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facilitated by applying the bevels to a box, the down bevel being laid on the side, and the bevel contained in the angle maf, on the top of the mitre-box.


Figure 1.

Shows the method of finding the lines on the face and butt-joints of a piece set to a given plane with the base, to mitre over an acute, obtuse, or right angle.

Let ABC be the plan of an acute angle; BD the line of intersection; Abc d the piece; ad the plane, and Af the base of the plane. To find the face-joint, make af equal to Ad. Draw dh parallel to AB, cutting the line в D in h. Perpendicular to dh draw hg. Join fg, making the line fg parallel to dh. Join вg; then the angle ABg, contains the bevel for the side Ad, of the piece to mitre over the acute angle, A B C.

To find the butt-joint, or the line over the edge dc, of the piece, make cy equal to cd; draw ya parallel to dh; produce gh to cut the line yx in x; join D, then yxD is the angle, and contains the bevel for the side dc, of the piece to mitre over the angle ABC, of the plan.

Figure 2.

Exhibits the application of the same principle to an obtuse angle, and a bare inspection of it will show that the process is precisely the same as in Fig. 1.


Figure 1.

Shows the method of mitering together a splayed box, also to find the cuts in a mitre box, by which to cut a moulding that a bevel cannot be applied to conveniently. Let AA be the sides, в the bottom, and cdaf, the plan of the box. With one point of the compass on h and the other on v, describe the arc, bvsk, and from the points bvsk, perpendicular to bk, draw the lines bq, vb, sm, and kn.

Join ba, and produce ba to m; then ba is the mitre, or intersecting line on the plan. Draw mn perpendicular to sm; join pn and pq; then the angle opn contains the bevel at which to cut the upper edge of the box at hi; and the angle opq contains the bevel for the out and inside of the piece, a. The other side of the figure shows the application of the same principle where the sides are not of the same bevel, the operation being performed on both sides of the box to find the lines for the two different bevels.

Figure 2.

Shows the same principle applied to an octagon figure, the mitre box being used to cut the mitres, either for facilitating the work, or when the surface of the moulding is such that a bevel cannot be employed. In such a case, ▲ will be the bevel to apply to the top of the box, and в the bevel applied to its side. In placing the piece in the box, lay the side A, downwards.


Shows the method of finding the lines for mitering together a grain-mill hopper at its angles; also to find the angle pieces that are required to secure the pieces together at their intersection.

Let ABCD, Fig. 1, be the plan of the angular box, and Fig. 2 the elevation. In Fig. 2 place one point of the compass on a, and with the radius ab, describe the dotted curve bd; then ad will be equal to ab. Draw the dotted line df perpendicular to a d, to the line fg on the plan, Fig. 1, cutting the line fg in f. Join Bf and of; then вof is the development of one side of the figure, and consequently contains all the angles for constructing a box of any size.

It is usual to place the sides of all mill-hopper boxes at an angle of 45°. It may, however, be desirable to make the elevation of the sides greater or less than 45°, and it is therefore proper to give the method of finding the butt-joint for any angle, from a horizontal to a perpendicular.

To find the mitre, or butt-joint across the edge of one of the sides of the box: Produce da to i. Make ai equal to ah. Produce the diagonal line Ac, until it cuts the line hm in m. Through the point m, perpendicular to hn, and parallel to Do, draw nm h. Join ok; then вck will be the angle at which to cut the edge of the board at a, Fig. 2; and Bof, Fig. 1, the side of the board, ba, Fig. 2. H, Fig. 1, is the angle piece, and is found by making og equal to of, and joining Dg; then the angle at H, will be the internal angle of the hopper box, and, if it is required to make the box at a greater or less angle than 45°, the angular piece at h will be, invariably, one half of the angle piece desired.


Shows the method of cutting buffer boards in a circular, or gothic-headed opening. Fig. 1, represents an elevation of one half of an equilateral gothic-headed window; No. 1, No. 2, and No. 3, being boards, to be placed against the curve, as shown at Fig. 2.

To find the points at which to cut the boards in order to fit against the circular jamb: Divide the board, No. 1, Fig. 2, into any number of equal parts, and from the points of division, 1, 2, 3, &c., on both sides, draw lines over the elevation of the board, parallel to its edges. With one point of the compass on &c., on the line ad equal to a 1,

ad, and from the

a, describe the arcs, 1.1, 2.2, &c., making a 1, &c., on the line ab; then ad will be the development of ab. Perpendicular to points, 1, 2, 3, &c., draw the parallel lines over the board to be cut at Fig. 3. Draw any line perpendicular to the edge of the board, Fig. 3, and transfer the distances between the vertical and curve lines, Fig. 1, to their corresponding line at Fig. 3. From the vertical line, and through the points thus found, trace the form of the boards required.

To make a perfect joint against the jamb, both sides will require to be found in the same manner.

Is a plan of a circular seat.

Figure 4.

Figure 5.

Is an elevation of the same, with an inclining back, and shows the method of finding the curved pieces bent around and forming a continuous back to the circular seat, in such a manner that the edges shall be parallel to the plan of the seat, and to the surface upon which the seat shall rest. In explanation, it is simply

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necessary to say that the principle is precisely the same as that upon which the covering for a circular roof is found.

Figure 6.

Shows the manner of finding the centre, or radius of a circle whose centre is lost.

Let ab be the curve. Take any distance, gg, and with the same radius describe arcs, ff. Through the points of intersection of the arcs draw the line, ffk; then at any other point in the circle repeat the same operation, and the radiating lines kff, will intersect each other in the centre of the circle ab, and can be proved by application of the compass.



Shows the method of finding the section of an inclined moulding that will miter with a horizontal moulding. Also, of finding the bevels to cut a box for mitering the horizontal and raking mouldings at any angle.

Let the moulding at A, be the given horizontal moulding, to find the section of the raking moulding B. Divide the surface of the moulding A, into any number of parts; and from the points thus obtained, and from all the angular points of the moulding, draw parallel lines to the given inclination, and to the vertical line, o a. From the point a, at A, draw ab, perpendicular to the vertical line, o a. All that is necessary in order to find the sections в and c, will be to transfer the distances 1, 2, 3, etc., on the line ab, at A, to the line ab at в and c; and, through the points of intersection with the parallel line, trace the section required to miter with the horizontal moulding at A.

To find the bevels to cut the mitre at the intersection of the horizontal and raking mouldings.

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Place one point of the compass on a, and, with the radius ab, describe the semicircle, fdbc. Draw ad at right angles with ab, and from the points a and b, drop the lines an and by, at right angles with ac. Draw the line

yns, which

is the plan of the miter. Draw ch, ds, and fv parallel to an; and from the

points and y, draw yh and sv. Join hn and vn. s h and sv. Join hn and vn. Then the angle anh, and

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