United States Standard Dimensions Of Bolts And Nuts
Bolt. 
Bolt Head and Nut. 


Diameter. 
Standard Number of threads per inch 
Long diameter, I, or diameter across corners 
Short diameter of hexagon and square, or width across J 
Depth of Nut, H 
Depth of bolt head, K 
Nominal. D. 
Effective. * 

Hexagon 
Square 



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 
13/32 
25/32 
7/16 
25/64 
1/2 
.400 
13 
1 
11/4 
7/8 
1/2 
7/16 
9/16 
.454 
12 
11/8 
13/8 
31/32 
9/16 
31/64 
5/8 
.507 
11 
17/32 
11/2 
11/16 
5/8 
17/32 
3/4 
.620 
10 
17/16 
13/4 
11/4 
3/4 
5/8 
7/8 
.731 
9 
121/32 
21/32 
17/16 
7/8 
23/32 
1 
.837 
8 
17/8 
25/16 
15/8 
1 
13/16 
11/8 
.940 
7 
23/32 
29/16 
113/16 
11/8 
29/32 
11/4 
1.065 
7 
25/16 
227/32 
2 
11/4 
1 
13/8 
1.160 
6 
217/32 
33/32 
23/16& 
13/8 
13/32 
11/2 
1.284 
6 
23/4 
311/32 
23/8 
11/2 
13/16 
15/8 
1.389 
51/2 
231/32 
35/8 
29/16 
15/8 
19/32 
13/4 
1.491 
5 
33/16 
37/8 
23/4 
13/4 
13/8 
17/8 
1.616 
5 
313/32 
45/32 
215/16 
17/8 
115/32 
2 
1.712 
41/2 
319/32 
413/32 
31/8 
2 
19/16 
21/4 
1.962 
41/2 
41/32 
415/16 
31/2 
21/4 
13/4 
21/2 
2.176 
4 
415/32 
515/32 
37/8 
21/2 
115/16 
23/4 
2.426 
4 
429/32 
6 
41/4 
23/4 
21/8 
3 
2.629 
31/2 
511/32 
617/32 
45/8 
3 
25/16 
31/4 
2.879 
31/2 
525/32 
71/16 
5 
31/4 
21/2 
31/2 
3.100 
31/4 
67/32 
719/32 
53/8 
31/2 
211/16 
33/4 
3.317 
3 
65/8 
81/8 
53/4& 
33/4 
27/8 
... 
3.567 
3 
71/16 
821/32 
61/8 

31/16 
41/4 
3.798 
27/8 
71/2 
93/16 
61/2 
41/4 
31/4 
41/2 
4.028 
23/4 
715/16 
923/32 
67/8 
41/2 
37/16 
43/4 
4.256 
25/8 
83/8 
101/4 
71/4 
43/4 
35/8 
5 
4.480 
21/2 
813/16 
1025/32 
75/8 
5 
313/16 
51/4 
4.730 
21/2 
91/4 
115/16 
8 
51/4 
4 
51/2 
4.953 
23/8 
911/16 
1127/32 
83/8 
51/2 
43/16 
53/4 
5.203 
23/8 
103/32 
123/8 
83/4 
53/4 
43/8 
6 
5.423 
21/4 
1017/32 
1229/32 
91/8 
6 
49/16 
* 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 onehalf times the diameter plus 1/8 inch for rough or unfinished bolts and nuts, and one and onehalf 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.
Fig. 155.
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 bolthead, 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 squareheaded 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 bolthead, 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.
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.
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 fortyfive 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.
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.
Fig. 160.
If the view showing a full side of the head of a squareheaded 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 onehalf 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.
Fig 161.
Fig. 162.
Fig. 163.
Fig. 164.
To draw a squareheaded 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.
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.
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.
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 centreline 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.
Fig. 168.
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.
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.
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.
Fig. 172.
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 fortyfive 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, straightedge, or whatever the triangle is rested on, is always a corner of a polygon having twelve sides.
Fig. 174.
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.
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 chuckplate 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.
Fig. 177.
For a fivesided figure it is best to step around the circumference of the circle with the compasses, but for a threesided 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.
Fig. 178.
Fig. 179.
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.
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 
" 
" 
521/2 
" 
" 
8 
" 
" 
551/2 
" 
" 
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 21/2 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 21/2 inches is 11/4 inches. Then look along the line O B for 11/4, 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 11/4 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 11/4 on O B to a, and the 11/4 is the radius, which, multiplied by 2, gives 21/2 inches, which is the required diameter across corners.
For any other number of sides the process is just the same. Thus: A body is 31/2 inches in diameter, and is to have 5 sides: what will be the length of each side? Now half of 31/2 is 13/4; hence from 13/4 on the line O B to the point C, where the diagonal line crosses the 13/4 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 1inch 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.
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 11/2 is shown a sevensided figure, whose diameter is 3 inches, and radius 11/2 inches, and if from the point at 11/2 (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 ninesided polygon, of 2inch radius, and the length of one of its sides will be found to equal the distance from the diagonal line marked 521/2 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 sixsided 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 sixsided figures, is along line 1 (Figure 182), and the radius of the circle enclosing the sixsided 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 ninesided figure, of 2inch radius, the radius across corners of the figure being 2 inches.
Fig. 183.
The dotted line 21/2 shows the length of the side of a ninesided polygon, having a radius across corners of 21/2 inches. The dotted line 1 shows the diameter, across corners, of a square whose sides measure an inch, and so on.
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.
Fig. 185.
A line at 75° for polygons having 12 sides.
" 72° " " 10 "
" 671/2° " " 8 "
" 60° " " 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.
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 532nds inches: hence the hexagon measures twice that, or 2 516ths 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.
Fig. 187.
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.
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.
Fig. 190.
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.
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.
Fig. 192.
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.
Fig. 194.
Fig. 195.
Fig. 196.
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.
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.
Fig. 199.
Fig. 200.
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.
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