In dealing with the trajectory of the howitzer’s projectile through airless space we have no concern with its diameter nor with its weight. We use the little diagram, Fig. 81, to represent the motions,—_c_ being a horizontal line, _a_, a vertical one, the angle at B is therefore a right angle, and we assume that at A to be 20°. Now, the most elementary geometry teaches us that every triangle having these angles will have the lengths of its sides in the same invariable proportions one to another whatever may be the size of the triangle itself, and it has been found convenient to calculate these proportions once for all, not merely for angle 20°, but for every angle up to 90°. Besides this, distinct names have been given to the proportions of every side of the triangle to each of the other two sides. Thus in the triangle before us, if we take _a_, _b_, and _c_ to represent the numbers expressing the lengths of the sides against which they are placed, _a_ divided by _b_, that is _a_ ÷ _b_, or _a_/_b_, is called the _sine_ of angle 20°, while _c_/_b_ is named the _cosine_ of that angle, etc. These therefore are _numbers_ which are given in mathematical tables, and we find by these that _sine_ 20° = 0·3420201, and _cosine_ 20° = 0·9396926, and these with the initial velocity give us all the data we require. We may first find the _time_ the projectile would take to reach the ground level, or strictly that of the muzzle of the gun at B. Taking _t_ to stand for this time, we know that AC = 1,120 × _t_, but CB will be the distance that a body would fall from rest at C by the influence of gravity in that same time, _t_, and it is known by experiment that this distance is 16·1 feet multiplied by the _square_ of the time from rest in seconds. We have now therefore the length of the line CB, and put _a_/_b_ = CB/AC = (16·1 × _t_^2)/(1,120 × _t_) = _sine_ 20° = ·3420201, and dividing numerator and denominator by _t_ and multiplying the above 3rd and 5th expressions by 1,120, we have
1,120 × ·3420201 and therefore _t_ = ———————————————— = 23·7927 secs. 16·1
Having obtained the time, it will be easy to work out the lengths _b_ and _a_ as 26,648 ft. and 9114·1 ft. respectively; and as _c_/_b_ = _cosine_ 20°, we have _c_ = 26,648 × ·9396926 = 25040·8 ft., which is the _range_. The trajectory will be a curve (parabola) symmetrical on each side of a vertical line half-way between A and B, and the length of this line within the triangle will be equal to half of _a_, and in half of 23·7927 seconds the projectile, supposed to move only along the line AC, would reach the point where this vertical axis intersects AC. If during this half-time it had been falling from rest at the same intersection, it would have reached a point below by a space just one quarter of CB (the spaces fallen through being as the _squares_ of the times), and therefore at this its highest point its distance above AB would also be one quarter the length of _a_ = 2278·525 ft., which distance is called the _height_ of the trajectory; and the descending curve being in every respect symmetrical to the ascending branch, the angle at which this would be inclined to AB would be 20°, but in the opposite direction to BAC, while the velocity would be the same as at A. We may now compare these results with those calculated when the air resistance is taken into account:—