. Stereotomy : Problems in stone cutting. In four classes. I. Plane-sided structures. II. Structures containing developable surfaces. III. Structrues containing warped surfaces. IV. Structures containing double-curved surfaces. For students of engineering and architecture . xample, substituted : — The half span 0B = S = ft. OBO = a = 30°. Then 00 = S. tan a = ft. and 0B = E = VS2 + 002 = ft. ( ft.) Again, let the angle of skew, EAC, = 0 then EAG = 90° — 0,whence BG = 2S. tan (90° — 0) (1) and AB = h/AG2+BG2= */l32 + (2S. tan (90 — Of oo or, AB = 2S. sec. (90 — 0) = 2S. cosec
. Stereotomy : Problems in stone cutting. In four classes. I. Plane-sided structures. II. Structures containing developable surfaces. III. Structrues containing warped surfaces. IV. Structures containing double-curved surfaces. For students of engineering and architecture . xample, substituted : — The half span 0B = S = ft. OBO = a = 30°. Then 00 = S. tan a = ft. and 0B = E = VS2 + 002 = ft. ( ft.) Again, let the angle of skew, EAC, = 0 then EAG = 90° — 0,whence BG = 2S. tan (90° — 0) (1) and AB = h/AG2+BG2= */l32 + (2S. tan (90 — Of oo or, AB = 2S. sec. (90 — 0) = 2S. cosec 6 = -~ .v J sin 0 Also CC, = AEB = I = R X .0174533 X 120° (2) = ft.; where AEB = 120°, and .0174533 is the length of 1° where R= 1°. Likewise, putting 0Ba = Runda2 = AjFB; = l1 = Rlx .0174533 X 120°, (3) where Rx = + = 10, whence nsa2= ft. Further ; let C2y4, perpendicular to AB, = L = 14 ft. Then, as A2C2^ = (90 — 0.) AC = BA = A2C2 = L. sec (90 — 0) = L. cosec 0 = ^ And, see (1) and (2), AB: = CDX = ^AGf+B^l (4) Now suppose CDX to be divided into n equal parts of whichDiiv are m parts. Then D,iv = — CDi. or, in this example, see (4), D,iv=|. CDX. Let ABxGx = B1D1C = ft then ^ = tan (3. Or, as by (1)B1G1 = BG = 2S. cot0. 5T- — . 2r0 28 | - - - - — |-}c- Jfc^f n 24 82 STEREOTOMY. 4°. Avoiding the acute diedral angles, between the face andthe intrados, near the acute angles, as B, of the abutments, by-terminating the cylindrical intrados by a plane parallel, andnear to AB ; through r5 for example, and then completing theintrados between this plane and AB by a conoid, whose direc-trices should be the ellipse in the plane r5, and a vertical line onBD, as at Q, and equal to 0E; and whose plane directorshould be horizontal. 5°. The substitution of brick for stone; either wholly, or ex-cept in the face ring of stones. 6°. The substitution of a circular for an elliptic face. Thiswill make the right section an arc of an ellipse whose
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