Let A B D C in Fig. 283 be the elevation of the face of a keystone, and G E2 F2 K of Fig. 284 a section of the same on its center line.

Sometimes problems occur which are so simple that it is not apparent that their solution is an exemplification of any rule. That this, with others in which plain surfaces form the largest factors, may be so designated, will be sufficient excuse for a brief reference to first principles. This problem is generally referred to as finding the "true face" of the keystone, because, the face being inclined, the elevation ABC D does not represent the "true face" or "true" dimensions of the face. To state the case, then, in conformity with the rule, A B and C D are the upper and lower lines of a molding, of which E2 F2 of Fig. 284 is the profile, and A C and B D are the surfaces against which it miters, or the miter lines. Therefore, to lay out the pattern, draw any line, as E1 F1, at right angles to A B for a stretchout line, upon which lay off the stretchout taken from the profile E2 F2, Fig. 284, which in this case consists of only one space, as shown by E1 E1; through the points E1 and F1 draw the horizontal lines A1 B1 and C1 D1, which are none other than the measuring lines. Then, with the T-square placed parallel with the stretchout line, drop the points from the miter lines A C and B D into lines of corresponding letter, which connect, as shown by A1 C1 and B1 D1, which completes the pattern.

Problem 10 Patterns Of The Face And Side Of A Plai 286

Fig. 283.

Problem 10 Patterns Of The Face And Side Of A Plai 287

Fig. 284.

Patterns of the Face and Side of a Plain Tapering Keystone.

In developing the pattern for the side, E2 G and F2 K are the lines of the molding, B D of Fig. 283 its profile and E2 F2 the miter line. Hence upon any vertical line, as L K1, lay off the stretchout of profile B D, locating the points M1 and H2, all as shown by LMH1 K1, through which points draw the measuring lines; then, with the T-square placed parallel to L K1, drop the points E2 and F2 into lines of corresponding letter, as shown by E3 F3. As the vertical lines at Grand K represent the position of surfaces against which the side is required to fit at the back, bring the T-square against each, thus locating them in the pattern at G1 and K1, as shown.

As the side must also fit over the molding of the arch an opening must be cut in it corresponding in shape to the profile of the arch molding N, which is given in the sectional view. It is therefore only necessary to transfer this profile to the pattern, placing the top at the measuring line M and the bottom at the measuring line H1, all as shown at N1.