790. Government Lands are usually surveyed into rectangular tracts, bounded by lines conforming to the cardinal points of the compass.

A Base-line on a parallel of latitude, and a Principal Meridian intersecting it, are first established. Other lines are then run six miles apart, each way, as nearly as possible.

The tracts thus formed are called Townships, and contain, as near as may be, 23040 acres.

A line of townships extending north and south is called a Range.

The ranges are designated by their number east or west of the principal meridian.

The townships in each range are designated by their number north or south of the base-line.

Since the earth's surface is convex, the principal meridians converge as they proceed northward. This tends to throw the townships and sections out of square, and necessitates occasional lines of offset, called "correction lines."

Townships are subdivided into Sections, and sections into HalfSections, Quarter-Sections, Half-Quarter-Sections, Quarter-QuarterSections, and Lots.

Diagram No. 1 shows the sub-divisions of a Township into Sections, and how they are numbered, commencing at the N. E. corner.

Diagram No. 2 shows the sub-divisions of a Section, on an enlarged scale, and how they are named.

A Lot is a subdivision of a section, usually of irregular form, on account of bordering upon a navigable river or lake-containing as near as may be the area of a Quarter-Quarter-Section, and described as lot No. 1, 2, 3, etc., of a particular section.

City and village plats are usually sub-divided into Blocks, and these into Lots.

PROBLEMS.

1. If a township of land is equally divided among 288 families, how many acres does each receive? What part of a section?

2. What number of rails will enclose a quarter-section of land with a fence 6 rails high, and 3 lengths for every 2 rods; and what will be the cost of the rails, at $40 per thousand?

3. A man bought the S. and afterward sold the E. What was his gain?

of a section of land at $21 an acre, of what he bought at $4.37 an acre. Ans. $700.

4. If I buy the N. E. and the E. land, how many acres do I purchase?

of N. W. of a section of

What part of a whole sec

tion? How are the parts located in respect to each other?

5. A speculator bought of the Ill. Central R. R. Co., the S. ¿ of Section 4, township 10 north, range 6 cast, at $2 an acre. He afterward sold the E. † of S. E. † at $2.75 an acre; the N. E. † of S. E. at $34 an acre; and the N. of S. W. † at $3.84 an acre. How many acres has he left? What was his gain on the purchase price of the whole? Draw diagram. Ans. $27.20.

6. A man having purchased a section of land from the U. S. Government at $1.25 an acre, sold the S. of S. W. † at $2.50 an acre; the N.W. of N.W. 1 at $1.75 an acre; the W. of S.E. 1 at $2 an acre; and the W. of S.W. † of N.E. † at $3 an acre. How many acres has he remaining, and what is his gain, provided the remainder is sold at $21 an acre? Draw diagram and explain.

BOARDS AND TIMBER.

791. A Board Foot is 1 ft. long, 1 ft. wide, and 1 inch thick. Hence 12 board feet make 1 cubic foot.

Board feet are changed to cubic feet by dividing by 12, and cubic feet are changed to board feet by multiplying by 12.

1. In Board Measure all boards are assumed to be 1 in. thick.

2. Lumber and sawed timber, as plank, scantling, etc., are usually estimated in board measure, hewn and round timber in cubic measure.

When lumber is not more than 1 inch thick:

RULES. 1. Multiply the length in feet by the width in inches, and divide the product by 12.

When more than 1 inch thick:

2. Multiply the length in feet by the width and thickness in inches, and divide the product by 12.

PROBLEMS.

1. What must be the width of a board 16 ft. long to contain. 12 board feet?

OPERATION.-16 ft. = 192 in.; 144 × 12 ÷ 1929 in., the width.

2. What must be the width of a piece of board 5 ft. 3 in. long, to contain 7 square feet?

3. Find the cost of 8 pieces of scantling, 3 in. by 4 in. and 14 ft. long. at $9.50 per thousand board feet.

4. A piece of timber is 10 in. by 16 in.; what length of it will contain 15 cubic feet? Ans. 131 ft.

5. How many board feet in a stick of timber 36 ft. long, 10 in. thick, 12 in. wide at one end and 9 in. wide at the other end? How many cubic feet? Ans. 261 cu. ft. 6. A rectangular field, 16 ch. long and 8 ch. wide, is enclosed by a post and board fence; the posts are set 8 ft. apart, the boards are 16 ft. long, and the fence is 5 boards high. The bottom board is 12 in. wide, the top board 6 in., and the other three 9 in. wide. The posts cost $25 per C., and the boards $14.80 per M. Required the number of posts, the amount of lumber, and the cost of both.

MASONRY.

792. Masonry is estimated by the cubic foot, and by the perch; also by the square foot and the square yard (287).

1. Brickwork is generally estimated by the thousand bricks; sometimes in cubic feet.

2. When stone is built into a wall, 22 cubic feet make a perch, 2 cubic feet being allowed for mortar and filling.

3. Philadelphia bricks are 8 in. × 43 × 2§; and Milwaukee bricks 8 in. × 43 × 28.

793. To find the number of bricks in a cu. ft. of masonry.

PROBLEMS.

1. How many Milwaukee bricks in a cubic foot of wall 12 in. wide, laid in courses of mortar of an inch thick?

OPERATION.

8.5+ (.25 × 2÷2) = 8.75 in. = length of brick and joint.
2.375+(.25 × 2÷2) = 2.625 in.
8.75 × 2.625 = 22.96875 sq. in.

=

thickness of brick and joint.
area of its face.

12.75÷3 (number of bricks in width of wall) = 4.25 in.=width of brick and mortar.

1728-97.617+ = 17.7+ = number of bricks in a cubic foot.

RULES. I. Add to the face dimensions of the kind of bricks used, one-half the thickness of the mortar or cement in which they are laid, and compute the area.

II. Multiply this area by the quotient of the thickness of the wall divided by the number of bricks of which it is composed, the product will be the volume of a brick and its mortar in cubic inches.

III. Divide 1728 by this volume, and the quotient will be the number of bricks in a cubic foot.

2. How many bricks, 8 in. x 4 x 2, will be required to build a wall 42 ft. long, 24 ft. high, and 16 in. thick, laid in courses of mortar of an inch thick? Ans. 31278

3. How many perches of stone, laid dry, will build a wall 60 ft. long, 16 ft. high, and 18 in. thick?

Ans. 60 Pch.

RULES. 1. Multiply the number of cubic feet in the wall, or work to be done, by the number of bricks in a cubic foot; the product will be the number of bricks required.

2. Divide the number of cubic feet in the work to be done by 24.75; the quotient will be the number of perches.

4. How many perches of masonry in a wall 120 ft. long, 6 ft. 9 in. high, and 18 in. thick?

5. At $.56 a cubic yard, what will it cost to remove an embankment 240 ft. long, 38 ft. wide, and 8.5 ft. high?

6. Find the cost of digging and walling the cellar of a house, whose length is 41 ft. 3 in., and width 33 ft.; the cellar to be 8 ft. deep, and the wall 1 ft. thick. The excavating will cost $.50 a load, and the stone and mason work $3.75 a perch. Ans. $4713.

7. What will be the cost of building a wall 60 feet long, 213 feet high, and 17 inches thick, of Philadelphia bricks, laid in courses of mortar of an inch thick, at $12 per M.? Ans. $423.53.

8. How many cubic feet of masonry in the wall of a cellar 37 feet long, 26 feet wide, and 9 feet deep, the wall being 2 feet thick, allowing one-half for the corners; and what will be the cost, at $3.85 a perch? Ans. 2214 cu. ft.; $344.40.

CAPACITY OF BINS, CISTERNS, ETC.

794. The Standard Bushel of the United States contains 2150.42 cu. in., and is a cylindrical measure 18 inches in diameter and 8 inches deep (311).

1. Measures of capacity are all cubic measures, solidity and capacity being measured by different nnits, as seen in the tables.

2. Grain is shipped from New York by the Quarter of 480 lb. (8 U. S. bu.), or by the ton of 331 U. S. bushels.

3. It is sufficiently accurate in practice to call 5 stricken measures equal to 4 heaped measures.

795. To find the exact capacity of a bin in bushels.

RULES. 1. Divide the contents in cubic inches by 2150.42; the quotient will represent the number of bushels.

Since a standard bushel contains 2150.42 cu. in., and a cubic foot contains 1728 cu. in., a bushel is to a cubic foot nearly as 4 to 5; or a bushel is equal to 1 cu. ft., nearly. Hence for all practical purposes,