1492

THEORY AND PRACTICE OF ENGINEERING.

Book II.

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point H on the axis B C prolonged ; the two points D, H are the centres of two extremearcs: to havethose of the fourintermediatearcs, draw IDparallel to P C,which gives thefirst arc AI of27 degrees: bycalculation wefind IK will be22 degrees 12

minutes, KL 17 degrees 24 \ **•.,

minutes, EM 12 degrees 36 \\\

minutes, MN 7 degrees 48 \\

minutes, and NB 3 degrees, mak- \\\

ing up 90 degrees for the whole. \\

From C as a centre, and with \

C A as a radius, describe thequarter circle A R; make V Requal to one-fifth, or 18 degrees: \\\

this arc, divided into six, gives \v,

t R 3 degrees, forming the first \ i’\

term of the arithmetical progres- \\ \

sion: to have the difference, the \i '

arc AV is divided into fifteen, 3

which gives s Y for this value; /A

D H being then divided into tenparts, four are given to DE,three to E F, two to F G, andone to GH. From the point Hwith radius equal to C R, havingdescribed an arc, iR is set offfrom 1 to 2; draw H 2 N,which forms an angle of 3 de-grees with H B. From the pointG, with CR for a radius, de-scribe a second arc, on which set Tig. 2712.

off from 3 to 4 the measure ofthe two arcs BN and NM taken together, equal to twice tr and s V, and draw G4M,which forms with H B an angle of 10 degrees 48 minutes, and with G N one of 7 degrees48 minutes.

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Fig. 2713.

From the point F, with the same radius, a third arc is described, on which measure offthe three arcs BN,NM, and ML, which is equal to three times iR and three times sV,from 5 to 6, and draw F 6 L, forming with H B an angle of 23 degrees 24 minutes, andwith MG an angle of 12 degrees 36 minutes.

From the point E a fourth arc is described, as are other arcs, from F and G : the pointswhere the lines intersect, are the centres of the four intermediate arcs.

Fig. 2711. is formed of eleven equal arcs, to draw which the centres d, H, are determinedas in the preceding figures: after drawing the diagonal PC, the angle A Pd is made equaltooPQ, and Pd is prolonged to H: from the point C, with the radius AC, draw thequarter circle A R, and divide it into six equal parts: from the point d raise a perpendi-cular to the intersection q , on the radius C 1, drawn from the first division of the quartercircle: from the point q draw qr parallel to AC, which cut the radii drawn from thedivisions 2, 3, 4, 5, in the points 6, 7, 8, 9. After setting out the parts $6, 6 7, 7 8, 8 9,and 9r, from C to m, m to 7, l to k, k to i, and i to d, from these points d, i, k , 2, m, with a-radius equal to C A, draw the arcs of circles, on which set off the arc A 1 of 15 degrees,once from 10 to D, twice from 11 to E, thrice from 12 to F, four times from 13 to G, andfive times from 14 to 15, through the points Dd, E i, Fk, G l, H m, draw the lines whoseprolongation makes the arcs A I, IK, KL, LM, MN, and NB, of the same number ofdegrees; the intersection of these lines gives the centres d, e, f g, k } and H, to describe theseveral arcs.

Fig. 2712. An arch of the same diameter and height as the preceding may have thequarter circle AR divided into six equal parts at 1, 2, 3, 4, 5, through which lines aredrawn parallel to R C: from the point C, with C B as a radius, another quarter circle isdescribed, also divided into six equal parts, in the points'6, 7, 8, 9, 0, through which lines