~
~
~
~
Qa &o N
~
Fig. 172.
ment of acone. ‘The pitch is measured parallel to the axis, as in the preceding case.
106 SPIRALS, HELICES, SCREW-THREADS AND BOLT-HEADS
132. Screw-Threads. In order to make the drawing of a screw-thread, it is necessary to
know the diameter, pitch and section of the thread.
If the thread to be drawn has a V section, as in Fig. 174, and the diameter and pitch are
given, begin by drawing the section of the screw as shown in dotted lines. Next, construct a
templet as follows: Draw on a rather light piece of cardboard two helices of the same pitch
as the required thread, one being of a diameter
equal to the outside of the thread and the other
equal to the diameter at the root of the thread.
These may be drawn separately, or together, as
in Fig. 178, and are to be carefully cut out so
as to be used as a pattern in pencilling the curves.
Having drawn the helices, mark the points of
tangency B, G, and A, F, as well as the centers
Fig. 173, L and K, in the manner indicated. Also repeat
| these marks on the opposite side after cutting
out. This will enable the templet to be used
either side up, and be readily set to the drawing. Observe that the curve is not to be cut off
abruptly at its termination, but continued a little beyond, so that in tracing the outline the
pencil-point may not injure the extreme point of the templet curve. This may now be used
for the drawing of all the helices on this screw, as from A to F, B to G, C to H, etc., Fig. 174.
If the pitch is small in proportion to the diameter, the drawing of the screw may now be
considered finished; but the contour line does not coincide with that of the section of the
thread, and in order to illustrate correctly the projection of a V thread we must consider the
L— | K
SCREW-THREADS
character of the surface and apply a correction to the
drawing. The surface which is being drawn is a
helicoid, and is generated by the motion of a line,
AB, which is made to revolve about the axis of the
screw and at the same time move in the direction of
the axis. ‘This will generate the upper half of the
surface of the screw. Every point of the line will
describe a helix, the diameters differing, but the pitch
remaining constant. The helices generated by the
extremities of the line have already been drawn and
the curves 1 2 3 and 4 5 6, described by two other
points, 1 and 4, are shown by the fine dotted lines.
The curved line M5 2N, drawn tangent to these
helices, will be the visible outline of the surface .
instead of the dotted line which is concealed. As
the labor of describing these helices would be great,
it is customary to draw the outlines of the screw as
follows: Having described the helices AF, BG,
etc., reverse the templet and draw a small portion of
the continuation of the helix on the opposite side of
the screw, as at BP and CO. Then, draw the line
MN tangent to the two helices, and in the other
direction the tangent OP, a part of which is invisible.
107
108 SPIRALS, HELICES, SCREW-THREADS AND BOLT-HEADS
Be sD
Fig. 176.
133. Conventional V Threads. In order to facilitate the
drawing of screw-threads it is customary to omit the drawing
of the helix, substituting therefor a right line, as in Fig. 175.
In most cases, however, even this would involve too much labor
as well as complication on the drawing, and the V’s are likewise
omitted, the representation shown in Fig. 177 being adopted.
When this is done, no care is taken to make the screw of the
required pitch, and the spaces between the fine lines which
represent the outer helices are estimated by the eye, as in
section lining. But it is imperative for the proper representa-
tion of a single thread, that the point C be over the middle of
the space BD. Having drawn the line AB, make the space
AC double that of EB and draw parallels. Afterwards, draw
the heavy lines to represent the root of the thread. As it is
difficult to make these of equal length without the aid of special
lines, the method illustrated in Fig. 178 is frequently used.
134. The Double Thread. As the use and character of the
double thread are generally misunderstood, Figs. 175 and 176
have been drawn to explain this problem more clearly. Fig.
175 illustrates a screw, the diameter and pitch of which are
supposed to have been given; but, as the pitch is excessive for
a screw of this diameter, the diameter at the root of the thread is small, and the screw propor-
U.S. STANDARD V THREADS 109
tionally weak. The only way to strengthen the thread at this
point without changing the angle of the V’s is by partly filing
the V’s at the root, as shown by the dotted line in Fig. 175
and the left-hand portion of the complete screw in Fig. 176,
thus increasing the diameter at the root. While overcoming
one weakness we have introduced a second by lessening the
section of the thread, so that with a nut of a given length the
tendency of the thread to be stripped from the body is doubled.
This last difficulty may be overcome by supposing an inter-
mediate thread wound between the present threads, as shown
by the right-hand portion of Fig. 176, which is a representa-
tion of a double thread having the same diameter and pitch
as the single thread of Fig. 175, but of increased strength. It
must be noted that the threads indicated by AB and CD, of
Fig. 176, are entirely independent of each other and that the
point C of one is diametrically opposite a point B in the par-
allel thread. This must be carefully observed-in the practical
representation of a double thread, as shown in Fig. 179. Fig.
180 represents a left-hand single thread.
_ 135. U.S. Standard V Threads. ‘The form of thread com-
monly used is that of the U.S. Standard, also known as the
Franklin Institute Standard and illustrated by Fig. 181.
RIGHT HAND SINGLE THREADS
Fig. (77. Fig. 178.
ANY 0H
DOUBLE THREAD LEFT HAND THREAD
Fig. 179. Fig. 180.
P= 0.24VD + 0.625 — 0.175.
S = 0.65 P.
110 SPIRALS, HELICES, SCREW-THREADS AND BOLT-HEADS
Although the pitch in a single-threaded screw is the:
distance between consecutive threads,. the term is often
applied to the number of threads per inch. Thus a screw
having eight threads to an inch is frequently spoken of as
8 pitch. This is obviously wrong, but leads to no confu-
sion, since the pitch and the number of threads per inch are
reciprocals of each other. The flattening of the thread, as
indicated in Fig. 181, is for the purpose of preventing injury
by the bruising of the otherwise sharp V.
136. Square Threads. Fig. 184 represents a thread of
square section. ‘The size of this square is equal to one-half
the pitch in a single thread, and one-quarter of the pitch
in a double thread. The construction is similar to that
described for the V thread. The outline of the section is
drawn first, and a templet is prepared for the inner and outer
helices. The threads are drawn as in Fig. 184 for a single
right-hand thread, and as in Fig. 185 for a nut in section.
Figs. 186 and 187 illustrate a double square thread and nut.
The conventional representation of the square thread is
similar to that of the V thread, the right line being substituted
for the helix. Fig. 182 is a correct drawing of the conventional square thread, and Fig. 183
a modification more commonly used.
<
Ek
=
(ea
Corea
Pars
OF 30 WITH G.L.
of GL, and the traces of P. Art. 96, page 66.
The drawing of the projection lines may be SEEN CHORD Akron Vine
omitted also. Use care to project all of the
invisible lines of the object. |
Pros. 20. It is required to represent three views of the preceding object when turned
around on its base. The construction is similar to that of Prob. 18. In this case the center
lines will no longer serve for the GL and traces of P. Use care to project all of the invisible
lines of the object.
132
DRAW THREE VIEWS | DRAW THREE VIEWS
OF THE OBJECT OF THE OBJECT
WHEN REVOLVED WHEN REVOLVED
PROBLEMS
21 TO 32
FROM THE POSITION | FROM THE POSITION
OF FIG. 1, 30° TO THE | OF FIG. ee 25° ABOUT
LEFT ABOUT AN AXIS|AN AXIS PERPENDIC-
DRAW THREE VIEWS |PERPENDICULAR TO V
ULAR TOH
FIG. 3.
DRAW THREE VIEWS | DRAW THREE VIEWS | DRAW THREE VIEWS
OF THE OBJECT OF THE OBJECT OF THE OBJECT
WHEN REVOLVED WHEN REVOLVED WHEN REVOLVED
FROM THE POSITION |FROM THE POSITION FROM THE POSITION
OF FIG.|I, 20 FORWARD OF FIG.2, 15 FORWARD] OF FIG. 5, 35° ABOUT
ABOUT AN AXIS PER- | ABOUT AN AXIS PER- | AN AXIS PERPENDIC-
PENDICULAR TO P PENDICULAR TO P ULAR TO H
FIG. 4. FIG. 5. FIG. 6.
ka Peor. 24%
altitude of Tie
a PROB. 22:
altitude of 13!’.
TOP VIEW OF FIG. Is
ob
RE
PROB? ?23.
altitude of 17! if
TOP VIEW OF. FIG. I.
1
1
F
Pros. 24.
altitude of 13".
TOP VIEW OF FIG. |.
PROBLEMS
145. Objects oblique to the coordinate planes.
A careful study of Art. 97, page 67, must be
made previous to the solution of Problems 21
to 32 inclusive. The plate will be divided into
six rectangles as in the accompanying figure,
the division lines being drawn in pencil only.
Six positions of three views each will be re-
quired in each case. Omit the drawing of pro-
jection and shade lines. If difficulty is found
with the problems, number the points as in Fig.
107, page 65, but use care to retain the same
number for each point throughout the problem.
In the required positions draw a rectangular pyramid having an
In the required positions draw a rectangular prism having an
In the required positions draw a triangular pyramid having an
The base is an equilateral triangle.
In the required positions draw a triangular prism having an
The bases are equilateral triangles.
ORTHOGRAPHIC PROJECTION
Prop. 25. In the required positions draw a pentagonal pyramid having an
altitude of 13’. Diameter of circumscribing circle of base 13’.
Pros. 26. In the required positions draw a pentagonal pyramid having an
altitude of 13’... Diameter of circumscribing circle of base 13'’.
Pros. 27. In the required positions draw a pentagonal pyramid having
an altitude of 13’. Diameter of circumscribing circle of base 12”.
Pros. 28. In the required: positions draw a hexagonal pyramid having an
altitude of 12’.
Pros. 29. In the required positions draw a hexagonal pyramid having an
altitude of 17’.
Pros. 80. In the required positions draw a wedge having an altitude
rot 12.
Pros. 31. In the required positions draw a wedge having an altitude
of 14”.
Pros. 32. In the required positions draw the frustum of. a rectangular
pyramid having an altitude of 12”.
4
138
TOP VIEW OF FIQ. I,
‘TOP VIEW OF FIQ. |.
TOP VIEW OF FIG. |
TOP VIEW OF FIG. I.
a Fs
ora
i
TOP VIEW OF FIG. |,
oe
ae
A
TOP VIEW OF FIG, I.
k- 13
Das) i
TOP VIEW OF FIG. I.
ca be
yanmar,
ee] +
5 aE 7
TOP VIEW OF FIQ. I.
134 PROBLEMS
PROB. 33.
OBTAIN THE PROJECTION
ON THE AUXILIARY PLANE
PROB. 35
HEIGHT OF, 3”
PRISM
8
REVOLVE CONE 30 ABOUT
VERTICAL AXIS
OBTAIN 3 VIEWS
PROBLEMS 33 to 37 require a knowledge of
the use of auxiliary planes. Art. 98, page 70.
Pros. 33. The auxiliary plane being par-
allel to the plane of the right-hand end of the
object, the projection on that plane will be a
true representation of that surface. Other lines
and surfaces will be foreshortened.
Pros. 34. In this problem the auxiliary
plane makes an angle of 30° with H, and the
projection on this plane is to be made in place ©
of the top view.
Pros. 35. Observe that the prism has a
triangular hole extending through it.
Pros. 36. This is similar to the preceding,
save that the prism is hexagonal instead of tri-
angular. A circular hole, 14!’ diameter, extends
from base to base.
Pros. 87. Study Arts. 99 and 100, pages
72 and 73, before solving this problem. Having
obtained the base of the cone in the manner
directed, locate the vertex of the cone. The
tangents to the base drawn from the vertex will
be the contour lines of the cone.
ORTHOGRAPHIC PROJECTION Veh
146. Special problems in projection. ‘These problems require a space of 5!’ x 7!’. Three
views are required in each case, and all invisible as well as visible lines should be shown on
each view. Leave all construction lines in pencil. All polygons are regular polygons.
Pros. 388. Draw the frustum of an octagonal pyramid having its base parallel to H and
two of its edges making an angle of 30° with V. The diameter of the circumscribing circle of
its lower base is 14’, and of the upper base, 14". The altitude is 17’.
Pros. 39. Revolve the pyramid of Prob. 88, 30° to right about an axis perpendicular
to V.
Pros. 40. Draw a pentagonal prism resting on one of its faces and having its lateral
edges at an angle of 223° with V. Diameter of circumscribing circle of base 1}/’.. Length of
prism 24’.
Pros. 41. Draw an equilateral triangular prism resting on one of its faces, and its lateral
edges making an angle of 15° with V. The edges of the base are 13’’, and the length of the
prism is 24’. There is a triangular hole extending through the bases and making the thickness
of the sides 4’.
Pros. 42. Draw a cylinder with its axis parallel to V and at an angle of 60° with H.
The diameter of the base is 13’’, and the length of cylinder, 2}/’.. Obtain the ellipses by the
method of trammels.
Pros. 43. Draw an equilateral triangular pyramid having an altitude of 24’, and the
edges of the base 17/’.. The base makes an angle of 30° with H and one of its edges is per-
pendicular to V.
Pros. 44. Revolve the pyramid of Prob. 43, 45° forward about an axis perpendicular
toRk:
136 PROBLEMS
Pros. 45. Draw a box having the following outside dimensions. Length 2!, width 13”,
depth, including cover, 1/’.. Thickness of material 4’... The long edges of the box are parallel
to H and make an angle of 30° with V. The cover is hinged on long edge and opened 30°.
Pros. 46. Draw a pyramid formed of four equilateral triangles having 22” sides. The
base is parallel to H and one of its edges makes an angle of 30° with V.
Pros. 47.. Draw a rectangular surface, 14!’ x 28’, in the following positions : The short
edges parallel to H and making an angle of 75° with V; the long edges making angles of 15°,
80° and 45° with H. Art. 102, page 76.
Pros. 48. Revolve the surface from the positions required in Prob. 47, 15° forward.
Pros. 49. Draw an isosceles triangle in three positions as follows : The base lying on V
and inclined at an angle of 30° with H. ‘The altitude making angles of 90°, 30° and 15° with V.
The base of the triangle is 14’’, and the altitude 23. Art. 102, page 76.
Pros. 50. Draw the same triangle revolved from the positions in Prob. 49, 30° in either
direction about a vertical axis. |
Pros. 51. Draw an isosceles triangle in the following positions: The base parallel to H
and making an angle of 60° with V. The altitude making angles of 45° and 60° with H. The
base of triangle is 2’, and the altitude 21’’.. Art. 102, page 76.
Pros. 52. Revolve the same triangle from the positions in Prob. 51, 30° backward.
Pros. 53. Draw an octagonal surface inclined at an angle of 60° with H, two of its edges
being parallel to H and making angles of 15° with V. ‘The diameter of circumscribing circle
is: 24"! Arte 102; page 76.
Pros. 54. Draw a hexagonal surface inclined at an angle of 45° with V, two of its
edges being parallel to V and making angles of 30° with H. The long diameter of the hexagon
is 24". Art. 102, page 76.
ISOMETRIC PROJECTION s¥6
Pros. 55. Draw the projections of a line located as follows: The left-hand extremity
of the line is 3 behind V and 7’ below H. The right-hand extremity is 14/’ behind. V, and
13" below H. The H projection of the line makes an angle of 30° with V. Find its length by
revolving it parallel to H, Vand P. Art. 101, page 74.
Pros. 56. Draw the projections of a lne of which the left-hand extremity is 1'’ behind
V and 12” below H. The right-hand extremity is 3’’ behind V and 3"’ below H. The H pro-
jection makes an angle of 15° with V. Find the length of the line by revolving it into the
planes of projection by the second method. Art. 101, page 75.
147. Isometric projection problems. Study Chapter VI, page 78. The problems require
a space of 5’ x 7'’.. Omit the invisible lines in inking. :
Pros. 1. Make the isometric drawing of a 2’ cube. Art. 110, page 80. Inscribe circles
on the upper and right-hand faces, the former by the exact method and the latter by the
approximate method. Art. 112, page 81. From the left-hand lower corner of the left-hand.
face draw lines making angles of 30°, 45° and 75° with the lower edge. Art. 113, page 82.
Pros. 2. Make the isometric drawing of the frustum of a pyramid, the lower base being 2”,
and the upper base 13/’ square. Height 14’’. Inscribe a circle on the upper base using the ap-
proximate method. Locate the front lower corner in the center and 1'’ from lower margin.
Pros. 3. Make the isometric drawing of a pentagonal plinth surmounted by a cylinder.
The sides of the pentagon are 2" and the height of the plinth is 3’’.. Art. 71, page 47. The
cylinder is 2!’ in diameter and 1” high. Art. 111, page 80.
Pros. 4. Make the isometric drawing of a box with cover opened through an angle of
120°. The outside dimensions are: length 21/’, width 13/’, depth #/’.. Thickness of material
is 4’. Locate the front lower corner in the middle of the space and 4/’ from lower margin.
138
PROBLEMS
or icpan
ska
ea!
PROB.
6
Pros. 5. Make the isometric
drawing of the bearing illustrated,
locating the upper corner at point A.
Pros. 6.. Make the isometric
drawing of a hexagonal bolt. ‘The
center line may be parallel with either
of the isometric axes. Art. 115, p. 83.
Pros. 7. Make the isometric
drawing of the pieces illustrated, and
a second isometric drawing of the
upright block showing the cuts neces-
sary for making the required fits.
The lowest portion of the upright
block will be located at A in the first
ease and B in the second.
Pros. 8. Make the isometric drawing of the connecting-rod strap, the scale to be 3/’=1 ft.
In drawing the curves observe the directions of Art. 115, page 83.
Pros. 9. Make the isometric drawing of the framing details illustrated. The dimensions
of the materials are given below. The space required is 10’ x 14’’. Draw toa scale of 13/’=1 ft.
Sills, 613".
Post, AS Bll
Brace, AM Di,
Window studs, 4x4".
Studs, 2" x4", 12" on centers.
Floor joists, 2x8", 12" on centers.
Under flooring, 7" x10".
Upper flooring, 7x 6".
Grounds, #3!'x 2".
Laths, 3" 11" x 48", 3" space.
Plaster, 3" thick.
Baseboard, 7x10", including cap.
‘ ISOMETRIC PROJECTION 139
eariieat= cial 9
md ;
=
Bia
BASE-BOARD
A, g
a2
v/,
Z
FLOOR JOIST 2X6. -
i
LLLLLLL LION a Roope FLOOR ho epee te
Cpoeneeneee _ UNDER FLOOR
|
: FLOOR JOIST
FLOOR JOIST 2x8
i
SILL.
WINDOW STUD
LAYOUT OF SHEET.
Aen
ET FT a
LABBawaa’
DETAILS FOR ISOMETRIC DRAWING
V|_ FLOOR voisT. OF FRAMING PLANS
SSX y
EY)
140
WE/EH7S
PULLEY
NS TN
QQ e{} 2
a AN Sy
Nt
PROBLEMS
Pros. 10.
detailed sketches.
Make the isometric drawing of a window from the
The dimensions of the materials are given below.
Show all that is given in the sections. Locate the drawing as in the
lay-out sketch, observing that the short edge of the plate is hori-
zontal.
LAYOUT OF SHEET
Art. 115, page 83.
Draw to scale of 3! =1 ft.
DIMENSIONS OF MATERIALS
Studs,
Apron,
Ground easing,
Inside architrave,
Outside architrave,
Outside casing,
Outside boarding,
Pocket,
Pulley style,
Parting bead,
Stop bead,
Laths,
Plaster,
Al x 4",
"x33",
3" x 4a",
gil x 5B
13"x 3h"
se ga
4.
Qi 4
i" thick.
BIN yc BI,
a" x14".
3" x 1i"x 48",
spaced 8",
3" thick:
Clapboards, 4" at thick edge, x 54!" x 48",
laid 34"' to the weather.
DEVELOPMENT OF SURFACES 141
148. Development problems.
ter VII, page 88.
plete development will be required in each case.
In general, begin to develop the surface on the
shortest edge. It will assist the student to a
better understanding of this subject if the de-
veloped surface be copied on stiff paper and
afterwards cut and folded. Observe the order
prescribed for performing these problems. Art.
119, page 90.
Pros. 1. This isa prism having a pentag-
onal base. ‘The vertical edges being parallel
to V are seen in their true length on that plane.
The length of the edges of the lower base can
be obtained from the top view, and the true shape
of the upper base will be found by projecting it
on to an auxiliary plane, as in Art. 98, page 70.
Pros. 2. Solve as for Prob. 1.
Pros. 8. In this and the following prob-
lems do not ink that portion of the surface lying
above the cutting plane. Art. 119, page 90.
Pros. 4. Observe that the slant edges of
this pyramid are not of equal length.
Study Chap-
Three views and the com-
BOB Me oe viEWion PROB. 2
| PENTAGON
- HEXAGONAL PRISM
— INSCRIBED Nat | REVOLVED
| 2 CIRCL SECTION
easy. aon
SIDE VIEW
| OBTAIN
igharanannns
DEVELOPMENT DEVELOPMENT
PROB. 3 PROB. 4
|
=| REVOLVED
SECTION | SECTION
REVOLVED
OEVELOPMENT DEVELOPMENT
142 PROBLEMS
PROB. 5 ’ PROB. 6
TOF VIEW OF TOP VIEW OF
s QUARE PYRAMID SQUARE tf atl
REVOLVED REVOLVED
SECTION ! errr
SIDE VIEW
ex
|
|
DEVELOPMENT
DEVELOPMENT DEVELOPMENT
Pros. 5. This is a square pyramid, the
diagonal of the base being 2’’.. Observe that the
base is cut by the plane, and one of the inclined
edges will not appear on the completed develop-
ment. In developing the surface open it on
this line.
Pros. 6. This differs from the preceding
in that the base is not cut by the plane, and the
position of the pyramid with respect to V is
changed.
Pros. 7. The surface cut by the plane
will not be symmetrical with respect to the
center line as in the preceding problems. Care
must be used to take no dimensions from the
top view that may not be represented by lines
parallel to H, the vertical projection of which
will be parallel to GL.
Pros. 8. This is similar to Prob. T, only
the position of the pyramid with respect to V
being changed.
DEVELOPMENT OF SURFACES
Pros. 9. The cutting plane makes an
angle of 40° with H, and the auxiliary plane
must be at the same angle, the projecting lines
being drawn perpendicular to it.
Pros. 10. This is similar to Prob. 5, the
base being cut by the cutting plane. In devel-
oping the surface open it on the uncut edge.
Pros. 11. Theslant edges of this pyramid
being of unequal length must be obtained sepa-
rately. One of these edges, AD, is shown in
its revolved position at AD’. One-half of the
development is also shown, and the method and
order for drawing the lines indicated by the
numbers. Thus, the line AB is drawn first,
and then arcs 3 and 2 described from its ex-
tremities, A and B, with radii equal to the true
lengths of AC and BC: this determines the
point C. In like manner the remaining points
and lines are found. «
Pros. 12. The development of a cylin-
der is required. Arts. 120 and 121, page 92.
Employ twenty-four elements in obtaining the
curve.
° 148
PROB. 9 te PROB. 10
TOP VIEW OF ,
HEXAGONAL PYRAMID
| PENTAGON
INSCRIBED IN
i 2-\"CIRCLE
as
REVOLVED
| SECTION
+
H REVOLVED
SECTION
ee
oy
nN
| SIDE VIEW SIDE VIEW
|
d lo
DEVELOPMENT DEVELOPMENT
PROB. I2
REVOLVED
SECTION
DEVELOPMENT
144
PROBLEMS
PROB. |3 PROB. 14
REVOLVED aa REVOLVED
SECTION . . SECTION
SIDE VIEW
DEVELOPMENT DEVELOPMENT
PROB. I5
Pros. 4. Solve without auxiliary planes.
Pros. 138. Use twenty-four elements for
obtaining the development. Art. 122, page 94.
Pros. 14. This being an elliptical cone,
the elements will be of unequal length.
Pros. 15. The cone is cut by four planes,
CG, CB, CF, CE. Determine the top view of
each and their projection on planes to which
they are parallel. ‘Test the latter as follows:
Determine the axes and foci of the ellipse, and
test eight points by the first method, Art. 80,
page 52. Test by trammels also. Test the
parabola by the second method, Art. 87, page
58. The cutting plane, CI, is parallel to an
element of the cone. Test the hyperbola by
the second method, Art. 91, page 59. The ver-
tex of the cone bisects the transverse axis.
149. Intersection problems. Chap. VIII.
Pros. 1. Assume twenty-four equidis-
tant elements on the small cylinder.
Pros. 2. The cylinders are tangent.
Use care in obtaining the limiting points.
Pros. 8. Use auxiliary planes. Page 98.
Art. 125, p. 100. Develop oblique cylinder.
INTERSECTION
DEVELOPMENT OF CYLINDER A
SIDE VIEW
DEVELOPMENT OF CYLINDER B
PROB. 2
DEVELOPMENT OF CYLINDER A
| SIDE VIEW
’
DEVELOPMENT OF CYLINDER B
OF SURFACES
TOP VIEW
DEVELOPMENT
OF CYLINDER A
DEVELOPMENT
OF CYLINDER
145
146 PROBLEMS
DEVELOPMENT OF DEVELOPMENT OF
EQUILATERAL TRIANGULAR EQUILATERAL TRIANGULAR
PRISM
TOP VIEW
DEVELOPMENT OF
HEXAGONAL PRISM HEXAGONAL PRISM A
Pros. 5. It is required to find the inter-
section of a cylinder and prism without using a
side view. Use cutting planes parallel with V.
The base of the prism will have to be revolved
in order to complete the top view and enable the
intersection of the cutting planes and prisms to
be determined.
Pros. 6. To determine the intersection
of a hexagonal and a triangular prism. This is
similar to Prob. 5, save that a prism is substi-
tuted for the cylinder. In all cases of inter-
section between prism and prism, it is only
necessary to find the point of intersection of
each edge of both prisms with a face of the
other prism. Art. 126, page 101.
Pros. 7. The lines of intersection on the
top view are to be completed, and the front
view with its lines of intersection are required.
Develop the prism only.
Pros. 8. Determine the intersection of
the oblique and vertical hexagonal prisms and
develop the latter. Art. 126, page 101.
SPIRALS AND HELICES 147
150. Spirals, screw-threads and bolt-head problems. Study Chapter IX. Page 102.
Problems 1 to 9 inclusive require a space of 43’. A graphic statement is made for the prob-
lems on screw-threads and bolt-heads.
Pros. 1. Draw an equable spiral with a pitch of 11/. Describe 13 revolutions of the
radius vector. Art. 128, page 102. Sketch the
curve, and ink as directed in Art. 15, page 12. Lea aad
Pros. 2. Draw the involute of a 3” square. --3 }-—>
j i! : SECTION OF
Art. 130, page 104. 2 SQUARE THREAD 14 3 Sa es
Pros. 3. Draw the involute of an equi- aoa ute at [THREAD |
lateral triangle having 3!’ sides. Art. 180,
page 104.
Pros. 4. Draw the involute of a right Branckaghiy enon
line 4’ long. Art. 130, page 104. pally ; ANGLE OF, V 90"
Pros. 5. Draw the involute of a hexagon
having sides of #/’. Art. 130, page 104.
Pros. 6. Draw the involute of a circle
having a diameter of 1’... Art. 130, page 104.
Pros. 7. Draw a right-hand helix of 14/’ pitch, 2’’ diameter and 23’ length. Art. 131,
page 104. |
Pros. 8. - Draw a left-hand double helix of 13/’ pitch, diameter 13’, length 21".
Pros. 9. Draw a right-hand conical helix. Pitch 1}!’, diameter of cone 2’, height 17/’.
Pros. 10. The diameter and pitch being the same in both cases, but two templets are re-
quired, one for the outer and one for the inner helix. Art. 132, page 106, and Art. 186, page 110.
148 PROBLEMS
R.H. THREAD
24 aye
5 °
3 PITCH 60 V
{—_ -
. R.H. DOUBLE
. | . . CA . sQ! it P.
Noo
— gt =e a
2 : itp
R.H.SINGLE y
,
haa S iciecae Same
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Ce oe oe
Pros. 11. The first four examples are of
conventional V threads, Art. 183, page 108.
The second four are conventional square
threads. Art. 186, page 110. The last four
are to be drawn by the methods illustrated by
Figs. 177 to 180, page 109. Make the pitches
about the same as those in the illustrations,
estimating the spaces by the eye. Distinguish
clearly between the single and double threads.
Pros. 12. Study Arts. 187, 188 and 139
before attempting this problem. The propor-
tion and character of the heads and nuts should
be so well understood that reference to the text
will be unnecessary. ‘The diameters are given,
and the sketch shows the character of the bolt,
whether rounded or chamfered. Observe every
detail and see that the dimensions are standard.
Draw the rounded heads and nuts before the
chamfered type.
For the further consideration of this subject,
the student is referred to the chapter on Bolts
and Screws, in ‘Machine Drawing” of this
series.
MISCELLANEOUS PROBLEMS
151. Miscellaneous problems. These
examples of a practical character are de-
signed for pencilling and inking practices
which may be used early in the course.
Problems 1 and 2 may be substituted for
some of the “examples for practice,” page
117 and following. They will afford an
excellent practice in lettering and figuring.
Problems 3 and 4 hkewise require very
little knowledge of projection and serve
as an excellent practice in penmanship.
Full instruction is given with each sketch
and the proper scale is specified for a
10" x 14" plate. The shading of the
drawing is left to the discretion of the
instructor. ‘The drawings should be cor-
rectly figured, and the title in proper form.
The author does not recommend the
making of machine drawings, save as above,
until the student has acquired a thorough
working knowledge of projection, and has
been taught something of the notation and
idiomatic use of applied graphics.
149
PROB.|
ECCENTRIC. eae
DRAW TWO FULLVIEWS.
SHADE AND DIMENSION LINES REQUIRED,
150 PROBLEMS
TWO COMPLETE SECTIONS OF THE FIRST
AND SECOND FORMS OR THE \FIRST ANDO
THIRD FORMS MAY BE Ba alsen INA
fA C 10/4 SPACE. |
STANDARD, SECTION OF
Z£OC al D SEWER
Ca ae YQ:
Ww
480
. ~ :
AN SI
RII; |
3
aX ID
So:
18 |
DRAW COMPLETE SECTION DRAW COMPLETE SECTION DRAW COMPLETE, SECTION
SCALE /°=/F7. SCALE 22/FT. SCALE 2 =/FT.
a
MISCELLANEOUS PROBLEMS 15
ONE WROUGHT JPON.
BoRDER £/NE
HAN) LEVER FOR CONTROLLING CYL/NDER
VALVES U.S. COAST DEFENSE VESSEL MONTEREY
152 PROBLEMS
2
Wa
ra
SO BARS %
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Pros. 4. A sketch is given for the Commutator of a 10 K.W.D.C. Dynamo, and
it is required to make a full-size sectional drawing of the complete commutator.
te
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-URBANA
3 0112 123898675
UNIVERSITY OF ILLINOIS
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