occupied in the fluid by the body if the body were removed. When the fluid is inelastic and incompressible, i.e., a liquid, as water, its density is the same throughout, and bodies placed in it either rise to the surface and float there partially immersed, or sink to the bottom. Thus, suppose a body only one-third as heavy as water (in other words, whose specific gravity is one-third) was floating on the surface of water, then, as the weight of the body must be equal to that of the water it displaces, it is clear that one third of the body must be immersed. In the case, however, of an elastic or gaseous fluid, such as air, the density gradually decreases as we recede from the surface of the earth, for each layer has to support the weight of all above it, and as air is elastic or compressible, the layers near the earth are more pressed upon, and therefore denser than those above. Thus, if a body lighter than the air it displaces be set free in the atmosphere, it rises to such a height that the air there is so attenuated that the weight of it displaced is equal to that of the body, when equilibrium takes place, and the body ascends no higher. In all cases, therefore, a body floating in the air is totally immersed, and it can never get beyond the atmosphere, and float, as it were, upon its surface.

To find therefore, how high any body (lighter than the air it displaces), such as a balloon, of given capacity and weight, will rise, it is only necessary to calculate at what height the volume of a quantity of air equal to the given capacity will be equal in weight to the given weight. Leaving temperature out of the question, the law of the decrease of density in the atmosphere is such that the density at a height x is equal to FORMULA x the density at the earth's surface, g being the measure of gravity, and k also a constant; the value of FORMULA is called the height of the homogeneous atmosphere, viz, it is equal to what would be the height of the atmosphere if it were homogeneous throughout, and of the same density as at the earth's surface. Its value may be taken at about 26,000 feet. Thus, let V be the volume of a balloon and its appurtenances, car, ropes, &c., (viz., the number of cubic feet, or whatever the unit of solidity may be, that it displaced), and let W be its weight (including that of the gas), then it will rise to a height x such that

FORMULA

G being the value of the force of gravity, and xo being the density of the air at the surface of he earth. This equation is not quite accurate, for several reasons -- (1) because the decrease of temperatures that results from increase of elevation has not been taken into account; (2) because of has been taken to measure the force of gravity on the earth's surface, whereas it should represent this force at a height x; this is easily corrected by replacing g by g, where of = FORMULA a being the radius of the earth, but as a is about 4000 miles, and x is never likely in any ordinary question to exceed 10 miles, we can replaced g by g without introducing sensible error, for the correction due to this cause would be much less than other uncertainties that must arise; and (3) because W and V could not both remain constant. If the balloon be not fully inflated on leaving, so that the gas contained in it can expand, then V, the volume of air displaced, will increase, while, if the balloon be full at starting, the envelope must either be strong enough to resist to resist the increased pressure of the gas inside, due to removal of some of the pressure outside (owing to the diminished density of the air), or some of the gas must be allowed to escape. The former alternative of the second case could not be complied with, as the balloon would burst; some of the gas must therefore escape, and so W is diminished. The weight of gas of which the balloon is thus eased cannot properly be omitted from the calculation, if x be considerable; but a good approximation is obtained without it, as the weight of the gas that escapes will generally bear a small proportion to the weight of balloon, car, grapnel, passengers &c. The true equation (except as regards temperature) is therefore, for a balloon full at starting-

FORMULA

vo denoting the volume actually occupied by the gas, g denoting FORMULA viz., gravity at height x, and po being the density of the gas on the ground. It will generally be sufficient especially when temperature is omitted, to take the formula in the approximate form written previously. As the volume of air displaced by the car, ropes, passengers, &c., is usually trifling compared to that displaced by the balloon itself, no great error can arise from taking vo = Vo. As an example, let us find now high a balloon of 1000,000 cubic feet capacity would rise if inflated with pure hydrogen gas, carrying with it a weight of 3000 lb (this including the weight of the balloon itself and appurtenances). A cubic foot of air, at temperature 32o Fahr., and under a pressure of 29.922 in., weighs 080728 lb, and a cubic foot of hydrogen weighs 005592 lb, so that 9supposing the barometer reading on the earth to be 29.922, and the temperature of the air to 32o) at the surface of the earth the balloon, &c., weighs 3559 lb, and the weight of the air displaced is 8073 lb. The balloon will therefore approximately rise to such a height x that 100,000 cubic feet of air shall there weigh 3559 lb; and x is given in feet by the equation.

FORMULA

Or x = 26,000 (log 8073- log 3559),

The logarithms being hyperbolic; if common or Briggian logarithms be used, the result must be multiplied by 2-30258. . . (the reciprocal of the modulus). In the above case we find x = about 21,000 feet, and a at this height rather more than half the gas will have escaped (it having been supposed that the balloon was full at starting). This only reduces the value 3559 by about 300, and the result of taking it into account is only to increase the height just found by about 200 feet. If 2000 lb out of the 3000 were thrown away during the ascent, the balloon would reach a height of about 10 miles; the weight of the gas gas that escapes is here important, as if it be not taken into account, the height given by the formula is only about 8 miles.

In actual aerostation, as at present practiced, ordinary coal gas is used, which is many times heavier than hydrogen, being, in fact, usually not less than half the specific gravity of air. Even when balloons are inflated with hydrogen, generated by the action of sulphuric acid on zinc filings, the gas is very far from pure, and its density is often double that of pure hydrogen, and even greater.

The hydrostatic laws relating to the equilibrium of floating bodies were known long previous to the invention of the balloon in 1783, but it was only in the latter half of the 18th century that the nature of gases was sufficiently understood too enable these principles to have been acted on. As we have seen, both Black and Cavallo did make use of them on a small scale, and if they had thought it possible gas they could have easily anticipated the Montgolfiers. As it was, so sooner was the fire-balloon invented, than Charles at once suggested and practically carried out the idea of the hydrogen or inflammable air balloon.

**Read the rest of this article:**

Aeronautics - Table of Contents