DOM

284

DOM

side of nA; or draw lines through the several points of sec-tion in the diagonal a i, parallel to i s, tc cut the lines perpen-dicular to h s, and the points of intersection will form thedirection of the line for the edge of the covering h s a. Inlike manner, by describing the quadrant of the ellipsis mqn,and by drawing lines through the points z, y, x, to cut thecurve n Q, by proceeding as before, we shall obtain the cover-ing a b t. The three coverings srb,hsa,atb, cover morethan one-quarter of the whole, by the half coverings grk,and l t b, of the side and end. The covering of each side sofound answers to the opposite side, by turning the back forthe front. Each covering, except that upon the side, andthat upon the end, will cover four different sides: thecovering upon the sides and ends only answer in two oppo-site places.

The angle-ribs of this dome are found as usual. If the cir-cumference of each quadrant or rib perpendicular to the side,were ascertained by calculation, then the boarding could easilybe laid out without the use of a plan, upon the same prin-ciple as in Figure 5, as is obvious from what has alreadybeen said

To construct the ribs of a spherical dome, with eight axalribs, and one purlin in the middle.

Plate II. Figure 1, No. 1. —Let abcdefgh, be thesemi-plan, which is supposed to be divided into four equalparts, and let a h be the diameter, terminating the semi-plan;divide the semi-circumference into four equal parts, from theextremity a, to the other extremity h, of the diameter a h ;and the points of division will mark the middle of the back,or convex sides of the ribs. This being the case, let b c c b,d e ed, f G gf be the plans of the intermediate ribs, B C, D E,and fg, having the points of division in the middle ; the linesn b, c c, d d, e e, vf g g, being the places of the verticalsides, and parallel to the lines drawn from the middle ofb c, d e, f G, to the centre. Draw v x, No. 2, parallel to A h,No. 1 ; from the side A a, b c c b, d e e d, &c., draw lines cut-ting v x. perpendicular to a h ; then taking v x for the underside of the kirb or wall-plate, draw its proper thickness. Inthe elevation, No. 2, of the dome, the front ribs are quadrants,forming a semi-circle with the upper side of the wall-plate,which is of course the diameter ; the curves of the sides ofeach of the other ribs are the quadrants of an ellipsis of thesame height with the front rib, and their projected places,from the plan upon the kirb v X, gives the lesser-semi-axis.To form the purlins, place the section of one of them in itssituation, and, circumscribing its angles, draw the squaremnop: draw m q, and n r, parallel to v x, and the lines fromthe several angles of the purlins, also parallel to v x ; then,where they cut the opposite rib y x form the section of thepurlin, and then the circumscribing square q r st. First forma ring, whose greater diameter is m q, or n r, and whose inner,or less diameter, is jo or os, and whose thickness is mn, or po ;the ring being thus formed, gauge lines from each of the sides,as is shown by the section ; then cut off the angles made bythe horizontal and perpendicular suffices, between each twolines, on each two adjoining sides, and the purlin will beformed. The ribs of this dome are not complete quadrants, asthey abut upon the upper kirb W Y at the top.

The method of covering this dome is, to suppose the sur-face polygonal; the principle is the same as is shown inPlate III.

Figure 2.—The ribs of an elliptic dome are formed in thesame manner as in Figure 1, and the covering as in the pre-ceding Plate ; the covering of one quarter being found,answers for the whole, as has been observed. It may benoticed, once for all, as a general rule to cover any dome,divide a quadrant of the plan into as many equal parts as

from

there are to be boards in each quarter, then draw l* n ^the points of section, to the centre of the plan, a”^illchords by joining each two adjacent points, so that ^ th ebe as many triangles formed as there are boards •

dil

'” J “'“‘v --, f t i.e cho ru '

centre of the plan, draw a perpendicular to each ot 11 n ^j c ul» rmeeting each chord; make the length of each P e ^P^| ie don 5 ®the base of a rib, and take the common height ot ^ atas the height of each rib, and place it at the ext f e in t of *right angles to each base; and describe the < l? a , t m ay b 6circle or ellipsis, according as the base and hog 1 ' jjvid®equal, or unequal. To find the covering for any ** . ’ jt0 s n/the curve of the upper side of the rib, so found,number of equal parts, and draw lines perpendicu ■base to intersect therewith, and the whole will be cas shown in Plate III. _ j N #j

id®

pose d _ the diameter, equal to 2 r, and a = r,then will -^1 = that iSj the g 0 iidity of the -

equal to two-thirds of the circumscribing rectan ° U ^Id l 1 ®® 11-It in the above dome, all the horizontal sections B h e-

circles instead of squares, the dome would have be® tl>®

— -v- .rea of a L ‘l srl >^

for

the

soli^

P t i,et>

'***Vivo

diameter of which?, p ~ - 78 H the area oi ~

um y> then in the case of the se &

° f ‘ he Sphere > We have p^rP-±f)

2 r = d, and wben less than a hemisphere; or,q t0 the a ltitude of the

( 4 ^-~ r )= P (2 da, - t he he^

n nressioa 30 7

solidity is p ■

The same form of exp re '

shown for all polygonal domes whatever : J^ c ircc ia3 'proper multiplier for p, when the radius o 1

ontf3>

circle is unity. . , „ „ , the golidW

Suppose it were required to nna w

of

to tbf]ete d '

t SMIUYVU 111 Jt lute XJl±, r J.JJ -

No. 6, shows the covering over loif; No. 7 ’^„[dc al1 ®No. 4, the middle rib; No. 5, the rib between theend ribs. , section

To find the solidity of a square dome, the ctxthrough the middlh of the sides being semi-circles. . sU pp° s ?

Let the radius of the circle be represented by parad,then, in the vertical section, that we draw any l' n w j,icb ! 3to the base, for the section of the generating pla ne > ( | ie »x l3equal to the side of the generating square. Supp° s ^. g pot °to be drawn upon the section, and the part of the x and H’the summit to the generating line, to be denoted J j^vCito denote the half generating line ; we shall t' ien ^ c on 3 ®'the property of the circle, y = (2 r x — X 1 ) $> a <• t ijequently 2 y = 2 (2 r x — x 2 ) £ the whole length oof the generating square.

Therefore, 4 f - 4 (2 r x — x?) goli d.

and 4 if x = 4 x (2 r x — x 1 ) the fluxion of the

A 3 • f

fluent of 4 y 2 x ~ 4 r xfl -— for the solidity o

3 e t ie i yb0 .

ment of the dome. Now, therefore, if the solidity a a i to r>dome be required, it is only supposing x to becom e

. , 8^ W

4 kt 4 t _,

we then find 4r x 2 -— = 4 P - 5 ’ 3

3

, the

. ~,o. i*