_EXAMPLES IN BOLTS, NUTS, AND POLYGONS._
[Illustration: Fig. 149.]
[Illustration: Fig. 150.]
[Illustration: Fig. 151.]
Let it be required to draw a machine screw, and it is not necessary, and
therefore not usual in small screws to draw the full outline of the
thread, but to represent it by thick and thin lines running diagonally
across the bolt, as in Figure 149, the thick ones representing the
bottom, and the thin ones the top of the thread. The pencil lines would
be drawn in the order shown in Figure 150. Line 1 is the centre line,
and line 2 a line to represent the lower side of the head; from the
intersection of these two lines as a centre (as at A) short arcs 3 and
6, showing the diameter of the thread, are marked, and the arcs 5 and 6,
representing the depth of the thread, are marked. The arc 7,
representing the head, is then marked. The vertical lines 8, 9, 10, and
11 are then marked, and the outline of the screw is complete. The thick
lines representing the bottom of the thread are next marked in, as in
Figure 151, extending from line 9 to line 10. Midway between these lines
fine ones are made for the tops of the thread. All the lines being
pencilled in, they may be inked in with the drawing instruments, taking
care that they do not overrun one another. When the pencil lines are
rubbed out, the sketch will appear as in Figure 149.
[Illustration: Fig. 152.]
For a bolt with a hexagon head the lines would be drawn in the order
shown in Figure 152. At a right-angle to centre line 1, line two is
drawn. The pencil-compasses are then set to half the diameter of the
bolt, and from point A arcs 3 and 6 are pencilled, thus showing the
width of the front flat of the head, as well as the diameter of the
stem. From the point where these arcs meet line 2, and with the same
radius, arcs 5 and 6 are marked, showing the widths of the other two
flats of the head. The thickness of the head and the length of the bolt
head may then be marked either by placing a rule on line 1 and marking
the short lines (such as line 7) a cross line 1, or the pencil-compasses
may be set to the rule and the lengths marked from point A. In the
United States standard for bolt heads and nuts the thickness of the head
is made equal to the diameter of the bolt. With the compasses set for
the arcs 3 and 4, we may in two steps, from A along the centre line,
mark off the thickness of the head without using the rule. But as the
rule has to be applied along line 1 to mark line 7 for the length of the
bolt, it is just as easy to mark the head thickness at the same time.
The line 8 showing the length of the thread may be marked at the same
time as the other lengths are marked, and the outlines 9, 10, 11, 12, 13
may be drawn in the order named. We have now to mark the arcs at the top
of the flats of the head to show the chamfer, and to explain how these
arcs are obtained we have in Figure 153 an enlarged view of the head. It
is evident that the smallest diameter of the chamfer is represented by
the circle A, and therefore the length of the line B must equal A. It is
also evident that the outer edge of the chamfer will meet the corners at
an equal depth (from the face of the nut), as represented by the line C
C, and it is obvious that the curves that represent the outline of the
chamfer on each side of the head or nut will approach the face of the
head or nut at an equal distance, as denoted by the line D D. It follows
that the curve must in each case be such as will, at each of its ends,
meet the line C, and at its centre meet the line D D, the centres of the
respective curves being marked in the figure by X.
[Illustration: Fig. 153.]
It is sufficiently accurate, therefore, for all practical purposes to
set the pencil on the centre-line at the point A in Figure 152 and mark
the curve 14, and to then set the compasses by trial to mark the other
two curves of the chamfer, so that they shall be an equal distance with
arc 14 from line 9, and join lines 10 and 13 at the same distance from
line 9 that 14 joins lines 3 and 4, so that as in Figure 153 all three
of the arcs would touch a line as C, and another line as D.
[Illustration: Fig. 154.]
The United States standard sizes for forged or unfinished bolts and nuts
are given in the following table, Figure 154 showing the dimensions
referred to in the table.
UNITED STATES STANDARD DIMENSIONS OF BOLTS AND NUTS.
KEY:
A: Nominal. D.
B: Effective.[*]
C: Standard Number of threads per inch.
------------------------+----------------------------+---------------------
BOLT. | BOLT HEAD AND NUT. |
------------------------+-----------------+----------+----------+----------
Diameter. | | Long diameter, | Short | |
-------+------| | I, or diameter | diameter | |
| | | across corners. | of | |
| | |--------+--------| hexagon | |
A | B | C | Hexa- | Square.| and | |
| | | gon. | | square, | Depth of | Depth of
| | | | | or width | nut, | bolt
| | | | | across J.| H. | head, K.
-------+------+---------+--------+--------+----------+----------+----------
1/4 | .185 | 20 | 9/16| 23/32| 1/2 | 1/4 | 1/4
5/16 | .240 | 18 | 11/16| 27/32| 19/32 | 5/16 | 19/64
3/8 | .294 | 16 | 25/32| 31/32| 11/16 | 3/8 | 11/32
7/16 | .345 | 14 | 29/32| 1-3/32 | 25/32 | 7/16 | 25/64
1/2 | .400 | 13 | 1 | 1-1/4 | 7/8 | 1/2 | 7/16
9/16 | .454 | 12 | 1-1/8 | 1-3/8 | 31/32 | 9/16 | 31/64
5/8 | .507 | 11 | 1-7/32 | 1-1/2 | 1-1/16 | 5/8 | 17/32
3/4 | .620 | 10 | 1-7/16 | 1-3/4 | 1-1/4 | 3/4 | 5/8
7/8 | .731 | 9 | 1-21/32| 2-1/32 | 1-7/16 | 7/8 | 23/32
1 | .837 | 8 | 1-7/8 | 2-5/16 | 1-5/8 | 1 | 13/16
1-1/8 | .940 | 7 | 2-3/32 | 2-9/16 | 1-13/16 | 1-1/8 | 29/32
1-1/4 |1.065 | 7 | 2-5/16 | 2-27/32| 2 | 1-1/4 | 1
1-3/8 |1.160 | 6 | 2-17/32| 3-3/32 | 2-3/16 | 1-3/8 | 1-3/32
1-1/2 |1.284 | 6 | 2-3/4 | 3-11/32| 2-3/8 | 1-1/2 | 1-3/16
1-5/8 |1.389 | 5-1/2 | 2-31/32| 3-5/8 | 2-9/16 | 1-5/8 | 1-9/32
1-3/4 |1.491 | 5 | 3-3/16 | 3-7/8 | 2-3/4 | 1-3/4 | 1-3/8
1-7/8 |1.616 | 5 | 3-13/32| 4-5/32 | 2-15/16 | 1-7/8 | 1-15/32
2 |1.712 | 4-1/2 | 3-19/32| 4-13/32| 3-1/8 | 2 | 1-9/16
2-1/4 |1.962 | 4-1/2 | 4-1/32 | 4-15/16| 3-1/2 | 2-1/4 | 1-3/4
2-1/2 |2.176 | 4 | 4-15/32| 5-15/32| 3-7/8 | 2-1/2 | 1-15/16
2-3/4 |2.426 | 4 | 4-29/32| 6 | 4-1/4 | 2-3/4 | 2-1/8
3 |2.629 | 3-1/2 | 5-11/32| 6-17/32| 4-5/8 | 3 | 2-5/16
3-1/4 |2.879 | 3-1/2 | 5-25/32| 7-1/16 | 5 | 3-1/4 | 2-1/2
3-1/2 |3.100 | 3-1/4 | 6-7/32 | 7-19/32| 5-3/8 | 3-1/2 | 2-11/16
3-3/4 |3.317 | 3 | 6-5/8 | 8-1/8 | 5-3/4 | 3-3/4 | 2-7/8
... |3.567 | 3 | 7-1/16 | 8-21/32| 6-1/8 | 4 | 3-1/16
4-1/4 |3.798 | 2-7/8 | 7-1/2 | 9-3/16 | 6-1/2 | 4-1/4 | 3-1/4
4-1/2 |4.028 | 2-3/4 | 7-15/16| 9-23/32| 6-7/8 | 4-1/2 | 3-7/16
4-3/4 |4.256 | 2-5/8 | 8-3/8 |10-1/4 | 7-1/4 | 4-3/4 | 3-5/8
5 |4.480 | 2-1/2 | 8-13/16|10-25/32| 7-5/8 | 5 | 3-13/16
5-1/4 |4.730 | 2-1/2 | 9-1/4 |11-5/16 | 8 | 5-1/4 | 4
5-1/2 |4.953 | 2-3/8 | 9-11/16|11-27/32| 8-3/8 | 5-1/2 | 4-3/16
5-3/4 |5.203 | 2-3/8 |10-3/32 |12-3/8 | 8-3/4 | 5-3/4 | 4-3/8
6 |5.423 | 2-1/4 |10-17/32|12-29/32| 9-1/8 | 6 | 4-9/16
-------+------+---------+--------+--------+----------+----------+----------
* Diameter at the root of the thread.
The basis of the Franklin Institute or United States standard for the
heads of bolts and for nuts is as follows:
The short diameter or width across the flats is equal to one and
one-half times the diameter plus 1/8 inch for rough or unfinished bolts
and nuts, and one and one-half times the bolt diameter plus, 1/16 inch
for finished heads and nuts. The thickness is, for rough heads and nuts,
equal to the diameter of the bolt, and for finished heads and nuts 1/16
inch less.
[Illustration: Fig. 155.]
[Illustration: Fig. 156.]
The hexagonal or hexagon (as they are termed in the shop) heads of bolts
may be presented in two ways, as is shown in Figures 155 and 156.
The latter is preferable, inasmuch as it shows the width across the
flats, which is the dimension that is worked to, because it is where the
wrench fits, and therefore of most importance; whereas the latter gives
the length of a flat, which is not worked to, except incidentally, as it
were. There is the objection to the view of the head, given in Figure
156, however, that unless it is accompanied by an end view it somewhat
resembles a similar view of a square head for a bolt. It may be
distinguished therefrom, however, in the following points:
If the amount of chamfer is such as to leave the chamfer circle (as
circle A, in Figure 153) of smaller diameter than the width across the
flats of the bolt-head, the outline of the sides of the head will pass
above the arcs at the top of the flats, and there will be two small flat
places, as A and B, in Figure 156 (representing the angle of the
chamfer), which will not meet the arcs at the top of the flats, but will
join the sides above those arcs, as in the figure; which is also the
case in a similar view of a square-headed bolt. It may be distinguished
therefrom, however, in the following points:
If the amount of chamfer is such as to leave the chamfer circle (A,
Figure 153) of smaller diameter than the width across the flats of the
bolt-head, the outline of the sides will pass above the arc on the
flats, as is shown in Figure 157, in which the chamfer A meets the side
of the head at B, and does not, therefore, meet the arc C. The length of
side lying between B and D in the side view corresponds with the part
lying between E and F in the end view.
[Illustration: Fig. 157.]
If we compare this head with similar views of a square head G, both
being of equal widths, and having their chamfer circles at an equal
distance from the sides of the flats, and at the same angle, we perceive
at once that the amount of chamfer necessary to give the same distance
between the chamfer circle and the side of the bolt (that is, the
distance from J to K, being equal to that from L to M), the length of
the chamfer N for the square head so greatly exceeds the length A for
the hexagon head that the eye detects the difference at once, and is
instinctively informed that G must be square, independently of the fact
that in the case of the square head, N meets the arc O, while in the
hexagon head, A, which corresponds to N, does not meet the arc C, which
corresponds to O.
When, however, the chamfer is drawn, but just sufficient to meet the
flats, as in the case of the hexagon H, and the square I, in Figure 157,
the chamfer line passes from the chamfer circle to the side of the head,
and the distinction is greater, as will be seen by comparing head H with
head I, both being of equal width, having the same angle of chamfer, and
an amount just sufficient to meet the sides of the flats. Here it will
be seen that in the hexagon H, each side of the head, as P, meets the
chamfer circle A. Whereas, in the square head these two lines are joined
by the chamfer line Q, the figures being quite dissimilar.
[Illustration: Fig. 158.]
It is obvious that whatever the degree or angle of the chamfer may be,
the diameter of the chamfer circle will be the same in any view in
which the head may be presented. Thus, in Figure 158, the line G in the
side view is in length equal to the diameter of circle G, in the end
view, and so long as the angle of the chamfer is forty-five degrees, as
in all the views hitherto given, the width of the chamfer will be equal
at corresponding points in the different views; thus in the figure the
widths A and B in the two views are equal.
[Illustration: Fig. 159.]
If the other view showing a corner of the head in front of the head be
given, the same fact holds good, as is shown in Figure 159. That the two
outside flats should appear in the drawing to be half the width of the
middle flat is also shown in Figure 158, where D and E are each half the
width of C. Let us now suppose, that the chamfer be given some other
angle than that of 45 degrees, and we shall find that the effect is to
alter the curves of the chamfer arcs on the flats, as is shown in Figure
160, where these arcs E, C, D are shown less curved, because the chamfer
B has more angle to the flats. As a result, the width or distance
between the arcs and line G is different in the two views. On this
account it is better to draw the chamfer at 45 degrees, as correct
results may be obtained with the least trouble.
If no chamfer at all is to be given, a hexagon head may still be
distinguished from a square one, providing that the view giving three
sides of the head, as in Figure 158, is shown, because the two sides D
and E being half the width of the middle one C, imparts the information
that it is a hexagon head. If, however, the view showing but two of the
sides and a corner in front is given, and no chamfer is used, it could
not be known whether the head was to be hexagon or square, unless an end
view be given, as in Figure 161.
[Illustration: Fig. 160.]
If the view showing a full side of the head of a square-headed bolt is
given, then either an end view must be given, as in Figure 162, or else
a single view with a cross on its head, as in Figure 163, may be given.
It is the better plan, both in square and hexagon heads, to give the
view in which the full face of a flat is presented, that is, as in
Figures 155 and 163; because, in the case of the square, the length of a
side and the width across the head are both given in that view; whereas
if two sides are shown, as in Figure 161, the width across flats is not
given, and this is the dimension that is wanted to work to, and not the
width across corners. In the case of a hexagon the middle of the three
flats is equal in width to the diameter of the bolt, and the other two
are one-half its width; all three, therefore, being marked with the same
set of compasses as gives the diameter of the body of the bolt, were as
shown in Figure 152. For the width across flats there is an accepted
standard; hence there is no need to mark it upon the drawing, unless in
cases where the standard is to be departed from, in which event an end
view may be added, or the view showing two sides may be given.
[Illustration: Fig 161.]
[Illustration: Fig. 162.]
[Illustration: Fig. 163.]
[Illustration: Fig. 164.]
To draw a square-headed bolt, the pencil lines are marked in the order
shown by figures in Figure 164. The inking in is done in the order of
the letters _a_, _b_, _c_, etc. It will be observed that pencil lines 2,
9, and 10 are not drawn to cross, but only to meet the lines at their
ends, a point that, as before stated, should always be carefully
attended to.
[Illustration: Fig. 165]
To draw the end view of a hexagon head, first draw a circle of the
diameter across the flats, and then rest the triangle of 60 degrees on
the blade _s_ of the square, as at T 1, in Figure 165, and mark the
lines _a_ and _b_. Reverse the triangle, as at T 2, and draw lines _c_
and _d_. Then place the triangle as in Figure 166, and draw the lines
_e_ and _f_.
[Illustration: Fig. 166.]
If the other view of the head is to be drawn, then first draw the lines
_a_ and _b_ in Figure 167 with the square, then with the 60 degree
triangle, placed on the square S, as at T 1, draw the lines _c_, _d_,
and turning the square over, as at T 2, mark lines _e_ and _f_.
[Illustration: Fig. 167.]
If the diameter across corners of a square head is given, and it be
required to draw the head, the process is as follows: For a view showing
one corner in front, as in Figure 168, a circle of the given diameter
across corners is pencilled, and the horizontal centre-line _a_ is
marked, and the triangle of 45 degrees is rested against the square
blade S, as in position T 1, and lines _b_ and _c_ marked, _b_ being
marked first; and the triangle is then slid along the square blade to
position T 1, when line _c_ is marked, these two lines just meeting the
horizontal line _a_, where it meets the circle. The triangle is then
moved to the left, and line _d_, joining the ends of _b_ and _c_, is
marked, and by moving it still farther to the left to position T 2, line
_e_ is marked. Lines _b_, _c_, _d_, and _e_ are, of course, the only
ones inked in.
[Illustration: Fig. 168.]
[Illustration: Fig. 169.]
If the flats are to lie in the other direction, the pencilling will be
done as in Figure 169. The circle is marked as before, and with the
triangle placed as shown at T 1, line _a_, passing through the centre of
the circle, is drawn. By moving the triangle to the right its edge B
will be brought into position to mark line _b_, also passing through
the centre of the circle. All that remains is to join the ends of these
two lines, using the square blade for lines _c_, _d_, and the triangle
for _e_ and _f_, its position on the square blade being denoted at T 3;
lines _c_, _d_, _e_, _f_, are the ones inked in.
[Illustration: 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, passing 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_.
[Illustration: 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.
[Illustration: Fig. 172.]
[Illustration: 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 compasses. 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.
[Illustration: Fig. 174.]
[Illustration: Fig. 175.]
In both of these examples we have assumed 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_.
[Illustration: 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, passing 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.
[Illustration: Fig. 177.]
For a five-sided figure it is best to step around the circumference of
the circle with the compasses, but for a three-sided one, or trigon, the
construction is as follows: It will be found that the compasses 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.
[Illustration: Fig. 178.]
[Illustration: Fig. 179.]
[Illustration: 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.
[Illustration: 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 compasses 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 pass 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 passes 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 compasses 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.
[Illustration: 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.
[Illustration: Fig. 183.]
The dotted line 2-1/2 shows the length of the side of a nine-sided
polygon, having a radius across corners of 2-1/2 inches. The dotted line
1 shows the diameter, across corners, of a square whose sides measure an
inch, and so on.
[Illustration: Fig. 184.]
This scale lacks, however, one element, in that the diameter across the
flats of a regular polygon being given, it will not give the diameter
across the corners. This, however, we may obtain by a somewhat similar
construction. Thus, in Figure 184, draw the line O B, and divide it into
inches and parts of an inch. From these points of division draw
horizontal lines; from the point O draw the following lines and at the
following angles from the horizontal line O P.
[Illustration: Fig. 185.]
A line at 75 deg. for polygons having 12 sides.
" 72 deg. " " 10 "
" 67-1/2 deg. " " 8 "
" 60 deg. " " 6 "
From the point O to the numerals denoting the radius of the polygon is
the radius across the flats, while from point O to the horizontal line
drawn from those numerals is the radius across corners of the polygon.
[Illustration: Fig. 186.]
A hexagon measures two inches across the flats: what is its diameter
measured across the corners? Now from point O to the horizontal line
marked 1 inch, measured along the line of 60 degrees, is 1 5-32nds
inches: hence the hexagon measures twice that, or 2 5-16ths inches
across corners. The proof of the construction is shown in the figure,
the hexagon and other polygons being marked simply for clearness of
illustration.
[Illustration: Fig. 187.]
[Illustration: Fig. 188.]
Let it be required to draw the stud shown in Figure 185, and the
construction would be, for the pencil lines, as shown in Figure 186; line
1 is the centre line, arcs, 2 and 3 give the large, and arcs 4 and 5
the small diameter, to touch which lines 6, 7, 8, and 9 may be drawn.
Lines 10, 11, and 12 are then drawn for the lengths, and it remains to
draw the curves in. In drawing these curves great exactitude is required
to properly find their centres; nothing looks worse in a drawing than an
unfair or uneven junction between curves and straight lines. To find the
location for these centres, set the compasses to the required radius for
the curve, and from the point or corner A draw the arcs _b_ and _c_,
from _c_ mark the arc _e_, and from _b_ the arc _d_, and where _d_ and
_e_ cross is the centre for the curve _f_.
[Illustration: Fig. 189.]
Similarly for the curve _h_, set the compasses on _i_ and mark the arc
_g_, and from the point where it crosses line 6, draw the curve _h_. In
inking in it is best to draw in all curves or arcs of circles first, and
the straight lines that join them afterward, because, if the straight
lines are drawn first, it is a difficult matter to alter the centres of
the curves to make them fall true, whereas, after the curves are drawn
it is an easy matter, if it should be necessary, to vary the line a
trifle, so as to make it join the curves correctly and fair. In inking
in these curves also, care must be taken not to draw them too short or
too long, as this would impair the appearance very much, as is shown in
Figure 187.
[Illustration: Fig. 190.]
[Illustration: Fig. 191.]
To draw the piece shown in Figure 188, the lines are drawn in the order
indicated by the letters in Figure 189, the example being given for
practice. It is well for the beginner to draw examples of common
objects, such as the hand hammer in Figure 190, or the chuck plate in
Figure 191, which afford good examples in the drawing of arcs and
circles.
In Figure 191 _a_ is a cap nut, and the order in which the same would be
pencilled in is indicated by the respective numerals. The circles 3 and
4 represent the thread.
[Illustration: Fig. 191 _a_.]
In Figure 192 is shown the pencilling for a link having the hubs on one
side only, so that a centre line is unnecessary on the edge view, as
all the lengths are derived from the top view, while the thickness of
the stem and height of the hubs may be measured from the line A. In
Figure 193 there are hubs (on both sides of the link) of unequal height,
hence a centre line is necessary in both views, and from this line all
measurements should be marked.
[Illustration: Fig. 192.]
[Illustration: Fig. 193.]
In Figure 194 are represented the pencil lines for a double eye or
knuckle joint, as it is sometimes termed, an example that it is
desirable for the student to draw in various sizes, as it is
representative of a large class of work.
These eyes often have an offset, and an example of this is given in
Figure 195, in which A is the centre line for the stem distant from the
centre line B of the eyes to the amount of offset required.
[Illustration: Fig. 194.]
[Illustration: Fig. 195.]
[Illustration: Fig. 196.]
[Illustration: Fig. 197.]
In Figure 196 is an example of a connecting rod end. From a point, as A,
we draw arcs, as B C for the width, and E D for the length of the block,
and through A we draw the centre line. It is obvious, however, that we
may draw the centre line first, and apply the measuring rule direct to
the paper, and mark lines in place of the arcs B, C, D, E, and F, G,
which are for the stem. As the block joins the stem in a straight line,
the latter is evidently rectangular, as will be seen by referring to
Figure 197, which represents a rod end with a round stem, the fact that
the stem is round being clearly shown by the curves A B. The radius of
these curves is obtained as follows: It is obvious that they will join
the rod stem at the same point as the shoulder curves do, as denoted by
the dotted vertical line. So likewise they join the curves E F at the
same point in the rod length as the shoulder curves, both curves in fact
being formed by the same round corner or shoulder. The centre of the
radius of A or B must therefore be the same distance from the centre of
the rod as is the centre from which the shoulder curve is struck, and at
the same time at such a distance from the corner (as E or F) that the
curve will meet the centre line of the rod at the same point in its
length as the shoulder curves do.
[Illustration: Fig. 198.]
Figure 198 gives an example, in which the similar curved lines show that
a part is square. The figure represents a bolt with a square under the
head. As but one view is given, that fact alone tells us that it must be
round or square. Now we might mark a cross on the square part, to denote
that it is square; but this is unnecessary, because the curves F G show
such to be the case. These curves are marked as follows: With the
compasses set to the radius E, one point is rested at A, and arc B is
drawn; then one point of the compass is rested at C, and arc D is drawn;
giving the centre for the curve F by a similar process on the other
side of the figure, curve G is drawn. Point C is obtained by drawing the
dotted line across where the outline curve meets the stem. Suppose that
the corner where the round stem meets the square under the head was a
sharp one instead of a curve, then the traditional cross would require
to be put on the square, as in Figure 199; or the cross will be
necessary if the corner be a round one, if the stem is reduced in
diameter, as in Figure 200.
[Illustration: Fig. 199.]
[Illustration: Fig. 200.]
[Illustration: Fig. 201.]
Figure 201 represents a centre punch, giving an example, in which the
flat sides gradually run out upon a circle, the edges forming curves, as
at A, B, etc. The length of these curves is determined as follows: They
must begin where the taper of the punch joins the parallel, or at C, C,
and they must end on that part of the taper stem where the diameter is
equal to the diameter across the flats of the octagon. All that is to be
done then is to find the diameter across the flats on the end view, and
mark it on the taper stem, as at D, D, which will show where the flats
terminate on the taper stem. And the curved lines, as A, B, may be drawn
in by a curve that must meet at the line C, and also in a rounded point
at line D.