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draw the line 14 d at the distance of three parts, then db isthe breadth of the hand-rail. Draw A n parallel to 9 b, atthe distance of b 9, then the point n is the centre of the eyeof the scroll. On the point a describe the quadrants b cand de, which is the length of the twisted part of the rail,the remaining part, to n, the eye, being level. On n describethe circle zxp, whose diameter, wp, must be equal to db,the breadth of the hand-rail. Divide the radius, np, intofour equal parts, and through the first part, at o, draw theline r l, cutting the line N F in x; on x describe the quadrantsc f and e g, make o t equal unto two parts of np, and drawthe line t s parallel to a n. On the point t describe thequadrants f h and g z, make n w equal to three parts of n p,and through the point w draw the line z k, parallel to r x ;on z describe the quadrant h v, and on w the quadrant v p,and then is the plan completed.”

This is a very anomalous attempt at the description of thescroll of a hand-rail with compasses, the first centre being ata, the second at x; the third centre is said to be at z, but itproves to be at o ; neither t nor w answer to the remainingcentres.

“ To describe the mould for the twist. —Continue b 9towards m, and m towards b, in Figure 141 ; also draw L iparallel to b N ; at the distance of N K, in any part of N b, asat c, draw the line af at right angles to b N ; and on cdescribe the semicircle ab f: make a d and f t each equal tothe rise of one stair, and draw the line dot. Make cn equalto ct; divide be into any number of equal parts, and drawthe ordinates 15, 1 ; 16, 2 ; k 3, &c.; divide cn into thesame number of equal parts as in b c, and make the ordinatesthereon equal to the ordinates on b c, and through theirextremes trace the curve vif which is the curve of the out-side of the mould. Make b k equal to the breadth of thehand-rail; and on c, with the radius c k, describe the innersemicircle. Make c h equal to r t. On k t, the semi-diameterof the inner semicircle, make ordinates, which transfer onc h, as before; and through their extremes trace the curveof the mould, which will complete the whole as required ;for as the outlines of the plan of the twisted part of the rail,be and d e, are quadrants, therefore the outer and inner curvesof the mould will be both a quarter part of two ellipses;because the twisted rail, strictly considered, is no other thanthe section of a cylinder, as imik, whose diameter, a f isequal to twice a b, in Figure 140, arid its transverse diameterequal to d t, and conjugate diameter to a f.

“ The twist of a rail over a circular base at a half pace, asa b, figure 142, is the very same thing as the preceding, asbeing the fourth part of an ellipsis, made by the section of acylinder, whose diameter is equal to twice a e.”

This method of forming the section is the same in effect asthat already shown by Price . The only difference betweenPrice and Langley is, that Price forms the face-mould of therail from the intersection of straight lines drawn from twolines of sines placed at right angles to each other, and tracinga curve through the diagonals of the rectangles, beginningat the extremity of one of the perpendiculars, and ending atthe extremity of the other. Whereas Langley divides thebreadth of the plan and the length of the face-mould of eachinto the same number of parts, and draws lines at rightangles through the points of division, and makes the respec-tive perpendicular of the face-mould equal to those of theplan, and then traces curves through the extremities for theconcave and convex sides ; and thus completes the face-mould of the twist.

Figures 143, 144.—“ To find the mould of a twisted railto a circular or elliptical staircase.—Figure 143. Let abcdbe the plan of a cylindrical staircase, whose base is a ci role, and

whose stairs wind about the cylinder ab d, &c. The pi 3 "of the stairs being divided, continue out the diametertowards the right hand, as to f of length at pleasure. Mak®a f equal to the girt of the semicircle abd , which divide in t0the same number of equal parts as there are stairs in tb®plan of the semicircle abd, as at the points 1,2, 3,4, & c 'from which erect perpendiculars, as 1 a, 2 a, 3 a, &c. 0length at pleasure. Consider the rise of a stair, and mak®the perpendicular, f g, equal to the rise of all the twel'' estairs that go round the semicircle ab d, and divide the p®j"'pendicular _/'<7 into twelve equal parts, as at the points L3, 4, &c. from which draw lines parallel to f d, continued outtowards the right hand at pleasure, which will intersect tb®perpendiculars on the line fad, in the points ac, ac, ac,and which are the breadths and heights of the treads anrisers of the twelve stairs at the side of the semi-cylin^ 61 "a b d; for were the whole of Figure g f a applied about tb®semi-cylinder, then the parts ac, ac, &c. would be in tb®respective place of each stair. Let a e represent the breadtj'of the hand-rail, and the semicircle e 10 c its base, over whi®its inside is to stand. Divide its diameter, e c, into a n /number of equal parts, as at 1 2 3 4, &c. and draw the ord*'nates 1, 6; 2, 7; 3,8; 4,9, &c. which continue upwards,as to meet the horizontal lines drawn from the perpendicul 3 *g f in the points 28, 27, 26, 25, &c. through which trace tb®ogee curve 28, 14, a, which is the sectional line of tb®cylinder over which it stands. Make the distances, 15, 21 ’

19, 14 ; 18, 13 ; 17, 12 ; and 16, 11, equal to the ordinate®10, 5 ; 9, 4 ; 8, 3 ; 7, 2 ; and 6, 1; and through the p°’ n !. S

20, 19, 18, 17, 16, to a on the line f d, trace the curve, 2^’16, a, which is the inside curve of the mould, and wb® seout-curve, 21a, being made concentric thereto, will be tb®mould required, whose end, 21,20, when set up in its pla®®’will stand perpendicular over its base 510.

‘ Nofe. This mould, though made but for one-fourth p 3 J’of the cylinder, will serve for the whole, by repeating 111

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same, or adding three or more others of the same kind toends of each other as often as there are revolutions incylinder.” . 9

It is not possible to conceive anything so void of truth ^the method here shown. Over and above the absenceprinciple, the description is contradictory to the diag I ' aII1 j.We are told to “ divide its diameter, e c (into any numberequal parts) as at 1 2 3 4, he. and draw the ordinates ^ ’2 7, 3 8, 4 9, &c. which continue upwards so as to meet thorizontal lines drawn from the perpendicular gf>” b .instead of being drawn through the diameter, they P 3through the divisions which divide the concave circuinfer® 11of the plan into equal parts. . fl

Langley has presumed to differ in method from Price finding the curvature of raking-mouldings ; but in tin®has been much mistaken, as may be observed infollowing

“ To find the curvature or mould of the raking ovoto 11shall mitre with the level ovolo. —Let np, Figure 145, k® ^part of the level cornice, and a n the points from which ^raking-cornice takes its rise; also, \etfa and gn repr® s ®.a part of the raking-cornice. On n erect the perpendic® c

nb, and continue l a to b; divide bn into any numbe*

equal parts at the points 1, 2, 3, &c. and from themthe ordinates, 1 2, 3 4, 5 6, &c. In any part of the ra

dr 3 ”

ki"?

- _ ‘ ' ’ * * . aU>

ovolo, as at c, draw the perpendicular c m, and makeequal to b a, the projection of the level ovolo. Divide cinto the same number of equal parts as are in b n, as atpoints 1, 3, 5, 7, he. from which draw ordinates equal toordinates in b n, arid through the points 2, 4, 6, he. trace .jcurve required. In th' same manner the curvature or < n °