A method of obtaining the bevels for this rafter is given in Fig. 50 where the steel square is shown laid on the pattern with the points 16 inches on the blade and 8 inches on the tongue applied to the edge of the stuff. The line HO on the blade gives the bevel for the foot of the rafter AC. The line OP, Fig. 50 gives the bevel for the top of the rafter or the plumb cut, as most workmen call it. Now, there is nothing in this diagram, which is from Bell’s Carpentry, an excellent work—from which the workman can get the length of his rafter, without complicating matters. Had the figures 12 inches and 6 inches on the square been employed instead of 16 and 8, then the distance across the diagonal from these two points would have equalled on the rafter, one foot on the base line or seat of the rafter, so that 15 times that length would have been the total length of the rafter. Better still, however, would have been the application of the square 15 times on the edge of the rafter pattern with the points 12 and 6 on gauge points, then both length and bevels would have been obtained at one operation.

Of course, the expert workman will often invent, or discover, methods of using the square in certain phases of roof framing, that can not be found in books, or that cannot be taught because of the peculiar circumstances of the particular case. Having a fair knowledge of the uses of the steel, the workman will seldom be overtaken by difficulties he cannot overcome if he studies the problems before him and then employs his knowledge of the square to their solution, as a little application on this line will remove all possible troubles.

Every carpenter knows, or ought to know, that the run and rise of the rafter taken on the square will give the seat and plumb cuts, but inasmuch as buildings are not all of the same width, it requires a different set of figures for each run, and as it requires an extra calculation to first find the run of the hip or valley, it is better to use the full scale for a one-foot run of the common rafter which answers for any run.

Referring to Fig. 51, we show a square bounded by A, B, C, D, the sides of which are 12 inches. E is at a point 5 inches from B, and C 12 inches from B. B-A represents the run of the common rafter. E-A represents the run of the octagon hip or valley, and C-A the same for the common hip or valley, their lengths, being 12, 13, and 17 respectively. Now since 12, 13, and 17 are fixed numbers, we take them on the tongue of the square, as shown in Fig. 52. Now suppose we want to find the lengths and cuts of the rafters for the ⅜ pitch. We take 9 on the blade. Why? Because the run being 12 inches, the span must be two times 12, which equals 24, and since the pitch is reckoned by the span, we find that ⅜ of 24 is 9, which represents the rise of the foot run. Then 12 and 9 give the seat and plumb cuts for the common rafter, 13 and 9 for the octagon hip or valley, and 17 and 9 gives the same for the common hip or valley. In Fig. 53 I show each separately.

The measurement line of hips and valleys is at a line along the center of its back, and just where to place the square on the side of the rafter so as to make the cuts and length come right at that point is a question that taxes the skill of most carpenters, especially so when the rafters are so backed. In Fig. 54 I have tried to make the above points clear.

First, I show the plan of the rafter. The cross lines on same represent an external corner for the hip and valley respectively. Above the plan is shown the elevation. The sections 1-2-3-4 represent the position of the rafters under the following conditions: No. 1 hip when not backed, No. 2 hip when backed, No. 3 valley when not backed, No. 4 valley when backed. No. 1 is outlined by heavy lines, and sets lower than the others. By tracing the bottom line of the sections down to the seat of No. 1, thence up to the second elevation will show just how deep the notching should be for each rafter. No. 1 cuts into the right hand vertical line from the plan, which would make it stand at the right height above the plate, but in order to make the seat cut clear the corner of plate, it is necessary to cut into the center line above the plan. No. 2 cuts into the same points as No. 1, but owing to its being backed, the seat cut drops accordingly. No. 3 cuts into the center vertical line, and in order to clear the edges of the plate must cut out at the sides to the left vertical line. No. 4 cuts in the same as the latter, but as much lower than No. 3 as No. 2 is below No. 1.

The outer vertical lines from the plan represent the width of the rafter. Therefore if the rafter be two inches thick, would be one inch apart, and this amount set off along the seat line (or a line parallel with it) will give the gauge point on the side of the rafter. To make this clearer refer to Fig. 53; 17 and 9 gives the cuts. Now leaving the square rest as it is, measure back from 17 one-half the thickness of the rafter, and this will be the gauge line point from which to remove the wood back to the center line of hip, and the measurement from the edge of the rafter taken vertically down to the gauge point set off on the plumb cut regulates how far apart the parallel lines of the seat cuts will be under the above conditions. This rule applies to any roof so long as the pitches are regular.

Proceed in like manner for the octagon hip, the variation, however, is practically one-half of the above results for the square cornered building.

Fig. 55 illustrates side cut of the jack, 12 on the tongue, and 15 (length of the common rafter) on the blade.

Fig. 56 illustrates side cut of the octagon jack, 5 on the tongue and 15 on the blade.

Fig. 57 illustrates the side cut of the hip or valley, 17 on tongue, 19¼ (length of the hip) on the blade giving the cut in each case.

The latter, however, is for the unbacked rafter. If it has been previously backed, then apply the square with the above figures on the lower edge at bottom of the plumb cut, or apply the square as for the jack, Fig. 56, to the backing line, which will give the same result as 17 and 19¼.

It is quite clear that when a workman cuts a common rafter, he is also cutting a timber that would answer for a hip for a building of less span having the same rise, only taking some adjustment of the top bevel to fit against a ridge. This is quite plain, and if we refer to Fig. 58, we find that the common rafter for a 1-foot run becomes a hip for an 8½-inch run, and that a hip for a 1-foot run of the building becomes a common rafter for a 17-inch run. Therefore, the rule that applies to the common rafter also applies to the hip rafter, i. e., the run and rise taken on the square will give the seat and plumb cuts. The run and length of the rafter taken on the square will give the side cuts, or taking the scale for a 1-foot run, Fig. 58, it is 12 on the tongue and the rise on the blade for the common rafter, and 17 on the tongue and rise on the blade for the hip. The tongue giving the seat cut and the blade the plumb cut. For the side cuts we take 12 on the tongue and 15⅝ inches on the blade, and the blade will give the side cut of the jack. Take 17 on the tongue and the length of the hip, 19¾ inches, on the blade and the blade will give the side cut of the hip. It would also be the side cut of the corresponding jack if it be a common rafter. Seventeen is used for a foot run of the hip rafter because the diagonal of a 12-inch square is practically 17 inches.

If we were to use 12 on the tongue for a foot run of the hip the rise to the foot would necessarily be less than 10 inches. In Fig. 59 I show what the difference is in rise to the foot.

From 12 to 12 is the length of the run of the hip would only have 10-17 of an inch to one run of the common rafter, and an equal rise of the common rafter, set off as at A, and a line from this to 12 on the tongue passes at 7 1-17 inches on the blade, because the common rafter having a rise of 10 inches to one foot, for one inch it would have 10-12 of an inch, while the hip would only have 10-17 of an inch to one inch and for 12 inches it would be 12 times 10-17 equals 120-17, or 7 1-17 inches. Therefore the figures given in the second illustration would give the same cuts as those in the first, but as the latter necessitates a calculation that ends in fractions—fractions not given on the square—and for that reason 17 is generally used for a foot run for the hips and valleys.

AN UNEQUAL PITCH.

In the matter of roofing over unequal pitches when there is no ridge and when all hips meet, the building being longer than it is wide, the backing of hips and their lengths and bevels would be a very easy matter if a drawing of the whole thing was made, but, to obtain these by the use of the square alone, is somewhat more difficult. Let us assume the building to be 18 feet wide and 28 feet long, and having a rise of 9 feet, then, by referring to Fig. 60, we show to one inch scale the length, run, rise, seat, and plumb cuts for the hip and common rafters as follows: The run of the long way of the building is 14, and 9 for the narrow way, which we take on the blade and tongue respectively, as shown on square No. 1, and to this apply square No. 2, as shown. AD equals the run of the hip. AE equals the rise and ED equals the length of the hip. The reader will notice that the letters A, B, C, D form a parallelogram, with side and ends equal to the runs of the common rafters. Therefore, by taking the runs on the tongue, as shown by the squares Nos. 3 and 4, will give their lengths, seat and plumb cuts.

In Fig. 61 is shown the intersection of the rafters at the peak and as the lengths of all rafters are scaled to run to a common center it is necessary that the common rafters must cut back so as to fit in the angle formed by the hips. The proper deduction for this is shown in Fig. 62 by placing two squares on the back of the rafter, with the heel or corner of the squares resting on the center line. The distance from the corner of the square to B measured square back (at right angles) from the plumb bevel, as shown in Fig. 61, will locate the point of the long common rafter at B in Fig. 61. Proceed in like manner for the short common rafter, taking the distance from the corner at C, and for the side cuts, take 14 on the tongue and the length of the short common rafter CE on the blade—the blade will give the cut at AC in Fig. 61. The reader will observe that this angle is the same as that for the side cut of the jack. Proceed in like manner for the long common rafter side, using 9 on the tongue and BE on the blade. These same figures will give the side cuts of the hip, provided hip has been previously backed. Taking the last for example the reader will observe that 9 on the tongue and BE on the blade the square would lay on the plane of the backing and the blade giving the cut along the line BB in Fig. 61, or these cuts may be found by measuring square back from a plumb bevel at points A and A, Fig. 62, the distance AC and AB, which will give the proper plumb cut at the sides and intersecting the line AA at the center. These same distances, AC and AB, but transferred to opposite sides, set off on the seat cut or a line parallel with it, will give the gauge points on the side of the hip for the backing.

The lengths of the jacks may be found by dividing the length of the common rafter by the number of the spacings for the jacks; the quotient will be the common difference.

END OF DIVISION B.

THE STEEL SQUARE AND ITS USES.
DIVISION C.
Introductory.

During my long experience as Editor of several of the leading building journals in the United States and Canada, I have been asked and have answered thousands of questions regarding matters concerning building construction, builders’ materials, tools and processes, and particularly regarding the “Steel Square and its Uses,” and I have concluded that the publication of a few of these questions and answers, along with other matter, in this division will be appreciated by my readers, and to this end I insert a number of the most useful items in this manual.

Besides these questions and answers, I also publish other up to date matter, all of which will make this volume one of the most useful little works to the American carpenter and wood-worker ever published.

I open this division with a few hints regarding the construction, or rather the laying out of a Hip-Roof where the design has been furnished by an architect, and which, of course, shows the pitch and the lay of the timbers. We suppose the roof to have a span of 18 feet and a rise of 6 feet, thus giving the roof a one-third pitch. The fence is used in this example to its full extent, and when placed on the square and fastened, the line of fence shows the slope or pitch of roof. Fig. 64 shows the square set to the pitch of the hip rafter. The squares as set give the plumb and level cuts. Fig. 65 is the rafter plan of a house 18 by 24 feet; the rafters are laid off on the level, and measure nine feet from center of ridge to outside of wall; there should be a rafter pattern with a plumb cut at one end, and the foot cut at the other, got out as previously shown. When the rafter foot is marked, place the end of the long blade of the square to the wall line, as in drawing, and mark across the rafter at the outside of the short blade, and these marks on the rafter pitch will correspond with two feet on the level plan; slide the square up the rafter and place the end of the long blade to the mark last made, and mark outside the short blade as before, repeat the application until nine feet are measured off, and then the length of the rafter is correct; remember to mark off one-half the thickness of ridge-piece. The rafters are laid off on part of plan to show the appearance of the rafters in a roof of this kind, but for working purposes the rafters 1, 2, 3, 4, 5 and 6, with one hip rafter, is all that is required.

To proceed, we first lay off common rafter, which has been previously explained; but deeming it necessary to give a formula in figures to avoid making a plan, we take 1-3 pitch. This pitch is 1-3 the width of the building, to point of rafter from wall plate or base. For an example, always use 8, which is 1-3 of 24, on tongues for altitude; 12, ½ the width of 24, on blade for base. This cuts common rafter. Next is the hip-rafter. It must be understood that the diagonal of 12 and 12 is approximately 17 in framing work, and the hip is the diagonal of a square added to the rise of roof; therefore we take 8 on the tongue and 17 on the blade; run the same number of times as common rafter which gives the length of hip and plumb and level bevels.

To cut jack rafters, divide the number of openings for common rafters. Suppose we have five jacks, with six openings, our common rafter 12 feet long, each jack would be 2 feet shorter. The first, next the hip, 10 feet, the second 8 feet, third 6 feet, and so on. The top down cut same as down cut for common rafter. For the bevel, cut against hip. Take half the width of building on tongue and length of common rafter on blade, and blade gives the bevel. Now find diagonal of 8 and 12, which is 14 7-16 in. Take this length on blade and 12 on tongue, blade gives bevels. If the hip-rafter is beveled or “backed” to suit jacks, then take height of hip on tongue, length of hip on blade, and tongue gives bevel. These figures will cover all bevels for cutting, cornice and sheathing. For bed moulds for gable to fit under cornice, take half width of building on tongue, length of common rafter on blade; blade gives cut. To cut planceer to run up valley, take height of rafter on tongue, length of rafter on blade; tongue gives bevel. For plumb cut, take height of hip on tongue, length of hip on blade; tongue gives bevel.

These figures were specially prepared for a hip roof having a one-third pitch, but will suit other pitches equally well if the difference in height of ridge is considered.

For a hopper the mitre is cut on the same principle. To make a butt joint, take the width of side on blade, and half the flare on tongue: the latter gives the cut. You will observe that a hip-roof is the same as a hopper inverted. The cuts for the edges of the pieces of a hexagonal hopper are found this way. Subtract the width of one piece at the bottom from the width of same at top, take remainder on tongue, depth of side on blade; tongue gives the cut. The cut on face of sides: Take 7-12 of the rise on tongue and the depth of side on blade; tongue gives cut. The bevel of top and bottom: Take rise on blade, run on tongue; tongue gives cut.

SOME QUESTIONS AND ANSWERS
FROM VARIOUS CORRESPONDENTS.

The following questions and answers from practical workmen are considered among the very best things regarding the use of the Steel Square, as they are from men who knew of what they were talking about. They are gathered from many sources, but chiefly from the columns of Technical Journals with which I have been connected, either as Editor or contributor.

Jas. Willis, Rochester, N. Y., asks: “How can I get the proper bevel for a butt joint on an obtuse or acute angle, by the use of the square only?”

Answer: Suppose Fig. 66 represents an obtuse angle formed by two parallel boards or timbers. To obtain the joint, A, space off equal distances from the point 1 to 3, 3, then square over from the lines, R, R, keeping the heel of the square at the points 3, 3. At the junction of the lines formed by the tongue of the square at 0 will be one point, and 1 will be the other by which the joint line, A, is defined.

To find the line of juncture for an acute angle, proceed as follows: Fig. 67 represents two parallel boards or timbers; 1 the extreme angle, 3, 3 equal distances from the angle 1 and are the points where the heel of the square must rest to form the lines 0, 3; 0 shows the junction of the lines formed by the blade of the square. Draw a line from 0 to 1, and the line, A, formed, is the bevel required.

It will be seen, by these two examples, that the bevel of a junction at any angle may be obtained by this method.

P. McVity, Milwaukee, asks: “How can I draw a circle with the Steel Square?”

Answer: A circle of any required diameter may be drawn by means of the square by using it as indicated in the accompanying sketch. Drive two pins or nails, A and B, Fig. 68, at whatever distance apart the circle is to have as its diameter. Bring the square against them, as shown, and use a pencil in the angle as indicated in the drawing. This rule is very convenient in many instances. Suppose A and B are two points through which a circle is required to be drawn. By bringing the square against pins or nails placed in the points, it may be described as indicated in the sketch.

A “Mechanic,” Tampa, Fla., asks: “Can the steel square be used in laying out a wreath for a handrail, and if so, please describe how?”

Answer: Some advance in this direction has been made, but not much, but the outlook is quite encouraging as many experts are trying to obtain all the lines required for forming circular handrails. It will be accomplished sooner or later. A few problems and solutions are given herewith: In getting out face-molds it has generally been considered necessary first to unfold the tangents and get the heights, and by construction get the bevels. The method shown is somewhat different, though results are the same, but are produced more rapidly. Take for illustration a side wreath mitered into a newel cap. This method will apply no matter where the newel is placed, or whether the easement is less or more than the one step of the example illustrated. What is meant by one step is, that the tangent of the straight rail continues to the point 2, Fig. 69. The tangent 2-1 is level.

To produce the face mould, lay the steel square in the position, indicated by the lines 1, 2, 3, 4, not the figure on the square at the points numbered, and transfer them to a piece of thin stuff, Fig. 70. Line 3-4 in Fig. 70 is indefinite. Now take the length of the long edge of the pitch board in the compasses, and with 2, Fig. 70, as a center, cut the line 3-4 in 4 and draw 2-4. Now 1-2 is the level, and 2-4 is the pitch tangent on the face mold.

To get the bevels and width of the face mold at both ends, take the distance 3-4 on the blade of the square, and the height of a riser on the tongue of the square, apply to the edge of a board and mark by the tongue; this gives the width of the mold at the lower end.

Next take the distance 4-X on the blade of the square, and the distance shown on the pitch board by the line squared from its top edge to the corner, on the tongue of the square; apply to the edge of a board and mark by the tongue; this gives the bevels for the top end of the wreath. Mark the width of the rail on the bevel, and this gives the width of the mold at the top end. An allowance of 6 inches is made at the top end to joint to the straight rail, and two inches at the bottom end to form the miter into the newel cap. The springing line is taken from the pitch board.

Fig. 69, in which are shown the bevels and the pitch board will help to make clear the methods used. The bevel at the back of the pitch board is for the bottom end of the wreath. The triangle has for its base the line 3-4 and for its height one riser. The hypothenuse is the length of 2-4, Fig. 70, and Fig. 70 stands over Fig. 69, level on the line 1-2-3, and inclined from it in this cast at an angle of nearly 45 degrees.

The top end bevel is shown below the pitch board. The angle has for its base the distance 4-x, and for its height not one rise, but the length of a line, from the corner of the pitch board squared from its top edge. This bevel will be understood better by placing the pitch board on the line 2-4 and applying the small triangle to it with its base on the line 4-x, and its point even with the top edge of the pitch board. It will then be at right angles to the top edge of the pitch board.

In practice, a parallel mold is generally used, and the wreath piece is cut out; both thickness of plank and width of molding being equal to the diameter of a circle that will contain a section of finished rail.

Jacques Demoux, Winnipeg, Man., wants to know how to lay out braces, regular and irregular by the use of the Steel Square.

Answer: Braces and trusses are something like rafters and when the run is known, there should be no difficulty in getting the lengths and proper bevels.

In the first place it is always best to make a pattern and then mark out the timber work from the pattern. Suppose we want braces having a “four-foot run”—that is, the brace is to form a diagonal from points four feet from the post and four feet from the girt. Take a piece of stuff already prepared, six feet long, four inches wide and half-inch thick, gauge it three-eighths from jointed edge.

Take the square as arranged at Fig. 71, and place it on the prepared stuff as shown at Fig. 72. Adjust the square so that the twelve-inch line coincide exactly with the gauge line o, o, o, o. Hold the square firmly in the position now obtained, and slide the fence up the tongue and blade until it fits snugly against the jointed edge of the prepared stuff, screw the fence tight on the square, and be sure that the 12-inch marks on both the blade and the tongue are in exact position over the gauge-line.

We are now ready to lay out the pattern. Slide the square to the extreme left, as shown on the dotted lines at x, mark with a knife on the outside edges of the square, cutting the gauge line. Repeat this process four times, marking the ends, and you have the length and bevels. Square over at each end from the gauge line and you have the toe of the brace. The lines ss, Fig. 72, show the tenons left on the end of the braces.

The cut at Fig. 73, shows the brace in position, on a reduced scale. The principle on which the square works in the formation of a brace can easily be understood from this cut, as the dotted lines show the position the square was in when the pattern was laid out.

It may be necessary to state that the “square,” as now arranged, will lay out a brace pattern for any length, if the angle is right, and the run equal. Should the brace be of great length, however, additional care must be taken in the adjustment of the square, for should there be any departure from truth, that departure will be repeated every time the square is moved, and where it would not affect a short run, it might seriously affect a long one.

To lay out a pattern for a brace where the run on the beam is three feet, and the run down the post four, proceed as follows:

Prepare a piece of stuff, same as the one operated on for four feet run; joint and gauge it. Lay the square on the left-hand side, keep the 12-inch mark on the tongue, over the gauge-line; place the 9-inch mark on the blade, on the gauge-line, so that the gauge-line forms the third side of a right angle triangle, the other sides of which are nine and twelve inches, respectively.

Now proceed as on the former occasion, and as shown at Fig. 74, taking care to mark the bevels at the extreme ends. The dotted lines show the position of the square, as the pattern is being laid out.

Fig. 75 shows the brace in position, the dotted lines show where the square was placed on the pattern. It is well to thoroughly understand the method of obtaining the lengths and bevels of irregular braces. A little study will soon enable any person to make all kinds of braces.

If we want a brace with a two-foot run, and a four-foot run it must be evident that, as two is the half of four, so on the square take 12 inches on the tongue, and 6 inches on the blade, apply four times, and we have the length, and the bevels of a brace for this run.

For a three by four foot run, take 12 inches on the tongue and 9 inches on the blade, and apply four times, because, as three feet is ¾ of four feet, so 9 inches is ¾ of 12 inches.

A young carpenter, Toronto, wants to know how to find the center of a circle by aid of the Square.

Answer: In Fig. 76 is shown how the center of a circle may be determined without the use of compasses; this is based on the principle that a circle can be drawn through any three points that are not actually in a straight line. Suppose we take A, B, C, D for four given points, then draw a line from A to D, and from B to C; get the center of these lines, and square from these centers as shown, and when the square crosses the line, or where the lines intersect, as at x, there will be the center of the circle. This is a very useful rule.

Ed. McDonald, Cincinnati, Ohio, says: “I want to know how much can be done with the square towards setting out stair railing?”

Answer: In a previous page a few remarks on this subject will be found and the following is further submitted:

Fig. 77 shows a plan of a stair well having three winders. The rail in this case will have two different pitches. These rails are a little more complicated than those having equal pitches, as in the latter the major axis is parallel off the diagonal line B D (Fig. 77). When the pitches differ the major axis ceases to be parallel; and the greater the difference in the pitches, the greater will be the difference in the axis and diagonal line. This fact can be easily demonstrated by cutting a model bed block out of 2-inch by 2-inch stuff to equal pitches. Procure a board, and draw a parallel line, say 8 inches off the edge. Now square over a line to cut the first line; set the bed block on, with the back corner touching the intersection of the lines. Lay a piece of cardboard on the inclined face of the bed block, and let it slide down until it touches the board. Make a mark along this edge, and it will be seen, on removing the card, that this line is equidistant from the corner (see Fig. 78). In Fig. 78 the cardboard is shown as though it were transparent. What has just been done is that the plane in which the rail lies has been projected to intersect the horizontal plane which contains the plan of the wreath. The name by which this line is generally known is the horizontal trace (shown at C, Figs. 78 and 79). The minor axis (Figs. 78 and 79) is always parallel off this, and always touches A as in Fig. 77. The major axis (Figs. 78 and 79) also touches this point A, and is always square off the minor axis and off the horizontal trace. It will be seen by this that the rail is pitched equally both ways; therefore the face mold will be of equal width at the ends.

When rails are cut by bandsaw on bed blocks, bevels are not necessary, as they can always be obtained by applying a bevel as shown at Fig. 80. The stock should lie solid on the block and square off the sides. When the block is thin it is best to apply the bevel near the corner, when a greater surface is obtained. These bevels are applied after the joint is squared off the tangent lines. To demonstrate a rail with unequal pitches, cut another piece of stuff 2x2 inches, as shown in Fig. 79, repeat the process with the cardboard as before. It will be found that the horizontal trace has departed from the angle of 45 degrees (see Fig. 79) and has approached nearer one corner and gone farther away on the other. The major axis B will have done likewise, as it is always square off the horizontal trace C. The wreath having two pitches, the face mold will obviously be wider at one end than at the other; and if bevels are required, they must be set off on the face of each side of the block. The width of the face mold is to be applied on the tangent line; this makes it slightly in excess on the joint, but it is better to have a little margin in thickness for working. Where thickness of stuff is a secondary consideration, it is preferable to take the rail out of stuff which is as thick as the diameter of a circle that will enclose the section of rail; the corners will then be left complete.

The following method shows the least thickness the rail can be cut out of, and also gives width of face molds on the joint. Set the bevel to the bed block as shown at Fig. 80, and apply at the side of the block. Draw a section of rail level; apply the bevel again, touching the bottom corner of the section. The distance between the marks is the thickness, a plumb line marked on shows the width of the face mold on the bevel line. Where the pitches are different the foregoing method has to be applied to each side of the bed block.

The bevels may be also obtained by the steel square. Take the width of prism face (shown by dotted lines) by laying the square with blade on the line C D (Fig. 82) and tongue, cutting at the center A. Note the length on the tongue of the square. Make a mark of this length on the edge of board. Now take the width of A to D (Fig. 84) which is 6 inches off the blade; keep this 6-inch mark fair at the end of this line made on the board (Fig. 81) and push on square until the tongue touches the end of the line; mark by the tongue, and this gives the bevel required.

To obtain an example of unequal pitches refer to Fig. 84. To set this out, run a line parallel off the edge of the board, and off this line square another. With the intersection A as a center, describe a semi-circle of 6 inches radius. This indicates the center of the rail. Run lines radial from A as shown; these are the riser lines. Draw the lines B C and C D, which are the tangents. Draw the diagonal B D. To make the bed blocks, procure a piece of one inch stuff; take it to the width shown at B C; square on a mark about 3 inches from the end (this is to allow for the shank to clear the saw table; the block is shown at Fig. 83 without the 3-inch allowance). Take on the steel square the rise on tongue and going on blade of the straight flight of stairs; mark on the inch board at tongue; this is pitch of the first tangent. Take the height at D, which is one and a half risers—10½ inches; deduct the height of the first tangent from 10½ inches; take the difference on the tongue and width from C to D on the blade; the tongue gives the cut for the second tangent. Mark the pitch of the first tangent on the edge of the second and cut to this; the pitch of the second tangent gives the edge cut of the first. Cut and fix together with stretcher as previously described.

To get out the face mold, procure a piece of thin stuff. Three-ply wood is excellent, as though it is liable to warp it does not shrink perceptibly. Shoot on edge and gauge on a center line; take the distance from B to D (Fig. 84) (the hypothenuse of 6 and 6) on the blade and the rise (10½) at D on the tongue; lay on the edge of the board to this. Lay off this length on the three-ply at B D (Fig. 82); take the width B C (Fig. 84) on the blade, rise at C on the tongue; find the hypothenuse, and apply with a pair of compasses at Fig. 82 with B as a center cutting at C. Then apply at D as a center, cutting at A. Now find the hypothenuse C to D (Fig. 84), and apply the compasses as before, with D and B as centers, cutting at A and C. Connect up the points where the arcs intersect to B and D; this is the face of the inclined prism, and contains the true shape of rail. Continue the tangent line C B, 3 inches or whatever is required for the shank, and square the joints of the lines B and D. In order to locate the major axis the horizontal trace is now required. Stand the bed block on the plan (see Fig. 83). Run the blade of the square down until it touches the board; mark this, and remove the block. It will be seen that the bed block has not got the 3 inches allowed at the bottom, but the horizontal trace is as easily found with as without the allowance; all that is required in the former case being to turn the blade to B (still keeping the heel at the top of the bed block), make a mark where the square touches and lay on the square as shown at Fig. 83. Mark at the blade, and slide back the square until the tongue touches at B, and also at the center A. This gives the true horizontal trace and major axis. Note the size indicated by the arrow lines on the tongue (from heel to B). Transfer the square to the three-ply board (Fig. 82), placing it as shown, with the blade touching A, and the distance of the arrow lines at B. Mark along the inside of the blade of the square and slide the square back until the tongue cuts at A. This gives the minor axis. Now continue this line downwards to guide the position of the square shown at Fig. 84. Describe a circle as wide as the rail on the minor axis (Fig. 82). The distance from A (Fig. 84) to the center of the rail is the distance to apply at Fig. 82 for the center of the rail, as this is the point where the center of the rail is fair with the plank. Obtain the width of the face molds, and apply at B and D; lay the square on the major and minor axis as shown at Fig. 84. Lay a lath on the square, with the point touching the outside of the circle at C; drive in a nail at the heel of the square; shift the lath until the point lies at B, and drive in another nail at the side of the square. This trammel is now ready to sweep the outside of the mold, which is done by reversing the square, as shown by dotted lines. Pull out the nails and repeat the process for the inside of the mold.