Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

EXAMPLE 2. The diameter of a circle is 24; required the diameter of a circle 4 times less.

Answer, 12. 24 x 24 = 576 = 4= 144 v of which is 12 the answer. Note 1.-To find the diameter and circumference of a circle

the one from the other. Multiply the diameter by 3.1416. the product will be the circumference:

Therefore, Divide the circumference by 3.1416 and the quotient will be the diameter. NOTE 2.When the diameter is given to find the area of a circle.

RULES. Rule 1.-Multiply the square of the diameter by .785398.

Or, RULE 2.-Multiply the square of the semidiameter by 3.1416.

Or, RULE 3.—Multiply half the circumference by half the diameter.

Or, Rule 4.—Multiply the square of the diameter by .7854, the product in either case will be the area. Note 3.-To find the area of an Ellipsis or Oval.

RULE. Multiply the transverse diameter by the conjugate and their product by .7854, the last product will be the area of the Ellipsis.*

LESSON 17 The conjugate and transverse diameters of an Ellipsis given to find the diameter of a circle equal in area.

RULE. Multiply the two diameters of the ellipsis together, and the square root of their product will be the diameter of a circle equal in area, to the ellipsis.

* Conjugate diameter is the longest line passing through the centre of an oval.

Transverse diameter is the shorter line passing through the centre of an an oval at right angles with the conjugate,

EXAMPLE. The area of an ellipsis is 1244 inches, the conjugate diameter is 44 and the transverse is 36; of what length must be the diameter of a circle equal in area? 39.8 inches nearly.

OPERATION. Conjugate, 44

15,84(39.8 nearly. Transverse, 36

9

[blocks in formation]

Product, 1584

788).6300*

6204
PROOF FOR LESSON 17.
Work by Note 2, and Rule 1, under Lesson 16.
Diameter, 72 39.8

39.8

3184 3582 1194

1584.04 square of diameter.

.7854 common multiplier.

633616 792020 1267232 1108828

1244.105016 area. Here note, if the diameter of the circle had been carried further in the decimal parts, say .79 instead of .8, the proof would have come out more exactly.

QUESTIONS FOR EXERCISE.

LESSON 18. If a pipe ; of an inch diameter, will fill a cistern in 4 hours, what must be the diameter of another pipe which will fill the same cistern in 2 hours.

Answer, 4" 2" 10" or .35355 + of an inch.

RULE.
As the time given,

is to the square of the given diameter;
So is the time required,
to the square of the required diameter.

OPERATION. As 4 hours are to 7 of 25, so are 2 hours to y of the diameter required. Inverse Proportion.

+ of an inch = .25

.25

125
50

.0625 square of the given diameter. Time given. Sg. Time required. 4 :.0625 : : 2

4

[blocks in formation]

7070). 377500

LESSON 19. If a pipe of an inch diameter will fill a cistern in 8 hours, in what time will a pipe whose diameter is 14 inch fill the same cistern?

Ans. 53 min. and 20 sec. When the answer consists of time, the second term will be in time; and the first and third terms in squares of given diameters.

į an inch =.5

.5

.25 square of the half inch pipe.
7 of 14 = 2.25

Diam. b. Diam.
.25 : 8. : : 2.25 inversely.

8

[blocks in formation]

LESSON 20. If 4 pipes of 3 inches diameter each, will be 4 hours in discharging a cistern of water, what time will 3 pipes of 4 inches diameter, require to discharge another cistern, which contains 5 times that quantity? Ans. 15 hours,

3 inches diam.
3

Diam. h. Diam.
9 square of diam.

36 : 4 :: 48 inversely. 4 number of pipes.

4 36 sum of the squares.

48)144(3 hours.

144
4 inches diam.
4

cis. h. cis.

1:3 :: 5 directly. 16 square of diam.

3 3

15 Answer 48 sum of the squares.

LESSON 21.
In the midst of a meadow well cover'd with

grass,
I will take just two acres to tether my horse ;
How long must the cord be, when he's feeding around,
To command him within these two acres of ground?

Answer, 252.313+ links : or, 166 ft. 6 in. 4 sec
See the rules under lesson 15.

Common multiplier, 1.273241
Links in 2 acres,

200000
✓ 25,46,48,20,00,00(504.626+

25

1004) 4648

4016

10086) 63220

60516

100922) 270400

201844

1009246). 6855600

6055476

. 800124 Diameter, 504.626+ Semidiameter, 252.313 or length of the cord.

i link = 7.92 inches. .313 of a link =24 in!

.313

2376

792 2376

2.47896 = 24 inches nearest to be ad

ded to the 252 links.
PROOF OF LESSON 21.
By Note 2, Rule 4 under Lesson 16.
Diameter,

7 504.626
504.626

See page 336.

« ΠροηγούμενηΣυνέχεια »