To find the Number of Balls in the complete Pile :

=

Number of balls for the whole pile 11480 Log. = 4.059942

Now, 11480-1330=10150 is the number of shot in the incomplete pile.

PROBLEM XVI.

To find the Number of Balls in an incomplete square Pile.

RULE.

Find the number of balls in the whole pile, considered as complete, by Problem XIII., page 567; and find also, by the same problem, the number of balls answering to the square pile, each side of whose base is represented by the number of shot in each side of the top course of the incomplete pile diminished by 1; then, the difference of the two results will be the number of shot remaining in the pile.

Example.

Required the number of shot in an incomplete square pile; each side of its bottom course containing 24 balls, and each side of its top course 8 balls?

To find the Number of Balls in the complete Pile :

Number of balls for the whole pile = 4900 Log.

= 3.690196

To find the Number of Balls deficient :

Balls in each side of top course = 8-1=7
Diminished course, or 7, increased by 1,=8
Twice ditto, increased by 1,

Log.=0.845098

Log.=0.903090

=

15

.

.

Log. 1. 176091

9.221849

Constant log. =

Number of balls wanting =

140

Log.=2.146128

Now, 4900-140 = 4760 is the number of shot in the incomplete pile.

PROBLEM XVII.

To find the Number of Balls in an incomplete rectangular Pile.

RULE.

Find the number of balls in the whole pile, considered as complete, by Problem XIV., page 567; and find also, by the same problem, the number of balls answering to the rectangular pile, whose sides are represented by the respective sides of the top course of the incomplete pile, the number of shot in each side being diminished by 1; then, the difference of the two results will be the number of shot remaining in the pile.

Example.

Required the number of shot in an incomplete rectangular pile; the length of its bottom course being 40 balls, its breadth 20, and the length of its top course 29 balls, and its breadth 9?

To find the Number of Balls in the complete Pile :-
Bottom course, 40 × 3 = 120

Number of balls for whole pile=7070 Log.=3.849419

To find the Number of Balls deficient :

Top row, 29-1 = 28 × 3 = 84

Now, 7070924 6146 is the number of shot in the incomplete pile.

Note. In triangular and square piles, the number of horizontal rows or courses is always equal to the number of balls in one side of the bottom row; and, in rectangular piles, the number of horizontal rows is equal to the number of balls in the breadth of the bottom row. In these piles, the number of balls in the top row or edge is always one more than the difference between the number of balls contained in the length and the breadth of the bottom row.

PROBLEM XVIII.

To find the Velocity of any Shot or Shell.

RULE.

From the logarithm of twice the weight of the charge of powder, in pounds, subtract the logarithm of the weight of the shot: to half the remainder add the constant logarithm 3.204120, and the sum (rejecting 5 in the index,) will be the logarithm of the velocity in feet, or the number of feet which the shot or shell passes over in a second.

Example 1.

With what velocity will a 24-pounds ball be projected by 8 lbs. of powder ?

Half the remainder =

Twice the charge =
Weight of the shot

= 1.204120.

4.911954

Constant log. =

3.204120

Log. 3.116074

Example 2.

Velocity of shot, in feet,=1306

With what velocity will a 13-inch shell, weighing 196 lbs., be discharged

Note. The constant logarithm made use use of in this problem is the logarithin of 1600 feet, which is the velocity acquired by a 1 lb. ball, when fired with 8 ounces of powder.

PROBLEM XIX.

To find the terminal Velocity of a Shot or Shell; that is, the greatest Velocity it can acquire in descending through the Air by its own Weight.

RULE.

For Balls. To half the logarithm of the diameter of the ball, in inches, add the constant logarithm 2. 244277; and the sum will be the logarithm of the terminal velocity of the ball.

And, for Shells.-To half the logarithm of the external diameter of the shell, in inches, add the constant logarithm 2. 168203; and the sum will be the logarithm of the terminal velocity of the shell.

Example 1.

Required the terminal velocity of a 24 lbs. ball, its diameter being 5. 6 inches?

Diameter of the ball = 5.6 Log. 0.748188

Required the terminal velocity of a shell weighing 196 lbs., its external

Note. The constant logarithms made use of in this problem are the respective logarithms of 175.5 and 147.3, the established multipliers for shot and shells. It is by this problem that the terminal velocities contained in Tables A and B, following, have been computed.

PROBLEM XX.

To find the Height from which a Body must fall, IN VACUO, in
order to acquire a given Velocity.

RULE.

Since the spaces descended by falling bodies are as the squares of the velocities, and as a fall of 16 feet produces a velocity of 32} feet,therefore, as the square of 32 feet, is to 16 feet; so is the square of any other given velocity, to the altitude from which it must fall, to acquire such velocity. Hence the following logarithmical expression :

To twice the logarithm of the given velocity, in feet, add the constant logarithm 8: 191564; and the sum, (abating 10 in the index,) will be the logarithm of the required altitude, or height.

Example 1.

From what height must à body fall, in order to acquire a velocity of 1340 feet per second?

=

Given velocity = 1340; twice its log. 6. 254210
Constant log. =

8. 191564

Altitude, or height, =27911

Log. 4.445774

Example 2.

From what height inust a body fall, in order to acquire a velocity of 1670

feet per second ?

Given velocity = 1670; twice its log. = 6.445434
Constant log. =

8. 191564

Altitude, or height, = 43352 Log. 4.636998

Note-It is by this problem that the altitudes in Tables A and B, following, have been computed; but, since the fractional parts beyond 16 and 32 were omitted, and the constant logarithm, in consequence thereof, assumed at 8. 193820, the respective altitudes, in these Tables, are something beyond the truth.