* This measure is used in reckoning the area of land in acres, rods, &c., or to find the surface of any thing, where length and breadth are considered without regard to thickness; as the annexed figure plainly shows:

3 feet or 1 yard.

D 3 feet or 1 yard. C

2

3

5

g

He

A SQUARE. The student already understands, that 3 feet in length make a yard in long measure. must now learn, that it requires 3 feet in length and 3 in breadth to make a yard in square measure.

And this is evident, from our Fi gure; each side of which measures a yard, consequently, it is a square yard. It is also plain, that it con tains 9 square feet; because the area or surface of a plain figure, is estimated by the number of little squares contained in it; each side of the little squares 1, 2, 3, and so on, being 1 foot, it is obvious, that there are in the whole Figure A B C D, as many square feet as there are squares in the figure. Or perhaps

1 foot.

7

5

9

A

1 foot.

E

B

may appear more plain to the young student, if the figure ABCD be measured by the little square, (E,) whose side is 1 foot; then as often as the little square (E) is contained in the whole square A B C D, so many square feet the whole Figure must contain; and it is evident, that the little square (E) is contained 9 times in the whole square A B C D. From the Figure, it is likewise plain, that the area or surface of any square may be found by multiplying the length and breadth together; thus, 3X3=9 square feet, the area of the Figure annexed; and it is also plain, from the Figure, that the area or surface may be obtained by multiplying the length and breadth together, where the length and breadth are unequal, as in the Parallelogram D f g C, for in that it will be seen that the side D C is 3 feet, and the side D f, 2 feet, and when multiplied, they give the area thus, 3X2-6 square feet, the area of the Parallelogram D fg C; and if the figure be 3 feet long and 1 foot wide, as the Figure D hi C, it is plain that the area or surface is 3 square feet; because 3X1=3 square feet, the area of the Figure D hi Ĉ.

EXAMPLES.

1. In 50 acres; how many roods, and perches?

50 acres.

4

200 roods.

40

Ans. 8000 perches.

Ans. 200 roods, 8000 perches. DEM.-We first multiply our fifty acres by 4, because we must have 4 times as many roods as acres, to equal our acres in quantity; we next multiply the product in roods by 40, because we must have 40 times as many rods or perches as we have roods, to equal our roods in quantity. 24000 rods?

Ans. 150 acres.

2. How many acres in 3. How many feet of boards will it take to lay a floor 20 feet long, and 18 feet wide?

Ans. 360ft. 4. If a field is 60 rods long, and 48 rods wide; how many acres does it contain? Ans. 18 acres.

48 rods
60

DEM.-We first multiply the length and breadth together, which gives the num ber of square rods in the whole field, be160)2880(18 Ans, cause the breadth is 48 rods, and counting

160

1280

1280

back one rod on the length, it is plain, we have 48 square rods; and counting back two rods, we have 96 square rods; it is then evident, by repeating the 48 rods, which may be considered as the front, by 60 rods, the distance back, the product must be the number of square rods in the whole lot; and when we divide these rods by 160, the number of rods in an acre, our quotient must express' acres, because our quotient shows how often 160 square rods are contained in the whole number of rods.

0

5. How many bricks will it take to lay a floor 20 feet long, and 18 feet wide; each brick being 8 inches square? 18×20=360×144=51840÷64=810. Ans. 810 bricks. 6. In a square mile; how many square rods, square yards, square feet?

and

Ans. 102400 rods, 3097600yds. 27878400 feet. SOLID OR CUBIC MEASURE.*

1728 Solid inches make

40 Feet of round timber or

50 Feet of hewn timber

128 Solid feet, or a pile 8 feet

Long, 4 feet wide, and 4 feet high

1 Solid foot.

1 Tun or load.

1 Cord of wood.

By this measure, things that have length, breadth, and thickness,

are measured.

H

1 yd. or 3 ft. thick.

1 yd. or 3 ft. long.

1 yd. or

ft. wide.

3

We have seen in square mea sure, that the product of the length and breadth gave the area or surface of a square yd., a long square, or Parallelogram.

It will now appear obvious on inspecting the annexed. figure, that a cubic yard is 3 feet in length, 3 feet in width, and 3 feet in thickness, that is, a square solid or cube, having six equal sides. Suppose the figure were 3 ft. in length, 3 ft. in breadth, and but 1 foot in thickness, it would then evidently contain (3X3X1-9) 9 cubiek feet, that is, 9 solid squares, each square containing 1 solid foot; because length, breadth and thickness, in feet, multiplied together, give cubick feet; if 2 feet in thickness, it would contain (3X3X2=18) 18 cubick feet; and, as it is 3 feet in thickness, it must contain (3×3×3=27,) 27 cubick feet.

It is also evident, that a cubick foot, that is, a solid square, 12 inches in length, 12 inches in breadth, and 12 inches in thickness, will contain (12X12X12-1728) 1728 solid or cubick inches.

f EXAMPLES.

1. In 9 tuns of round timber; how many feet and inches? Ans. 360 solid feet, 622080 solid inches.

9 Tuns.

40

360 solid feet.

1728

2880

720

2520

360

Ans. 622080 solid inches.

DEM.-We first multiply the tuns by 40, because 40 feet of round timber make a tun; we next multiply the 360ft. by 1728, because 1728 inches make a solid' foot.

2. In 9 tuns of hewn timber; how many solid feet, and inches?

Ans. 450ft. 777600in.

3. In 2688 solid feet; how many cords?
4. In 10 cords of wood; how many solid feet?

Ans. 21cds.

Ans. 1280 feet.

5. In a pile of wood, 32 feet long, 5 feet high, and 4 feet wide; how many cords? Ans. 5 cords. NOTE.-Multiply length, breadth, and thickness together, and the product will be the solidity, in the same denomination of the given numbers thus, 32ft. X5ft.X4ft.=640ft.÷128ft.-5 cords.

6. In 221184 solid inches; how many cords? Ans. lcd.

By the Calendar the year is divided into 12 months.-The number of days in each month, may be easily retained by the following verse:

"Thirty days hath September, April, June, and November; February twenty-eight alone, and all the rest have thirty-one." February has 29 days every bissextile, or leap year.

NOTE.—The civil or solar year contains 365 days, being short of the true year, according to the best computations, about 5 hours, 48 minutes, and 48 seconds; so that the time runs forward through the seasons nearly one day in four years. And it was on this account that Julius Cesar caused one day to be added to February every fourth year; reckoning the 24th day of this month twice, because the 24th day was the sixth (sextilis) before the kalends of March, there being in this year two of these sextiles, which gave the year the name of Bissextile year; but this has since been corrected and every 4th year February has 29 days, and in honour of Julius Cesar it is called the Julian year. The periodical year is the time which the earth takes in performing one revolution round the sun, in leaving any fixed point in the heavens and returning again to the same point. On account of certain points in the Ecliptick having a backward or retrograde motion, the Equinox will apparently meet the sun, so that the sun will arrive at the Equinox or first point of Aries, before his entire revolution is completed, and this time is called the tropical year.

EXAMPLES.

I. In 15 weeks; how many days, hours, minutes, and seconds? Ans. 105d. 2520h. 151200m. 9072000sec.

DEMONSTRATION.-Here, we first multiply by 7, because, we must have seven times the number of days that we have weeks to equal our weeks in time; we next multiply by 24, because we must have twenty-four times as many hours as days to equal the days; we then multiply by 60, because we must have 60 times as many minutes as hours to equal the hours in time; we lastly multiply by 60, because we must have 60 times as many seconds as minutes.

15 weeks.
7

105 days.

24

420

210

2520 hours.

Proof.

60

151200 minutes.

60

610)9072000 seconds.

60)151200
24)2520(7) 105

24

15

120

120

2. In 413 days; how many hours and minutes?

Ans. 9912h. 594720m. 3. In 413280 minutes

how many hours, days and weeks?

Ans. 6888h. 287d. 41w. 4. How many hours in 4 years, allowing 365 days and 6 hours to the year?

Ans. 35064h. 5. How many seconds in 60 years, allowing 365 days 6 hours, to the year?

Ans. 1893456000s.

6. From the 2d of March to the 19th of November; how many days? Ans. 262d.

7. From April the 19th to the 20th of January in the next

year; how many days?

Ans. 276d: NOTE.-In reckoning time, exclude the first day and reckon the last. * CIRCULAR MEASURE OR MOTION.

Minute, min. marked ' iDegree, deg.

1 Sign,

S.

12 Signs, or 360 degrees,

.1 Circle of the zodiack.

*This measure is used in measuring latitude and longitude, and in computing the revolutions of the earth and other planets round the sun. The great circle of a sphere, containing the 12 signs, through which the sun passes, is called the zodiack

NOTE.-Every, circle, great or small, contains 360 degrees.

The sun passes 1 degree on the earth in 4 minutes.

EXAMPLES.

1. How many minutes in the 12 signs of the zodiack?