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To find the length of any rafter, draw a line from the one extremity of the plan-line of that rafter, perpendicular to that line, and make the height of the perpendicular equal to the height of the roof; join the point of height and the other extremity of the plan-line, and the line thus joined is the length of the rafter as required.

Example 1; fig. 1.—To find the length of the common rafters standing upon IK; divide IK into two equal parts in the point F: draw FL perpendicular to IK; make FL equal to the height of the roof, and join IL or KL; then IL or KL is the length of each rafter.

Example 2; fig. 2.—To find the length of the hip-rafter standing upon AF. Draw FS, perpendicular to AF: make FS equal to the height of the roof; join AS, and AS is the length of the hip,

Example 3.—To find the bevels of the back of the hip-rafters. Let FD, fig. 2, be the plan of a rafter; and draw XW perpendicular to FD, to meet the lines of the wall-plates in X and W, and intersect FD in w. From the point w, as a centre, describe a circle to touch the elevation of the rafter DT; and from the point where the circle cuts the line FD, draw lines to X and W, then will the angle formed by these lines be the proper bevel for the back of the hip-rafter.

133. To find the bevels of a purlin against a hip-rafter, when the plan-line of a common rafter, that of the hip-rafter, and the angle which the common rafter makes with its plan are known.

Place the section of the purlin in its real position with respect to the common rafter. Produce that side of the section of the purlin, of which the bevel is required upon the hip, toward the plan of the rafter; from one extremity of the line thus produced, and, with the length of the said line as a radius, describe a circle. Draw three lines, parallel to the wall-plate, to meet the hipped line: viz. one from the centre of the circle, one from the point where the line meets the circle, and the third to touch or be a tangent to the circle. From the point in the plan of the hip-rafter, where the middle line meets the said plan, draw a line perpendicular to that middle line to meet the tangent; join the point, where this perpendicular meets the tangent, to the point where the line drawn from the centre meets the plan of the hip-rafter, and the angle formed by the line thus joining, and the line drawn from the centre of the circle, will be the bevel of the purlin

Example; plate XXIV, fig. 1.—Let AF be the plan of a hip-rafter, IF that of a commonrafter, and FIL the angle which the common-rafter makes with its plan, and abcd the section of the purlin.

Now suppose it were required to find the bevel of that side of the purlin represented by ad. Produce ad to any point, f; and from a, with the radius af, describe a circle, efgh. Parallel to the line of the wall-plate, AB, draw two lines to cut the plan, AF, of the hip; viz., from the centre, a, draw ai; and from the point f, where af meets the circle, draw fk, the former cutting AF in i, and the latter in k; also draw el to touch the circle. Draw kl perpendicular to ƒk, cutting el in l; and join il; then the angle lia is the bevel required.

In the same manner, by producing ab, we may find the angle formed upon the end of the side, of which the section is ab.

134. In order that the different inclined planes, which form the sides of a roof, may have an equal inclination to the horizon, the plan-lines of the hip-rafters ought to bisect the angles formed by the wall-plates.

When a roof is wider at one end than at the other, as in fig. 3, in order to prevent its winding, let IK and OP be the plans of the two common rafters, passing through each extremity of the

ridge-piece, and let the rafters IL and KL be found as before; divide OP into two equal parts, in E; draw ER perpendicular to OP. Make the angle EPR equal to the angle FKL; then FR will be the height of the roof at the point E.

If this should be objected to, because it makes the ridge higher at one end than at the other let E, fig. 4, be the end of the ridge next to the narrow end of the roof.

Bisect all the four angles of the roof by the straight lines AF, BE, CE, DF; and, through E, draw EG, parallel to AB, cutting AF in G; and draw EH, parallel to CD, cutting DF in H; and join GH: then GH will be parallel to AD. This is true, because, since all the angles are bisected, if we imagine perpendiculars drawn from E to the three sides, the three straight lines thus drawn will be equal: and because EG is parallel to AB, the perpendiculars drawn from the points E and G, to the straight line AB, are equal; from the same reason, because EH is parallel to CD, the perpendiculars drawn from the points E and H, to the straight line CD, are equal; therefore the perpendicular drawn from the point G, to the straight line AB, is equal to the perpendicular drawn from H to the straight line CD. And, since the angles BAD and CDA are bisected by the straight lines AG and DH, the two perpendiculars, drawn from G, to the sides AB and AD, are equal; as also the two perpendiculars from the point H to the sides DA and DC: but the perpendicular drawn from G, to the side AB, is equal to the perpendicular drawn from H to the side CD; therefore the perpendiculars, drawn from the points G and H, to the straight line AD, are equal to each other; but when the perpendiculars drawn between two straight lines are equal, these two straight lines are parallel: therefore the straight line GH is parallel to AD.

Whence, if all the angles of a roof be bisected, and if any point be taken in any one of the bisecting lines, and if a line be drawn through the point thus assumed, parallel to one of the adjacent sides, to meet the next bisecting line, and so on from one to another, till only one line remains to be drawn; then, if the point assumed be joined to the point where the parallel meets the last bisecting line, the line thus joining will be parallel.

GEOMETRICAL LINES FOR POLYGONAL ROOFS.

135. THE plans of these roofs are supposed to be regular polygons, and all the sections of the same roof, parallel to the plan, to be similar to the plan, and therefore all the parallel sections similar to one another. They may be conceived to be formed of a series of triangular prisms whose joining planes meet in the same point, and their exterior surfaces cut to the form of the roof. 136. In pl. XXV, fig. 1, the plan of the roof is denoted by the letters ABCDEFA. Then the centre of the polygon being the point I, draw the lines AI, BI, CI, &c. Bisect any of the sides, as AB, in the point L, and draw LI; then LI is perpendicular to AB.

Produce the line IL to M, and let ILN be the section applied upon IL. In the curve LN take any number of points, 1, 2, 3, at equal distances, and transfer these distances to the line LM, so that LM may be equal to the arc LN. Through the points 1, 2, 3, &c. in LM, draw lines 1g, 2h, 3i, &c. parallel to AB; and from the points 1, 2, 3, &c., in the arc LN, draw lines 1d, 2e, 3f, &c., parallel to AB, cutting LI at the points a, b, c, &c., and BI at the points d, e, f, &c.: Make 1g equal to ad, 2h equal to be, 3i equal to cf, &c. Through the points g, h, i, &c., draw a curve, which will be the edge of the covering which corresponds to the joirt over the mitre IB.

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