EXERCISES.

1. Find the terminal height for an iron ball 6 inches in diameter. Ans. 5400.

2. Find the terminal height for a 3-pound iron ball.

meter.

Ans. 2520. 3. Find the terminal height for a shell 12 inches in diaAns. 8640. 4. Find the terminal height for a leaden ball 2 inches in diameter. Ans. 2818.

232. In the following table, the first, third, and fifth columns, contain the actual ranges of projectiles expressed in terms of F, that is, the F for the ball in any particular case is the unit of measure; and the second, fourth, and sixth columns, contain the corresponding potential ranges, that is, with the same elevation and initial velocity, expressed in the same manner :—

Actual Potential

Range.

Range.

⚫0100

⚫0201

*0405

*0612

*0822

•1034

•1249

•1468

•1578

•2140

'3324

*4591

⚫5949

'7404

*8964 1.0638

1.2436

1.4366

1.6439

1.8669

Actual Potential
Range.

Range.

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2.0

2.1

2.2

2.3

2.4

2.5

2.6

2.7

2.8

2.9

3.0

3.1

3.2

2.1066

2.3646

2.6422

2.9413

3.2635

3.6107

3.9851

4.3890

4.8249

5.2955

5.8036

6.3526

6.9460

7.5875

8.2813

9.0319

9.8442

10.7237

11.6761

12.7078

Actual Potential

Range.

Range.

3.3

3.4

3.5

3.6

3.7

3.8

3.9

4.0

4.1

4.2

4.3

4.4

4.5

4.6

4.7

4.8

4.9

5.0

13.8258

15.0377

16.3517

17.7767

19.3229

21.0006

22.8218

24.7991

26.9465

29-2792

31.8138

34.5686

37.5632

40.8193

44.3605

48.2127

52.4040

56.9653

233. PROBLEM IV.-Given the actual range of a given projectile, at an angle not greater than 10°, and its original velocity, to find its potential range, and the elevation to produce the actual range.

Case I.-When the potential random does not exceed 39000 feet.

'Divide the actual range by the terminal height, and find the quotient in one of the columns of actual ranges in the table, and opposite to it in the next column of potential ranges is a number, which, multiplied by the preceding height, will give the potential range. The potential range and initial velocity being known, find the elevation by article 204. the terminal height in feet, r = the actual range in feet, R = the potential range in feet, r' = the actual range in the table, the potential range in the table, v = the initial velocity,

Let

F

R'

h

the impetus,

then

Then

Or,

e = the elevation,

d

=

2h

the diameter of the projectile in inches;
the potential random.

and R = FR',

F= 900 d, r' =

2

h = (3) nearly, or Lh = 2 (Lv — ·903090).

Lh=2Lv-1.806180.

R 32 R

Sin 2 e=

=

2h

v2

Or,

L sin 2e

10+ LR-L2h (art. 204).

234. In this case 2h does not exceed 39000, and v does not exceed 1050, for v=8√h=8 × 131, and e may be found without previously calculating h; by substituting in the last formula the value of L2 h, it becomes

L sin 2 e 11.505150+ LR-2 Lv.

=

EXAMPLE.-At what elevation must an 18-pounder be fired, with a velocity of 984 feet, in order that its actual range on a horizontal plane may be 2925 feet?

r =

F900d900 × 5·09 = 4581

r 2925

= =·64, and R=FR'=4581 ×·8028=3678

F 4581

Lh2 (Lv

903090) 2 (2·992995-903090) =2.089905 × 2=4∙179810, and h=15129.

=

=

L sin 2e 10+ L 3678-L2h 13.565612 - 4.480840 =9.084772, and 2 e 6° 59′, and e3° 29' 5.

235. If the potential random is required, it is easily found by the theorem in article 202, when e = 45°.

Or it is found more simply, when h is known, by article 200; for if R′′ the potential random, then R”"=2h.

Thus, for the above example,

R"

2h 2 × 15129 = 30258.

=

EXERCISES.

1. At what elevation must a 12-pound ball be fired, with a velocity of 700 feet, in order that it may reach an object 2000 feet distant? Ans. 4° 28'.

2. Find the elevation at which a ball 5 inches in diameter must be discharged, with a velocity of 800 feet, that its actual range may be of a mile. Ans. 2° 53'. When the potential random exceeds

236. Case II. 39000 feet.

'Find two mean proportionals between 39000 and the potential random; then the less of these means is to the potential random as the potential range, found by the former case, to the true potential range; then the elevation is found as before."

Find has in the preceding case, then, if

R" the potential range found by the preceding case,
Rthe true potential range,

then R00138 R′′ 3/ h2.

Or, LR 3.139977+ LR" + 3 Lh.

Instead of LR", LF +LR' may be used.

Then find e, as in the former case, or, by this formula,
L sin 2 e 6·838947 + LR” — } Lh,
which gives e at once, when h and R" are found.

=

EXAMPLE.-At what elevation must a 24-pounder be discharged, with a velocity of 1730 feet per second, in order that its actual range may be 7500 feet?

r 7500

=
F 5040

Hence, R" FR'5040 × 2·587 = 13048

=

Lh=2(Lv-903090) = 2 (3·238044-903090) 2 × 2·334956 = 4.669912, and h = 46764, and 2h93528, which exceeds 39000. L sin 2 e 6.8389469+ LR′′

Then,

=

• Lh
= 6·838947 +4·115280 — 1·556637
=9-397590 L sin 14° 28';
e=7° 14'.

=

F 900d900 × 5.6=5040, and r' =

―――

and therefore

y

Let a = 39000, then 2h being the potential random, let x and be two mean proportionals between a and 2 h; then a:x=x:y, and x: y=y: 2h; y2 X3

x2

and 2 h =

X

y

hence,

and

also,

therefore, R=

Or,

LR=3·139977+LR" + Lh.

Then e can be found for this potential range, and given initial velocity, as in the preceding case; or, Ꭱ 2h R" R" since sin 2 e= 2h2h2ath 72ah' therefore, L sin 2e = 10—3 L2 a2 + LR” — } Lh, L sin 2 e 6.838947+ LR"-Lh.

or,

2hR"

X

x=/2a2h;

=

x: 2h R": R;

2h R"

/2a2h

=

=1.48

=·00138 R′′ / h2.

=

EXERCISES.

1. At what elevation must a ball 4-5 inches in diameter be fired, with a velocity of 1200 feet per second, in order that its actual range may be 4500 feet? Ans. 4° 45'.

2. Required the elevation at which a 24-pounder must be fired, with a velocity of 1600 feet per second, that its actual range may be a mile. Ans. 4° 29'.

237. PROBLEM V.-Given the elevation not exceeding 45°, and the velocity with which a given projectile is discharged, to determine its actual range.

Case I.-When the potential random does not exceed 39000 feet.

'Reduce the terminal height F corresponding to the given projectile in the ratio of radius to the cosine of of the angle of elevation; find the potential range by article 202; divide this range by the reduced F, and find the quotient in the tabular column of potential ranges, and opposite to it in the preceding column of actual ranges is a number, the product of which, by the reduced F, will give the actual range.'

Let F the terminal height found by article 230,
F' the reduced height,

F: F',

the other letters denoting as before.
Then, to find F',
rad: cose: =
LF LF+L cos &e-10,
and h is to be found as in article 203.

or,

=

To find R, rad: sin 2e2h: R, and R = 2h · sin 2 e,
LR=L2h+L sin 2e-10.

or,

R

Then,

R' =
F

and r = F'r',

or,

Lr

LF+Lr'.

Or, R' may be found without previously calculating R and h, by substituting their values (234) thus:LR' LR-LF'=2 Lv + L sin 2 e-(10.505150+LF')

=

EXAMPLE.-What is the actual range of a musket bullet, of the usual diameter of of an inch, discharged at an elevation of 15%, with a velocity of 900 feet?

F1409 d 1409 × 1057 ... (by 230)

=

LF-L1057+ L cos 11° 25′-10-3-024075+9-991574 10=3·015649, and F' = 1036.

and (202), R =

Also, h=(3) = (8 )=12656, and 2h = 25312,

2 900, 2

=2hx=h=12656,

=

2 h sin 2 e
rad