Part 305

RULES.

1. To the square of the diameter of the excavation, add the square of double the line of least resistance, and reserve the said sum.

2. Multiply the square root of the reserved sum by double the line of least resistance, and subtract the product from the same sum.

3. Multiply half the remainder by the line of least resistance, and 1.57 times the product, will give the solidity of the excavation.

4. The charge will then be determined from the nature of the soil, as in the following example.

Example I.

It is required to make a mine in the second sort of soil, mentioned in the foregoing experiments, which shall have a line of least resistance of 10 feet, and the diameter of its excavation 20 feet; what will be the proper charge?

The nature of this soil, by the table, requires 10 pounds of powder to 216 cubic feet.

Calculation.

1. The diameter of the excavation is 20, and its square 400 Double the line of least resistance is 20, and its square 400 --- Therefore the sum to be reserved is 800

2. The square root of 800 is 28.3 Double the line of least resistance is 20 566 --- Which leaves the remainder 234 --- 3. Half the remainder is 117 Which multiplied by the line of least resistance, 10 --- Gives the product 1170 Which multiplied by 1.57 ------ Gives the solidity of the excavation feet 1836.9

feet. lb. feet. lb. 4. If 216 : 10 ∷ 1836.9 : 85 which is the charge required.

By Logarithms.

1. Diam. of excavation is = 20 1.301030 Diameter squared is 2.602060 400 Double the line of least resistance is = 20 and its square 400 --- The sum to be reserved is 2.903090 800 -------- 2. Square root of sum is 28.3 1.451545 Double the line of least resistance is = 20 1.301030 -------- Product to be subtracted is 2.752575 566 -------- --- Remainder is 2.369216 234 Line of least resist. = 10 1.000000 10 pounds of powder 1.000000 To 216 cubic feet, compl. arith. 7.665546 To which add the const. log. 9.894870 -------- And the sum is the logarithm charge required 1.929632 = 85 lb.

Problem II.

Given the nature of the soil, the line of least resistance, and the charge, to find the diameter of the excavation.

RULES.

1. Find the solidity of the earth to be raised, by a proportion from the nature of the soil, and multiply it by 1.27.--Divide the product by the line of least resistance, and to the quotient add the square of the line of least resistance: reserve the sum.

2. Multiply the square root of the sum reserved by twice the line of least resistance, and add the product to the said sum, and from the result subtract 3 times the square of the line of least resistance; so will the square root of the remainder be the diameter of the required excavation.

Example I.

Let a mine be charged with 100 pounds of powder in a soil which requires 11 pounds of powder to raise 216 cubic feet, and let its line of least resistance be 10 feet: what will be the diameter of the excavation?

By the nature of the soil 11lb. : 216 feet ∷ 100lb. : 1964 feet, which is the solidity of the earth to be raised.

1. Therefore multiply 1964 By 1.27 ------- The product is 2494.28 ------- Which divided by the line of least resistance, 10, is 249.428

To which add the square of the line of least resistance 100.000 ------- And the sum to be reserved is 349.428

2. The square root of 349.428 is 18.7, which multiplied by twice the line of least resistance, 20, gives 374. ------- This added to the sum reserved gives 723.428

From which subtract 3 times the square of least resistance 300. ------- And there will remain 423.428

The square root of which is, 20.5 feet, being the required diameter of the excavation.

By Logarithms.