278 



THE CIVIL ENGINEER AND ARCHITECT'S JOURNAL. 



[Sept. 



Dimensions and Calculations. 

 A B width of arch on the square 

 O E height of arch 

 A A' depth of arch at springing 

 E T ditto at crown 



* AB»28^ + 20E122 AB28 + 20E12 

 BFor AEB=:^/ ^ ''' i ' 



31416 

 2 

 \ C width of arch on the skew = sin obi. 60' 50' ; R : : A B 



28: AC =32-065 



B C = R : cosin obi. 60° 50' : : A C 32065 : B C =15-627 



C L external width of bridge including projection of arch-stones 



11 in. .. .. •■ •• 17-25 



^.ngle BFC = BF32-626 : BC 15-627 ::R: tan BFC .. 25° 35' 35" 

 CF = sin BFC 25° 35' 35" : R : : B C 15'627 : CF 

 Convenient number of voussoirs, 31 



Of 36-175 

 Thickness of ditto — rj — 



C D from face to face of springer = R : cosin obi. 60o 50" : : C L 



17-25: CD =19-755 



D K= 7 courses x the thickness 1-167 =8-169 



AngleDCKadjusted = CD 19-755 : DK8-169:: R :sinDCK 24° 25' 14" 

 C M axial length=R : cot D C K 24° 25' 14" : : B F or M I 



32-626: CM .. .. .. .. ri'85 



Development of half the Arch C, I, m, n, o,p, q, to obtain the bevels for 

 Heads of Arch-stones. 

 Divide BE into a convenient number of parts and lay them on B/, 

 dravf lines parallel to B C and the ordinates will be found by similar 

 triangles, as A B : BC : : A 6' ; 6'm' and F B : BC : : F 6 : 6m. 

 Then V m' — bh=zh m. 



= 32-626 



36-175 

 1-167 



Divisions ore B E or B /. 

 B a = l-80 

 a 4=1-81 

 6 = 3-22 

 e rf=3-20 

 d e = 3-15 

 e /=3-14 



Ordinates calculated on C g. 

 g ^=-561 

 /im = -736 

 i n = -689 

 j o = -478 

 k p = -238 



Fig. 2. Elliptical arch on the square divided into circular arcs. 

 B S = 1"25, second arc = M3, third = 1-05, fourth &c. = 1-0. 



Ordinates of Ellipse. 

 The one semi-axis is to the other as the circle to the ellipse, — 

 or BO : OE :: aa" : aa' (14 : 6 :: 13-645 : 5-848) = 5-848 

 4 i' (14 : 6 :: 12-527 : 5-368) = 5-368 

 e e' (14 : 6 :: 10-42 : 4-465) = 4-465 

 dd'(14:6:: 6-831 : 2-892) = 2-892 

 oe'(14:6:: 3-885 : 1-665) = 1-665 



To find the .Angles. 



1st. E'Eborgb'E. E'6 6-25 : EE' (6-5-368) -632 : : R : tan 

 E 6' E' 5° 46' 27" and 90° - E 6' E' 5° 46' 27" = E' E 6' or g 6' E = 

 84° 13' 33". 



2nd. h b' c' or h c' b'. t' 8 3-1 : s c' -903 : : R '. tan s V c' 16° 14' 

 25" and 180°- (^ 6' E 84° 13' 33" + E6'6" 84° 13' 33") = i"4'/ 

 11° 32' 54" and 90° - 6" b' t 11° 32' 54" =itb'a 78° 27 6", then 

 180°— (/6's 78° 27' 6" + s 6' c' 16° 14' 25") = A6'c' or hc'b' — 

 85° 18' 29". 



3rd. ABe'. Be-55 : ee' 1-665 : : R : Ian eBe' = 71° 43' 12". 

 To find the Radii. 



1st Rad. g v. 180° - 2 g- E 6' 84° 13" 33" = E g 6' 1 P 32' 54" and 

 sinE'E6'84° 13' 33" : R : : E' 6' 6-25 : E 6' 6-281. Sin Eg 6' IT 

 32' 54" : sin g E b' 84° 13' 33" : : E i' 6-281 : g i' = 31-219. 



To find the lines drawn from the centres of the several arcs perpendi- 

 cular to the semi-axis O B. 



Ist.i^andBZ. je' 5-927 -i c' 2-795 =j i 3-132 and R -.sinjkl 

 or B A e' 36° 33' 36" :: ^4 3-132 :^/=: 1-865. 



And R : cosin; A/ 36° 33' 36" : ; j k 3-132 : kl 2-564 and Bk 

 2-795 + k I 2-564 = B / = 5-359. 



2nd. !»jandBm. 90°- (f (i'2 or yij' 36° 30' 14" = (i2i' or 22m 

 53° 29' 46" and sin dl d' 53° 29' 46" : R : : c^i' 2-892 : d' 2 3-597 

 and id 12-067- 2 i' 3-597=2 « 8-470 R : sin »2m 53° 29' 46" : ; 2i 

 8-470 : zm=: 6-808. 



And R • cosin ilm 53° 29' 46" : ; 2 i 8-470 : 2 m 5-039 and 

 R - sin 2d' d 36° 30' 14" : : 2^' 3-597 : 2rf 2-14 and Be-55 + fid 

 1-23 + 2^ 2-14 + 2m 5-039 = Bm = 8-959. 



Fig. 3. Development for circular arcs. 



^"^ 



The angle D C K, fig. 1, or G E H is the angle of intrado, and G E I 

 the angle of extrado, and the difference of these, or H E I, is the twist 

 wrought on the beds of the arch-stones; a sectional view through the 

 crown of the arch to show the radiation of the courses would exhibit 

 the checks on the soffit to which the template for intrado applies, but 

 the calculations generally will be sufficient for practical purposes. 



