Descriptive Geometry II
LEARNING OBJECTIVES
After completing this material, you will be able to:
• Determine bearing, slope, and percent of grade.
• Establish the true angle between a line and a plane.
• Find the true distance between a plane and a point.
• Find the true distance between two skew lines.
• Find the piercing point of a line and a plane.
• Determine the visibility of lines.
• Find the true angle between intersecting planes.
• Graphically solve vector force problems.
T H E E N G I N E E R I N G D E S I G N A P P L I C AT I O N
going through the roof. This is the concept of a line
intersecting a plane. “Determine the shape of the roof
opening” explains itself. The true shape of the opening is asked for here. You are finding the true shape
of a plane to determine the angle between the pipe
and the roof. The concept of the angle between a
line and a plane is used to determine the slope of the
discharge pipe. Therefore, the problem requires you
to find:
A sample descriptive geometry problem solved by the
design process follows:
1. Identify and define the problem. Given an isometric and
partial detail drawing of a building with a chip blower
and its discharge pipe location, determine the location
and the shape of the roof opening shown in Figure 1.
Also, the angle between the pipe and the roof must be
determined to complete the design of the pipe support bracket. Finally, determine the slope of the discharge pipe. The partial plane view shows the location
of a chip blower in a 40' wide 3 100' long metal building. The building has 16' high walls and a 4/12 slope
on the roof. The blower point 1 is 4' above the floor,
and the discharge point 2 is 28' above the floor.
2. Research and generate solutions. To see how much can
be understood from the problem statement, take it
word by word. “Determine the location of the roof
opening” means there will be a pipe intersecting or
a. The intersection of a line and a plane.
b. The true shape of a plane.
c. The angle between a line and a plane.
d. The slope of a line.
3. Evaluate the possible solutions. Draw the top, front,
and side projection planes or, as in architectural drawing, the top, front elevation, and side elevation of the
given problem. This will help in the evaluation if there
12
4
10
C
L
16'-0"
0'-
0"
2
40'-0"
32'-0"
48'-0"
© Cengage Learning 2012
OPENING ?
1
FIGURE 1 Problem layout with partial detail.
1
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Descriptive Geometry II
is doubt about which part of the information is useful.
Only one solution for each part has been determined,
so this step is limited in its usefulness.
4. Use the best solution. Decide which views or projection planes will give the necessary information.
The front, top, and side views do not give the true
lengths, shapes, or angles needed. Auxiliary views
must be found and constructed, as in Figure 2. The
roof plane must be seen as a true shape in order to
see the true shape of the hole in it. You must see the
edge of the roof plane and the true length of the pipe
in the same view to determine where they intersect
and the angle between them. To find the slope of the
discharge pipe, you must see the true length of the
centerline of the pipe in an elevation view.
5. Evaluate the best solution. Check the answers by other
methods if possible. Does everything in the answer
make sense?
6. Finalize the solution. Present the solution to your employer or the instructor with the completed notations.
1h
C
L
2h
C
L
2f
H
F
1f
F
A
ROOF
UNDER
TRUE SHAPE OF
HOLE IN ROOF
1a
A
X
1x
C
L
2a
PARALLEL TO
ROOF EDGE
TRUE ANGLE BETWEEN
ROOF AND SHAFT
C
L
2x
TRUE LENGTH OF SHAFT
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FIGURE 2 Fold-line auxiliary view solution.
The procedure to locate lines and planes in space was discussed
in Descriptive Geometry I on the Student CD. To solve many
engineering problems, the direction of lines and planes must
also be known. The direction of lines and planes is identified in
space by a variety of ways, depending on their uses.
Horizontal Directions
Bearings are generally used for map directions and can only be
determined in a horizontal view or a top view. Knowing the
bearing angle of a line or plane is especially useful in mining
and civil engineering. The direction of north is always understood to be toward the top of a drawing layout unless otherwise
indicated by a north-pointing arrow pointing in another direction. South is always opposite north, west is to the left, and east
is to the right, forming the word WE as shown in Figure 3. A
bearing angle is always 90° or less and is identified either from
the north or south. Lines pointing directly north are identified
as due north. Also, lines pointing directly east, west, or south are
identified as due east, due west, and due south, respectively. The
bearings of several angles are shown in Figure 4. Notice that the
bearing is first identified by a north or south direction indicator.
The acute angle formed from the vertical direction and east or
west follows next. Lastly, the corresponding side of the acute
angle direction east or west is noted. Examples include N60°E,
N75°W, S30°E, and S45°W.
09574_DGII_p1-23.indd 2
Bearing Is Found
in the Horizontal View
The bearing of a line can only be found in the horizontal view.
The line does not have to be in true length to represent the
bearing.
There is really no such thing as the bearing of a plane, but
in mining applications it is common practice to locate a stratum by using the bearing of a horizontal line in the plane of a
stratum. Stratums in their simplest form are horizontal layers
© Cengage Learning 2012
DIRECTIONS OF LINES AND PLANES
FIGURE 3 Compass directions.
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DESCRIPTIVE GEOMETRY II
3
N
1500
N60°E
1400
N75°W
nh(1300')
ah
1000
1300
1200
dh
1100
900
800
W
oh(750')
mh(900')
E
600
700
930 920
910
S45°W
S30°E
S
bh
FIGURE 4 Bearing angles of lines on a contour map.
of rock. Within a specific area, it is reasonable to assume that
a stratum is uniform in thickness and that it lies between two
parallel planes. For the discussion in this text, these conditions
are assumed. The bearing of a horizontal line in a stratum is
called the strike. To find the strike of plane m–n–o in Figure 5,
you can use the same technique found in Descriptive Geometry
I on the Student CD by locating a horizontal line in the front
view and finding that same line in the horizontal view where it
appears as true length.
STEP 1 Draw horizontal line mf –sf in the F view.
STEP 2 Find line m–s in the H view. Line mh–sh shown in
true length is not important, because this is where
the bearing of the line is shown whether it is in true
length or not.
STEP 3 Measure the angle of line mh–sh from the north or south
direction. This bearing N65°E of line m–n as seen in the
horizontal view is the strike of the plane.
This discussion of the application of descriptive geometry
to mining problems is brief for the purposes of this material. It
will be necessary to study in more detail the concepts of contour lines, saddles, ridges, ravines, stratas, veins, faults, outcrops, cuts, and fills to adequately utilize descriptive geometry
for civil engineering applications. Many of the calculations to
determine cross sections for highways, and the designs and
drawings from many civil engineering projects, are currently
being made using computers.
Vertical Directions
The vertical direction of a line can only be found when the
following two conditions exist: The line must be in an elevation
09574_DGII_p1-23.indd 3
N
nh
STRIKE
5°
N6
E
sh
mh
oh
H
F
nf
1300'
sf
mf
HORIZONTAL LINE
(b)
1000'
of
700'
© Cengage Learning 2012
ch
© Cengage Learning 2012
(a)
FIGURE 5 Stratums: (a) horizontal view of a stratum; (b) finding the
strike of the stratum.
view, and the line must appear in true length. The vertical direction of a line is identified by the acute angle formed between
the vertical line and the horizon (ground). This angle varies
from true horizontal (0°) to vertical (190°). If the angle is between the horizon and 190°, or the line goes upward from its
origin, the angle is called inclination and is assigned a positive
value. If the angle is between the horizon and 290°, or the line
goes downward from its origin, it is called declination and is assigned a negative value (see Figure 6). The terminology used to
identify the inclination of a line is specified differently for various engineering fields. Each of the terms that follow are those
that express the relationship of a line to a horizontal plane. In
all cases, the same two conditions must exist for the true direction of a vertical line to be identified. The line must be in an
elevation plane where it appears in true length.
Rules for Finding the Slope
of a Line
When solving problems that require you to find the slope of
a line, always start from the horizontal view. For example,
when using the fold-line technique, place the fold line parallel
to the line where it exits on the horizontal view. When using
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Descriptive Geometry II
© Cengage Learning 2012
4
(a)
FIGURE 7 Slope angle.
the revolution method, revolve the line where it exists in the
horizontal view parallel to the existing fold line. Either method
results in showing the line in true length in an elevation view.
This is the only place you can measure the true slope of a line.
The slope angle or slope of a line is the angle in degrees that
the line makes with the horizontal (ground or level) plane. The
slope angle can be expressed by the following formula:
Rise
Slope angle (s) is the angle where tangent (arctan s) 5 ____
Run
(b)
© Cengage Learning 2012
© Cengage Learning 2012
FIGURE 6 Inclination and declination.
FIGURE 8 Calculating
the grade of a line: (a) formula one;
(b) formula two.
This is illustrated in Figure 8a. Observe also that the grade
is the tangent of the slope angle multiplied by 100. From trigonometry, you know that the tangent of an angle is defined as
dividing the length of the opposite side of a triangle of the included angle by the adjacent side of that angle as seen in Figure
8b. One of the most common ways of measuring run and rise
is to use an engineer’s scale, which has divisions in multiples of
10, or set the computer scale in CADD to the appropriate scale.
Percent of grade is simply another way to express slope; therefore, the conditions for finding the percent of grade of a line are
the same as for finding the true slope of a line (see Figure 9).
The term grade is usually used when referring to a line that has
an angle with less than a 45° inclination. Grade is usually, but
not always, associated with civil engineering. Slope is usually
applied to a line that has an angle with greater than 45° inclination, such as the slope of a roof on a house. Slope is used more
with architectural engineering but not always.
09574_DGII_p1-23.indd 4
© Cengage Learning 2012
Again, the true slope of a line can only be seen in an elevation view where the line appears in true length (see Figure 7).
The grade of a line is expressed in percent form and is found
by using the following formula:
Rise 3 100
Percent grade 5 ____
Run
FIGURE 9 Grade of a road.
In architectural engineering, the slope is also referred to as
the pitch. When describing the angle the roof plane has with
the horizon, it is expressed as the ratio of the vertical rise to 12'
of span. The graphical method of showing pitch on a drawing is
illustrated in Figure 10.
In structural engineering, slope is usually referred to as the
bevel of a line. Figure 11 shows the bevel of several beams.
In civil engineering, the terms grade and slope are frequently
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DESCRIPTIVE GEOMETRY II
LEVEL
12
SLOPE
1 ON 3
SLOPE
1 ON 1
10
10
3
1
SLOPE
1 ON 3
SLOPE
1 ON 4
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12' SPAN
(2)
CONCRETE
DAM
6
1:2
6
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12
ER
BA
TT
ER
TT
BA
1:
1
(3)
BA
TT
FIGURE 10 Pitch of a roof.
ER
2:3
:8
BATTER 1
PITCH 10:12
(a)
(b)
12
FIGURE 12 Terms in civil engineering: (a) slopes of earth dam; (b)
batter of concrete dam.
4
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FIGURE 11 Bevel of a beam.
used as shown in Figure 12a. In addition, batter is applied
when dealing with the slope of concrete footings as shown in
Figure 12b. Dip is used when finding the slope of a stratum
(plane). The dip is the slope of the strike. Figure 13 shows how
to find the dip of a stratum.
In review, slope, grade, pitch, bevel, batter, and dip are all representations of the angle between a true-length line and the
horizon. Finding this angle graphically is done with the same
procedure in each situation.
TRUE DISTANCE BETWEEN A POINT
AND A LINE
To find the true distance between a point and a line, you must
create a view that shows the line and the point appearing as two
points. This occurs when you are viewing the end of the line and
the point on one view. As with most problems, there are several
ways to solve this problem. Figure 14 shows how to find this
distance by the fold-line method.
09574_DGII_p1-23.indd 5
Finding the Distance Between Two
Points with the Fold-Line Method
The first step in the solution of this type of problem is always
to place a fold line parallel to the given line. The line is in true
length when it is projected onto the new view. The second step
is to place another fold line perpendicular to the true-length
line. The line will then appear as a point when it is projected
onto the next new view. After projecting the point onto the two
new views, it is just a matter of measuring the distance between
two points.
STEP 1 Find the true length of line r–s and the location of
point x on the A view (see Figure 14).
STEP 2 Find the end view of the true-length line r–s and the
location of point x on the I view. The distance measured between point x and the end view of line r–s
on the i view is the true distance between point x and
line r–s.
TRUE DISTANCE BETWEEN TWO SKEW
LINES
Skew lines are two nonparallel, nonintersecting lines. An example of two skew lines is shown in Figure 15. The true distance between two skew lines is the perpendicular between
them. To be able to determine the true distance between the
two skew lines, you must create a view where one line appears
as a point. (Note: One line appears as a point, but the other line
will not necessarily appear in true length.) The perpendicular
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Descriptive Geometry II
1500
1400
nh(1300')
1000
1300
1200
1100
900
800
600
oh(500')
mh(900')
300
T
L
© Cengage Learning 2012
500
(a)
H A
na
EDGE VIEW
OF PLANE
(EVP)
nh
FIGURE 14 True distance between a line and a point. TL represents
true length.
sa – ma
sh
N
oa
E
65°
mh
P
DI
oh
1300'
sf
mf
(b)
900'
of
500'
© Cengage Learning 2012
nf
© Cengage Learning 2012
H
F
FIGURE 13 Fold-line method of finding the dip of a stratum. EVP
represents the edge view of the plane. (a) Plan view with
contour lines. (b) Finding the dip of the stratum using the
fold-line method.
distance between the line appearing as a point and the other
line is the true distance between both lines (see Figure 16).
FIGURE 15 Skew lines.
STEP 1 Find the true length of one of the lines. In this exam-
ple, line r–s was selected to find the true length. The
true length of line r–s and the location of line m–n has
been found on the A view (see Figure 16a).
STEP 2 Create the view where line r–s appears as a point. Line
r–s appears as a point in the i view. Line m–n is also
carried onto this view (see Figure 16b).
STEP 3 On the I view, the true distance can be measured between line r–s and line m–n by constructing a perpendicular line from line r–s (appearing as a point) and
line m–n (see Figure 16c).
Sometimes a problem may require you to adjust the distance
between two skew lines or make sure there is enough clearance
09574_DGII_p1-23.indd 6
between the two lines. An application for determining the true
distance or clearance determination might be for the location
of a valve in a piping system or the clearance between power
lines and a nearby metal structure in an electrical power substation. When you have found the true distance between the
two skew lines as in Step 3 in the previous problem, adjust
the point (line r–s) in or out along the perpendicular distance
from line m–n, as shown in Figure 16d. You then can relocate
the new position of line r–s in all the previous views. If line
r–s cannot be moved in the original problem, then you need to
solve the problem by starting with line m–n rather than line r–s
as in this example.
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DESCRIPTIVE GEOMETRY II
(a)
(b)
POINT VIEW
(c)
TL
© Cengage Learning 2012
POINT VIEW
(d)
FIGURE 16 Finding the true distance between skew lines: (a) Step 1; (b) Step 2; (c) Step 3; (d) adjusting the true distance.
INTERSECTION OF A LINE
AND A PLANE
Important considerations for the designer in structural and
mining problems include the location and anchoring of guy
wires; the location of holes for shafts, cables, pipes, and wires;
and the angles that are created by these features.
Finding the Piercing Point
of a Line and a Plane
To find the point of intersection, the piercing point, of a line
and a plane, you must create a view where the plane appears as
an edge. You can then project this apparent point of intersection
back into the original views.
09574_DGII_p1-23.indd 7
Use the following steps if you have a problem that requires
finding the intersection of the roof and the circular shaft as
shown in Figure 17a (piercing point problem).
STEP 1 Find a view that shows the roof plane as an edge.
Figure 17b shows that the P view contains the roof
plane as an edge. When the centerline of the shaft
is projected onto the P view, the apparent point
of intersection, wp, of the centerline of the shaft and
the roof plane can be seen.
STEP 2 Project this apparent point of intersection, wp, back
onto the previous views. As Figure 17c shows, a projector is projected from the A view back to the F view
intersecting the centerline of the shaft. This is the actual point of intersection, point wf of the roof plane
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Descriptive Geometry II
(a)
(b)
© Cengage Learning 2012
8
(c)
FIGURE 17 Piercing point problem: (a) problem setup; (b) Step 1, connecting two points and finding the apparent point of intersection; (c) Step 2,
finding the actual piercing point.
shown on the F view. You can also show the proper
location of point W in the H view by the normal procedure of locating a point in space.
VISIBILITY
When two lines are shown in space, sometimes it is difficult to
know which line is on top of which or if they intersect. Problems dealing with the intersection of planes require you to
know which line is in front of the other in order to correctly
solve and complete the problem as you will see later in this material. For example, look at Figure 18a. Looking at the F view,
you cannot tell if lines a–b and r–s intersect or if line a–b is in
front of or behind line r–s. There is a simple way to determine
which situation exists.
STEP 1 On the F view shown in Figure 18b, project a projec-
tor from the apparent point of intersection of lines
a–b and r–s perpendicular to fold line F/H. Continue
projecting until it crosses one or both of the lines on
the H view.
STEP 2 If the projector passes through the apparent point of
intersection on the H view, these are intersecting lines
09574_DGII_p1-23.indd 8
and the real point of intersection. But this is not the
case in this example. The projector crosses line r–s in
the H view first and verifies that line r–s is closest to
you in the F view.
STEP 3 You can apply the same procedure when determining
the visibility of lines r–s and a–b on the H view. Figure
18c shows a projector being projected from the apparent point of intersection on the H view perpendicular
to the H/F fold line. The projector is then continued
on until it crosses one or both of lines a–b and r–s
in the F view. The projector crosses line a–b in the F
view first and verifies that line a–b is closest to you in
the H view.
INTERSECTION OF PLANES
Structural sheets, houses, sheet metal forms, and many other
features have intersecting surfaces. Determining the lines of
intersection is essential in completing each view of a drawing
and eventually determining the shape and size of each surface
for the construction of each part in the fabrication shop. If two
planes are not parallel, they intersect and form a straight line
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(a)
(b)
9
© Cengage Learning 2012
DESCRIPTIVE GEOMETRY II
(c)
FIGURE 18 Visibility: (a) two lines in space; (b) finding visibility of lines in the F view; (c) finding visibility of lines in the H view.
that is common to both planes. Only two points that lay on
both planes need to be found to fix the position of this line of
intersection, which is common to both planes. To find the line
of intersection of two planes, a view must be created that shows
both planes where one appears as an edge view. Figure 19a
shows two views in which the two planes appear to intersect.
STEP 1 Create a projection plane where one of the planes ap-
pears as an edge view. Figure 19b shows finding plane
a–b–c as an edge view in the A view. You already know
how to do this from Descriptive Geometry I.
STEP 2 Figure 19c shows the common points J and K of intersection of the two planes. These points can now be
projected back onto the F and H views and completed
as the line of intersection.
STEP 3 Identify the visibility of the intersecting lines of each
plane on the F and H views by the process previously
explained. You will need to do this in order to show
which lines are visible and which lines are hidden
on the two planes. Figure 19d shows this process
completed.
DIHEDRAL ANGLES
Working with structural sheets or various sheet metal projects
requires knowing the exact angle between surfaces or planes.
The true angle between two planes, or dihedral angle, is found
in a view in which both of the intersecting planes appear as an
edge (see Figures 20 and 21). Figure 22a shows the top and front
views of a sheet metal funnel on the H and F views. The problem here is to determine the dihedral angle between the front
side and the right side of the funnel.
09574_DGII_p1-23.indd 9
Steps in Finding the True
Dihedral Angle
To find the true dihedral angle between two planes, you need to
find the end view of the line that is created by the intersection
of the two planes. That first step is to place a fold line parallel
to this line of intersection. Then, place a fold line perpendicular
to this true-length line.
STEP 1 Create a view in which you show the true length of the
line of intersection of the two planes for which you are
trying to determine the dihedral angle. In Figure 22b,
the true length of line d–t, the line of intersection, has
been found in the i view. You already know how to find
the true length of a line from Descriptive Geometry I.
Then project only the points that identify the rest of
the two intersecting planes onto the i view as shown.
STEP 2 Create a view that shows the intersecting line d–t appearing as a point. In Figure 22c, line d–t has been
found on the A view. It appears as a point. Notice
also that when you project all the other points of the
two planes onto the A view, they form two intersecting straight lines. The angle formed between these two
planes is the true dihedral angle. This angle can be measured with a protractor or calculated on the computer.
AN INTRODUCTION
TO VECTOR GEOMETRY
Vector analysis is a branch of mathematics that includes the
manipulation of vectors. Vector geometry is included in this material because vectors are a vital part of many of the engineering
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Descriptive Geometry II
(a)
(b)
EV
P
J
EV
P
J
(c)
© Cengage Learning 2012
J
(d)
FIGURE19 Two planes that appear to intersect: (a) problem setup; (b) Step 1, finding the edge view of a plane; (c) Step 2, finding the
line of intersection of two planes; (d) Step 3, determining visibility and completing the visible and hidden lines of two planes.
EVP represents the edge view of the plane.
bp-ap
bf
EDGE
EDGE
rf
© Cengage Learning 2012
DIHEDRAL
ANGLE
TL
cf
sp-rp
sf
df
F P
DIHEDRAL
ANGLE
bp-cp
© Cengage Learning 2012
af
FIGURE 21 Descriptive
FIGURE 20 Pictorial representation of a dihedral angle.
09574_DGII_p1-23.indd 10
geometry flat layout showing the dihedral
angle of two planes.
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DESCRIPTIVE GEOMETRY II
(b)
(c)
© Cengage Learning 2012
(a)
FIGURE 22 Dihedral angle problem: (a) two views of a sheet metal funnel; (b) Step 1, finding the true length of the line of intersection;
sciences. Graphic solutions are usually sufficiently accurate for
engineering analysis of forces and other directional quantities
in structural design, mechanics, and other physical sciences.
Vectors can be manipulated in some very complex ways, but
the discussion here is limited to vector addition and its practical purposes. Because vectors are segments of straight lines, the
principles of descriptive geometry are directly applicable to the
graphic solution of problems involving vectors.
A vector quantity is one that has magnitude and direction,
such as displacement, acceleration, momentum, force, velocity,
and torque. The speed of a space shuttle, for example, describes
only the magnitude of its velocity, which is a physical or scalar
quality. The velocity of that space shuttle, on the other hand,
is a vector quality that describes its speed and its direction of
motion in space. A vector quantity is usually represented by a
straight line with an attached arrowhead pointed in the same
direction in which the quantity is acting. The length of the
09574_DGII_p1-23.indd 11
© Cengage Learning 2012
(c) Step 2, finding the end view of a true-length line and creating the dihedral angle of two planes.
FIGURE 23 Vector representation/vector diagram.
vector is drawn proportionally to the magnitude of that quantity. An example of a vector diagram representation of a force
of 10 lbs directed N45°E is shown in Figure 23. To solve problems using vector geometry, it is essential to know the difference between a vector diagram and a space diagram. Figure 24
shows the space diagram of the 10 lb vector. Notice that the
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Descriptive Geometry II
Addition of Concurrent
Coplanar Vectors
© Cengage Learning 2012
12
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FIGURE 24 Space diagram.
FIGURE 25 Space diagram of concurrent forces.
Resolution of Concurrent
Coplanar Vectors
Two or more vectors can be added to form a single resultant
vector. By reversing that process, a single vector can be resolved
© Cengage Learning 2012
space diagram is not drawn to scale. It shows sense and the
accurate direction. When two or more vectors act on an object, it is called a vector or force system. The discussion in
this material is limited to forces that are concurrent, because
they are the most common. In concurrent forces, vectors pass
through a common point, as seen in Figure 25. Forces with
a line of action lying in one plane are called coplanar forces.
Figure 26 shows five coplanar vectors pictorially and with their
corresponding vector diagram. When those forces lie in more
than one plane but still meet at one point, they are identified as
noncoplanar, as shown pictorially and in their corresponding
vector diagram in Figure 27.
To determine the net effect of a vector system, vector quantities
must be combined. The graphic addition of concurrent coplanar
vectors requires you to draw a vector diagram locating the vectors head to tail in their correct direction. The sum, or the space
that is left open, is called the resultant. The resultant has the
same effect as the total of the individual vectors added together.
In Figure 28a, the two vectors M and N shown in the space
diagram are to be added. The magnitude of vector M is 40 lb,
and vector N is 60 lb. Remember that the space diagram shows
the correct direction, so in the vector diagram, Figure 28b, the
direction of the vectors is duplicated and the vector length is
scaled to the same units. After placing each vector head to tail in
any order, the space left is connected with a straight line. This is
the resultant, and the direction arrow is in the same direction as
the added vectors. Measuring the length of the resultant to scale
gives a length of 80 lb. If the direction of the resultant is reversed
as shown in Figure 28c, it is called the equilibrant. An equilibrant is a force that balances all other forces through a point.
An equilibrant is the same as the resultant of zero. If you have
just two vectors to be added, the parallelogram method can be
used. This method requires more construction, in that the sides
of the parallelogram are drawn parallel to the given vectors.
Then the diagonal of the parallelogram is drawn, which is the
resultant of the two vectors, as shown in Figure 28d. When you
have three or more vectors, similar methods can be used. In Figure 29a, the vectors are drawn to scale, using the space diagram.
In Figure 29b, the correct direction is obtained as previously explained. The closing side of the vector polygon is the resultant.
As before, if the equilibrant is required, the direction of the arrowhead of the resultant is reversed, as shown in Figure 29c.
FIGURE 26 Concurrent coplanar vector system.
09574_DGII_p1-23.indd 12
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13
© Cengage Learning 2012
DESCRIPTIVE GEOMETRY II
FIGURE 27 Concurrent noncoplanar vector system.
M
40 LB
60 LB
N
60 LB
E = 80 LB
EQUILIBRANT
(a)
40 LB
60 LB
(HEAD TO
TAIL)
R = 80 LB
RESULTANT
40 LB
PARALLEL TO
VECTOR N
PARALLEL TO
VECTOR M
SCALE =
0 1 2 3
UNITS
M
40 LB
N
60 LB
(d) VECTOR ADDITION
© Cengage Learning 2012
R=8
0 LB
(b)
(c) VECTOR DIAGRAM
FIGURE 28 Addition of concurrent coplanar vectors: (a) space diagram; (b) vector diagram showing the resultant;
(c) vector diagram showing the equilibrant; (d) vector addition by parallelogram method.
into two or more component vectors. In the case of coplanar
vectors, only two can be unknown in magnitude or direction.
Most commonly, a vector must be resolved into two other vectors having specified directions. In Figure 30a, a given vector T
is to be resolved into two component vectors having direction
of OR and OS. The magnitude of each needs to be found. By
using the parallelogram method for the addition of two vectors,
the solution is obtained in reverse. Vector R is moved parallel until it crosses the arrow end of the given vector (see Figure 30b). The S vector is extended through the tail end of the
09574_DGII_p1-23.indd 13
given vector. Where the two moved vectors cross will determine
the lengths of each. The direction of each of the two vectors is
opposite that of the given vector.
Addition of Concurrent
Noncoplanar Vectors
The addition of noncoplanar vectors is the same as for coplanar
vectors, except that three dimensions are used now. Remember
that noncoplanar means in more than one plane, so the principles
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Descriptive Geometry II
B = 500 LB
C = 200 LB
R
O
=
S = 2500 LB
(a)
S
B = 500 LB
(a)
R
B
3
LB
50
© Cengage Learning 2012
0L
1
2
UNITS
T=
0
00
400
SCALE =
O
R
B
A = 300 LB
T
400 =
0L
14
O
(b)
S
FIGURE 30 Resolution of a single vector into two components: (a)
A = 300 LB
R = 350 LB
RESULTANT
space diagram; (b) vector diagram.
C = 200 LB
(b)
E = 350 LB
EQUILIBRANT
A = 300 LB
C = 200 LB
(c)
© Cengage Learning 2012
B = 500 LB
FIGURE 29 Addition of concurrent coplanar vectors: (a) space diagram;
(b) vector diagram showing the resultant; (c) vector diagram
showing the equilibrant.
of descriptive geometry are needed to solve these problems.
Figure 31a shows the space diagram on the F and H views for
the given vector quantities. To solve this problem, you must
create a projection plane that shows the true length of the resultant force.
STEP 1 Set up the space diagram on a frontal view and the
adjacent horizontal view as shown in Figure 31a.
STEP 2 Find the resultant vector on the H and F views just as
you would for coplanar vectors with one major exception: You will need to find the true lengths of those
vectors, which are not shown in true length on either
the H view or the F view. To do this, you can use either
the fold-line or the revolution method. Then project
the true lengths back onto the H and F views correctly. In Figure 31b, the S vector is already shown in
true length on the F view. Therefore, you can scale the
S vector on the F view. The T vector is shown already
in true length on the H projection plane. Therefore,
you can scale the T vector on the H view. However,
the R vector is not shown in true length on either the
H view or the F view. To find the true length of the R
vector, use the same procedure as you did when you
had to find the slope of a line when given only the
bearing of that line as explained in Descriptive Geometry I on the Student CD. In this case, line sh–rh was
established with an arbitrary length after the vector
09574_DGII_p1-23.indd 14
diagram was completed and placed in the correct direction on the H view. Then, the corresponding vector diagram was completed in the F view. Vector R
was placed in the correct direction as shown in the
space diagram on the F view with the corresponding
length as projected from the H view. The true length
of line sf –rf was found in the H view. Notice that the
line needed to be lengthened to equal the scaled 30
lbs. Line sh–rh2 now represents the true length of vector R on the H view. This length is then projected into
the vector diagrams as shown in Figure 31b.
STEP 3 Find the true length of the resultant force, the space
left in the vector diagrams. The resultant force is not
shown in true length on either the H or F views, so
the fold-line method was used this time to find the
true length of the resultant force on the I view. The
resultant force is measured at 15 lbs as shown on the
I view (see Figure 31b).
FORCES IN EQUILIBRIUM
Equilibrium is a condition in which the series of forces acting on
a body or structure equals zero and remains at rest. Graphically,
the vector polygon closes, with vectors in a continuous headto-tail arrangement (see Figure 32). When a body is known to
be in equilibrium under the action of certain known and unknown forces, these graphic conditions can be used to evaluate
the unknown forces. However, the number of unknown forces
is strictly limited. Coplanar vector systems can have as many
as three unknown forces. Vector diagrams can more easily be
constructed if you isolate the structure or part to which the
known forces are applied. This isolation means free from all
adjacent bodies and is referred to as free body. Construction of
a free-body diagram should be the first step in the analysis of
every equilibrium problem.
Finding the Equilibrium of Two
Unknowns in a Coplanar System
The following steps may be used to find the equilibrium of two
unknowns in a coplanar system when you are given a weight
that is supported by two ropes as shown in Figure 33a.
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DESCRIPTIVE GEOMETRY II
rh
15
SCALE =
0
1
2
UNITS
3
4
oh
T L = 30 LB
LB
rh3
R
rh
T
L
=
40
O
sh
th
th
H
H
F
F
rh2
rh1
sh
sf
rf1 rf2
O
tf
T
tf
L
=
LB
rf
rf3
R = T L = 15 LB
ri
of
oi
F I
rf
(a)
(b)
© Cengage Learning 2012
sf
50
FIGURE 31 Addition of concurrent noncoplanar vectors: (a) space diagram; (b) vector diagram.
STEP 1 Imagine the ropes are cut, making the knot an isolated
STEP 1 Draw the space diagram. The forces acting in the ca-
bles are known, but the direction of the vectors is not.
09574_DGII_p1-23.indd 15
A = 89 LB
B = 100 LB
C = 73 LB
VECTOR DIAGRAM
© Cengage Learning 2012
free body, as shown in Figure 33b. Draw a free-body diagram with the tension forces in the diagram being represented with force vectors pointing away from the knot.
STEP 2 Make a vector diagram. Start with the known vector
force of 100 lbs by drawing it to scale with the proper
direction. The direction of forces A and B are known
but not their magnitudes. It was previously explained
that for forces to be in equilibrium, the vector diagram
must close. As shown in Figure 33c, the vectors A and
B are drawn parallel to their corresponding vectors in
the free-body diagram shown in Figure 33b. When
the vectors are drawn properly, point L is established
in the vector diagram.
STEP 3 Scale the magnitudes of vectors A and B. In this example, both vectors A and B measure to be 60 lbs.
A similar type of problem occurs when two directions are
unknown. In this situation, you are given the space diagram in
Figure 34a, which shows a weight of 10 lbs supported by two
cables that pass over pulleys; these, in turn, support two other
weights of 6 and 8 lbs.
FIGURE 32 Vector diagram showing forces in equilibrium.
As shown in Figure 34b, start with the vector with the
force and direction known. The vector that represents
10 lbs is drawn in the down position. As shown, draw
the other two vector forces at any two different directions as shown in Figure 34b. The wavy lines indicate
that the real directions are unknown.
STEP 2 Draw the free-body diagram. This is solved by using
the method of creating a triangle when three sides of
the triangle are known. Start with the known force of
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16
Descriptive Geometry II
A
B
B = 60 LB
L
100 LB
100 LB
100
SCALE =
0 1 2 3 4 5 6
(a)
(b)
(c)
© Cengage Learning 2012
A = 60 LB
(b)
(a)
in equilibrium with two magnitudes unknown:
(a) space diagram; (b) free-body diagram; (c) vector diagram.
10 lbs. Draw it in its correct direction and magnitude.
Next, with a radius equal to 6 lbs, an arc is drawn with
A as its center (see Figure 34c). With a radius equal to
8 lbs, another arc is drawn with B as its center. These
arcs intersect at C, completing the triangle.
STEP 3 Determine the directions of the 6 lb and 8 lb vectors.
The directions are indicated by the angles shown and
are determined by placing the arrowheads in a headto-tail manner in the diagram (see Figure 34c).
(c)
© Cengage Learning 2012
FIGURE 33 Forces
FIGURE 34 Forces
in equilibrium with two directions unknown:
(a) space diagram; (b) free-body diagram; (c) vector diagram.
WEB SITE RESEARCH
The following Web sites can provide additional information for research or further study into topics covered in this material.
Address
Company, Product, or Service
www.asme.org
Find information and publications related to the American Society of Mechanical Engineers.
www.ansi.org
The American National Standards Institute. Information about national and international drafting standards.
www.adda.org
American Design Drafting Association. Information related to the drafting and design profession.
PROFESSIONAL PERSPECTIVE
In this material, many new concepts have been introduced.
But as you may have noticed, the techniques used to solve
each of the unknown angles and locations of the intersections are the same as those used in Descriptive Geometry
I on the Student CD with, for the most part, only the terminology changed. Again, accuracy in the construction
and drawing of lines is essential and natural with CADD.
In most cases, the problem situations are more complex, so
it is essential to understand how to break each one down
09574_DGII_p1-23.indd 16
into various simple elements. To solve vector problems, the
correct terminology must be learned for the different types
of vectors and diagrams. This will be especially true if there
has been no previous experience with the related physics.
Problem solving, although difficult for many, can be made
easier if a logical system is used, such as the one used at the
beginning of this material. This system is basically the best
one that most engineers or anyone else would use to solve
any type of problem.
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DESCRIPTIVE GEOMETRY II
17
Descriptive Geometry II Test
INSTRUCTIONS
Answer the following questions with short, complete statements or drawings as needed.
QUESTIONS
1. Identify the view where bearings can be found.
2. Define strike.
3. Give the two conditions that must exist to find the vertical
direction of a line.
4. Describe inclination.
5. Explain declination.
6. Describe the slope angle or slope of a line.
7. Give the formula for finding the grade of a line.
8. How is pitch specified in architectural engineering?
9. Define dip.
10. What are skew lines?
11. Define dihedral angle.
12. Name the branch of mathematics that includes the manipulation of vectors.
13. Identify the condition in which the series of forces acting
upon a body or structure equals zero and remains at
rest.
14. Briefly describe concurrent forces.
15. What are coplanar forces?
Descriptive Geometry II Problems
INSTRUCTIONS
Carefully study the problems before you begin working. Solve Problems 1 through 19 on graph paper or with CADD, showing and
identifying all projectors, brackets with distances used, and point identifications in the required view(s). If using graph paper, use
a separate sheet of graph paper for each problem and the graph lines included with each problem for proper setup. Use a scale of at
least three squares to one to obtain accuracy in the problem solutions.
Part 1: Problems 1 Through 19
Given the adjacent horizontal and front
views showing the cable r–s, find its bearing, slope, and
grade.
PROBLEM 1
sh
N
rh
rf
sf
09574_DGII_p1-23.indd 17
© Cengage Learning 2012
H
F
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18
Descriptive Geometry II
True length and
slope
A crane is mounted on the bow
of a boat as shown. Find the
true length and the slope of
the cable A–B and the support
structure B–C.
PROBLEM 2
B
A
C
B
STEEL CABLE
A
© Cengage Learning 2012
C
Bearing, slope, and true length
From the following given data, draw separate adjacent
horizontal and vertical views that contain the following
lines:
PROBLEM 3
Line
Bearing
Slope
True Length
B–C
R–S
M–N
N 22° W
S 75° E
N 60° E
130°
210°
115°
3 in.
4 in.
5 in.
Slope and grade
In this problem, two sections of pipe are connected
together. Using only the centerlines of each pipe, draw the
necessary views to answer the following questions. Leave all
construction lines on the drawing. Identify on the drawing
where the answers appear:
a. Identify and calculate the true length of pipe a–b.
b. Identify and calculate the slope in degrees of pipe b–c.
c. Identify and calculate the grade in percent of pipe a–b.
d. Identify and calculate the true angle between the
pipes.
09574_DGII_p1-23.indd 18
© Cengage Learning 2012
PROBLEM 4
4/4/11 9:51 PM
19
DESCRIPTIVE GEOMETRY II
True length, true angle
A designer’s sketch, below, shows the adjacent front
and horizontal views of the roof plane a–b–c–d and the
location where the antenna is to be mounted on the roof
plane. The top end of the antenna is to project 8’ above
the ridge of the house.
a. Identify and calculate the true length of the antenna
in feet and inches.
b. Find the place where the bottom of the antenna
touches the roof in the front view.
c. Create a view that will show the true angle made by
the roof plane and then calculate the true angle in
degrees.
d. Show, in the horizontal and front views, three
equally spaced guy wires anchored on the roof
plane. Two guy wires are attached 5' from the base
of the antenna along the roof plane, and the third
guy wire is attached at the point shown in the
horizontal view. All three guy wires are anchored
to the antenna at a distance of 6' above the ridge.
e. Identify and calculate the total length of guy wire
used to stabilize the antenna to the nearest foot.
True distance between point and a plane.
Find the shortest distance between line (cable) r–t and
point E.
PROBLEM 5
© Cengage Learning 2012
© Cengage Learning 2012
PROBLEM 6
09574_DGII_p1-23.indd 19
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20
Descriptive Geometry II
Skew lines. It is important to keep the clearance between the two line (cable) segments a–b and m–n
of a diameter of 1/2" each at a minimum of 1 in. and a
maximum of 2 in. Create a view where you can verify
how far apart they are in the setup. If they do not fit this
criteria, readjust cable a–b in all views so it will be met.
(Drawing scale: Each square 5 1/4")
True angles. Two parallel pipe corner segments, k–p and r–s, are shown in adjacent horizontal and
front views. Create a view that will show the true angle of
the pipe segments.
PROBLEM 9
© Cengage Learning 2012
PROBLEM 7
Piercing point. A vein of ore has been determined by plane d–h–g. A mine tunnel, r–s, is being dug toward this vein. Create a view that will show the true distance
of how much further the tunnel will need to be extended
from point R to reach the vein. Calculate the distance to the
nearest foot. (Drawing scale: Four squares 5 200')
© Cengage Learning 2012
PROBLEM 10
True distance between parallel lines. Two
parallel line (cable) segments m–n and r–s are shown in
adjacent horizontal and front views. Create a view that will
show the true distance between the cable segments.
09574_DGII_p1-23.indd 20
© Cengage Learning 2012
© Cengage Learning 2012
PROBLEM 8
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DESCRIPTIVE GEOMETRY II
Visibility. The adjacent front and horizontal
projection planes of three cable segments a–b, m–n, and
r–s are shown. Determine which cable segments are in
front of which in each projection plane. Then, correctly
complete the cable segment drawings by adding the
missing object lines in each projection plane. (Drawing
scale: One square 5 1/4")
PROBLEM 11
mh
21
HEIGHT ABOVE GROUND OF:
RIDGE 5 24'-0"
EAVES 5 17'-0"
45'-0"
4'-0"
10'-0"
5'-0"
bh
20'-0"
30'-0"
sh
ah
af
nh
mf
nf
rf
25'-0"
sf
bf
© Cengage Learning 2012
R
F
© Cengage Learning 2012
rh
3'-0"
True shapes and angles between planes. A
designer’s sketch of a roof plan of a residential house with
specifications for size of the roof and chimney flashing
detail is given. It is necessary to determine some information about the roof area and the amount of roof flashing
needed to keep the building job on schedule. Use the
sketch and detail information to develop the necessary
projection planes that will supply the information needed.
You need to:
Dihedral angle. Given are two adjacent
front and horizontal views of a laundry chute. Find the
true angle between side A and side B. (Drawing scale: One
square 5 1/4")
PROBLEM 13
a. Create the views that will show how many squares
of roofing will be needed to cover the entire roof
area. (100 ft.2 of roof surface 5 1 square.) Calculate
to the nearest square how much roofing will be
needed to cover the roof, with any extra to be left
for repairs.
b. Calculate how much 18" wide flashing will be
needed to flash all the ridges, valleys, and hips to
the nearest foot.
c. Create the views that will show all the pitches of
the roof. Calculate each of the pitches in the correct
pitch ratio.
d. Create the view that will show the ridge angle needed to bend the chimney flashing correctly. Calculate
the ridge angle to the nearest degree.
09574_DGII_p1-23.indd 21
© Cengage Learning 2012
PROBLEM 12
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Descriptive Geometry II
True shapes and dihedral angle. This sketch
of a sand collection hopper was given to you by an engineer. He asked you to do the following:
a. Create a view that will show the true shape of plane
A including the hinged door.
b. Create a view that will show how far the top of the
door, which is hinged at the bottom, will be from
the bottom of the hopper when the door is hanging
down in the open position. Show the measurement
of this distance in feet and inches.
c. Create a view that will show the true angle between
plane A and plane B. Show the measurement of this
angle to the nearest degree.
PROBLEM 14
6'-0"
Vectors. Draw the correct vector diagram
to determine the resultant force of the concurrent coplanar forces below. Measure the resultant force to the nearest 10 lb.
PROBLEM 15
4000 lb.
2000 lb.
5000 lb.
B
1'-0"
h
h
bh
dh
ah
ch
© Cengage Learning 2012
22
3000 lb.
2'-0"
H
Vectors. Given two adjacent front and horizontal views showing three concurrent noncoplanar forces,
find the resultant force of vectors a–b, a–c, and a–d to the
nearest pound.
PROBLEM 16
F
8'-0"
4'-0"
xf
f
af
cf
DOOR
bh
df
8'-0"
ch
5l
b.
7'-6"
f
f
3'-0"
© Cengage Learning 2012
A
6'-0"
dh
6 lb.
.
4 lb
H
F
af
cf
df
bf
09574_DGII_p1-23.indd 22
© Cengage Learning 2012
bf
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DESCRIPTIVE GEOMETRY II
Vectors. Given the front view showing the
space diagram of three forces, find the tensions in members A–B and A–C acted upon by the given force.
23
Given the following information on sloping
lines, determine their slope, slope angle, and percent of
grade.
1. Line A–B: run 5 50 ft.; rise 5 34 ft.
2. Line C–D: run 5 100 ft.; rise 5 25 ft.
3. Line D–E: run 5 148 yards; rise 5 100 yards
4. Line F–G: run 5 65 in.; rise 5 0 in.
5. Line H–J: run 5 100 m; rise 5 75 m
PROBLEM 19
© Cengage Learning 2012
PROBLEM 17
500 lb.
Vectors. Given the front view showing a
space diagram of three forces in equilibrium, find angles A
and B to the nearest degree.
PROBLEM 18
lb.
09574_DGII_p1-23.indd 23
© Cengage Learning 2012
lb
.
lb.
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