DOG

287

DOG

'■P

Ptticfent* 1 '/'^ 6 convex surface of the segment of a dome, inde-Lgj the diameter of the great circle.

~ the diameter of the great circle,rf * the altitude of the dome,

the diameter of the base of the dome ;

le ". b

y the property of the circle, d a ■

therefore

d*

d 4

out

d = a + —.4 a

& D a is equal to the convex surface ; therefore- vd 2

' P ° a + ~~ = the area of the convex surface

*0 , ..

Therefore,

< 0 yy

f l Iulti i area °f the segment of a dome :

3 nh Par the sum of the square of the altitude and the‘Ulg " ar *' °f the square of the diameter of the base, bythe product will be the superficies of the dome.

e '7~What is the superficies of the segment of a5l 8ht4p diameter of the base being 17.25 feet, and thefeet.

17.25 4.5

17.25 4.5

8625

3450

12075

1725

225

180

20.25

4 ) 297.5625

~~- [base’s diameter.

74.3906 the fourth part of the square of the20.25 square of the altitude.

94.6406 the sum.

3.1416

56784369464063785624

’'^^0896 feet, the surface required.

i-Hje 64 bet\v t ' le superficies of any portion of a sphereCt „ of th e 6e - n an l r tvvo P ara ^ e l planes, is equal to the^e r, a c ! rcum ference of a great circle into the dis-b > the ) Plaues ; that is, equal to the surface ofNlel ' p e, and baSe ° f which is equal to the great circle of^Wane s . t ^ e altitude equal to the distance of the

) of 1 Hiw® ^°th t^ 16 a ^’ tu( l e of the segment of the sphere,u He °f the ae a ^ltude of the segment wanting, and the^te^StUg^t Mallei planes ; also, let a equal the altitudeV^U 1 c ' r clp ' Va uting ; and let c be the circumference of

1

'Hi,

>tt;

k Hs „ 6 Paraji i 1 ^ 6 ^goaent include the part contained be-dfV so r anes ’ or t f‘ e segtuent cut off, the area

tk'V j e oitcumr ace still be expressed by the product‘ . if 1-62106 °f the great circle and the height

•s j n ’j aorefore, c A is equal to the superficies ofPofaUgi ) ^at °f the solid contained betweenPlanes: also, c a is the superficies of the

segment cut off; but the difference between the areas of thecurved surfaces of these two segments is equal to the curvedsurface of the solid contained between the two parallel planestherefore, c a — c a = c ( a — a) is the surface of the spherecontained between the parallel planes.

It would, however, be very desirable to have another sub-stitute in terms of the upper and lower diameter of the solid,in place of the diameter of a great circle : for this purposewe shall here give the following rules:

Given any two parallel chords in a circle, and their distance,to find the distance of the greater chord from the centre.

To the square of the distance between the chords add theSquare of half the lesser chord. The difference between thissum and the square of half the greater chord, divided bytwice the distance of the chords, gives the distance from thegreater chord to the centre.

Example. —Suppose the greater chord, c D, is 48 feet, andthe lesser, A b, 30 feet, and their distance, e g, 13 feet; whatis the distance, e f, from the centre to the greater chord, c d ?

13 30 = 15 48 = 24

13 ~~n 15 ft 24

169

[chord. -

225 square of the less169 square of distance.

9648

[chord.576 square of greater394

394 26 ) 182 ( 7 dist. required.

182

Given the chord of a circle, and its distance from the centre,to find the radius of the circle.

To the square of the half chord, add the square of thedistance from the centre, and the square root of the sum willbe the radius required.

Example. — Given the chord C d, 48 feet, and its distanceef from the centre, 7 feet, the radius of the circle isrequired.

7 X 7 = 49

l

«=24

24

625 ( 25 the radius required.

4

45)225225

Suppose, then, that we would wish to obtain the area ofthe curved surface of a dome, contained between two parallelplanes, the greater of which is 30 feet diameter, and thelesser 20 feet diameter, and the distance between them being5 feet.

15 X 15 = 225

10 X 10 = 1005 X 5 = 25

125

125

1,0 ) 10,0 the difference.10. feet,

the distance of the greater chord from the centre.