BE Civil Engineering (IOE, TU) Engineering Drawing I (IOE, ME 401) Question Paper 2076 Nepal

Given data

• True length $TL = 70\,\text{mm}$
• End $A$: $15\,\text{mm}$ above HP, $20\,\text{mm}$ in front of VP
• True inclination to HP: $\theta = 30^{\circ}$
• True inclination to VP: $\phi = 45^{\circ}$

Step 1 — Lengths of front view and top view

The top view (plan) length is the horizontal projection of the true length, governed by the inclination to the HP:

$$TV = TL\cos\theta = 70\cos 30^{\circ} = 70 \times 0.8660 = 60.62\,\text{mm}$$

The front view (elevation) length is the vertical-plane projection, governed by the inclination to the VP:

$$FV = TL\cos\phi = 70\cos 45^{\circ} = 70 \times 0.7071 = 49.50\,\text{mm}$$

Step 2 — Vertical extents (used to verify the apparent angles)

The difference in heights of the two ends (projection onto a vertical line):

$$\Delta h = TL\sin\theta = 70\sin 30^{\circ} = 70 \times 0.5 = 35.00\,\text{mm}$$

The difference in distances from VP (projection onto a line perpendicular to VP):

$$\Delta d = TL\sin\phi = 70\sin 45^{\circ} = 70 \times 0.7071 = 49.50\,\text{mm}$$

Step 3 — Apparent angles

Apparent angle of the front view $\alpha$ (front view makes this angle with the XY line):

$$\tan\alpha = \frac{\Delta h}{FV} = \frac{35.00}{49.50} = 0.7071 \quad\Rightarrow\quad \alpha = 35.26^{\circ}$$

Apparent angle of the top view $\beta$ (top view makes this angle with the XY line):

$$\tan\beta = \frac{\Delta d}{TV} = \frac{49.50}{60.62} = 0.8165 \quad\Rightarrow\quad \beta = 39.23^{\circ}$$

(Check: a consistency relation for an oblique line is $\sin^2\theta + \sin^2\phi = \sin^2 30^\circ + \sin^2 45^\circ = 0.25 + 0.50 = 0.75 \le 1$, so the configuration is physically valid.)

Step 4 — Construction procedure

        b'                          (elevation)
         \                          a'b' = 49.50 mm at alpha=35.26 deg to XY
          \
  a'-------\----------  apparent elevation
   |        \
 ---+---------+---------------- XY line
   |          \
  a ---------- b   (plan) a-b = 60.62 mm at beta=39.23 deg to XY

1. Mark $a'$ at $15\,\text{mm}$ above XY and $a$ at $20\,\text{mm}$ below XY on the same projector.
2. Auxiliary (true-length) step for FV: From $a'$ draw a line $70\,\text{mm}$ long at $30^{\circ}$ to XY; project its end down to fix the horizontal travel; the rotated final elevation $a'b'$ has length $49.50\,\text{mm}$ at $\alpha = 35.26^{\circ}$.
3. Auxiliary step for TV: From $a$ draw a line $70\,\text{mm}$ long at $45^{\circ}$ to XY; the rotated final plan $ab$ has length $60.62\,\text{mm}$ at $\beta = 39.23^{\circ}$.
4. Both views share common projectors at $A$ and at $B$.

Step 5 — Traces

Horizontal Trace (HT): point where the line, produced, meets the HP — found where the produced front view $a'b'$ crosses XY, then projected down onto the produced top view.

Using the elevation, the FV rises $35\,\text{mm}$ over a horizontal run of $49.50\,\text{mm}$ from $a'$ (which is $15\,\text{mm}$ above XY). Extending downward from $a'$ to reach XY ($0$ height) needs a run of:

$$x_{HT} = \frac{15}{\tan\alpha} = \frac{15}{0.7071} = 21.21\,\text{mm}$$

measured back from $a'$ along the line. The HT lies on the XY line in the elevation and is then projected onto the produced plan; HT distance in front of VP follows the plan geometry.

Vertical Trace (VT): point where the line, produced, meets the VP — found where the produced top view $ab$ crosses XY, then projected up onto the produced front view. From $a$ ($20\,\text{mm}$ in front of VP), extending toward XY:

$$x_{VT} = \frac{20}{\tan\beta} = \frac{20}{0.8165} = 24.50\,\text{mm}$$

measured back from $a$ along the plan. The VT lies on XY in the plan and is projected up to the elevation.

Final Answers

• Front view length $FV = 49.50\,\text{mm}$
• Top view length $TV = 60.62\,\text{mm}$
• Apparent angle $\alpha = 35.26^{\circ}$, $\beta = 39.23^{\circ}$
• HT obtained by producing $a'b'$ to XY (about $21.2\,\text{mm}$ from $a'$); VT obtained by producing $ab$ to XY (about $24.5\,\text{mm}$ from $a$).

Given data

• Pentagonal prism, base edge $a = 25\,\text{mm}$, axis length (prism height) $= 60\,\text{mm}$
• Resting on a rectangular face on HP
• Axis inclined $40^{\circ}$ to VP

Key dimensions of the regular pentagon (base)

For a regular pentagon of side $a = 25\,\text{mm}$:

• Circumradius $R = \dfrac{a}{2\sin 36^{\circ}} = \dfrac{25}{2 \times 0.5878} = 21.27\,\text{mm}$
• Inradius (apothem) $r = \dfrac{a}{2\tan 36^{\circ}} = \dfrac{25}{2 \times 0.7265} = 17.20\,\text{mm}$
• Width across (point-to-opposite-side) $= R + r = 38.47\,\text{mm}$

Stage 1 — Simple position (axis perpendicular to VP, a face on HP)

Because a rectangular face rests on the HP and the axis is initially perpendicular to the VP, the front view shows the true pentagon and the top view shows a rectangle.

• Draw the pentagon in the FV with one edge (the resting face's edge) horizontal at the bottom, side $25\,\text{mm}$. Its overall height is $R + r = 38.47\,\text{mm}$.
• Project the top view: a rectangle of length $= 60\,\text{mm}$ (axis) and width $= 25\,\text{mm}$ (the resting face seen edge-on width corresponds to the pentagon's base edge along XY). The five corners of the pentagon give five horizontal lines in the plan.

  Stage 1 FV (true pentagon)      Stage 1 TV (rectangle)
         *                         +-----------------+
        / \                        |                 |
       *   *                       |   5 face lines  |  width 25
       |   |                       |                 |
       *---*                       +-----------------+
   base edge 25                       length 60

Stage 2 — Tilt the axis to $40^{\circ}$ with the VP

The inclination to the VP is controlled in the top view:

1. Redraw the top view rectangle so that the axis (its long centre line, $60\,\text{mm}$) makes $40^{\circ}$ with the XY line, keeping the resting face on HP (so the plan keeps the same shape, just rotated).
2. From each corner of the rotated plan, draw vertical projectors upward.
3. From the Stage-1 front view, draw horizontal projectors (heights are unchanged because the solid still rests on the HP — the tilt is about a vertical axis).
4. The intersections of corresponding projectors give the final front view, an inclined pentagonal prism shape.

  Stage 2 (final)
   FV: pentagon faces seen foreshortened, prism leaning
   ___________ XY ___________
   TV: rectangle rotated 40 deg to XY
        \
         \  axis at 40 deg
          \________

Why heights are preserved

Tilting the axis to the VP is a rotation about a vertical line. Vertical heights above the HP do not change, so all FV heights are carried straight across from Stage 1; only the horizontal (left-right) spread changes according to the rotated plan.

Final Answer

• Pentagon dimensions: $R = 21.27\,\text{mm}$, $r = 17.20\,\text{mm}$, height $= 38.47\,\text{mm}$.
• Stage 1: true pentagon in FV, $60\times25\,\text{mm}$ rectangle in TV.
• Stage 2: rotate the plan so the $60\,\text{mm}$ axis is at $40^{\circ}$ to XY, re-project upward keeping Stage-1 heights to obtain the final inclined views.

Interpreting the object

• Base block: $80 \times 50 \times 20\,\text{mm}$ (L × W × H)
• Top square prism: $30 \times 30\,\text{mm}$ section, $40\,\text{mm}$ tall, centred on the base.
• Through hole: $\varnothing 20\,\text{mm}$ vertical, full depth of the square prism ($40\,\text{mm}$).
• Total height $= 20 + 40 = 60\,\text{mm}$.

In first-angle projection the views are arranged: TOP view below the FRONT view, and the LEFT side view is placed on the right of the front view (object → plane → observer order; "view falls beyond the object").

Centring of the top prism

The $30\,\text{mm}$ square sits centred on the $80\times50$ top, so margins are:

• Along length: $(80-30)/2 = 25\,\text{mm}$ each side.
• Along width: $(50-30)/2 = 10\,\text{mm}$ each side.

Front View (looking along the width / Y-direction)

Width seen = $80\,\text{mm}$, height seen = $60\,\text{mm}$.

   <------------ 80 ------------>
   .          ______             .   ^
   .         |      |            .   |  40 (prism)
   .         | ...  |  hole dashed.   |  (Ø20 shown as 2 dashed lines, 20 wide centred)
   ._________|______|___________.   v
   |                            |   ^ 20 (base)
   |____________________________|   v
   <------------ 80 ------------>

• Base rectangle $80 \times 20$.
• Square prism rectangle $30 \times 40$ centred ($25\,\text{mm}$ from each end).
• Hole: two dashed vertical lines $20\,\text{mm}$ apart, centred, running the $40\,\text{mm}$ height of the prism (hidden in this view).

Top View (projected BELOW the front view in first angle)

Length seen = $80\,\text{mm}$, depth seen = $50\,\text{mm}$.

   <------------ 80 ------------>
    ____________________________   ^
   |        +----------+        |  |
   |        |   (O)    |        |  50   (O) = Ø20 hole, solid circle
   |        +----------+        |  |   square 30x30 centred
   |____________________________|  v

• Outer rectangle $80 \times 50$.
• Inner square $30 \times 30$ centred ($25$ from ends, $10$ from front/back edges).
• Circle $\varnothing 20$ at centre with centre lines (solid circle — the hole opening is visible from the top).

Left Side View (projected to the RIGHT of the front view in first angle)

Depth seen = $50\,\text{mm}$, height seen = $60\,\text{mm}$.

   <----- 50 ----->
      ___________      ^
     |   ....    |     |  40   (hole hidden -> dashed)
    _|___________|_    v
   |               |   ^ 20
   |_______________|   v

• Base rectangle $50 \times 20$.
• Prism rectangle $30 \times 40$, centred across the $50$ depth ($10\,\text{mm}$ each side).
• Hole shown as dashed lines (hidden), $20\,\text{mm}$ apart.

Dimensioning (overall)

Final Answer

• Three first-angle views drawn as above, total height $60\,\text{mm}$, top prism centred with $25\,\text{mm}$ and $10\,\text{mm}$ margins, $\varnothing20$ hole shown solid in top view and dashed (hidden) in front and side views.

Given data

• Cylinder: diameter $D = 50\,\text{mm}$ (radius $R = 25\,\text{mm}$), height $H = 70\,\text{mm}$
• Central square through-hole, side $20\,\text{mm}$, vertical axis

Step 1 — Isometric scale

In a true isometric projection, edges parallel to the isometric axes are foreshortened. The ratio is:

$$\text{Isometric length} = \text{True length} \times \frac{2}{\sqrt{3}}^{-1}\;=\;\text{True length}\times 0.8165$$

More precisely the isometric scale factor is

$$k = \sqrt{\frac{2}{3}} = 0.8165.$$

Thus an isometric drawing (using true lengths directly, scale $1$) is larger than a true isometric projection (using $k=0.8165$). The drawn figure shape is identical; only the size differs.

Step 2 — Foreshortened height (true isometric projection)

$$H_{iso} = 70 \times 0.8165 = 57.16\,\text{mm}$$

Similarly the diameter in true isometric projection $= 50 \times 0.8165 = 40.82\,\text{mm}$ and the square side $= 20 \times 0.8165 = 16.33\,\text{mm}$.

(If an isometric drawing is asked, use the true values $70$, $50$, $20\,\text{mm}$ directly.)

Step 3 — Constructing the circle (cylinder) in isometric — four-centre method

A circle of diameter $50\,\text{mm}$ appears as an ellipse inscribed in the isometric rhombus of the bounding $50\times50$ square.

1. Draw the isometric rhombus of the top face: sides $50\,\text{mm}$ (drawing) at $30^{\circ}$ to horizontal.
2. Mark mid-points of the four sides.
3. The two obtuse-angle corners are the centres of the larger arcs; lines from these corners to the opposite mid-points intersect to give the centres of the smaller arcs.
4. Draw the four arcs to complete the ellipse (the isometric circle).

Step 4 — Build the solid

        ____
      /      \        top ellipse (Ø50)
     |  []    |       square hole (20) as a rhombus on top face
     |        |
      \      /        verticals 70 mm (or 57.16 mm in true iso)
     |        |
      \______/        bottom ellipse

1. Draw the top ellipse by the four-centre method.
2. Drop vertical edges of length $70\,\text{mm}$ (drawing) at the extreme left and right tangent points.
3. Draw the bottom ellipse (only the visible lower half/front portion).
4. Square hole: on the top face draw the central isometric square (rhombus) of side $20\,\text{mm}$. Drop vertical lines from its corners; since the hole is through, the lower opening is hidden — show visible inner walls of the hole as seen from above.

Final Answer

• Isometric scale factor $k = \sqrt{2/3} = 0.8165$.
• Foreshortened height $= 70 \times 0.8165 = 57.16\,\text{mm}$ (use $70\,\text{mm}$ if an isometric drawing is required).
• Top/bottom circles drawn as ellipses by the four-centre method; central square hole drawn as a rhombus with vertical walls.

Given data

• Cone: base diameter $60\,\text{mm}$ ($R = 30\,\text{mm}$), axis height $75\,\text{mm}$
• Section plane $\perp$ VP, inclined $45^{\circ}$ to HP, crossing the axis at $35\,\text{mm}$ height

Step 1 — Identify the conic

The semi-apex angle of the cone:

$$\tan\alpha = \frac{R}{H} = \frac{30}{75} = 0.4 \quad\Rightarrow\quad \alpha = 21.80^{\circ}.$$

The angle the generator makes with the base (HP) is

$$\gamma = 90^{\circ} - \alpha = 68.20^{\circ}.$$

The cutting plane is at $45^{\circ}$ to the HP. Since the cutting-plane angle ($45^{\circ}$) is less than the generator's inclination to the base ($68.20^{\circ}$), i.e. the plane is steeper than a generator is allowed for a parabola but cuts only one nappe and does not equal the generator angle, the plane crosses all generators on one side: the section is an ELLIPSE.

Rule recap: If plane angle $\theta$ to base satisfies $\theta < \gamma$ (the base angle of the generator) → ellipse; $\theta = \gamma$ → parabola; $\theta > \gamma$ → hyperbola. Here $45^{\circ} < 68.20^{\circ}$ → ellipse.

Step 2 — Locate the cutting line in the front view

In the FV the cone is a triangle, base $60\,\text{mm}$ on XY, apex $75\,\text{mm}$ up. The cutting line passes through the axis point at height $35\,\text{mm}$, inclined $45^{\circ}$.

The radius of the cone at height $y$ is

$$r(y) = R\left(1 - \frac{y}{H}\right) = 30\left(1-\frac{y}{75}\right).$$

The cutting line: $y = 35 + (x)\tan45^{\circ}$ where $x$ is horizontal distance from the axis... Let us find the two extreme cut points along the axis plane of symmetry.

Lower end (toward base): moving from the axis a horizontal distance $x_1$ where the line meets the left generator. Left generator (in FV) goes from $(-30,0)$ to $(0,75)$: $y = 75 + \frac{75}{30}x = 75 + 2.5x$ for $x<0$. Cutting line through $(0,35)$ at $45^{\circ}$ going down-left: $y = 35 + x$ (taking down-left as negative $x$, the $45^\circ$ line $y-35 = (x-0)$ with slope $+1$, so toward $x<0$, $y<35$). Set $35 + x = 75 + 2.5x \Rightarrow -40 = 1.5x \Rightarrow x = -26.67\,\text{mm}$, giving $y = 35 - 26.67 = 8.33\,\text{mm}$.

Upper end (toward apex): right generator from $(30,0)$ to $(0,75)$: $y = 75 - 2.5x$ for $x>0$. Cutting line $y = 35 + x$. Set $35 + x = 75 - 2.5x \Rightarrow 3.5x = 40 \Rightarrow x = 11.43\,\text{mm}$, $y = 46.43\,\text{mm}$.

So the cut line in the FV runs from $(-26.67,\,8.33)$ to $(11.43,\,46.43)$.

Length of the cut in the FV (the major-axis projection / apparent length):

$$L = \sqrt{(11.43-(-26.67))^2 + (46.43-8.33)^2} = \sqrt{38.10^2 + 38.10^2} = 38.10\sqrt2 = 53.88\,\text{mm}.$$

This $53.88\,\text{mm}$ is the true major axis of the ellipse (it lies in the VP-perpendicular plane, seen in true length in the FV because the plane is $\perp$ VP).

Step 3 — Section points by cutting-plane / horizontal sections

Divide the base circle (top view) into $12$ equal generators. Each generator in the FV is a line from apex to a base point. Where each generator meets the cut line gives a section point; project these to the top view and then to a true-shape auxiliary view.

The minor axis equals the chord of the cone at the mid-height of the cut. Mid-height $y_m = (8.33+46.43)/2 = 27.38\,\text{mm}$, where the cone radius is

$$r(27.38) = 30\left(1-\frac{27.38}{75}\right) = 30 \times 0.6349 = 19.05\,\text{mm}.$$

So the minor axis $\approx 2 \times 19.05 = 38.10\,\text{mm}$ (the full width of the cone at the cut's mid-height).

Step 4 — True shape

Project the cut points perpendicular to the cutting line (auxiliary view) to get the true shape — an ellipse with:

• Major axis $= 53.88\,\text{mm}$
• Minor axis $\approx 38.10\,\text{mm}$

  FV (triangle, cut at 45 deg)        True shape
        /\                              ___
       /  \   cut line                /     \
      /----\----                     |       |   ellipse
     /      \                         \_____/
    /________\                      major 53.88, minor 38.10

Final Answer

• The section is an ELLIPSE (plane at $45^{\circ}$ < base-generator angle $68.20^{\circ}$).
• Cut in FV from $(-26.67, 8.33)$ to $(11.43, 46.43)\,\text{mm}$, length $53.88\,\text{mm}$.
• True shape: ellipse, major axis $= 53.88\,\text{mm}$, minor axis $\approx 38.10\,\text{mm}$.

Lettering proportions (single-stroke vertical, BIS / IS 9609)

Two standard families are defined by the ratio of line thickness $d$ to height $h$:

General capital width is about $\tfrac{6}{10}h$ (most letters) with $h:w \approx 10:6$ for type B.

Computation for type B, $h = 14\,\text{mm}$

Stroke thickness:

$$d = \frac{h}{10} = \frac{14}{10} = 1.4\,\text{mm}$$

Spacing between two adjacent letters:

$$a = 2d = 2 \times 1.4 = 2.8\,\text{mm}$$

Minimum spacing between two words:

$$e = 6d = 6 \times 1.4 = 8.4\,\text{mm}$$

Final Answer

• Type B: $d = h/10$, letter spacing $= 2d$, word spacing $= 6d$.
• For $h = 14\,\text{mm}$: stroke $d = 1.4\,\text{mm}$, letter spacing $= 2.8\,\text{mm}$, word spacing $= 8.4\,\text{mm}$.

Given data

• Major axis $2a = 100\,\text{mm} \Rightarrow a = 50\,\text{mm}$
• Minor axis $2b = 60\,\text{mm} \Rightarrow b = 30\,\text{mm}$

Concentric-circles method — steps

1. Draw the major axis $AB = 100\,\text{mm}$ and minor axis $CD = 60\,\text{mm}$, intersecting at centre $O$.
2. With centre $O$, draw the major auxiliary circle of radius $a = 50\,\text{mm}$ and the minor auxiliary circle of radius $b = 30\,\text{mm}$.
3. Divide both circles into a number of equal parts using radial lines (e.g. every $30^{\circ}$, giving $12$ divisions).
4. From each point on the outer circle, draw a line parallel to the minor axis (vertical).
5. From the corresponding point on the inner circle, draw a line parallel to the major axis (horizontal).
6. Their intersection is a point on the ellipse. Repeat for all radial lines and join smoothly.

Coordinates at radial angle $\theta = 30^{\circ}$

The parametric equations are:

$$x = a\cos\theta, \qquad y = b\sin\theta.$$

At $\theta = 30^{\circ}$:

$$x = 50\cos 30^{\circ} = 50 \times 0.8660 = 43.30\,\text{mm}$$ $$y = 30\sin 30^{\circ} = 30 \times 0.5 = 15.00\,\text{mm}$$

Verification (point lies on the ellipse):

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{43.30^2}{50^2} + \frac{15^2}{30^2} = \frac{1874.9}{2500} + \frac{225}{900} = 0.75 + 0.25 = 1.00 \;\checkmark$$

        C
     ___|___        outer R=50, inner R=30
   /    |    \   . P(43.30, 15.00)
  A-----O-----B
   \    |    /
     ---|---
        D

Final Answer

• Steps of concentric-circles method as listed above.
• Point at $30^{\circ}$: $(x, y) = (43.30,\ 15.00)\,\text{mm}$, verified on the ellipse.

Recall — the four quadrants

Convention: front view (elevation) distance above/below XY = distance above/below HP; top view (plan) distance below/above XY = distance in front of / behind VP. In the standard layout, "above HP" → $p'$ above XY, "in front of VP" → $p$ below XY.

Point P — $25\,\text{mm}$ above HP, $40\,\text{mm}$ behind VP

Above HP and behind VP → Second quadrant (II).

• Front view $p'$: $25\,\text{mm}$ above the XY line.
• Top view $p$: behind VP, so in the rotated layout the plan also goes above XY, at $40\,\text{mm}$ above XY.
• In the second quadrant both views lie above the XY line (a characteristic check).

Point Q — $30\,\text{mm}$ below HP, $20\,\text{mm}$ in front of VP

Below HP and in front of VP → Fourth quadrant (IV).

• Front view $q'$: $30\,\text{mm}$ below the XY line.
• Top view $q$: in front of VP → plan lies below XY, at $20\,\text{mm}$ below XY.
• In the fourth quadrant both views lie below the XY line.

            p (40 above)
            p'(25 above)
  __________|__________ XY
            q'(30 below)
            q (20 below)

Final Answer

• P is in Quadrant II: $p'$ is $25\,\text{mm}$ above XY; $p$ is $40\,\text{mm}$ above XY (both above).
• Q is in Quadrant IV: $q'$ is $30\,\text{mm}$ below XY; $q$ is $20\,\text{mm}$ below XY (both below).

Given data

• Regular hexagon, side $a = 30\,\text{mm}$, resting on an edge on HP
• Surface inclined $50^{\circ}$ to HP

Key dimensions of the hexagon

• Width across flats (edge to opposite edge) $= a\sqrt3 = 30 \times 1.7321 = 51.96\,\text{mm}$
• Width across corners $= 2a = 60\,\text{mm}$

Resting on an edge, the dimension perpendicular to that edge (across flats) $= 51.96\,\text{mm}$.

Stage 1 — Surface parallel to HP

The lamina lies flat on the HP:

• Top view: true-shape hexagon, side $30\,\text{mm}$ (across flats $51.96\,\text{mm}$), one edge on XY.
• Front view: a straight horizontal line on XY (the lamina seen edge-on), length $= 60\,\text{mm}$ (across corners) along XY.

Stage 2 — Tilt the surface to $50^{\circ}$ with the HP

Inclination to the HP is set in the front view:

1. Redraw the Stage-1 front view (a line) tilted at $50^{\circ}$ to XY, keeping the resting edge on XY.
2. Project each corner up from the new (tilted) front view and across from the Stage-1 top view.
3. Intersections give the new top view — a foreshortened hexagon.

Foreshortened width of the top view

The dimension perpendicular to the hinge edge (across flats, $51.96\,\text{mm}$) is foreshortened in the plan by $\cos 50^{\circ}$:

$$w_{TV} = 51.96 \times \cos 50^{\circ} = 51.96 \times 0.6428 = 33.40\,\text{mm}$$

The dimension along the hinge edge (the $60\,\text{mm}$ across-corners direction parallel to XY) is unchanged at $60\,\text{mm}$ in the plan.

 Stage 1 TV (true hexagon)   Stage 2 FV (line at 50 deg)   Stage 2 TV (foreshortened)
      ____                          /                          ____
    /      \                       /  50 deg                  /     \   width 33.40
    \______/                  ____/____ XY                    \_____/
  across flats 51.96                                       along XY still 60

Final Answer

• Stage 1: true hexagon plan + edge-line elevation. Stage 2: tilt elevation to $50^{\circ}$, re-project to get foreshortened plan.
• Apparent top-view width (across flats) $= 51.96\cos 50^{\circ} = 33.40\,\text{mm}$; the along-XY dimension stays $60\,\text{mm}$.

Sectioning conventions

1. Cutting-plane line

A long chain line (long dash – short dash) thick at the ends and at changes of direction, thin in between, with arrows at the ends showing the direction of viewing. Labelled with capital letters, e.g. $A\text{–}A$.

2. Section (hatching) lines

• Drawn as thin continuous lines over the cut material.
• Standard inclination is $45^{\circ}$ to the principal outline / main axis of the part.
• They are uniformly spaced; typical spacing $1\,\text{mm}$ to $3\,\text{mm}$ depending on the size of the sectioned area (the larger the area, the wider the spacing).

3. Adjacent parts in an assembly

• When two adjacent parts are sectioned, the hatching of one is drawn at $45^{\circ}$ one way and the adjacent part at $45^{\circ}$ the opposite way (or at a different angle such as $30^{\circ}/60^{\circ}$), or with different spacing, so the boundary between parts is clear.
• The same part keeps the same direction and spacing of hatching in all its sectioned areas across all views.

4. Thin sections

Very thin sections (gaskets, sheet metal) may be filled solid (blacked in).

Parts NOT sectioned (even if the plane passes through them)

Conventionally shown un-hatched because sectioning gives no useful information and may mislead:

• Shafts, bolts, nuts, screws, studs, rivets (solid fasteners along their axis)
• Keys, cotters, pins
• Ribs and webs (when the plane passes longitudinally through them)
• Spokes of wheels/pulleys, gear teeth, balls and rollers of bearings

Final Answer (two required examples)

• Cutting-plane line: chain line thick at ends, arrows = viewing direction, lettered $A\text{–}A$.
• Hatching: thin lines at $45^{\circ}$, uniform $1\text{–}3\,\text{mm}$ spacing; adjacent parts hatched in opposite directions/different spacing; same part identical in all views.
• Two parts NOT sectioned: a bolt (or shaft) and a rib/web (also keys, nuts, rivets).

Part (a) — Regular polygon in a circle

General method (general/central-angle method)

1. Draw the circle of given radius, mark the centre $O$.
2. The chord (side) subtends a central angle $= \dfrac{360^{\circ}}{n}$ at $O$.
3. Starting from any point on the circle, step off this central angle $n$ times with radial lines; the intersections with the circle are the polygon's vertices.
4. Join consecutive vertices.

Heptagon ($n = 7$)

Central (subtended) angle:

$$\frac{360^{\circ}}{7} = 51.43^{\circ}$$

Interior angle:

$$\frac{(n-2)\times 180^{\circ}}{n} = \frac{5 \times 180^{\circ}}{7} = \frac{900^{\circ}}{7} = 128.57^{\circ}$$

Check: interior + exterior $= 128.57^{\circ} + 51.43^{\circ} = 180^{\circ}\ \checkmark$ (exterior angle = central angle = $51.43^{\circ}$).

Part (b) — Plain scale, RF $= 1:50$, up to $6\,\text{m}$

Length of the scale (length of the scale line on paper)

$$L = \text{RF} \times \text{max length} = \frac{1}{50} \times 6\,\text{m} = \frac{1}{50} \times 6000\,\text{mm} = 120\,\text{mm} = 12\,\text{cm}$$

Main divisions

Measure up to $6\,\text{m}$ → divide the $120\,\text{mm}$ line into 6 equal main divisions, each representing $1\,\text{m}$:

$$\text{one main division} = \frac{120\,\text{mm}}{6} = 20\,\text{mm} \;\;(=1\,\text{m on the object}).$$

Sub-divisions (decimetres)

Divide the first main division into 10 equal parts, each representing $1\,\text{dm} = 0.1\,\text{m}$:

$$\text{one sub-division} = \frac{20\,\text{mm}}{10} = 2\,\text{mm}\;\;(=1\,\text{dm}).$$

Mark $0$ at the start of the second main division; metres to the right, decimetres to the left.

  dm |        metres ->
  10 0   1   2   3   4   5
  |||||___|___|___|___|___|
   <-20-> each main div = 1 m = 20 mm
  total length 120 mm, range 0..6 m

Final Answer

• Heptagon: central angle $= 51.43^{\circ}$, interior angle $= 128.57^{\circ}$.
• Plain scale length $= 120\,\text{mm}$ (12 cm); 6 main divisions of $20\,\text{mm} = 1\,\text{m}$ each; first division split into 10 parts of $2\,\text{mm} = 1\,\text{dm}$ each.