Page:Encyclopædia Britannica, Ninth Edition, v. 1.djvu/223

resistance appears from experiments to vary pretty nearly as the velocity; and when the motion is very swift, as in the case of a rifle-bullet, as the cube of the velocity; but when the motion is neither very rapid nor very slow, the law of the square of the velocity probably represents the truth very fairly. By ${\displaystyle g^{\prime }}$ is denoted the value of gravity at the height ${\displaystyle x,}$ so that

${\displaystyle a}$ being, as above, the radius of the earth. In the exponential term, we shall replace ${\displaystyle g^{\prime }}$ by ${\displaystyle g,}$ as no sensible error can result therefrom. The value of ${\displaystyle \sigma v}$ is constant, as by Boyle's and Marriotte's law it always ${\displaystyle =\sigma _{0}v_{0}.}$ Writing, therefore, for brevity—

the equation of motion takes the form

whence, following the usual rule for the integration of linear differential equations of the first order, and writing ${\displaystyle {\text{X}}}$ for ${\displaystyle e^{-nx},}$ for convenience of printing,

Herein put ${\displaystyle x=0,}$ so that ${\displaystyle u=u_{0},}$ and we have

whence, by subtraction,

therefore

in which ${\displaystyle {\text{Ei}}x}$ is used to denote the exponential integral of ${\displaystyle x,}$ viz.: ${\displaystyle \int _{-\infty }^{x}{\frac {e^{x}}{x}}dx,}$ according to a recognised notation. The values of the integral ${\displaystyle {\text{Ei}}x,}$ which may be regarded as a known function, have been tabulated (see Philosophical Transactions for 1870, pp. 367–388).

We thus have, except for temperature, the complete solution of the problem of the motion of the balloon so far as velocity and height are concerned; it would not be possible to connect the time and the height except by the performance of another integration, for the practicability of which it would be necessary to submit to some loss of generality, viz., we should have to regard ${\displaystyle x}$ as small as compared to ${\displaystyle a,}$ and take ${\displaystyle \lambda }$ as small, and so on. The equation last written gives the motion until the height (say ${\displaystyle h}$) is attained at which the balloon becomes quite full, after which the gas begins to escape, and we have the second case of the problem.

Before proceeding, however, to the discussion of this second case, it is worth while to examine the solution more carefully, leaving out of consideration quantities that make no very great difference in the practical result, for the sake of simplicity. Supposing, then, gravity to be constant at all heights, and ${\displaystyle \lambda }$ to be zero, the equation of motion takes the simple form

and we see, what is pretty evident from general reasoning, that if a balloon, partially filled, rises at all, it will at least rise to such a height that it will become completely full.

The letters meaning the same as before, the equation of motion of a balloon completely filled at starting is

or substituting for ${\displaystyle \rho }$ and ${\displaystyle \sigma }$ their values

The integral of this differential equation could be obtained in series as before, only that the resulting equations would be more complicated. As we do not propose to discuss the formulæ obtained, it will be sufficient for our purpose to deduce an approximate solution by neglecting ${\displaystyle {\text{V}}_{0}\rho _{0}(1-e^{nx})}$ compared to ${\displaystyle {\text{M}},}$ viz., neglecting the mass of the gas that has escaped during the ascent compared to the mass of the whole balloon and appurtenances. It must be borne in mind, however, that when coal gas is used, and the ascent is to a great height, the mass of gas that escapes is by no means insensible. The equation thus becomes

or

${\displaystyle \gamma }$ being ${\displaystyle {\frac {2g}{\text{M}}}.}$ This is an equation which can be integrated in exactly the same way as that previously considered, viz., by multiplying by a factor ${\displaystyle e^{-m{\text{X}}},}$ and integrating at once; thus,