The relation of models to machines, as to strength, deserves the particular attention of the mechanic. A model may be perfectly proportioned in all its parts as a model, yet the machine, if constructed in the same proportion, will not be sufficiently strong in every part; hence, particular attention should be paid to the kind of strain the different parts are exposed to; and from the statements which follow, the proper dimensions of the structure may be determined.

If the strain to draw asunder in the model be 1, and if the structure is 8 times larger than the model, then the stress in the structure will be 83 equal 512. If the structure is 6 times as large as the model, then the stress on the structure will be 63 equal 216, and so on; therefore, the structure will be much less firm than the model; and this the more, as the structure is cube times greater than the model. If we wish to determine the greatest size we can make a machine of which we have a model, we have,

The greatest weight which the bean of the model can bear, divided by the weight which it actually sustains equal a quotient which, when multiplied by the size of the beam in the model, will give the greatest possible size of the same beam in the structure.

Example

If a beam in the model be 7 inches long, and bear a weight of 4 lbs., but is capable of bearing a weight of 26 lbs., what is the greatest length which we can make the corresponding beam in the structure? Here 26 / 4 = 6.5, therefore, 6.5 X 7 = 45.5 inches.

The strength to resist crushing increases from a model to a structure in proportion to their size, but, as above, the strain increases as the cubes; wherefore, in this case, also, the model will be stronger than the machine, and the greatest size of the structure will be found by employing the square root of the quotient in the last rule, instead of the quotient itself; thus,

If the greatest weight which the column in a model can bear is 3 cwt., and if it actually bears 28 lbs., then, if the column be 18 inches high, we have √ (336\ 28) = 3.464; 3.464 x 18 = 62.362 inches, the length of the column in the structure. TABLE OF MANILLA ROPE.

Diam.

Ins.

Circ.

Ins.

Wt. per foot. lbs.

Breaking load.

Tons.

lbs.

•239

¾

•019

•25

560

•318

1

•033

•35

784

•477

1½

•074

•70

1,568

•636

2

•132

1.21

2,733

•795

2½

•206

1.92

4,278

•955

3

•297

2.73

6,115

1.11

3½

•404

3.81

8,534

1.27

4

•528

5.16

11,558

1.43

4½

•668

6.60

14,784

1.59

5

•825

8.20

18,368

1.75

5½

•998

9.80

21,952

Diam. Ins.

Circ. Ins.

Wt. per foot, lbs.

Breaking load.

Tons.

lbs.

1.91

6

1.19

11.4

25,536

207

6½

1.39

130

29,120

2.23

7

1.62

14.6

32,704

2.39

7½

1.86

16.2

36,288

2.55

8

211

17.8

39,872

2.86

9

2.67

21.0

47,040

3 18

10

3.30

24.2

54,208

3.50

11

3.99

27.4

61,376

3.82

12

4.75

30.6

68,544

414

13

5.58

33.8

75,712

4.45

14

6.47

37.0

82.880

The strength of Manilla ropes, like that of bar iron, is very variable; and so with hemp ones. The above table supposes an average quality. Ropes of good Italian hemp are considerably stronger than Manilla; but their cost excludes them from gen-eral use.

The Tarring of ropes is said to lessen their strength; and, when exposed to the weather, their durability also. We believe that the use of it in standing rigging is partly to diminish contraction and expansion by alternate wet and dry weather.

The common rules for finding the strength of rope by multiplying the square of the diameter or circumference by a given coefficient are entirely erroneous.

Prices in Philadelphia, in 1873 : Manilla 17 to 18 cents per pound; Italian hemp, 25 cents; American hemp, 15 cents; Sisel hemp, 16 cents; jute (East Indies), 10 cents.