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March 1,1868.] THE CIVIL ENGINEER AND ARCHITECT’S JOURNAL. 85 I find that N increases with 0, and is greatest at the point B on the springing. Thus when 0 is 60°, N=8. <j> (13’694), and when 0=tan- 1 2, which is the case when E is at B, and 0 is 63°. 26'T, then N=8 . <j> (18’25). Hence we must consider that the greatest thrust is at B. We can now compare the maximum thrust in this dome 'with that of the spherical of the same dimensions and material; for in the parabolic dome N = 8. <f> (18’25) = 79’63 lbs., and in the spherical N=92 lbs. Hence it appears that the thrust is much less in the parabolic than in the spherical dome. Also, since N acts lower down in the former than in the latter, its moment about the outer edge (S) of the base of the pier will be proportionally reduced ; the greatest thrust (N) in the sphere acting at a point 20° above the springing line, while in the paraboloid it is at the springing line. We can now calculate the thickness which must be given to the “ drum,” or pier, in the foregoing example, by forming the Equation of Stability, which in this case will be 2N8=P . a+Q . q in which Z> = H, a=t, Q . q being the’same as before. Then, putting H = 50 ft., we have Q . 5=43’63275 (f-f-SOt 5 ), P . a=384-05t, 2N . 8=7962’9 Whence the Equation of Stability is t 3 +30t s +8’8t—182’5=0, from which we obtain t=2’244 ft., and in the sphere of equal span we found t=2’45 ft. It is evident that if a belt is placed round this dome to assist in counteracting the outward thrust, the position of it must be at the springing in order to have the greatest effect. I will next consider the dome whose surface is formed by the revolution of an ellipse about its major axis. In Fig. 6 OD is the semi-major axis (a), and O B the semi-minor axis (J) of the quarter-ellipse BED. Let A B or C D be called X, the thick ness of the dome, which will be very nearly equal throughout if we consider A F C to be the quarter of another ellipse whose semi-axes are a+X and Z>+X; so that the domical shell is the difference between the two semi-ellipsoids generated by the revolution of these quadrants about the axis O C. If we attempt to find the centre of gravity of any portion of the shell by the method used in the parabolic dome, we obtain an integral which cannot be expressed in a finite algebraical form. I therefore take E F, one of the joints of the masonry, as horizontal instead of being in the normal to the surface as in the former cases. This will not produce any material error as we have only to consider the joints near the springing, which are nearly horizontal. Let g be the centre of gravity of the portion of the wedge cut out of the dome whose section is C D E F, and formed by its revolution about 0 C through the angle </>; P the weight of this portion of the wedge. Then if P x is the weight of the slice F K C cut out of the outer ellipsoid; z, the distance from the axis of its centre of gravity; P 2 and z„ the same for the inner ellipsoid; we have by the principles of mechanics, z i—Pi z i P» z 2 P " Pi-P 2 ' If we call x, and the current co-ordinates of the outer ellipse, x„ and of the inner ; taking 0 as the origin, 0 A being axis of y, and 0 C of x, we have yf=V (at-xf), S rA rt + ^ P^-^r/y 2 ^, 2 xi Z x 2 _2fy^ x ~sfy^ 7 2_/j/ 2 3 <fc 2 ^'3/^ p *- y (Sr)’! H ’ ■ f 1 ci - j- X a/(«+X) 2 -«i 2 (| (a+X) 2 -^ 2 )) } P 2 X 2 = 3 \a) 18 \2 P i—~2^ {(®+ x ) 2 («+X—Ki)— | ((a+X) 3 —a/) | From these expressions we can find the position of the centre of gravity of any part of the shell of any given elliptical dome. In order to calculate the value of N=P-, I take the ellipse in y which a=12 ft., 5=10 ft., X=1 ft. Substituting those values in the equations I find that the greatest value of N is obtained when oq=a: 2 =4 ft. This gives us P A -P 2 c 2 =3 . <f> (608’1) Pj—P 2 =8 • <f> (81’03)=P GE=a:=i/ 2 —x=^y7 t 2Z^ o 2—.-=9’4282—7’5=1’9282 ft. NE=y=a+X—ft. N=P? = 8 . d> (81’03) X -2?®?=90’867 lbs. y 9 This is the maximum thrust, and is rather less than that of the spherical dome, which I found to be 92 lbs. I have supposed N in this dome to act at C as in the spherical dome. If we call F the weight of the portion of the wedge below E F, z' the distance from 0 C of its centre of gravity, we find »' in the in the same way as we found z, namely, k'=51AzSJ_=10’719 ft. Fi—P 2 F, being the weight of the whole slice A F K 0, z\ the distance of its centre of gravity from 0 C ; F 2 , ss' 2 the same for the slice B E K 0. We can now form the Equation of Stability by taking the moments of all the forces about S, the forces N and P being transposed to the point E. If the height (H) of pier is 50 ft., S 2 /=6=H+O K=54 ft. Se=a=t+OB—KE=t+-5718 P=P 2 —P 2 =358-56 lbs., F=F 1 —F 2 =175’7 lbs. c=t+OB—«'=t—-719 2N . 5=9813’636 P . <(.=358’561+205.024 F . c=175-7t—126’33 Q . 5=43’63275 (t 3 +30« 2 ) Equating and reducing, we obtain the Equation of Stability, < 3 +30t 2 +12’25t—223’11=0; hence t=2.44 ft., which is almost exactly the same as was obtained for the thick ness of pier in the spherical dome of same span. Hence it appears that in any elliptical dome of the above proportions, namely, where the internal height from the springing is to the half-span as 6 : 5, and the thickness of the shell one-twentieth of the span, the weakest part of the dome is at a distance of one-third of the internal height above the spring ing line ; and therefore at this part the dome may be advan tageously strengthened by a surrounding belt. There is another kind of dome commonly used in Eastern countries, the section of which is represented by Fig 7. It is sometimes called the “ Ogival ” dome, the contour being a curve having a point of “ contrary flexure.” The simplest mathematical curve which lias this form is that called the “ curve of sines,” the equation of which is y'=r. sin (?) In Fig. 7 let W be the origin of co-ordinates of the outer contour, WL=aq, LF=i/ 1( then AT and V C are each equal to r n and W T=W y=^- v Let M be the origin of co-ordinates of the inner contour, M l=x„, I E=i/, ( , then B H and D It are each equal to r 2 , and 22