[Ill.u.s.tration: Fig. 170.]

For a hexagon head we have the processes, Figures 170 and 171. The circle is struck, and across it line _a_, Figure 170, pa.s.sing through its centre, the triangle of sixty degrees will mark the sides _b_, _c_, and _d_, _e_, as shown, and the square blade is used for _f_, _g_.

[Ill.u.s.tration: Fig. 171.]

The chamfer circles are left out of these figures to reduce the number of lines and so keep the engraving clear. Figure 171 shows the method of drawing a hexagon head when the diameter across corners is given, the lines being drawn in the alphabetical order marked, and the triangle used as will now be understood.

[Ill.u.s.tration: Fig. 172.]

[Ill.u.s.tration: Fig. 173.]

It may now be pointed out that the triangle may be used to divide circles much more quickly than they could be divided by stepping around them with compa.s.ses. Suppose, for example, that we require to divide a circle into eight equal parts, and we may do so as in Figure 172, line _a_ being marked from the square, and lines _b_, _c_ and _d_ from the triangle of forty-five degrees; the lines to be inked in to form an octagon need not be pencilled, as their location is clearly defined, being lines joining the ends of the lines crossing the circle, as for example, lines _e_, _f_.

Let it be required to draw a polygon having twelve equal sides, and the triangle of sixty is used, marking all the lines within the circle in Figure 173, except _a_, for which the square blade is used; the only lines to be inked in are such as _b_, _c_. In this example there is a corner at the top and bottom, but suppose it were required that a flat should fall there instead of a corner; then all we have to do is to set the square blade S at the required angle, as in Figure 174, and then proceed as before, bearing in mind that the point of the circle nearest to the square blade, straight-edge, or whatever the triangle is rested on, is always a corner of a polygon having twelve sides.

[Ill.u.s.tration: Fig. 174.]

[Ill.u.s.tration: Fig. 175.]

In both of these examples we have a.s.sumed that the diameter across corners of the polygon was given, but suppose the diameter across the flats were given, and the construction is a little more complicated.

Circle _a_, _a_, in Figure 175, is drawn of the required diameter across the flats, and the lines of division are drawn across with the triangle of 60 as before; the triangle of 45 is then used to draw the four lines, _b_, _c_, _d_, _e_, joining the ends of lines _i_, _j_, _k_, _l_, and touching the inner circle, _a_, _a_. The outer circle is then pencilled in, touching the lines of division where they meet the lines _b_, _c_, _d_, _e_, and the rest of the lines for the sides of the polygon may then be drawn within the outer circle, as at _g_, _h_.

[Ill.u.s.tration: Fig. 176.]

It is obvious, also, that the triangle may be used to draw slots radiating from a centre, as in Figure 176, where it is desired to draw a chuck-plate having 6 slots. The triangle of 60 is used to draw the centre lines, _a_, _b_, _c_, etc., for the slots. From the centre, the arcs _e_, _f_, _g_, _h_, etc., are marked, showing where the centres will fall for describing the half circles forming the ends of the slots.

Then half circles, _i_, _j_, _k_, _l_, etc., being drawn, the sides of the slots may be drawn in with the triangle, and the outer circle and the slots inked in.

If the slots are not to radiate from the centre of the circle the process is as follows:

The outer circle _a_, Figure 177, being drawn, an inner one _b_ is drawn, its radius equalling the amount; the centres of the slots are to point to one side of the centre of circle _a_. The triangle is then used to divide the circle into the requisite number of divisions _c_ for the slots, and arcs _i, j_, are then drawn for the lengths of the slots. The centre lines _e_ are then drawn, pa.s.sing through the lines _c_, and the arcs _i, j_, etc., and touching the perimeter of the inner circle _b_; arcs _f, g_, are then marked in, and their sides joined with the triangle adjusted by hand. All that would be inked in black are the outer circle and the slots, but the inner circle _b_ and a centre line of one of the slots should be marked in red ink to show how the inclination of the slot was obtained, and therefore its amount.

[Ill.u.s.tration: Fig. 177.]

For a five-sided figure it is best to step around the circ.u.mference of the circle with the compa.s.ses, but for a three-sided one, or trigon, the construction is as follows: It will be found that the compa.s.ses set to the radius of a circle will accurately divide it into six equal divisions, as is shown in Figure 178; hence every other one of these divisions will be the location for a corner of a trigon.

The circle being drawn, a line A, 179, is drawn through its centre, and from its intersection with the circle as at _b_, here a step on each side is marked as _c_, _d_, then lines _c_ to _d_, and _c_ and _d_ to _e_, where A meets, the circle will describe a trigon. If the figure is to stand vertical, all that is necessary is to draw the line _a_ vertical, as in Figure 180. A ready method of getting the dimension across corners, across the flats, or the length of a side of a given polygon, is by means of diagrams, such as shown in the following figures, which form excellent examples for practice.

[Ill.u.s.tration: Fig. 178.]

[Ill.u.s.tration: Fig. 179.]

[Ill.u.s.tration: Fig. 180.]

Draw the line O P, Figure 181, and at a right angle to it the line O B; divide these two lines into parts of one inch, as shown in the cut, which is divided into inches and quarter inches, and from these points of division draw lines crossing each other as shown.

[Ill.u.s.tration: Fig. 181.]

From the point O, draw diagonal lines, at suitable angles to the line O P. As shown in the cut, these diagonal lines are marked:

40 degrees for 5 sided figures.

45 " " 6 " "

49 " " 7 " "

52-1/2 " " 8 " "

55-1/2 " " 9 " "

But still others could be added for figures having a greater number of sides.

1. Now it will be found as follows: Half the diameter, or the radius of a piece of cylindrical work being given, and the number of sides it is to have being stated, the length of one side will be the distance measured horizontally from the line O B to the diagonal line for that particular number of sides.

EXAMPLE.--A piece of work is 2-1/2 inches in diameter, and is required to have 9 sides: what will be the length of the sides or flats?

Now the half diameter or radius of 2-1/2 inches is 1-1/4 inches. Then look along the line O B for 1-1/4, which is denoted in the cut by figures and the arrow A; set one point of the compa.s.ses at A, and the other at the point of crossing of the diagonal line with the 1-1/4 horizontal line, as shown in the figure at _a_, and from A to _a_ is the length of one side.

Again: A piece of work, 4 inches in diameter, is to have 9 sides: how long will each side be?

Now half of 4 is 2, hence from B to _b_ is the length of each side.

But suppose that from the length of each side, and the number of sides, it is required to find the diameter to which to turn the piece; that is, its diameter across corners, and we simply reverse the process thus: A body has 9 sides, each side measures 27/32: what is its diameter across corners?

Take a rule, apply it horizontally on the figure, and pa.s.s it along till the distance from the line O B to the diagonal line marked 9 sides measures 27/32, which is from 1-1/4 on O B to _a_, and the 1-1/4 is the radius, which, multiplied by 2, gives 2-1/2 inches, which is the required diameter across corners.

For any other number of sides the process is just the same. Thus: A body is 3-1/2 inches in diameter, and is to have 5 sides: what will be the length of each side? Now half of 3-1/2 is 1-3/4; hence from 1-3/4 on the line O B to the point C, where the diagonal line crosses the 1-3/4 line, is the length of each of the sides.

2. It will be found that the length of a side of a square being given, the size of the square, measured across corners, will be the length of the diagonal line marked 45 degrees, from the point O to the figures indicating, on the line O B or on the line O P, the length of one side.

EXAMPLE.--A square body measures 1 inch on each side: what does it measure across the corners? Answer: From the point O, along diagonal line marked 45 degrees, to the point where it crosses the lines 1 (as denoted in the figure by a dot).

Again: A cylindrical piece of wood requires to be squared, and each side of the square must measure an inch: what diameter must the piece be turned to?

Now the diagonal line marked 45 degrees pa.s.ses through the 1-inch line on O B, and the inch line on O P, at the point where these lines meet; hence all we have to do is to run the eye along either of the lines marked inch, and from its point of meeting the 45 degrees line, to the point O, is the diameter to turn the piece to.

There is another way, however, of getting this same measurement, which is to set a pair of compa.s.ses from the line 1 on O B, to line 1 on O P, as shown by the line D, which is the full diameter across corners. This is apparent, because from point O, along line O B, to 1, thence to the dot, thence down to line 1 on O P, and along that to O, encloses a square, of which either from O to the dot, or the length of the line D, is the measurement across corners, while the length of each side, or diameter across the flats, is from point O to either of the points 1, or from either of the points 1 to the dot.

[Ill.u.s.tration: Fig. 182.]

After graphically demonstrating the correctness of the scale we may simplify it considerably. In Figure 182, therefore, we have applications shown. A is a hexagon, and if one of its sides be measured, it will be found that it measures the same as along line 1 from O B to the diagonal line 45 degrees, which distance is shown by a thickened line.

At 1-1/2 is shown a seven-sided figure, whose diameter is 3 inches, and radius 1-1/2 inches, and if from the point at 1-1/2 (along the thickened horizontal line), to the diagonal marked 49 degrees, be measured, it will be found exactly equal to the length of a side on the polygon.

At C is shown part of a nine-sided polygon, of 2-inch radius, and the length of one of its sides will be found to equal the distance from the diagonal line marked 52-1/2 degrees, and the line O B at 2.

Let it now be noted that if from the point O, as a centre, we describe arcs of circles from the points of division on O B to O P, one end of each arc will meet the same figure on O P as it started from at O B, as is shown in Figure 181, and it becomes apparent that in the length of diagonal line between O and the required arc we have the radius of the polygon.

EXAMPLE.--What is the radius across corners of a hexagon or six-sided figure, the length of a side being an inch?

Turning to our scale we find that the place where there is a horizontal distance of an inch between the diagonal 45 degrees, answering to six-sided figures, is along line 1 (Figure 182), and the radius of the circle enclosing the six-sided body is, therefore, an inch, as will be seen on referring to circle A. But it will be noted that the length of diagonal line 45 degrees, enclosed between the point O and the arc of circle from 1 on O B to one on O P, measures also an inch. Hence we may measure the radius along the diagonal lines if we choose. This, however, simply serves to demonstrate the correctness of the scale, which, being understood, we may dispense with most of the lines, arriving at a scale such as shown in Figure 183, in which the length of the side of the polygon is the distance from the line O B, measured horizontally to the diagonal, corresponding to the number of sides of the polygon. The radius across corners of the polygon is that of the distance from O along O B to the horizontal line, giving the length of the side of the polygon, and the width across corners for a given length of one side of the square, is measured by the length of the lines A, B, C, etc. Thus, dotted line 2 shows the length of the side of a nine-sided figure, of 2-inch radius, the radius across corners of the figure being 2 inches.

[Ill.u.s.tration: Fig. 183.]