The distance AB between the piece of ordnance and the object, is called the and also the Let tl e angle EAB contained between the vertical and the direction of the object be called the ANGLE OF POSITION = p.
And let the angle DAB, contained between that direc tion and the axis of the piece, be called the direction of the mortar = d, and let .7. express the zenith distance or angle LAD, contained between the axis of the mortar and the vertical line AL.
The leading problem from which almost all the others may be derived, is the following.
Let a shell be thrown from A, (Fig. 3, 4.) with the ve locity required by Falling through the vertical FA, so as to hit an object B. Required the direction AD of the pro jection.
Let All be a horizontal line, and AB the line of po sition of the object. In the vertical Al:, take AE= 4 AF, and cn EA describe an arch of a circle ED d A, which shall touch the line of position AB. Draw through the oh jest the vertical line BD, cutting the circle in D and d, and join Al) and A d. I say that All or A d are the di rections required. Join ED and E d.
For, because AB touches the circle in A, the angle ADE is equal to the exterior angle EA a, or DBA, and the alternate angles EAD, ADB are equal. The trian gles ADB and EAD are therefore similar, and I)B : DA = DA ; AE, and DB x EA. Therefore B is in a parabola, of which the vertical AI is a diameter, AD a tangent in A, and AE the parameter of that diameter. If, therefore, the body be projected from A in the direc tion AD, with the velocity acquired by falling through FA, the fourth part of this parameter, it will describe a para bola AVB which passes through B.
By the same reasoning, it is demonstrated that the body will hit the mark B, if projected in the direction A d with the same velocity, describing the parabola A v B.
From this very simple construction, we may draw se veral very instructive ccitollaries.
Cor. 1. When the vertical line passing through B cuts the circle EDA, it always cuts it in two points D and d, giving two directions AD and A d, either of which will solve the problem.
Car. 2. But if the vertical through b only touch the circle, as it touches it in one point only, it gives but one direction, along which the body must be projected to hit the mark b. This direction is AG.
Car. 3. The direction AG evidently bisects the angle LAB, and the directions AD and A d are equidistant from the middle direction AG.
Cor 4. If the vertical passing through B do not meet the circle described on AE, according to the conditions specified, the object is too remote to be struck by a body projected from A with the velocity acquired by _ falling from F. There is no direction that will enable it to go so far on the line AB. The distance A b is the greatest possible with this velocity, and it is attained by taking the elevation AG which bisects the angle EAB. We may therefore call Ab the maximum range on the line AB, and AG the middle direction.
Cor. 5. The distances on a given line of position to which a body will be projected in a given direction AD, are proportional to the squares of the velocities of pro jection. For the figure being similar, the range AB has the same proportion to AF, the fall necessary for acquir ing the velocity. Now the falls are in the duplicate ratio of the velocities required by falling. Therefore, &c.
The converse of this problem is solved with the same facility of construction.
Let a body be projected in the direction AD, with the velocity acquired by falling through FA, it is required to find to what distance it will reach on the line AB.
Describe, as before, on AE,=4 AF, the circle EDA, touching AB, and cutting AD in D. Through D draw the vertical DB, cutting AB in B. Then B is the point to which the projectile will reach. The proof is too evi dent to need discussion.
Lastly, suppose the object B to be given, and also the line of direction AD (which is a very common case, see ing that our mortars are often so fixed in their beds that their elevation can be very little altered) it is required to determine the velocity that must be given to the projectile.
Draw through the object the vertical BD, meeting the direction in D. Draw the vertical AE, and make it a third proportional to DB and DA, that is, make AE= D z — —A D E and take FA Then FA is the fall which B, — 4 will generate the velocity required for the projection. The demonstration of this is also very evident.