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area.

Examples.

1. What is the superficial content of a board, that is 1] feet long, and 12 in. wide ?

An..

11 ft.

NOTE.The board being 11 ft

. lung, I look for the line numbered 11 at the end, and apply that line to the middle of the board, and the width extends to that division numbered 11, which is its content.

Ans. 11 ft. 111.12 13 14 15 16

CASE VI. To measure boards with Gunter's sliding rule, RULE.---Bring the width of the board in inches, on the slider, against 12 on the line above; then

look along on the line above for the number expressing the length of the board ; and against that number, (on the slider) stands the number expressing the area of the board.

Example.

1. What is the content of a board that is 18 feet. long and 10 inches wide ?

Ans. 15. feet.

Note.--I first brought the width of the board in inches, against 12, on the line above; I then looked on the line above for the length of the board, vis; 18; and against the length (on the slider) stood 15 the number, answering to the area of the board,

CASE VII,

To measure boards with Gunter's scale and dividers.

Rule. On the line of numbers extend from 1 to the width, and that extent will reach from the length to the superficial content or area ; any right angled parallelogram may be measured by the same rule.

Example 1. What is the superficial content of a board that is 20 fcet long and id feet wide ?

Ans. 25 feet. NOTE. On the line of oumbers I extend from 1, to 2} tenths, or of the distance from 1 to 2; and that extent reaches from 20, (the length of the board) to the superficial content 25.

CASE VIII.

To measure joist, plank, &C. DEFINITION.--Joists are of different dimensions, sometimes 3 by 3, or 3 by 4, &c. plank are also of different thickness ; and both plank and joists are reduced to board, or superficial measure, and are measured thus,

RULE. Find the area, of one side of the joist, or plank, by any of the preceding rules in superficies ; multiply that area by the thickness in inches, the last product will be the superficial content of the joist, or plank.

Eramples. 1. What is the superficial area, or board measure of a joist, that is 20 feet, long 4 wide, and 3 thick? 20 ft. Oxo ft. 46,. ft. X3' thick equal 20 ft. Ans,

2. What is the area, or board measure of a plank that is 23 fț. long, and 16 in..wide, and 3. in. thick!

Ans. 100 feet. CASE IX. To measure any irregular plane surface. RULE.-Divide the whole surface into triangles and measure each triangle separately, as taught in case third, superficial measure.

Examples 1. What is the superficial content of a plat of ground, in form of the following figure ; diagonal A B 50 ft. ; perpendiculars, 24 and 12?

Ans. 900

24

12.

B CASE X. To measure the surface of a circle. DEFINITION. Circles are round figures bounded every where by a circular line called the periphery, or arch, or sometimes the circumference; the line passing through the centre is called the diameter ; half the diameter, or a line proceeding from the centre to the periphery is called the semidiam

eter, or radius.

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Rule 1.--As 7 iş to 22 so is the diameter to the circumference.

RULE As 113 is to 355,50 is the diam.to the circum..

Examples.
1. What is the circumference of a circle, which
has a diameter of 14 rods?.

By Rule 1, As 22 : :,14 to 44 rds. Ans.
By Rule 2. As 113 : 355:: 19 to 43'') Ans.*

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PROBLEM II.

The circumference given to find the diameter. RULE 1.Aiş 22 is to: 7. so is the cicum. to the diam.

Rule 2.-As.355 is to 113 so is the cicum. to the diam.

RULE 3.-Annex two cyphers to the circum. and divide by 3.14 the quotient is the diameter nearly.

2

*The two methods do not exactly agree, the last method is nearest to the truth; an exact proportion between the diame. ter and circumference of a circle bas not yet been discovered.

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Examples.
1. What is the diameter of a circle whose cir.
cumference is 44 rods ?

diam.
By Rule 1.-As 22 : 7:: 44 ; 14. Ans.
By Rule 2.--As 355 : 113 :: 44 : 14717 Ans.
By Rule 3.-As 44.00---3•14=1471. Ans.

PROBLEM III.
The diameter and circumference given to find the area.
RULE.-Multiply half the diameter by half the
circumference, the product is the answer; or multi-
ply the diameter and circumference together and
the product is the area.

Examples. 1. What is the superficial content of a circle whose diameter is 14 rds. and circumference 44 rds.?

rds. rds.

147*X44 =154 square rods, Ans. The same question done by rule second.

14x44=616+1=154 sq. rds. Ans.

PROBLEM IV.

The diameter given to find the area. RULE.--Multiply the square of the diameter by •7854 the product is the area.

Example. 1. What is the area of a circle whose diameter is 14 rods?

rds.

14x14x•7854–153.9384, Ans.

PROBLEM V.
The circumference given to find the area.
Rule.-Multiply the square of the circumfer-
ence by .07958, the product is the area.

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