296

And conversely, to find what velocity will remain after

passing over any space, as 4000 feet, the first velocity being v=200. Here the hyp. log. of ———

80

=

107

is

5350

=

4000

5350

=

400

535

74766, the natural number of which is 2·1120,

V

that is, 2·112 =; therefore v =

velocity.

i

Again, for the time t: since =

32cd

νυ

-So, for example,

if v = 300, and v = 100; then

300 velocity to 100, or of passing over the space 5878 feet.

And, reversing, to find the velocity v, answering to any.

as above, is brought out for small velocities, will also serve for the higher ones, if we employ the medium resistance between the two proposed velocities, as was done in prob. 5. Thus, as in the first example of this problem, where the two velocities are 1780 and 1500, the resistance due to the velocity 1700, in the first table of resistances, being 74-13, say as 1700: 17802:: 74·13: 81.27, the resistance due to the velocity 1780; then the mean between 81.27 and 57.25, due to 1500 velocity, is 69-26, or rather take 694. Again, as 65769: 1600 1646, the velocity due to the medrum resistance 691. Hence, as in prob. 5, as 16462: 02:: 69: 00002565v2 = suppose av, the resistance due to any velocity v, between 1780 and 1500, for the 105 lb ball. And, as 1965: 5.62 :: av2 : 8·124αp2 =·00020838v2 = bv2;

resistance due to the same velocity with the 24lb 24f, and 32f4bux, and

ball. Therefore

3

= the correct fluent of which is x h. 1.÷= //

460

x h. 1. 150

46

46

178

3

x h. l.

46

89 75

3600 × 171148 616 the

velocity sought.

PROBLEM XI.

To determine the Ranges of Projectiles in the Air.

To determine, by theory, the trajectory a projectile describes in the air, is one of the most difficult problems in the whole course of dynamics, even when assisted by all the experiments that have hitherto been made on this branch of physics; and is indeed much too difficult for this place, in the full extent of the problem: the consideration of it must therefore be reserved for another occasion, when the nature of the air's resistance can be more amply discussed. Even the solutions of Newton, of Bernoulli, of Euler, of Borda, &c, &c, after the most elaborate investigations, assisted by all -the resources of the modern analysis, amount to no more than distant approximations, that are rendered nearly useless, even to the speculative philosopher, from the assumption of a very erroneous law of resistance in the air, and much more so to the practical artillerist, both on that account, and from the very intricate process of calculation, which is quite inapplicable to actual service. The solution of this problem requires, as an indispensable datum, the perfect determination by experiment of the nature and laws of the air's resistance at different altitudes, to balls of different sizes and densities, moving with all the usual degrees of celerity. Unfortunately however, hardly any experiments of this kind have been made, excepting those which on some occasions have been published by myself, as in my Tracts of 1786, as well as in my Dictionary, some few of which are also given in the 2d vel. of this course, art. 105, with some practical inferences. And though I have many more yet to publish, of the same kind, much more extensive and varied, I cannot yet undertake to pronounce that they are fully adequate to the purpose in hand.

All that can be here done then, in the solution of the present problem, besides what is delivered in the 2d volume, is to collect together some of the best practical rules, founded partly on theory, and partly on practice. 1. In the first place then, it may be remarked, that the initial or first velocity of a ball may be directly computed by prob. 17, near the end of our 2d volume; having given the dimensions of the piece,

the

24

the weight of the ball, and the charge of powder. Or other wise, the same may be made out from the table of experimented ranges and velocities in pa. 161 of that volume, by this rule, that the velocities to different balls, and different charges of powder, are as the square roots of the weights of the powder directly, and as the square roots of the weights of the balls inversely. Thus, if it be enquired, with what velocity a 24lb ball will be discharged by 8 lb of powder. Now it appears in the table, that 8 ounces of powder discharge the 1lb ball with 1640 feet velocity; and because 8 lb are = 128 ounces; therefore by the rule, as ::: 1640: 1640 1640/= 1339, the velocity sought. Or otherwise, by rule 1 p. 162 of the 2d vol. as 24 : √16 : : 1600 1306, the same velocity nearly. But when the charges bear the same ratio to one another as the weight of the balls, that is when the pieces are said to be alike charged, then the velocities will be equal. Thus, the 1lb ball by the 2 oz charge, being the 8th part of the weight, and the 24lb ball, with 3lb of powder, its 8th part also, will have the same velocity, viz, 860 feet. In like manner, the 1230 tabular velocity, answering to 4 oz of powder, the 4th part of the ball, will equally belong to the 24lb ball with 6 lb of powder, being its 4th part, and the tabular velocity 1640, answering to the 8 oz charge, which is the weight of ball, will equally belong to the 24 lb ball with 121b of powder, being also the of its weight.

2. By prob. 9 will be found what is called the terminal velocity, that is, the greatest velocity a ball can acquire by descending in the air; indeed a table is there given of the several terminal velocities belonging to the different balls, with the heights, in an annexed column, due to those velocities in vacuo, that is the heights from which a body must fall in vacuo, to acquire those velocities.

3. Given the initial velocity, to find the elevation of the piece to have the greatest range, and the extent of that range. These will be found by means of the annexed table, altered

A

from

9, and look for the quotient in the first column here annexed. Against this, in the 2d column will be found the elevation to

give the greatest range; and the number in the 3d column. multiplied by a, the altitude due to the terminal velocity, also found in the table in problem 9, will give the range, nearly.

1640

Ex. 1. Let it be required to find the greatest range of a 241b ball, when discharged with 1640 feet velocity, and the corresponding angle to produce that range. By the table in prob. 9, the terminal velocity of the 241b ball is 415, and its producing altitude 2691: hence 3.95, nearly equal to 3.9865 in the 1st column of our table, to which corresponds the angle 34° 15', being the elevation to produce the greatest range; and the corresponding number 2.9094, in the 3d column, multiplied by 2691, gives 7829 feet, for the greatest range, being nearly a mile and a half.

415

Exam. 2. In like manner, the same ball discharged with the velocity 860 feet, will have for its greatest range 3891 feet, or nearly 2 of a mile, and the elevation producing it 39° 55'.

These examples, and indeed the whole table in the 9th

problem,

problem, are only adapted to the use of cannon balls. But it is not usual, and indeed not easily practicable, to discharge cannon shot at such elevations, in the British service, that practice being the peculiar office of mortar shells. On this account then it will be necessary to make out a table of terminal velocities, and altitudes due to them, for the different sizes of such shells. The several kinds of these in present use, are denominated from the diameters of their mortar bores in inches, being the five following, viz, the 4'6, the 5.8, the 8, the 10, and the 13 inch mortars, as in the first column of the following table. But the outer diameters of the shells are somewhat smaller, to leave a little room or space as windage, as contained in the 2d column.

Table of dimensions, &c, of Mortar Shells.

The 3d column contains the weight of each shell when the hollow part is filled with powder: the diameter of the hollow is usually of that of the mortar: the weight of the shells empty and when filled, with other circumstances, may be seen at Quest. 53, pa. 265, vol. 2. On account of the vacuity of the shell being filled only with gunpowder, the weight of the whole so filled, and contained in column 3, is much less than the weight of the same size of solid iron, and the corresponding weights of such equal solid balls are contained in col. 4. The ratio of these weights, or the latter divided by the former, occupies the 5th column.

Now because the loaded or filled shells are of less specific gravity, or less heavy, than the equal solid iron balls, in the ratio of 1 to 142, as in column 5, the former will have less power or force to oppose the resistance of the air, in that same proportion, and the terminal or greatest velocity, as determined in the 9th prob. will be correspondently less. Therefore, instead of the rule there given, viz, 175.5✅✅d, for that velocity, the rule must now be 175°5 / =147·3√d=v,

d

the