184.9.1 



THE CIVIL ENGINEER AND ARCHITECT'S JOURNAL. 



•147 



pearance is that of nakedness, blankness, and poverty: the con- 

 sequence of which is, that the more there is of ornament bestowed 

 on other parts of such an interior, all the meaner do the windows 

 show themselves, especially as the glazing itself is invariably of a 

 very mean description, and has a cold and dingy look. Yet, surely it 

 would not be difficult to remedy the last-mentioned defect by em- 

 ploying metal-work instead of lead, and making it of various 

 ornamental patterns that should be in conformity with the style of 

 the architecture. Internally, the metal-work might be either en- 

 tirely gilded, or partly gilt and partly bronzed, accordingly as a 

 greater degree of enrichment might be found suitable. Hardly 

 necessary is it to observe, that both stained and diapered glass 

 might also be employed with excellent effect; although, of course, 

 it would require to be treated in quite a different manner from the 

 painted windows in Gothic churches, so that the resemblance should 

 be only that of material, and not of style and ideas. And why 

 should the use of coloured glass be confined to one particular style, 

 any more than other materials are? — "Gentle Sliepherd, tell me 

 why." 



ON THE OBLIQUE BRIDGE. 

 By F. Bashfokth, Esq. 



Spiral Courses. 



Mr. Buck, in his "Essay on Oblique Bridges," directs the 

 "twisting rules" used in working the beds of tlie arch-stones, 

 to be placed at a greater distance apart at the extrados than at 

 the intrados. There could be no objection to such an arrange- 

 ment, provided the dimensions of the "twisting rules" were de- 

 termined accordingly ; but it has long appeared to me that the 

 method by which Mr. Buck has arrived at the difference of the 

 widths of the two ends of one rule (=/tan 5), supposes that they 

 are to be applied in parallel directions on the bed of the stone, at 

 a distance / apart. 

 Let R, r, be the radii of the extrados and intrados respectively; 



*, (p, the angles of the extrados and intrados; 



/, I', the distances apart of the intradosal and extradosal ends 

 of the twisting rules; 



* -(p:=5; and R — r = e=the thickness of the arch. 



™. , sec* 



Tlien I' — I . 



secip 



In Examples, page 25, R = 20-32 ft.; »-=17-32ft.; e = R-r = 

 3ft.; /= 42 in. = 35 ft. 



* = Zof the extrados (Mr. Buck's <),) = 59° 23' 7"; 

 <fi = / of the intrados (Mr. Buck's 0„) = 55° 13' 49"; 

 5 = * — (p = 4° 9' 18"; 

 / tan 5 = 42 tan 4° 9' 18" = 3-05 inches; 



/' — 



sec * 



/ T = 47-03 



sec <p 



47 inches nearly. 



J3 



Mr. Buck directs two strips of wood to be provided (AB, C D, 

 fig. 1), each in length R- ?-=3ft. 

 The opposite sides of one, AB, 

 are to be parallel, and of the 

 other the end D is to be greater 

 than the end C by / tan 5, or 3-05 

 inches in this case. A, C, are to Fv. l. 



be applied at the intrados at a distance I, or 42 inches, apart; 

 B, D, are to be at the extrados, and at a distance := 47 inches 

 apart. 



The twisted beds of the arch-stones may be supposed to be 

 generated in the following manner :— Let the straight line AC 

 (fig. 2) have an arm BDE rigidly attached at right angles to it; 



Fig. 2. 



and let D, E, be two points, such that BD = r, B E = R. Suppose 

 AC to coincide with the axis of the intrados of the oblique arch, 

 and have a uniform motion in the direction of its length AC ; and 

 also let the arm BE revohe uniformly. Suppose these uniform 

 velocities to be so adjusted that BE may revolve through 180° 



whilst any point B in AC describes a space L = the axial length. 



Then DE will describe the twisted surface proper for the bed's of 



the arch-stones, and the equations to the path of D will be 



L 

 r = rcose, x = rnne, y=—B =. fi9 (1) 



IT ^ ' 



the horizontal axis of the cylinder being the axis of y, and a verti- 

 cal line with which BD coincides when 6=0 being the axis of z. 

 Also, the equations to the path of E will be 



Z = Kcosfl, X = Ksine, Y = fi9 (2) 



And if, in equations (1) and (2), we give 6 the same value a, the 

 corresponding values of .Tyz,XYZ, will be the co-ordinates of 

 the points D, E, respectively, for one position of the arm BE; and 

 these points may be considered to be coincident with two corners of 

 the bed of an arch-stone. Also, by giving to 6 another value, 0, "e 

 may determine the co-ordinates of the remaining two corners of 

 the same bed. 



Suppose that P,p, Q,7, are the four corners of the bed of an 

 arch-stone, yj 7 being the intradosal, and PQ the extradosal cours- 

 ing Joints. Let X' Y'Z', ,r ^z'; X"Y"Z", x"y"x", be the co-ordi- 

 nates of P,p, Q,f/, respectively 

 and }. 



Fur P. Fur p. 



r'=.r sill al 

 (3) z' =rcu»a 1(4) 

 y'= Mil J 



Let z =z Ax+'By+c{7) be the equation to a plane passing through 

 the three points P,/>, 7. 



Then, since (3) and (4) are points in (7), substituting, we get 

 Z' = AX'-)-BY'-t-c; i'^Ax'-f-By'+c ; 



or, (Rcoso) =: A(Rsina)-fBMa+c (8) 



rcosa = Ar sin a -t-B^a-f-c (9) 



Subtracting, 



U cos a — rcosa = A (Rsin a — r sin o) ; 

 or, (R — r)co>a = A(R — rjs.no; or, cot a = A. 



And therefore equation (9) becomes 



rcosa — r uutasiu a-|-Li^aT-c = r cos a-j-B^a+c ; 

 or, = B^a-j-c; or, c = —lifia. 

 Equation (7) now becomes 



z = jrcola-|-By— Umi = X col a+R(!/ — iia) (10) 



And the point (6) is situated in this plane; hence, 

 • " = x" cot a-^ [5{i/" — fiay, 

 or, r cos;8 — rcot asin3 — B(/»|8 — ;ua) = B/i(e — a); 



or, . — Bu(B — ii): nr R — S_'" 



Sin a 



X =Riin a] 

 Z' = R CDsa y ( 

 Y'= ^.a J 



— a for P and p ; 6 = /3 for (j 

 For Q. For q. 



.(o) 2" = rcos;3 1(0) 

 i"= m3 J 



X"=RsinB] 

 Z" = Rcos/8 W 

 V'= ^S J 



- Bn{B-a); 



or, B 



M (3 — n) sin a 



Substituting in (10), we get the equation to the required plane 

 passing through P, p, and q, 



r 6ina(— /3) 



z = xcota -i- ■ 



{y-h") (11) 



/i(/3 — a) sin a 



Suppose now that the distance p 5 (fig. 3), measured along the 

 intradosal arris of tlie arch-stone, =: /; and that 

 the angle 7/) TO = <p. Let fall the perpendicular 

 pm on qm. Then mpq may be considered to be a 

 portion of the development of the soffit, pm being 

 parallel to the top of the abutment. 

 pm =y"-y' = Y"-Y', in (3), (4), (5), & (6), = ;x(B-a) 



= ;;y cos ip = / cos (p (12) 



Also, /sin<p = my = r((3— a) (13) 



Now, the length of a perpendicular let fall from 

 a point X" Y" Z", on a plane z = Ax+^y+c, is 



Z'-AX"-BY'-c 



V» = + ^ — == Hymers' Geom. of 3 Dimensions. Art. (.'.;) 



And substituting from equation (11) the values of A B and c 

 we get length of the perpeudicular let fall from X", Y" 2"' on tlie 

 plane (11), 



nil .i-n f- sina — S 



Z"-X"cota -)- - , _ ^,_ g- (Y"-Ma) 



Qra = -I- 



V 



H (g — fljsin a 



1 -(- col- a -(- 



ir sin(n-)3) \ 

 \l). una {a— 0)1 



= + 



R (sin g cos g— cos a sin $) 



bill a 



+ 



r sin(g — /3) 

 ^ (g — fl)sing 



M (e-g) 



V 



cosec'' a -j- 



r sing — 3^" 



(r sing — g\ 

 M {a-0)) 



20* 



