D O M
285
DOM
Wore ^kpheric dome, let x = the altitude of the dome, as>* a; ^ len > by the nature of the circle, we have^ equal to one quarter of the generating square.
i ff* ~~ 4 x 3 for the generating square.
~~ 4 7-2 i — 4 x 2 x the fluxion of the solid.
The
die
in.
® Ue nt of 4 y 5 i r: 4 r s a: —
4 X s
now let d — the
n et er
e qual 2 r, and a equal the altitude of the dome, then~~s— = d* a equal to the solidity of the
tr uncated dome, and p (d? a - ec l ua l the
^Preision c ' rcu ' ar dome. Let us suppose the same
• es i -‘c circular aome. ivet us suppose ir;u to be applied to a hemispheric dome, then
P(<P„
3
<P r
"P(*H
,) =
')
2 <P r
P — 5 — =P
3
2 d 3 a
■, the same for-
as ab 0
ve.
•P
Tqj. Tactlca l application of the preceding rules.
I cf^are f iB so ^dity of a dome less than a hemisphere, uponVpiej, P lari , when its axal sections, parallel to the sides, are' ' ^^0^
!i t0 g W ' Ce the diameter of the vertical section multipliedd|? es the if 6 o4> t^ le altitude, subtract the third part of fourV of t i cu ° e of the altitude, and the remainder is the soli-
*af*® % j»fc S<1U G are se S mental dome -
of 5 the ^oppose the altitude of the dome to be 4 feet,2o^ e dome^ ter *^ ie vert ' c:d section 20 feet, the solidity
required.
<°tv
Ir *ice tl, . —
t e diameter of the vertical 16
i 6^ Action. =
85a 4
sfJ 4
square of the[altitude.
8 °Udi t y
16
^ired of the square 4
64
4
To find the solidity of a truncated square dome, the axaisection being that part of a semi-circle left by cutting off asegment parallel to the diameter.
From the square of the side of the base, multiplied into thealtitude of the dome, subtract the third part of four timesthe cube of the altitude, and the remainder is the solidityrequired.
Example .—Suppose the side of the base 20 feet, and thealtitude 6 feet, what is the solidity ?
20
6
20
6
400
36
6
6
2400
216
288
4
2112 solidity of the dome required.
3)864
288
But if the horizontal sections had been circles instead ofsquares, we should then have the solidity of the circulardome as follows:
2112
.7854
8448
10560
16896
14784
1658.7648 the solidity of the domewhen the horizontal sections are circular.
It may here be remarked, that the segment and truncateddomes make together a complete square dome, each side ofthe base being 20 feet, and the altitude 10 feet.
Now the solidity of the segment dome is 554§ feet,and that of the truncated dome . . . 2112
Therefore the whole square dome whose -
vertical section is a semi-circle, is . . 26661
Now, the rule for measuring a square dome with semi-cir-cular vertical sections parallel to the sides of the base, is to taketwo-thirds of the area of the base, multiplied into the height:
20
20
3)256
' 0 S<j
COty
851
a circular dome of the same dimensions: Then
.7854
554£
31416
39270
39270
43 5.1116
2618
2618
43 5.6352 feet, the solidity required.
400 area of the base.
10
3)4000
13331
13331
2666f the solidity, as before.
But as it may be objected by many, that, when a square orcircular dome, whose vertical section is the segment of a cir-cle, is required to find its solidity, it is difficult to find thediameter of the vertical section, and that it would be moreeligible to find the solidity from the side or diameter of itsbase, and the altitude of the section: in order to save thetrouble of finding the diameter, we shall here show dieinvestigation of another rule, independent of any foreign oradventitious dimension.