Among the readers of this book, without a doubt, there are those who wish themselves to check the calculations mentioned above. Here these calculations. They are only approximately correct, as it is based on the assumption that the channel of the gun, the projectile is moving uniformly-accelerated (in fact, the increased speed is uneven).

For calculations have to use the following two formulas uniformly-accelerated motion:

the speed *v* after *t*-th second equal *at*where *and* acceleration:

*v = at*;

the path *S*, past *t* seconds, is determined by the formula

*S = at*^{2}/2.

These formulas define a first acceleration of the projectile when it is slid in the channel "the columbiad".

From the novel the known length of the cannon, not occupied by the charge - 210 m; it is traversed by the projectile path *S*.

We know the final velocity: *v* =16 000 m/s. Data *S* and *v* allow to determine the value of *t* is the duration of the movement of the projectile in the gun (looking at this movement as a uniformly-accelerated). In fact:

*v* = *at* = 16000,

where

t = 210/8000 = about 1/40 sec.

The projectile, as it turns out, slipping inside the gun would only 1/40 of a second! Substituting

t = 1/40 in the formula* v = at*, we have:

16 000 = 1/40 and where *and* = 640 000 m/s^{2}.

Hence, the acceleration of the projectile in motion in the channel is equal to 640 000 m/s^{2}, i.e. in 64 000 times the acceleration of gravity!

What length should the gun to get the acceleration of the projectile was only 10 times greater than the acceleration of a falling body (i.e., equal to 100 m/s^{2})?

This is a task, the reverse of which we are now decided. Data:

a = 100 m/s^{2},

*v* =11 000 m/s (in the absence of atmospheric drag this speed is sufficient).

From the formula *v = at* are:

11000 = 100*t*, where t = 110 sec.

From the formula *S = at*^{2}/2 we get that the length of the gun should be

i.e. round score of 600 km.

Such calculations obtained the figures, which destroy tempting plans of the heroes of Jules Verne.