Engineering Graphics and Computer Aided Drawing. Module 1, introduction, projection of points and lines.

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Engineering drawing is how engineers describe a solid object on flat paper. The trick is to draw it as seen from two directions at once: from the front (the front view) and from above (the top view). Together the two views fix where every point of the object is.

This module builds that skill from the ground up:

1. Drawing basics: the lines and dimensions every drawing uses.
2. Planes, quadrants and points: the two planes the views are drawn on, and how to draw a point.
3. Lines: how a straight line shows up in the two views, and how to find its true length and its angles to the planes.
4. Traces: where a line pierces the two planes.
5. Problems, from easy to hard. Each one lists what is given, what to find, the steps, and the finished drawing.

How to use the figures

Every line problem has a figure under its steps.

• Drawing (2D) builds the construction in four steps. Press the step buttons to see what each step adds.
• In space (3D) shows the same line between the two planes. Drag to turn it. Move the slider to fold the horizontal plane down into the vertical plane: the picture in space becomes the flat drawing you make on paper.

Try it on this line before you read on.

Drawing basics: lines and dimensions

Syllabus note. The BIS code of practice for technical drawing is not asked in the end-semester examination. Use this page to draw neatly.

Types of lines

Draw construction lines light so the final lines stand out. Keep every construction line: it shows how you got the answer.

Dimensions

• A dimension line is thin, with an arrowhead at each end, and is drawn outside the shape.
• Extension lines are thin lines that run from the shape to the dimension line. They stop just short of the shape and go slightly past the dimension line.
• Write the number above the dimension line, in the middle, so it reads from the bottom or the right.
• Give each dimension once, in millimetres, without writing "mm" on every number. A note on the sheet says "all dimensions in mm".

Planes, quadrants and points

The two planes

Two flat planes meet at right angles:

• the Horizontal Plane (HP), lying flat like a floor,
• the Vertical Plane (VP), standing upright like a wall.

They meet in a line called the XY line. The two planes cut space into four quadrants.

This course uses first-angle projection: the object is placed in the first quadrant, between you and the planes.

From space to paper

You look at the object from the front. What you see is projected on the VP: that is the front view. You look from above. What you see is projected on the HP: that is the top view.

To draw both on one sheet, the HP is turned down through 90° until it lies flat in the VP. After the turn:

• the front view is drawn above XY,
• the top view is drawn below XY (for a point in front of the VP),
• the front view and top view of one point lie on one straight line at right angles to XY, called the projector.

Letters: write the front view with a dash (a′, b′) and the top view without (a, b).

Projecting a point

Two distances place a point:

Above the HP or in front of the VP: the view goes to the usual side of XY (front view up, top view down). Below the HP or behind the VP: the view goes to the other side.

Example: four points

Draw the views of A (20 above HP, 30 in front of VP), B (25 above HP, 35 behind VP), C (30 below HP, 25 behind VP) and D (15 below HP, 40 in front of VP).

1. Draw XY and four projectors across it, one for each point.
2. A, first quadrant: mark a′ 20 above XY and a 30 below XY.
3. B, second quadrant: mark b′ 25 above XY and b 35 above XY.
4. C, third quadrant: mark c′ 30 below XY and c 25 above XY.
5. D, fourth quadrant: mark d′ 15 below XY and d 40 below XY.

A point in the HP has its front view on XY. A point in the VP has its top view on XY.

Lines: true length and inclinations

A straight line AB is fixed by its two end points, so its views are the lines a′b′ and ab joining the views of A and B.

Positions of a line

Rule. A line parallel to a plane shows its true length in its view on that plane.

Angles and lengths

Four angles are used. Learn them first.

For a line of true length TL:

• Top view length: ab = TL cos θ
• Front view length: a′b′ = TL cos φ
• Height difference of the ends: TL sin θ
• Distance difference of the ends from the VP: TL sin φ
• θ + φ is never more than 90°.
• A view is never longer than TL, and the apparent angles α and β are never smaller than θ and φ.

Finding the true length

When a line is inclined to both planes, neither view shows TL. Turn the line until it is parallel to one plane:

1. Keep end A fixed. Turn the top view ab about a until it is parallel to XY. Its length does not change.
2. Project the new end up to the height of b′. The line from a′ to that point is the true length, and the angle it makes with XY is θ.
3. In the same way turn the front view a′b′ about a′ until it is parallel to XY and project down. The new line in the top view is the true length, and its angle with XY is φ.

The figure under every problem draws exactly this.

Traces of a line

A line that is inclined to a plane, if extended, meets that plane in a point called a trace.

• HT (horizontal trace): where the line meets the HP. It lies on the top view and its front view is on XY.
• VT (vertical trace): where the line meets the VP. It lies on the front view and its top view is on XY.

Finding the traces

1. HT: extend the front view a′b′ until it meets XY at h′. Draw a projector down from h′ to meet the extended top view ab. That point is HT.
2. VT: extend the top view ab until it meets XY at v. Draw a projector up from v to meet the extended front view a′b′. That point is VT.

HT is in front of the VP when it falls below XY, and behind it when it falls above XY. VT is above the HP when it falls above XY, and below it when it falls below XY.

Remember. A line parallel to the HP has no HT. A line parallel to the VP has no VT.

Reading a problem

A line is fixed by five facts. A line problem always gives five, in some mix of these:

How to start:

1. Write down the five facts. If a word like "assume" appears, the statement is giving you one.
2. Draw XY and the views of the end you know.
3. Use any fact that fixes a height or a distance as a horizontal locus line parallel to XY. The other end must lie on it.
4. Use true length and true angles by turning the line parallel to a plane, as on the previous pages.

The problems below start with the simplest data and end with applications.

Line Rotation Method: True Length and True Inclinations or Loci

Now that you understand how rotating a line parallel to a reference plane reveals its true length and angle, we will apply this forward and backward to draw projections across different quadrants. By drawing the true-length line at its true inclination to establish the endpoint's horizontal locus, you can project and swing arcs of the foreshortened lengths to lock in the final front and top views.

The idea

A line inclined to both the Horizontal Plane (HP) and the Vertical Plane (VP) looks shorter than its true length ($TL$) in both orthographic views. In the Line Rotation Method, we imagine pivoting the line around one endpoint until it becomes parallel to one of the reference planes while keeping its inclination to the other plane unchanged.

When we rotate a line $AB$ about end $A$ until it is parallel to the VP, its inclination $\theta$ to the HP stays constant. Therefore, its new front view $a'b_1'$ shows the true length $TL$ at the true angle $\theta$, and the height of $b_1'$ defines the horizontal locus of $b'$ (the path of constant height for end $B$). Simultaneously, its new top view $ab_1$ lies parallel to the $XY$ line and reveals the exact apparent length of the top view ($ab = ab_1 = TL \cos\theta$). Symmetrically, rotating the line parallel to the HP keeps its angle $\phi$ to the VP constant: the top view $ab_2$ shows $TL$ at angle $\phi$, establishes the horizontal locus of $b$, and projects to $a'b_2'$ to give the apparent front view length ($a'b' = a'b_2' = TL \cos\phi$). Swinging $ab_1$ and $a'b_2'$ onto their respective loci completes the views.

Finally, the points where the extended line (if necessary) pierces the HP and VP are the Horizontal Trace ($HT$) and Vertical Trace ($VT$). To locate $HT$, extend the front view $a'b'$ to meet the $XY$ line at $h'$, erect a perpendicular to $XY$, and intersect it with the extended top view $ab$. To locate $VT$, extend the top view $ab$ to meet the $XY$ line at $v$, erect a perpendicular to $XY$, and intersect it with the extended front view $a'b'$.

Worked example

Given: Line $AB$ is 90 mm long, inclined at $\theta = 30^\circ$ to the HP and $\phi = 40^\circ$ to the VP, with end $A$ 20 mm above the HP and 30 mm in front of the VP (first quadrant). Find: The front view $a'b'$, top view $ab$, and the traces $HT$ and $VT$.

1. Draw the reference line $XY$ and mark $a'$ 20 mm above $XY$ and $a$ 30 mm below $XY$ on a common vertical projector, because $A$ is in the first quadrant.
2. From $a'$, draw $a'b_1'$ of length 90 mm at $30^\circ$ above the horizontal to represent the line rotated parallel to the VP, and draw a horizontal line through $b_1'$ (65 mm above $XY$) as the locus of $b'$.
3. Project $b_1'$ vertically down to meet the horizontal line through $a$ at $b_1$, which gives the apparent top view length $ab_1 = 77.9$ mm.
4. From $a$, draw $ab_2$ of length 90 mm at $40^\circ$ below the horizontal to represent the line rotated parallel to the HP, and draw a horizontal line through $b_2$ (87.9 mm below $XY$) as the locus of $b$.
5. Project $b_2$ vertically up to meet the horizontal line through $a'$ at $b_2'$, which gives the apparent front view length $a'b_2' = 68.9$ mm.
6. With center $a$ and radius $ab_1$, swing an arc down to the locus of $b$ to find $b$, and join $ab$; with center $a'$ and radius $a'b_2'$, swing an arc up to the locus of $b'$ to find $b'$, and join $a'b'$ (check that $b'$ and $b$ lie on the same vertical projector).
7. Extend $a'b'$ to meet $XY$ at $h'$ and project vertically to meet extended $ab$ at $HT$; extend $ab$ to meet $XY$ at $v$ and project vertically to meet extended $a'b'$ at $VT$.

Answer: Front view $a'b' = 68.9$ mm, top view $ab = 77.9$ mm, $HT$ is 4.3 mm from the $XY$ line, and $VT$ is 3.3 mm from the $XY$ line.

More problems

Problem 2: Finding inclinations and views from endpoint loci

Given: $AB = 75$ mm; $A$ is 25 mm above the HP and 20 mm in front of the VP; $B$ is 50 mm above the HP and 65 mm in front of the VP. Find: Front view $a'b'$, top view $ab$, inclinations $\theta$ and $\phi$, and traces $HT$ and $VT$.

1. Mark $a'$ 25 mm above $XY$ and $a$ 20 mm below $XY$ on a vertical projector.
2. Draw the horizontal locus of $b'$ 50 mm above $XY$ and the locus of $b$ 65 mm below $XY$.
3. Swing an arc of 75 mm from $a'$ to cut the locus of $b'$ at $b_1'$, measuring $\theta = 19.5^\circ$.
4. Project $b_1'$ down to the horizontal line through $a$ to get $b_1$, then swing $ab_1$ to the locus of $b$ for $ab = 70.7$ mm.
5. Swing an arc of 75 mm from $a$ to cut the locus of $b$ at $b_2$, measuring $\phi = 36.9^\circ$.
6. Project $b_2$ up to the horizontal through $a'$ to get $b_2'$, then swing $a'b_2'$ to the locus of $b'$ for $a'b' = 60$ mm.
7. Extend $a'b'$ and $ab$ to the $XY$ line at $h'$ and $v$, and project vertically to locate $HT$ and $VT$.

Answer: $a'b' = 60$ mm (printed: 47 mm), $ab = 70.7$ mm (printed: 60 mm), $\theta = 19.5^\circ$ (printed: $23^\circ$), $\phi = 36.9^\circ$ (printed: $44^\circ$), $HT$ is 25 mm from $XY$ (printed: 24 mm), and $VT$ is 13.9 mm from $XY$ (printed: 14 mm).

Watch out. The printed values on the drawing ($a'b' = 47$ mm, $ab = 60$ mm, $\theta = 23^\circ$, $\phi = 44^\circ$) do not match the given coordinates; a 75 mm line rising 25 mm and advancing 45 mm yields $a'b' = 60$ mm, $ab = 70.7$ mm, $\theta = 19.5^\circ$, and $\phi = 36.9^\circ$.

Problem 5: Line in the third quadrant

Given: $AB = 80$ mm, $\theta = 30^\circ$, $\phi = 45^\circ$; $A$ is 20 mm below the HP and 25 mm behind the VP (third quadrant). Find: Front view $a'b'$, top view $ab$, and traces $HT$ and $VT$.

1. Mark $a$ 25 mm above $XY$ and $a'$ 20 mm below $XY$ because $A$ is in the third quadrant.
2. From $a$, draw $ab_2 = 80$ mm at $45^\circ$ above the horizontal to set the locus of $b$ 81.6 mm above $XY$.
3. Project $b_2$ down to the horizontal through $a'$ at $b_2'$, finding the front view length $a'b_2' = 56.6$ mm.
4. From $a'$, draw $a'b_1' = 80$ mm at $30^\circ$ below the horizontal to set the locus of $b'$ 60 mm below $XY$.
5. Project $b_1'$ up to the horizontal through $a$ at $b_1$, finding the top view length $ab_1 = 69.3$ mm.
6. Swing $a'b_2'$ down to the locus of $b'$ for $b'$, and swing $ab_1$ up to the locus of $b$ for $b$.
7. Extend $a'b'$ and $ab$ to meet $XY$ at $h'$ and $v$, and project vertically to intersect the views at $HT$ and $VT$.

Answer: $a'b' = 56.6$ mm, $ab = 69.3$ mm, $HT$ is 3.3 mm from $XY$, and $VT$ is 2.3 mm from $XY$.

Problem 4: Line ending in the second quadrant

Given: $PQ = 100$ mm, $\theta = 30^\circ$, $\phi = 20^\circ$; $P$ is on the HP and 20 mm in front of the VP; $Q$ is in the second quadrant. Find: Front view $p'q'$ and top view $pq$.

1. Mark $p'$ on $XY$ and $p$ 20 mm below $XY$ on a common vertical projector.
2. From $p'$, draw $p'q_1' = 100$ mm at $30^\circ$ above $XY$ to set the locus of $q'$ 50 mm above $XY$.
3. Project $q_1'$ down to the horizontal through $p$ at $q_1$ to get the top view length $pq_1 = 86.6$ mm.
4. From $p$, draw $pq_2 = 100$ mm at $20^\circ$ above the horizontal (going behind the VP) to set the locus of $q$ 14.2 mm above $XY$.
5. Project $q_2$ to $XY$ (the horizontal through $p'$) at $q_2'$ to get the front view length $p'q_2' = 94$ mm.
6. Swing $p'q_2'$ from $p'$ to the locus of $q'$ to locate $q'$, and swing $pq_1$ from $p$ to the locus of $q$ to locate $q$.
7. Join $p'q'$ and $pq$; $HT$ is at $p$ (20 mm from $XY$) and $VT$ is 29.2 mm from $XY$.

Answer: $p'q' = 94$ mm and $pq = 86.6$ mm (printed: 86 mm).

Problem 6: Line crossing into the second quadrant

Given: $AB = 100$ mm, $\theta = 30^\circ$, $\phi = 45^\circ$; $A$ is 20 mm above the HP and 25 mm in front of the VP; $B$ is in the second quadrant. Find: Front view $a'b'$, top view $ab$, and traces $HT$ and $VT$.

1. Mark $a'$ 20 mm above $XY$ and $a$ 25 mm below $XY$ on a vertical projector.
2. From $a'$, draw $a'b_1' = 100$ mm at $30^\circ$ above the horizontal to set the locus of $b'$ 70 mm above $XY$.
3. Project $b_1'$ down to the horizontal through $a$ at $b_1$ to get the top view length $ab_1 = 86.6$ mm.
4. From $a$, draw $ab_2 = 100$ mm at $45^\circ$ above the horizontal (crossing $XY$ into the second quadrant) to set the locus of $b$ 45.7 mm above $XY$.
5. Project $b_2$ down to the horizontal through $a'$ at $b_2'$ to get the front view length $a'b_2' = 70.7$ mm.
6. Swing $a'b_2'$ to the locus of $b'$ for $b'$, and swing $ab_1$ upwards across $XY$ to the locus of $b$ for $b$.
7. Extend $a'b'$ and $ab$ to meet $XY$ at $h'$ and $v$, and project vertically to locate $HT$ and $VT$.

Answer: $a'b' = 70.7$ mm (printed: 70 mm), $ab = 86.6$ mm (printed: 86 mm), $HT$ is 53.3 mm from $XY$ (printed: 53 mm), and $VT$ is 37.7 mm from $XY$ (printed: 38 mm).

Problem 3: Line spanning first and second quadrants from loci

Given: $AB = 100$ mm; $A$ is 25 mm above the HP and 25 mm in front of the VP; $B$ is 50 mm above the HP and 50 mm behind the VP. Find: Front view $a'b'$, top view $ab$, inclinations $\theta$ and $\phi$, and traces $HT$ and $VT$.

1. Mark $a'$ 25 mm above $XY$ and $a$ 25 mm below $XY$ on a vertical projector.
2. Draw a single horizontal line 50 mm above $XY$ to serve as the locus for both $b'$ and $b$ (since $B$ is in the second quadrant).
3. Swing an arc of 100 mm from $a'$ to cut the 50 mm locus at $b_1'$, measuring $\theta = 14.5^\circ$.
4. Project $b_1'$ down to the horizontal through $a$ at $b_1$ ($ab_1 = 96.8$ mm), and swing $ab_1$ up to the 50 mm locus to find $b$.
5. Swing an arc of 100 mm from $a$ up across $XY$ to cut the 50 mm locus at $b_2$, measuring $\phi = 48.6^\circ$.
6. Project $b_2$ down to the horizontal through $a'$ at $b_2'$ ($a'b_2' = 66.1$ mm), and swing $a'b_2'$ up to the 50 mm locus to find $b'$ (coinciding with $b$).
7. Extend $a'b'$ and $ab$ to meet $XY$ at $h'$ and $v$, and project vertically to find $HT$ and $VT$.

Answer: $a'b' = 66.1$ mm (printed: 66 mm), $ab = 96.8$ mm (printed: 97 mm), $\theta = 14.5^\circ$ (printed: $14^\circ$), $\phi = 48.6^\circ$ (printed: $49^\circ$), $HT$ is 100 mm from $XY$, and $VT$ is 33.3 mm from $XY$.

Watch out. Because $B$ is 50 mm above the HP and 50 mm behind the VP, the loci of $b'$ and $b$ coincide 50 mm above $XY$, so the endpoints $b'$ and $b$ land on the exact same spot.

Line Rotation Method: Working from Apparent Lengths and Angles

Having used true lengths and inclinations to find apparent views, you can now run the exact same rotation arcs in reverse when the problem gives you the apparent view lengths or angles first. Any known apparent view (front or top) can be rotated until it is parallel to the XY line and then projected to the other view's locus to recover the missing true length and true inclination.

The idea

When a line is inclined to both reference planes, its front view $a'b'$ and top view $ab$ are foreshortened and make apparent inclinations with the XY line: $\alpha$ in the front view and $\beta$ in the top view. Because the line leans away from the plane, an apparent angle is always larger than the corresponding true inclination ($\alpha > \theta$ and $\beta > \phi$).

In the forward method, you started with a true-length line $a'b_1'$, dropped a vertical projector to a horizontal line through $a$ to get the foreshortened length $ab_1$, and swung an arc of radius $ab_1$ to find the top view $ab$. Reverse line rotation runs that exact path backward. Whenever you know the length of a view—say, the front view $a'b'$—you can lay off that length horizontally from $a'$ to get $a'b_2'$, then project straight down to the locus of $b$ (or intersect with a true-length arc from $a$) to recover the true-length position $ab_2$ and the true angle $\phi$.

Similarly, if a problem gives an apparent angle ($\alpha$ or $\beta$), draw that angle ray first and intersect it with either the endpoint's horizontal locus or a rotation arc to lock in the endpoint. Finally, a complete projection requires five parameters, including the starting point's coordinates. If a problem gives only three geometric parameters and omits the endpoint location, make the standard simplifying assumption: place endpoint A on both the HP and the VP so that $a'$ and $a$ sit directly on the XY line.

Worked example

Given: Line AB is 80 mm long, inclined at $\theta = 30^\circ$ to the HP, with front view $a'b' = 60$ mm; end A is 15 mm above the HP and 20 mm in front of the VP. Find: The top view $ab$ and the inclination $\phi$ with the VP.

1. Draw the XY line and mark $a'$ 15 mm above XY and $a$ 20 mm below XY on the same vertical projector.
2. From $a'$, draw the true-length line $a'b_1' = 80$ mm at $\theta = 30^\circ$ to the horizontal to establish the horizontal locus of $b'$ at 55 mm above XY.
3. Drop a vertical projector from $b_1'$ to the horizontal line through $a$ to find $b_1$; the segment $ab_1 = 69.3$ mm is the true top-view length because $a'b_1'$ is parallel to the VP.
4. With center $a'$ and radius equal to the given front view 60 mm, cut the locus of $b'$ to locate $b'$ and draw the front view $a'b'$.
5. Drop a vertical projector from $b'$ and intersect it with an arc of radius $ab_1 = 69.3$ mm centered at $a$ to locate $b$, completing the top view $ab$ and establishing the locus of $b$ at 72.9 mm below XY.
6. To find $\phi$ by reverse rotation, mark $b_2'$ on the horizontal line through $a'$ at $a'b_2' = 60$ mm, drop a projector to the locus of $b$ at $b_2$, and measure $\phi$ on $ab_2$ (or cut the locus of $b$ directly with an 80 mm arc from $a$).

Answer: Top view $ab = 69.3$ mm, true inclination with the VP $\phi = 41.4^\circ$.

More problems

Problem 191: True length, $\theta$, and plan angle $\beta$ given

Given: Line PQ is 80 mm long, $\theta = 30^\circ$, top-view angle $\beta = 60^\circ$; P is on the HP and 15 mm in front of the VP. Find: The projections of PQ and its inclination $\phi$ with the VP.

1. Mark $p'$ on the XY line and $p$ 15 mm below XY on a common vertical projector.
2. From $p'$, draw $p'q_1' = 80$ mm at $\theta = 30^\circ$ to XY and draw the horizontal locus of $q'$ through $q_1'$ at 40 mm above XY.
3. Project $q_1'$ down to the horizontal line through $p$ to obtain the top-view length $pq_1 = 69.3$ mm.
4. From $p$, draw a ray at $\beta = 60^\circ$ below the horizontal and swing an arc from $q_1$ centered at $p$ to cut this ray at $q$, locating the locus of $q$ 75 mm below XY.
5. Project $q$ vertically up to the locus of $q'$ to locate $q'$ and join $p'q'$ ($52.9$ mm).
6. With center $p$ and true length radius 80 mm, cut the horizontal locus of $q$ at $q_2$ and measure $\phi$ between $pq_2$ and XY.

Answer: Front view $p'q' = 52.9$ mm, top view $pq = 69.3$ mm, inclination with the VP $\phi = 48.6^\circ$.

Problem 9: Only three parameters given (assume A on XY)

Given: Line AB is 70 mm long, $\theta = 40^\circ$, and plan inclination $\beta = 30^\circ$. Find: The projections of AB and its inclination $\phi$ with the VP.

1. Because endpoint coordinates are omitted, assume end A lies on both the HP and the VP so that $a'$ and $a$ coincide on the XY line.
2. From $a'$, draw the true-length line $a'b_1' = 70$ mm at $\theta = 40^\circ$ above XY and draw the horizontal locus of $b'$ at 45 mm above XY.
3. Drop a vertical projector from $b_1'$ to the XY line at $b_1$ to get the top-view radius $ab_1 = 53.6$ mm.
4. From $a$, draw a ray at $\beta = 30^\circ$ below XY and swing an arc from $b_1$ centered at $a$ to intersect it at $b$, establishing the locus of $b$ at 26.8 mm below XY.
5. Project $b$ vertically upward to the locus of $b'$ to find $b'$ and draw the front view $a'b' = 64.7$ mm.
6. With center $a$ and radius 70 mm, cut the horizontal locus of $b$ at $b_2$ and measure the true angle $\phi$ between $ab_2$ and XY.

Answer: Front view $a'b' = 64.7$ mm, top view $ab = 53.6$ mm, inclination with the VP $\phi = 22.5^\circ$.

Problem 8: Both apparent inclinations $\alpha$ and $\beta$ given

Given: Front view angle $\alpha = 35^\circ$, top view angle $\beta = 45^\circ$; A is 20 mm above the HP and 25 mm in front of the VP; B is 40 mm above the HP. Find: The projections, true length, $\theta$, and $\phi$.

1. Mark $a'$ 20 mm above XY, $a$ 25 mm below XY, and draw the horizontal locus of $b'$ 40 mm above XY.
2. From $a'$, draw a ray at $\alpha = 35^\circ$ to the horizontal to cut the 40 mm locus at $b'$, giving the front view $a'b' = 34.9$ mm.
3. From $a$, draw a ray at $\beta = 45^\circ$ below the horizontal and drop a vertical projector from $b'$ to intersect it at $b$, giving the top view $ab = 40.4$ mm and the locus of $b$ at 53.6 mm below XY.
4. Rotate $ab$ centered at $a$ up to the horizontal through $a$ at $b_1$, project $b_1$ up to the locus of $b'$ at $b_1'$, and measure true length $a'b_1'$ and angle $\theta$.
5. Rotate $a'b'$ centered at $a'$ down to the horizontal through $a'$ at $b_2'$, project $b_2'$ down to the locus of $b$ at $b_2$, and measure angle $\phi$ on $ab_2$.

Answer: True length $AB = 45.1$ mm, front view $a'b' = 34.9$ mm, top view $ab = 40.4$ mm, $\theta = 26.3^\circ$, $\phi = 39.3^\circ$.

Problem 204: Top view, $\beta$, and locus of Q given

Given: Top view $pq = 70$ mm at $\beta = 45^\circ$ to XY; P is in the VP and 15 mm above the HP; Q is 30 mm above the HP in the first quadrant. Find: True length, elevation length $p'q'$, inclinations $\theta$ and $\phi$, and traces.

1. Mark $p'$ 15 mm above XY, $p$ on the XY line, and draw the horizontal locus of $q'$ 30 mm above XY.
2. From $p$, draw the top view $pq = 70$ mm at $\beta = 45^\circ$ below XY, establishing the locus of $q$ at 49.5 mm below XY.
3. Project $q$ vertically up to the 30 mm locus line to locate $q'$ and join $p'q'$ to get the elevation length.
4. Rotate $pq$ centered at $p$ up to the XY line at $q_1$, project $q_1$ up to the locus of $q'$ at $q_1'$, and measure true length $p'q_1'$ and $\theta$.
5. With center $p$ and radius equal to the true length $71.6$ mm, cut the locus of $q$ at $q_2$ and measure $\phi$ between $pq_2$ and XY.
6. Note that $p'$ is already the Vertical Trace (VT) 15 mm above XY since $p$ is on XY; extend $p'q'$ back to XY at $h'$ and project to the extended line $qp$ to locate the Horizontal Trace (HT) 49.5 mm from XY.

Figure: Worked solution step 204 for projection of line PQ

Orthographic projection diagram labelled 204 showing line PQ with reference line xy. Point p(v) lies on the xy line, and p'(VT) is 15 mm above the xy line on the same vertical projector. Locus of q' and q1' is a horizontal line 30 mm above xy. The top view pq has length 70 mm and is inclined at beta = 45 degrees below xy, ending at q (which coincides with q2). An arc goes from q(q2) up to q1 on the xy line (with upward arrow), and a vertical projector goes from q1 up to q1' on the 30 mm locus (with upward arrow), giving true length p'q1'. A vertical projector goes from q up to q' on the 30 mm locus (with upward arrow), giving front view p'q'. Extending p'q' backwards to the left meets the xy line at h'. Extending the top view pq backwards above the xy line meets the vertical projector through h' at HT above xy. Box 'GIVEN': [1] TV - 70mm, [2] beta - 45°, [3] P - in VP, [4] P - 15 aHP, [5] Q - 30 aHP. Box 'ANSWERS': [1] p'q' - 52mm, [2] TL - 72mm, [3] phi - 44°, [4] theta - 12°, [5] h'-HT - 49mm, [6] v-VT - 15mm.

Answer: True length = 71.6 mm, elevation $p'q' = 51.7$ mm, $\theta = 12.1^\circ$, $\phi = 43.7^\circ$, HT = 49.5 mm from XY, VT = 15 mm from XY.

Problem 193: Both view lengths and front-view angle $\alpha$ given

Given: Front view $p'q' = 50$ mm at $\alpha = 45^\circ$ to XY, top view $pq = 65$ mm; P is on the HP and 20 mm in front of the VP. Find: The true length, inclinations $\theta$ and $\phi$, and the traces.

1. Mark $p'$ on the XY line and $p$ 20 mm below XY on the same vertical projector.
2. From $p'$, draw the front view $p'q' = 50$ mm at $\alpha = 45^\circ$ above XY and draw the horizontal locus of $q'$ through $q'$ at 35.4 mm above XY.
3. Drop a vertical projector from $q'$ and cut it with an arc of radius 65 mm centered at $p$ to locate $q$, establishing the locus of $q$ at 74.5 mm below XY.
4. Mark $q_1$ on the horizontal line through $p$ at $pq_1 = 65$ mm, project $q_1$ up to the locus of $q'$ at $q_2'$, and measure true length $p'q_2'$ and $\theta$.
5. Project $q'$ down to the horizontal through $p'$ (on XY) or cut the locus of $q$ with the true length radius 74 mm from $p$ to measure $\phi$.
6. Point $p$ is the HT (20 mm from XY) since $p'$ is on XY; extend $qp$ to XY at $v$ and project down to the extended $q'p'$ to find the VT 13 mm from XY.

Figure: Projection of line PQ (solution step 193)

Orthographic projection of line PQ with reference line xy. Point p' is on the xy line. Point p is 20 mm below xy. Front view p'q' has length 50 mm and makes an angle alpha = 45 degrees with xy. Top view pq has length 65 mm. Point q is projected up to the xy line to get q1', and rotated around p' to the locus of q' to get q2', giving the true length in the front view p'q2' inclined at theta to xy. Point q' is projected down to the horizontal line through p, and rotated around p to the locus of q to get q1, giving the true length in the top view pq1 (labelled TL) inclined at phi to the horizontal. Box '193'. Box 'GIVEN': [1] P - on HP, [2] P - 20 fVP, [3] alpha - 45 deg, [4] FV - 50mm, [5] TV - 65mm. Box 'ANSWERS': [1] TL - 74mm, [2] theta - 29 deg, [3] phi - 48 deg.

Answer: True length = 74 mm, $\theta = 28.5^\circ$, $\phi = 47.5^\circ$, HT = 20 mm from XY, VT = 13 mm from XY.

Problem 10: True length and both apparent view lengths given

Given: Line AB has true length 75 mm, top view $ab = 65$ mm, and front view $a'b' = 50$ mm; end A is on both the HP and the VP. Find: The projections of AB and its inclinations $\theta$ and $\phi$.

1. Mark both $a'$ and $a$ at the same point on the XY line.
2. Along the XY line from $a$, mark $ab_1 = 65$ mm (the un-rotated top view) and erect a vertical projector upward from $b_1$.
3. With center $a'$ and true-length radius 75 mm, cut the vertical projector from $b_1$ at $b_1'$ to find $\theta$ and draw the horizontal locus of $b'$ at 37.4 mm above XY.
4. With center $a'$ and radius $a'b' = 50$ mm, cut the locus of $b'$ at $b'$ to complete the front view $a'b'$.
5. Drop a vertical projector from $b'$ below XY and intersect it with an arc of radius 65 mm centered at $a$ to locate $b$ and its locus at 55.9 mm below XY.
6. With center $a$ and true-length radius 75 mm, cut the horizontal locus of $b$ at $b_2$ and measure $\phi$ between $ab_2$ and XY.

Watch out. You cannot draw $a'b'$ or $ab$ directly at the start because neither locus is known; you must lay off $ab_1 = 65$ mm along the horizontal through $a$ and intersect its projector with the 75 mm true-length arc from $a'$ to find the locus of $b'$ first.