Ropes,cables, and all other descriptions of cordage are distinguished by their circumference, thus a two-inch rope means a rope two inches in circumference.
To find the weight of a rope.
First method.—Multiply the length in fathoms by the square of the circumference, and divide the product by 480 for the weight in cwts.
Example.—Required the weight of 110 fathoms of 3-inch rope.
3 × 3 × 110 = 990, which divided by 480, gives 2 cwt. 7 lb. Weight required.
Second method.—Divide the square of the circumference by 4, the quotient will give the weight, in pounds, per fathom.
Example.—What is the weight of a 3-inch rope per fathom?
32÷ 4 = 2¼ lb. Weight required.
To find the strength of a rope, or the weight it will support.
First method.—Square the circumference, and divide by 5, for the number of tons which it will bear suspended from it.[46]
Example.—What weight will 3-inch rope of the best description support?
(3 × 3)5= 9/5 = 1⅘ ton, or 4030 lb. Weight required.
Second method.—Multiply the square of the circumference by 2, the product will give thepractical weight in cwts. that may be lifted by it, or about half the breaking weight.
Example.—What number of cwts. may be lifted by a 3-inch rope?
32× 2 = 18 cwts. Weight required.
To find the weight of chains.
The square of the diameter of the link, measured in eighths of inches, will give the weight of the chain, per fathom, in pounds.
Example.—What is the weight per fathom of a ¾-inch chain?
¾-inch =68; 62= 36 lb. Weight per fathom.
Or, the weight per foot of the chain, multiplied by 24, will give the weight per fathom of the chain,nearly. A chain cable with a stay across the links will weigh about one-twelfth more than the foregoing examples.
To find the weight that may be safely lifted by a chain.
Divide the square of the diameter of the links, taken in eighths of an inch by 8, and the quotient will give the number of tons that may be lifted by the chain.
Example.—What number of tons will a chain made of ¾-inch iron carry with safety?
¾-inch =6862= 36368= 4½ tons. Weight required.
The safe strainis equal to about 8 tons, per square inch, of the iron of which the chain is made.
The stay across the link of a chainincreases its strength about one-sixth.
When the chain is of great length, a deduction, from the above rules, must be allowed for the weight of it.
To find the weight of round iron rods.
Divide the square of the diameter, in quarter inches, by 2, and the quotient will give the weight in pounds, per yard.
Example.—What is the weight of a yard of 1-inch round iron.
1 inch = 4 quarters 42= 16162= 8 lb. Weight required.
To find the weight of square rods.
The weight of round rods, of similar diameter, divided by ·7854 will give the weight of the square rods.
To find the weight that may be sustained, or lifted by round iron rods.
Find the weight in pounds, per yard; two-thirds of which will give the safe load, in tons.
A round iron rod of average quality of iron, one inch in diameter, will be torn asunder by 16 tons; it will be perceptibly damaged by half this strain, or 8 tons; its safe load will be one-third, or 5·33 tons.
To find the area, or superficial content of a plank.
Multiply the length by the mean breadth.[47]
Example.—Required the content of a board whose length is 11 feet 2 inches, and breadth 1 foot 10 inches
To find the solid content of squared, or four-sided timber.
Multiply the mean breadth by the mean thickness, and the product by the length, for the content, nearly.
Note 1.—If the tree taper regularly from the one end to the other, either take the mean breadth, and thickness in the middle, or take the dimensions at the two ends, and half their sum will be the mean dimensions; which, multiplied as by the above rule, will give the content, nearly.
Note 2.—If the piece do not taper regularly, take several different dimensions, add them all together, and divide their sum by the number of them, for the mean dimensions.
Example.—Required the content of a piece of timber 16 feet long, and side of square 14 inches.
To find the solidity of round, or unsquared timber.
1. Multiply the square of the quarter girt (or the square of ¼ of the mean circumference), by the length, for the content.
Note.—When the tree is tapering, take the mean dimensions, either by girting it in the middle for the mean girt, or at the two ends, taking half the sum of the two; or by girting it in several places, then adding all the girts together, and dividing the sum by the number of them for the mean girt. But when the tree is very irregular, divide it into several lengths, and find the content of each part separately.
Example.—Required the content of a tree, whose mean girt is 3·15 feet, and length 14½ feet.
3.154= ·7875 ·7875 × ·7875 = ·62015625.
·62015 × 14·5 = 8·9922 feet of solid timber. The content required.
2. Find the mean area of a round tree, and multiply it by the length for the content.
To find the weight of a tree.
Find its content in feet, and multiply that by the specific gravity of the wood.
(VideGravity,and Table of Specific Gravities.Page318.)
Example.—Required the weight of an elm-tree; whose mean girt is 5 feet, and length 60 feet.
54= 1·25 1·25 × 1·25 = 1·5625.
1·5625 × 60 = 93·75. Content in feet.