Orthographic Projection

(i) In orthographic projection, the projectors (lines of sight) are perpendicular to the plane of projection.

(ii) In I angle projection, the object comes between the observer and the plane of projection.

(iii) In III angle projection, the plane of projection comes between the observer and the object.

Given/Concept: In orthographic projection, two principal planes are used — the Vertical Plane (V.P.) and the Horizontal Plane (H.P.). When the H.P. is rotated 90° downward (clockwise) to lie in the same plane as the V.P., the line of intersection of V.P. and H.P. — called the reference line XY — appears on the drawing sheet.

Explanation:
- The reference line XY represents the line of intersection of V.P. and H.P.
- Any point above XY on the drawing represents a projection on the V.P. (Front View).
- Any point below XY on the drawing represents a projection on the H.P. (Top View).
- Thus, the single line XY simultaneously acts as the base line of V.P. and the top edge of H.P. after the H.P. is rotated open.
- The perpendicular distance of a view from XY gives the distance of the object from the respective plane.

Conclusion: The reference line XY represents both principal planes — distances above XY relate to V.P. projections and distances below XY relate to H.P. projections.

First Angle Projection Symbol:
- A frustum of a cone is drawn with the apex on the left and the base on the right.
- The left view (smaller circle) represents the view from the left side (apex end).
- The right view (larger truncated shape) represents the front view.
- This symbol indicates that the object is placed between the observer and the plane (First Angle / European method).

$$\text{Symbol: } \bigcirc \!\!\!\!\!\supset \quad \text{(Truncated cone — apex left, base right, with end views)}$$

Third Angle Projection Symbol:
- A frustum of a cone is drawn with the apex on the right and the base on the left.
- The left view (larger truncated shape) represents the front view.
- The right view (smaller circle) represents the view from the right side.
- This symbol indicates that the plane is placed between the observer and the object (Third Angle / American method).

$$\text{Symbol: } \subset\!\!\!\!\!\bigcirc \quad \text{(Truncated cone — base left, apex right, with end views)}$$

Note: These symbols are placed in the title block of the drawing sheet to indicate which projection method has been used.

Reason:

When an object is placed in the II quadrant (above H.P. and behind V.P.) or the IV quadrant (below H.P. and in front of V.P.), and the H.P. is rotated 90° downward to open the glass box:

- In the II quadrant: Both the Front View (on V.P.) and the Top View (on H.P.) fall on the same side (above XY) after rotation. This causes the two views to overlap, making the drawing confusing and unreadable.

- In the IV quadrant: Both the Front View and the Top View fall below XY after rotation, again causing overlapping of views.

Conclusion: Since the views overlap and cannot be clearly distinguished, the II and IV quadrants are not used in practice. Only the I quadrant (First Angle Projection) and the III quadrant (Third Angle Projection) are used, as they produce non-overlapping, clearly readable views.

Given:
- Top View is 40 mm above XY.
- Front View is 20 mm below the Top View.

Analysis using Table 4.2:
- Top View above XY → the point is behind V.P.
- Front View: 20 mm below the Top View means the Front View is at $40 - 20 = 20$ mm above XY → Front View is above XY → the point is above H.P.

Conclusion: The point is above H.P. and behind V.P. → It lies in the Second (II) Quadrant.

Given:
- Both Front View and Top View coincide and are 40 mm below XY.

Analysis using Table 4.2:
- Front View below XY → the point is below H.P.
- Top View below XY → the point is in front of V.P.

Conclusion: The point is below H.P. and in front of V.P. → It lies in the Fourth (IV) Quadrant.

Given: Point D is 25 mm from H.P. and 30 mm from V.P.

Case 1 — First Quadrant (above H.P., in front of V.P.):

- Draw reference line XY.
- The Front View $d'$ is above XY at a distance of 25 mm (distance from H.P.).
- The Top View $d$ is below XY at a distance of 30 mm (distance from V.P.).
- Both $d'$ and $d$ lie on the same projector (perpendicular to XY).

Case 2 — Third Quadrant (below H.P., behind V.P.):

- The Front View $d'$ is below XY at a distance of 25 mm.
- The Top View $d$ is above XY at a distance of 30 mm.
- Both $d'$ and $d$ lie on the same projector perpendicular to XY.

Summary Table:

| Quadrant | Front View ($d'$) | Top View ($d$) |
|---|---|---|
| I | 25 mm above XY | 30 mm below XY |
| III | 25 mm below XY | 30 mm above XY |

Given:
- Point P: 15 mm above H.P., 20 mm in front of V.P. → First Quadrant
- Point Q: 25 mm behind V.P., 40 mm below H.P. → Third Quadrant

Projections of Point P:
- Front View $p'$: 15 mm above XY (above H.P.)
- Top View $p$: 20 mm below XY (in front of V.P.)
- $p'$ and $p$ are on the same projector ⊥ to XY.

Projections of Point Q:
- Front View $q'$: 40 mm below XY (below H.P.)
- Top View $q$: 25 mm above XY (behind V.P.)
- $q'$ and $q$ are on the same projector ⊥ to XY.

Drawing Steps:
1. Draw reference line XY.
2. Mark $p'$ 15 mm above XY and $p$ 20 mm below XY on the same vertical projector.
3. Mark $q'$ 40 mm below XY and $q$ 25 mm above XY on another vertical projector.
4. Label all points clearly.

Given: Four points to be projected on the same XY line.

Analysis of each point:

| Point | Position | Quadrant | Front View | Top View |
|---|---|---|---|---|
| B | 20 mm above H.P., 25 mm in front of V.P. | I | 20 mm above XY | 25 mm below XY |
| D | 25 mm below H.P., 15 mm behind V.P. | III | 25 mm below XY | 15 mm above XY |
| E | 15 mm above H.P., 10 mm behind V.P. | II | 15 mm above XY | 10 mm above XY |
| F | 20 mm below H.P., 25 mm in front of V.P. | IV | 20 mm below XY | 25 mm below XY |

Drawing Steps:
1. Draw a horizontal reference line XY.
2. For Point B: Draw projector; mark $b'$ 20 mm above XY and $b$ 25 mm below XY.
3. For Point D: Draw projector; mark $d'$ 25 mm below XY and $d$ 15 mm above XY.
4. For Point E: Draw projector; mark $e'$ 15 mm above XY and $e$ 10 mm above XY (both above XY — II quadrant).
5. For Point F: Draw projector; mark $f'$ 20 mm below XY and $f$ 25 mm below XY (both below XY — IV quadrant).
6. All projectors are perpendicular to XY. Label all views clearly.

Given:
- Line CD, length = 40 mm
- Line is in V.P. (on the V.P.)
- Line is parallel to H.P.
- End C is 30 mm above H.P.

Analysis:
- Since the line lies in V.P., its Top View will be on XY (i.e., the top view is a point or line on XY).
- Since the line is parallel to H.P., the Front View will be a true-length line parallel to XY.
- End C is 30 mm above H.P., so the Front View is 30 mm above XY.

Drawing Steps:
1. Draw reference line XY.
2. Draw the Front View $c'd'$: a horizontal line 30 mm above XY, 40 mm long (true length, since line is parallel to H.P. and in V.P.).
3. Draw the Top View $cd$: a line on XY (since the line is in V.P., its H.P. projection lies on XY), 40 mm long, directly below the front view.
4. Project $c$ and $d$ vertically down from $c'$ and $d'$ to XY.

Result: Front View = 40 mm line, 30 mm above XY; Top View = 40 mm line on XY.

Given:
- Line EF, length = 40 mm
- Line is in H.P. (lies on H.P.)
- Line is parallel to V.P. and 25 mm in front of V.P.

Analysis:
- Since the line lies in H.P., its Front View will be on XY.
- Since the line is parallel to V.P., the Top View will be a true-length line parallel to XY.
- The line is 25 mm in front of V.P., so the Top View is 25 mm below XY.

Drawing Steps:
1. Draw reference line XY.
2. Draw the Top View $ef$: a horizontal line 25 mm below XY, 40 mm long (true length).
3. Draw the Front View $e'f'$: a line on XY, 40 mm long, directly above the top view.
4. Project $e'$ and $f'$ vertically up from $e$ and $f$ to XY.

Result: Top View = 40 mm line, 25 mm below XY; Front View = 40 mm line on XY.

Given:
- Line GH, length = 40 mm
- Line lies in both H.P. and V.P. simultaneously.

Analysis:
- A line lying in both planes must lie along the XY line (the line of intersection of H.P. and V.P.).
- Both the Front View and Top View will coincide with XY.

Drawing Steps:
1. Draw reference line XY.
2. Both the Front View $g'h'$ and Top View $gh$ coincide and lie on XY, each 40 mm long.

Result: Both Front View and Top View are 40 mm lines lying on XY.

Given:
- Line JK, length = 40 mm
- Line is ⊥ to H.P.
- Line is 20 mm in front of V.P.
- Nearest point to H.P. is J, 15 mm above H.P.
- Therefore K is $15 + 40 = 55$ mm above H.P.

Analysis:
- Since the line is perpendicular to H.P., its Top View is a point.
- The Front View is a vertical line (true length) perpendicular to XY.

Drawing Steps:
1. Draw reference line XY.
2. Top View $jk$: a single point (both J and K project to the same point) 20 mm below XY (20 mm in front of V.P.).
3. Front View $j'k'$: a vertical line perpendicular to XY, directly above the top view point.
- $j'$ is 15 mm above XY.
- $k'$ is 55 mm above XY.
- Length of front view = 40 mm (true length).

Result: Top View = a point 20 mm below XY; Front View = vertical line from 15 mm to 55 mm above XY.

Given:
- Line UV, length = 40 mm
- Line is ⊥ to V.P.
- Farthest end V is 65 mm in front of V.P. and 20 mm above H.P.
- Therefore, nearer end U is $65 - 40 = 25$ mm in front of V.P. and 20 mm above H.P.

Analysis:
- Since the line is perpendicular to V.P., its Front View is a point.
- The Top View is a horizontal line (true length) perpendicular to XY.

Drawing Steps:
1. Draw reference line XY.
2. Front View $u'v'$: a single point (both U and V project to the same point) 20 mm above XY (20 mm above H.P.).
3. Top View $uv$: a horizontal line perpendicular to XY (i.e., a line going away from XY), directly below the front view point.
- $u$ is 25 mm below XY.
- $v$ is 65 mm below XY.
- Length of top view = 40 mm (true length).

Result: Front View = a point 20 mm above XY; Top View = horizontal line from 25 mm to 65 mm below XY.

Given:
- Line AB, length = 40 mm
- Line is parallel to both H.P. and V.P.
- 25 mm behind V.P. (above XY in top view)
- 30 mm below H.P. (below XY in front view)

Analysis:
- Since the line is parallel to both planes, both Front View and Top View are true-length lines parallel to XY.

Drawing Steps:
1. Draw reference line XY.
2. Front View $a'b'$: horizontal line 30 mm below XY, 40 mm long.
3. Top View $ab$: horizontal line 25 mm above XY, 40 mm long.
4. Projectors from $a'$ to $a$ and $b'$ to $b$ are perpendicular to XY.

Result: Front View = 40 mm line, 30 mm below XY; Top View = 40 mm line, 25 mm above XY.

Given:
- Line LM, length = 40 mm
- Line is ⊥ to H.P.
- 30 mm behind V.P.
- Nearest point L is 10 mm above H.P.
- Therefore M is $10 + 40 = 50$ mm above H.P.

Analysis:
- Since the line is perpendicular to H.P., its Top View is a point.
- The Front View is a vertical line (true length).

Drawing Steps:
1. Draw reference line XY.
2. Top View $lm$: a single point 30 mm above XY (30 mm behind V.P.).
3. Front View $l'm'$: a vertical line directly above the top view point.
- $l'$ is 10 mm above XY.
- $m'$ is 50 mm above XY.

Result: Top View = a point 30 mm above XY; Front View = vertical line from 10 mm to 50 mm above XY.

Given:
- Line NP, length = 40 mm
- Line is ⊥ to V.P.
- 30 mm below H.P.
- Nearest point P is 10 mm in front of V.P.
- Therefore N (farthest) is $10 + 40 = 50$ mm in front of V.P.

Analysis:
- Since the line is perpendicular to V.P., its Front View is a point.
- The Top View is a horizontal line (true length).

Drawing Steps:
1. Draw reference line XY.
2. Front View $n'p'$: a single point 30 mm below XY (30 mm below H.P.).
3. Top View $np$: a horizontal line directly below the front view point.
- $p$ is 10 mm below XY.
- $n$ is 50 mm below XY.

Result: Front View = a point 30 mm below XY; Top View = horizontal line from 10 mm to 50 mm below XY.

Given:
- Line QR, length = 40 mm
- Line is ⊥ to V.P.
- 10 mm below H.P.
- Farthest point Q is 65 mm behind V.P.
- Therefore nearer point R is $65 - 40 = 25$ mm behind V.P.

Analysis:
- Since the line is perpendicular to V.P., its Front View is a point.
- The Top View is a horizontal line (true length).
- Both points are behind V.P., so top view is above XY.

Drawing Steps:
1. Draw reference line XY.
2. Front View $q'r'$: a single point 10 mm below XY (10 mm below H.P.).
3. Top View $qr$: a horizontal line directly above the front view point (behind V.P. → above XY).
- $r$ is 25 mm above XY.
- $q$ is 65 mm above XY.

Result: Front View = a point 10 mm below XY; Top View = horizontal line from 25 mm to 65 mm above XY.

Given:
- Line ST, length = 40 mm
- Line is ⊥ to H.P.
- Behind V.P.
- Nearest point S is 20 mm from V.P. (behind) and 15 mm below H.P.
- Therefore T is $15 + 40 = 55$ mm below H.P.

Analysis:
- Since the line is perpendicular to H.P., its Top View is a point.
- The Front View is a vertical line (true length).
- Both points are below H.P. → Front View is below XY.
- Behind V.P. → Top View is above XY.

Drawing Steps:
1. Draw reference line XY.
2. Top View $st$: a single point 20 mm above XY (20 mm behind V.P.).
3. Front View $s't'$: a vertical line directly below the top view point.
- $s'$ is 15 mm below XY.
- $t'$ is 55 mm below XY.

Result: Top View = a point 20 mm above XY; Front View = vertical line from 15 mm to 55 mm below XY.

Given:
- Pentagonal plate, side = 35 mm
- Inclined at 30° to H.P., perpendicular to V.P.
- One edge ⊥ to V.P., 20 mm above H.P.
- Nearer end of that edge is 30 mm in front of V.P.

This is a two-step problem (plane inclined to H.P., ⊥ to V.P.):

Step 1 — Assume the plate is parallel to H.P. (⊥ to V.P.):
- Draw the Top View as the true shape: a regular pentagon with 35 mm sides, with one edge perpendicular to XY (i.e., parallel to the direction of projection).
- The Front View will be a straight line (edge view) parallel to XY, at 20 mm above XY.

Step 2 — Tilt the plate at 30° to H.P.:
- In the Front View, redraw the edge view (straight line) at 30° to XY.
- The edge that is ⊥ to V.P. and 20 mm above H.P. with nearer end 30 mm in front of V.P. fixes the position.
- Project the new Top View from the tilted Front View.

Drawing Procedure:
1. Draw XY. Mark the front view as a line 20 mm above XY (Step 1).
2. Draw the true-shape pentagon in the top view with one edge ⊥ to XY, nearer end 30 mm below XY.
3. Tilt the front view line at 30° to XY, keeping the nearer end fixed at 20 mm above XY.
4. Project all corners vertically from the tilted front view and horizontally from the Step 1 top view to get the new top view.
5. Join the projected points to complete the top view (foreshortened pentagon).

Result: Front View = a line inclined at 30° to XY; Top View = foreshortened pentagon.

Given:
- Equilateral triangular lamina, side = 30 mm
- Side AB lies in V.P.
- Surface inclined at 60° to V.P.

This is a two-step problem (plane inclined to V.P.):

Step 1 — Assume the surface is parallel to V.P. (⊥ to H.P.):
- Draw the Front View as the true shape: equilateral triangle with 30 mm sides, with side AB on XY (in V.P.).
- The Top View will be a straight line on XY (edge view).

Step 2 — Tilt the surface at 60° to V.P.:
- In the Top View, redraw the edge view (line on XY) at 60° to XY.
- Project the new Front View from the tilted Top View.

Drawing Procedure:
1. Draw XY. Draw the true-shape equilateral triangle as Front View with AB on XY.
2. Project the Top View as a line on XY (length = 30 mm, the length of AB).
3. Tilt the top view line at 60° to XY, keeping AB end fixed on XY.
4. Project all three corners vertically from the tilted top view and horizontally from the Step 1 front view to get the new front view.
5. Join the projected points to complete the front view (foreshortened triangle).

Result: Top View = line at 60° to XY; Front View = foreshortened triangle.

Given:
- Square plate, side = 35 mm
- Inclined at 45° to V.P., ⊥ to H.P.
- One corner in V.P.
- Two sides containing that corner are equally inclined to V.P.

This is a two-step problem (plane ⊥ to H.P., inclined to V.P.):

Step 1 — Assume the plate is parallel to V.P. (⊥ to H.P.):
- Draw the Front View as the true shape: square with 35 mm sides.
- The Top View is a straight line on XY (edge view), with one end on XY (corner in V.P.).
- Since two sides are equally inclined to V.P., the diagonal of the square is perpendicular to V.P. in Step 1.

Step 2 — Tilt the plate at 45° to V.P.:
- In the Top View, redraw the edge view line at 45° to XY, with the corner on XY.
- Project the new Front View.

Drawing Procedure:
1. Draw XY. Draw the square as Front View with one corner on XY and diagonal vertical.
2. Project the Top View as a line on XY (length = $35\sqrt{2}$ mm, the diagonal length).
3. Tilt the top view line at 45° to XY, keeping the corner on XY.
4. Project all four corners from the tilted top view and from the Step 1 front view to get the new front view.
5. Join the projected points to complete the foreshortened front view.

Result: Top View = line at 45° to XY; Front View = foreshortened square (rhombus shape).

Given:
- Regular hexagonal plate, side = 30 mm
- Resting on one side on H.P. (ground)
- That side is parallel to V.P.
- Surface inclined at 45° to H.P.

This is a two-step problem (plane inclined to H.P., ⊥ to V.P.):

Step 1 — Assume the plate is parallel to H.P.:
- Draw the Top View as the true shape: regular hexagon with 30 mm sides, with one side parallel to XY (parallel to V.P.).
- The Front View is a straight line (edge view) parallel to XY, on XY (resting on H.P.).

Step 2 — Tilt the plate at 45° to H.P.:
- In the Front View, redraw the edge view line at 45° to XY, with the resting side on XY.
- Project the new Top View.

Drawing Procedure:
1. Draw XY. Draw the true-shape hexagon as Top View with one side on XY.
2. Project the Front View as a line on XY (length = distance across the hexagon perpendicular to the resting side = $30\sqrt{3}$ mm).
3. Tilt the front view line at 45° to XY, keeping the resting side on XY.
4. Project all six corners from the tilted front view and from the Step 1 top view to get the new top view.
5. Join the projected points to complete the foreshortened top view.

Result: Front View = line at 45° to XY; Top View = foreshortened hexagon.

Given:
- Rectangular lamina, 25 mm × 20 mm
- Parallel to H.P. and 15 mm above H.P.
- One longer edge (25 mm) makes 30° to V.P.

Analysis:
- Since the lamina is parallel to H.P., the Front View is a straight line parallel to XY, 15 mm above XY.
- The Top View is the true shape (rectangle), with the longer edge at 30° to XY.

Drawing Steps:
1. Draw reference line XY.
2. Draw the Top View: a rectangle 25 mm × 20 mm with the longer edge (25 mm) at 30° to XY.
3. Draw the Front View: a straight horizontal line 15 mm above XY, with length equal to the extent of the rectangle in the direction perpendicular to V.P. (projected length).
- The projected length = $25\cos30° + 20\cos60°$ ... (depends on orientation; draw by projecting corners).
4. Project all four corners of the top view vertically up to the front view line to determine the extent.

Result: Top View = true-shape rectangle with longer edge at 30° to XY; Front View = horizontal line 15 mm above XY.

Given:
- Circle, diameter = 30 mm (radius = 15 mm)
- Plane is vertical (⊥ to H.P.) and inclined at 30° to V.P.
- Centre is 25 mm above H.P. and 20 mm in front of V.P.

This is a two-step problem (plane ⊥ to H.P., inclined to V.P.):

Step 1 — Assume the plane is parallel to V.P. (⊥ to H.P.):
- Front View: true shape — a circle of 30 mm diameter, centre 25 mm above XY.
- Top View: a straight line (edge view) parallel to XY, 20 mm below XY, length = 30 mm.

Step 2 — Tilt the plane at 30° to V.P.:
- In the Top View, redraw the edge view line at 30° to XY, with centre 20 mm below XY.
- Project the new Front View.

Drawing Procedure:
1. Draw XY. Draw the circle (Front View) with centre 25 mm above XY.
2. Divide the circle into 12 equal parts; project each point down to the top view line (20 mm below XY).
3. Tilt the top view line at 30° to XY.
4. Project all 12 points from the tilted top view vertically up and horizontally from the Step 1 front view to get the new front view points.
5. Join the points with a smooth curve — the new front view will be an ellipse.

Result: Top View = line at 30° to XY; Front View = ellipse with major axis = 30 mm.

Given:
- Square prism: base edge = 35 mm, height = 50 mm
- Resting on H.P. (base on H.P.)
- Two vertical rectangular faces parallel to V.P.
- Axis is perpendicular to H.P.

Step 1 — Start with Top View (axis ⊥ to H.P.):
- Top View is the true shape of the base: a square of 35 mm sides with sides parallel and perpendicular to XY.
- Label corners as $a, b, c, d$.

Step 2 — Project Front View:
- Front View is a rectangle: width = 35 mm (one side of square), height = 50 mm.
- The rectangle is placed with its base on XY.
- Visible edges are shown as continuous lines; hidden edges as dashed lines.

Drawing Steps:
1. Draw XY.
2. Draw the square top view (35 mm × 35 mm) with sides parallel to XY, below XY.
3. Project corners vertically up to XY and draw the rectangle 50 mm high above XY.
4. The front view shows two visible rectangular faces (front face) as a rectangle 35 mm wide and 50 mm tall.

Result: Top View = 35 mm square; Front View = 35 mm × 50 mm rectangle with base on XY.

Given:
- Triangular prism: base edge = 40 mm, height = 60 mm
- Standing on base on H.P.
- One vertical rectangular face is at the rear, parallel to V.P.
- Axis ⊥ to H.P.

Step 1 — Top View:
- True shape of base: equilateral triangle with 40 mm sides.
- One side (rear face) is parallel to XY (parallel to V.P.).
- The apex of the triangle points toward the observer (toward V.P.).

Step 2 — Front View:
- The front view shows the rear rectangular face as a rectangle: width = 40 mm, height = 60 mm.
- The apex of the triangle projects as a vertical line at the centre, in front.
- The front view is a rectangle 40 mm wide and 60 mm tall, with a vertical centre line (apex edge) shown as a visible line.

Drawing Steps:
1. Draw XY.
2. Draw the equilateral triangle (top view) with one side parallel to XY (rear), 40 mm sides.
3. Project all three corners up; draw the front view rectangle 40 mm × 60 mm.
4. The apex edge appears as a vertical line at the front centre of the rectangle.

Result: Top View = equilateral triangle (one side parallel to XY); Front View = 40 mm × 60 mm rectangle with apex edge visible.

Given:
- Cylinder: base diameter = 30 mm, axis length = 40 mm
- Resting on base on H.P.
- Axis ⊥ to H.P.

Step 1 — Top View:
- True shape of base: a circle of 30 mm diameter.

Step 2 — Front View:
- A rectangle: width = 30 mm (diameter), height = 40 mm (axis length).
- Base on XY.
- Two vertical lines represent the extreme generators.

Drawing Steps:
1. Draw XY.
2. Draw the circle (top view) of 30 mm diameter, tangent to XY (or with centre at appropriate distance below XY).
3. Project the extreme points of the circle up to XY; draw the rectangle 30 mm wide and 40 mm tall.

Result: Top View = circle of 30 mm diameter; Front View = 30 mm × 40 mm rectangle.

Given:
- Hemisphere: diameter = 60 mm (radius = 30 mm)
- Resting on H.P. with its flat circular face on top (curved surface on H.P.).

Analysis:
- The curved surface rests on H.P., so the flat circular face is at the top.
- The axis is vertical (⊥ to H.P.).

Step 1 — Top View:
- Looking from above, we see the flat circular face: a circle of 60 mm diameter.

Step 2 — Front View:
- The front view shows a semicircle of 30 mm radius on top of a rectangle...
- Actually: the curved surface is below, flat face is on top. The front view is a semicircle with the diameter (flat face) at the top and the curved part below.
- The overall height = 30 mm (radius).
- The front view is a semicircle of 60 mm diameter with the flat edge at the top (30 mm above XY) and the curved part touching XY.

Drawing Steps:
1. Draw XY.
2. Draw the circle (top view) of 60 mm diameter.
3. Draw the front view as a semicircle: flat diameter 60 mm wide at the top, curved part resting on XY. Centre of the diameter is 30 mm above XY.

Result: Top View = circle of 60 mm diameter; Front View = semicircle with flat edge at top.

Given:
- Hexagonal pyramid: base edge = 25 mm, height = 70 mm
- Cut at mid-height (35 mm) parallel to base
- Frustum: base edge = 25 mm, top edge = $25 \times \frac{35}{70} = 12.5$ mm, height = 35 mm
- Resting on H.P. (base on H.P.)
- Axis ⊥ to H.P.

Step 1 — Top View:
- Two concentric regular hexagons:
- Outer hexagon: 25 mm sides (base)
- Inner hexagon: 12.5 mm sides (top cut face)
- Both with sides parallel to each other.

Step 2 — Front View:
- A trapezium: base width = $2 \times 25 = 50$ mm (across flats or across corners depending on orientation), top width = $2 \times 12.5 = 25$ mm, height = 35 mm.
- For a hexagonal pyramid with two sides parallel to V.P., the front view shows the outline as a trapezium.

Drawing Steps:
1. Draw XY.
2. Draw the two concentric hexagons (top view) with one pair of sides parallel to XY.
3. Project the corners up to draw the trapezoidal front view, 35 mm tall.
4. Show slant edges and hidden lines appropriately.

Result: Top View = two concentric hexagons; Front View = trapezium, 35 mm tall.

Given:
- Hollow cylinder (pipe): outer diameter = 50 mm, inner diameter = 40 mm, length = 50 mm
- Resting on H.P., axis ⊥ to V.P.

Since axis ⊥ to V.P., start with Front View:

Front View:
- Two concentric circles:
- Outer circle: diameter = 50 mm
- Inner circle: diameter = 40 mm
- Centre is at 25 mm above XY (resting on H.P., so bottom of outer circle touches XY).

Top View:
- A rectangle showing the length and outer diameter: 50 mm wide × 50 mm tall (length × outer diameter).
- Two horizontal lines at the top and bottom represent the outer surface.
- Two dashed horizontal lines inside represent the inner bore (40 mm apart from centre).
- The end circles project as vertical lines.

Drawing Steps:
1. Draw XY.
2. Draw the front view: two concentric circles (outer ∅50, inner ∅40), with the outer circle tangent to XY at the bottom.
3. Project the top and bottom of the outer circle down to XY; draw the rectangle 50 mm wide and 50 mm tall below XY.
4. Show the inner bore as dashed lines 5 mm from each horizontal edge (since inner radius = 20 mm, outer radius = 25 mm, difference = 5 mm).

Result: Front View = two concentric circles; Top View = rectangle with outer and inner boundary lines.

Given:
- Triangular pyramid: base edge = 50 mm, axis = 50 mm
- Resting on a base corner on H.P.
- Upper edge of base is horizontal.
- Base is on the rear and parallel to V.P.
- Axis is parallel to V.P. (since base is parallel to V.P.).

Since axis is parallel to V.P. and base is parallel to V.P., start with Front View:

Front View:
- The base (equilateral triangle, 50 mm side) appears as its true shape since it is parallel to V.P.
- The pyramid rests on one corner of the base on H.P., with the opposite edge (upper edge) horizontal.
- The apex is in front of the base.
- Draw the equilateral triangle with one vertex at the bottom (on XY) and the opposite side horizontal at the top.
- The apex of the pyramid projects as a point in front.

Top View:
- The base appears as a horizontal line (edge view) on XY (since base is parallel to V.P. and perpendicular to H.P. in this position... actually the base is vertical since it is parallel to V.P.).
- The apex projects below XY.
- The top view shows the base as a line and the apex as a point.

Drawing Steps:
1. Draw XY.
2. Draw the front view: equilateral triangle (50 mm sides) with one vertex on XY and opposite side horizontal at top. The apex of the pyramid is shown as a point in front (coinciding with the centroid projected).
3. Project the top view: base appears as a vertical line on XY; apex appears as a point below XY at the appropriate distance.

Result: Front View = equilateral triangle with apex point; Top View = line (base edge view) and apex point.

Given:
- Pentagonal prism: base edge = 30 mm, length = 60 mm
- Long edges (axis) ⊥ to V.P.
- Rectangular face on top is parallel to H.P.

Since axis ⊥ to V.P., start with Front View:

Front View:
- True shape of the pentagonal base: regular pentagon with 30 mm sides.
- One side (top rectangular face) is horizontal (parallel to H.P.).
- The pentagon is oriented with one side at the top.
- The bottom vertex rests on H.P. (or the bottom side — need to determine from context; since top face is parallel to H.P., one side is at top and the opposite vertex is at the bottom).

Top View:
- A rectangle: width = 60 mm (length of prism), height = distance from top face to bottom vertex of pentagon.
- The top edge represents the top rectangular face.
- Hidden lines show the bottom vertex edge.

Drawing Steps:
1. Draw XY.
2. Draw the front view: regular pentagon (30 mm sides) with one side horizontal at the top, resting on the bottom vertex on XY.
3. Project the top view: rectangle 60 mm wide. The top edge is 30 mm wide (top face), and the overall height equals the distance from the top side to the bottom vertex of the pentagon.
4. Show appropriate hidden lines.

Result: Front View = regular pentagon (one side at top); Top View = rectangle 60 mm × (height of pentagon).

Given:
- Frustum of square pyramid: base edge = 40 mm, top edge = 20 mm, axis = 50 mm
- Resting on H.P. on a base edge
- Axis horizontal and ⊥ to V.P.
- Cut face (top) is in front

Since axis ⊥ to V.P., start with Front View:

Front View:
- The cut face (square, 20 mm sides) is in front, parallel to V.P.
- The front view shows the cut face as a square: 20 mm × 20 mm.
- The base edges project behind.
- The overall front view is a square (cut face) with the base outline visible.

Top View:
- The frustum is lying on a base edge on H.P.
- The top view shows the frustum from above: a trapezium shape.
- The base edge on H.P. is on XY; the cut face is at the front (below XY in top view).

Drawing Steps:
1. Draw XY.
2. Draw the front view: the cut face square (20 mm × 20 mm) centred on the axis, with the base square (40 mm × 40 mm) behind it. The front view shows a square within a square (or the outline of the frustum).
3. Project the top view: trapezium with the base edge on XY and the cut face edge in front.
4. Show slant edges and hidden lines.

Result: Front View = square cut face with base outline; Top View = trapezium.

Given:
- Hexagonal prism: base edge = 25 mm, axis = 60 mm
- Resting on one base edge on H.P.
- Axis ⊥ to V.P.

Since axis ⊥ to V.P., start with Front View:

Front View:
- True shape of hexagonal base: regular hexagon with 25 mm sides.
- The prism rests on one edge, so one side of the hexagon is on XY.
- The hexagon is oriented with one side at the bottom (on XY).

Top View:
- A rectangle: length = 60 mm (axis length), height = distance from the bottom edge to the topmost point of the hexagon.
- For a regular hexagon with one side at the bottom, the height = $25 + 25\sin60° \times 2$...
- Actually, for a regular hexagon of side 25 mm with one flat side at the bottom:
- Height of hexagon = $25\sqrt{3}$ mm ≈ 43.3 mm
- Top view rectangle: 60 mm × 43.3 mm.
- Hidden lines show the bottom edge and internal edges.

Drawing Steps:
1. Draw XY.
2. Draw the front view: regular hexagon (25 mm sides) with one side on XY.
3. Project the top view: rectangle 60 mm wide and 43.3 mm tall (height of hexagon), with the bottom edge on XY.
4. Show appropriate hidden lines for the bottom edge.

Result: Front View = regular hexagon (one side on XY); Top View = rectangle 60 mm × 43.3 mm.

Given:
- Cylinder: base diameter = 50 mm, height (length) = 70 mm
- Resting on H.P. (on its curved surface — generator on H.P.)
- Axis ⊥ to V.P.

Since axis ⊥ to V.P., start with Front View:

Front View:
- True shape of circular base: a circle of 50 mm diameter.
- The cylinder rests on H.P. on its curved surface, so the bottom of the circle touches XY.
- Centre of circle is 25 mm above XY.

Top View:
- A rectangle: length = 70 mm (axis), width = 50 mm (diameter).
- The bottom edge of the rectangle is on XY (generator on H.P.).

Drawing Steps:
1. Draw XY.
2. Draw the front view: circle of 50 mm diameter with bottom tangent to XY.
3. Project the top view: rectangle 70 mm × 50 mm with bottom edge on XY.

Result: Front View = circle of ∅50 mm; Top View = 70 mm × 50 mm rectangle.

Given:
- Cone: base diameter = 30 mm, axis = 65 mm
- Axis ⊥ to V.P.
- Vertex (apex) is in front.

Since axis ⊥ to V.P., start with Front View:

Front View:
- True shape of circular base: a circle of 30 mm diameter.
- The cone rests on H.P. on its curved surface (generator on H.P.).
- Centre of base circle is 15 mm above XY.
- The apex is in front (toward the observer).

Top View:
- An isosceles triangle: base = 30 mm (diameter), height = 65 mm (axis).
- The apex point is at the front (below XY in top view, toward observer).
- The base of the triangle is at the rear.
- The bottom of the triangle (apex) is on XY (generator on H.P.).

Drawing Steps:
1. Draw XY.
2. Draw the front view: circle of 30 mm diameter with bottom tangent to XY.
3. Draw the top view: isosceles triangle with base 30 mm at the rear and apex at the front, height 65 mm. The apex touches XY.

Result: Front View = circle of ∅30 mm; Top View = isosceles triangle (base 30 mm, height 65 mm) with apex on XY.

Given:
- Square prism: base edge = 40 mm, axis = 70 mm
- Lying on one rectangular face on H.P.
- Axis ⊥ to V.P.

Since axis ⊥ to V.P., start with Front View:

Front View:
- True shape of square base: a square of 40 mm sides.
- The prism lies on one rectangular face, so one side of the square is on XY (bottom).
- The square is oriented with one side horizontal at the bottom.

Top View:
- A rectangle: length = 70 mm (axis), width = 40 mm (base edge).
- The bottom edge of the rectangle is on XY.

Drawing Steps:
1. Draw XY.
2. Draw the front view: square (40 mm × 40 mm) with one side on XY.
3. Project the top view: rectangle 70 mm × 40 mm with bottom edge on XY.

Result: Front View = 40 mm square (one side on XY); Top View = 70 mm × 40 mm rectangle.

Given:
- Frustum of triangular pyramid: base edge = 50 mm, top edge = 20 mm, axis = 60 mm
- Resting on H.P. on a base edge
- Axis parallel to H.P. and ⊥ to V.P.
- Cut face (top) is in front

Since axis ⊥ to V.P., start with Front View:

Front View:
- The cut face (equilateral triangle, 20 mm sides) is in front, parallel to V.P.
- The front view shows the cut face as an equilateral triangle: 20 mm sides.
- The base triangle outline is visible behind.

Top View:
- The frustum rests on a base edge on H.P.
- The top view shows a trapezium: the base edge (50 mm) at the rear and the cut face edge (20 mm) at the front.
- The height of the trapezium = 60 mm (axis length).

Drawing Steps:
1. Draw XY.
2. Draw the front view: equilateral triangle (20 mm sides) for the cut face, with the base triangle (50 mm sides) outline behind it.
3. Project the top view: trapezium with 50 mm base at rear and 20 mm top at front, height 60 mm.
4. Show appropriate hidden lines.

Result: Front View = triangular cut face with base outline; Top View = trapezium.

Given:
- Pentagonal pyramid: base edge = 25 mm, height = 60 mm
- Axis ⊥ to V.P.
- Base parallel to V.P.

Since axis ⊥ to V.P., start with Front View:

Front View:
- The base (regular pentagon, 25 mm sides) is parallel to V.P., so it appears as its true shape in the front view.
- The apex is behind the base (away from observer).
- Draw the regular pentagon with one side at the bottom (resting on H.P. or positioned appropriately).
- The apex projects as a point at the centroid of the pentagon.

Top View:
- The top view shows the pyramid from above.
- The base appears as a line (edge view) since the base is parallel to V.P. and perpendicular to H.P.
- The apex projects as a point in front of (or behind) the base line.
- The top view shows an isosceles triangle: base = 25 mm (one side of pentagon), height = 60 mm (axis).

Drawing Steps:
1. Draw XY.
2. Draw the front view: regular pentagon (25 mm sides) with one side on XY, with the apex point at the centroid.
3. Project the top view: the base appears as a line on XY; the apex appears as a point 60 mm away from XY (below XY, in front of V.P.).
4. Draw lines from the apex to the ends of the base line.

Result: Front View = regular pentagon with apex point; Top View = triangle with base on XY and apex 60 mm below XY.

Given:
- Hexagonal prism: base edge = 25 mm, axis = 60 mm
- Lying on one rectangular face on H.P.
- Axis parallel to both V.P. and H.P.

Since axis is parallel to both planes, start with Side View (or use the standard approach):

Step 1 — Assume axis ⊥ to V.P. (start with Front View):
- Front View: regular hexagon (25 mm sides) with one side on XY (lying on H.P.).

Step 2 — Since axis is parallel to V.P., the Front View remains the hexagon and the Top View is a rectangle:

Front View:
- Regular hexagon with 25 mm sides, one side on XY.

Top View:
- Rectangle: length = 60 mm (axis), width = distance across the hexagon (from the bottom face to the top face) = $25\sqrt{3}$ ≈ 43.3 mm.
- Bottom edge on XY.

Drawing Steps:
1. Draw XY.
2. Draw the front view: regular hexagon (25 mm sides) with one flat side on XY.
3. Project the top view: rectangle 60 mm × 43.3 mm with bottom edge on XY.
4. Show hidden lines for the bottom face.

Result: Front View = regular hexagon; Top View = 60 mm × 43.3 mm rectangle.

Given:
- Triangular pyramid: base edge = 25 mm, axis = 50 mm
- Resting on one base edge on H.P.
- Axis parallel to both V.P. and H.P.

Front View:
- The pyramid rests on one base edge on H.P. with axis parallel to V.P.
- The front view shows the true shape of the triangular face visible from the front.
- The base edge is on XY; the apex is above.
- The front view is a triangle.

Top View:
- The top view shows the pyramid from above.
- The base edge (25 mm) is on XY; the axis extends parallel to XY.
- The top view shows the base edge as a line on XY and the apex as a point above XY (behind V.P.) or the outline of the pyramid.

Drawing Steps:
1. Draw XY.
2. Draw the front view: equilateral triangle (25 mm base on XY, height = $25\frac{\sqrt{3}}{2}$ ≈ 21.65 mm... but this is the base triangle; the axis = 50 mm gives the height of the pyramid).
- The front view shows the base edge on XY and the apex at the top.
3. Project the top view: the base edge is on XY; the apex projects as a point at the appropriate distance.

Result: Front View = triangle (base on XY, apex at top); Top View = outline showing base edge and apex.

Given:
- Cylinder: base diameter = 40 mm, axis = 60 mm
- Lying on a generator on H.P.
- Axis parallel to both V.P. and H.P.

Front View:
- A rectangle: length = 60 mm (axis), height = 40 mm (diameter).
- Bottom edge on XY.

Top View:
- A rectangle: length = 60 mm (axis), width = 40 mm (diameter).
- Bottom edge on XY.
- (Both views are rectangles since the axis is parallel to both planes.)

Drawing Steps:
1. Draw XY.
2. Draw the front view: rectangle 60 mm × 40 mm with bottom edge on XY.
3. Draw the top view: rectangle 60 mm × 40 mm with bottom edge on XY (directly below the front view).
4. The two end circles project as vertical lines at the ends.

Result: Front View = 60 mm × 40 mm rectangle; Top View = 60 mm × 40 mm rectangle.

Given:
- Frustum of hexagonal pyramid: base edge = 20 mm, top edge = 10 mm, axis = 50 mm
- Resting on H.P. on a base edge
- Axis horizontal and parallel to V.P.
- Cut face (top) is in front

Front View:
- The cut face (regular hexagon, 10 mm sides) is in front, parallel to V.P.
- The front view shows the cut face as a regular hexagon (10 mm sides) with the base hexagon outline behind.
- The frustum rests on a base edge, so the base hexagon has one side on XY.

Top View:
- The top view shows the frustum from above.
- The base edge (20 mm) is on XY; the cut face edge (10 mm) is at the front.
- The top view shows a trapezium: base 20 mm at rear, top 10 mm at front, height = 50 mm (axis).

Drawing Steps:
1. Draw XY.
2. Draw the front view: regular hexagon (10 mm sides) for the cut face, with the base hexagon (20 mm sides) outline behind it. One side of the base hexagon is on XY.
3. Project the top view: trapezium with 20 mm base at rear (on XY) and 10 mm top at front, height 50 mm.
4. Show appropriate hidden lines.

Result: Front View = hexagonal cut face with base outline; Top View = trapezium (50 mm × 20 mm/10 mm).

Given:
- Triangular prism: base edge = 25 mm, axis = 60 mm
- Resting on one base edge on H.P.
- Axis parallel to V.P. and inclined at 30° to H.P.

Two-step method:

Step 1 — Assume axis ⊥ to H.P. (axis vertical):
- Top View: equilateral triangle (25 mm sides) with one side on XY (resting edge).
- Front View: rectangle 25 mm wide and 60 mm tall, with base on XY.

Step 2 — Tilt the axis at 30° to H.P. (axis inclined to H.P., parallel to V.P.):
- Redraw the Front View with the axis inclined at 30° to XY, keeping the resting edge on XY.
- The new front view is a parallelogram/rectangle tilted at 30°.
- Project the new Top View from the tilted front view.

Drawing Procedure:
1. Draw XY. Complete Step 1 views.
2. Redraw the front view: tilt the rectangle at 30° to XY, keeping the bottom edge (resting edge) on XY.
3. Project all corners of the tilted front view vertically down to XY.
4. Draw horizontal lines from the Step 1 top view corners to intersect with the projectors from Step 3.
5. Join the intersection points to get the new top view.

Result: Front View = tilted rectangle at 30° to XY; Top View = foreshortened triangular prism outline.

Given:
- Pentagonal pyramid: base edge = 25 mm, axis = 50 mm
- Axis ⊥ to V.P. and 50 mm above H.P.
- One edge of base inclined at 30° to H.P.

Since axis ⊥ to V.P., start with Front View:

Front View:
- True shape of pentagonal base: regular pentagon (25 mm sides).
- The pyramid is 50 mm above H.P., so the centre of the base is 50 mm above XY.
- One edge of the base is inclined at 30° to H.P. (to XY).
- Draw the pentagon with one side inclined at 30° to XY.
- The apex is behind the base (a point at the centroid).

Top View:
- The top view shows the pyramid from above.
- The base appears as a line (edge view) since the base is parallel to V.P.
- The apex projects as a point.

Drawing Steps:
1. Draw XY.
2. Draw the front view: regular pentagon (25 mm sides) with centre 50 mm above XY, one side inclined at 30° to XY.
3. Project the top view: base appears as a line 50 mm above XY (parallel to XY); apex as a point.

Result: Front View = regular pentagon (one side at 30° to XY, centre 50 mm above XY); Top View = line and apex point.

Given:
- Frustum of square pyramid: top edge = 20 mm, base edge = 40 mm, axis = 50 mm
- Side surface (face) inclined at 45° to H.P.
- Axis parallel to V.P.

Two-step method:

Step 1 — Assume axis ⊥ to H.P.:
- Top View: two concentric squares (outer 40 mm, inner 20 mm), with sides parallel to XY.
- Front View: trapezium with base 40 mm, top 20 mm, height 50 mm, on XY.

Step 2 — Tilt so that one side face is at 45° to H.P.:
- In the Front View, tilt the trapezium so that one slant face makes 45° with XY.
- Redraw the Front View accordingly.
- Project the new Top View.

Drawing Procedure:
1. Draw XY. Complete Step 1 views.
2. Tilt the front view (trapezium) so that one slant edge makes 45° with XY, keeping the base edge on XY.
3. Project all corners of the tilted front view down to get the new top view.
4. Use horizontal projectors from Step 1 top view to complete the new top view.

Result: Front View = tilted trapezium at 45°; Top View = foreshortened frustum outline.

Given:
- Frustum of cone: base diameter = 90 mm, top diameter = 30 mm
- Axis parallel to V.P.
- Case (i): axis inclined at 30° to H.P.
- Case (ii): axis inclined at 60° to H.P.

Two-step method for each case:

Step 1 — Assume axis ⊥ to H.P.:
- Top View: two concentric circles — outer circle ∅90 mm (base), inner circle ∅30 mm (top).
- Front View: trapezium with base 90 mm, top 30 mm, height = axis length (not given; assume $h$ mm), base on XY.

Step 2(i) — Tilt axis at 30° to H.P.:
- Redraw the Front View with the axis inclined at 30° to XY, keeping the base on XY.
- The new front view is a tilted trapezium.
- Project the new Top View from the tilted front view.
- Divide the circles into 12 parts; project each point to get the elliptical outlines in the top view.

Step 2(ii) — Tilt axis at 60° to H.P.:
- Same procedure as above but tilt the front view at 60° to XY.
- The top view will be more foreshortened.

Drawing Procedure (for each case):
1. Draw XY. Complete Step 1 views.
2. Redraw the front view tilted at the required angle (30° or 60°) to XY.
3. Divide the base and top circles into 12 equal parts in Step 1 top view.
4. Project each point from the tilted front view down and from the Step 1 top view horizontally to get new top view points.
5. Join the points with smooth curves (ellipses) for the new top view.

Result:
- (i) 30° to H.P.: Front View = trapezium at 30° to XY; Top View = frustum outline with elliptical ends.
- (ii) 60° to H.P.: Front View = trapezium at 60° to XY; Top View = more foreshortened frustum outline with elliptical ends.

Given:
- Pentagonal prism: base edge = 25 mm, axis = 50 mm
- Resting on base on H.P.
- One edge of base inclined at 30° to V.P.
- Axis is perpendicular to H.P. (resting on base).

Two-step method (axis ⊥ to H.P., base edge inclined to V.P.):

Step 1 — Assume one base edge parallel to V.P. (one side parallel to XY):
- Top View: regular pentagon (25 mm sides) with one side parallel to XY.
- Front View: rectangle showing the height (50 mm) and width of the prism.

Step 2 — Rotate the top view so that one base edge makes 30° to XY (V.P.):
- Redraw the Top View with one edge of the pentagon at 30° to XY.
- Project the new Front View from the rotated top view.

Drawing Procedure:
1. Draw XY. Complete Step 1 views.
2. Redraw the top view: rotate the pentagon so that one side makes 30° to XY.
3. Project all corners of the rotated top view vertically up to XY.
4. Draw horizontal lines from the Step 1 front view corners to intersect with the projectors from Step 3.
5. Join the intersection points to get the new front view.

Result: Top View = pentagon with one side at 30° to XY; Front View = foreshortened prism outline.

Given:
- Triangular pyramid: base edge = 50 mm, axis = 60 mm
- One base corner touching V.P.
- Axis parallel to H.P. and inclined at 45° to V.P.

Two-step method (axis parallel to H.P., inclined to V.P.):

Step 1 — Assume axis ⊥ to V.P. (axis horizontal, perpendicular to V.P.):
- Front View: true shape of base — equilateral triangle (50 mm sides) with one corner on XY (touching V.P.).
- Top View: triangle showing the axis length (60 mm) and base.

Step 2 — Rotate the top view so that the axis makes 45° to XY (V.P.):
- Redraw the Top View with the axis at 45° to XY.
- Project the new Front View from the rotated top view.

Drawing Procedure:
1. Draw XY. Complete Step 1 views.
2. Redraw the top view: rotate so that the axis makes 45° to XY, keeping the base corner on XY.
3. Project all corners of the rotated top view vertically up to XY.
4. Draw horizontal lines from the Step 1 front view corners to intersect with the projectors from Step 3.
5. Join the intersection points to get the new front view.

Result: Top View = pyramid outline with axis at 45° to XY; Front View = foreshortened pyramid.

Given:
- Frustum of cone: base diameter = 40 mm, top diameter = 20 mm, axis = 50 mm
- Resting on H.P. (on curved surface)
- Axis parallel to H.P. and inclined to V.P. at 30°

Two-step method (axis parallel to H.P., inclined to V.P.):

Step 1 — Assume axis ⊥ to V.P.:
- Front View: two concentric circles — outer ∅40 mm (base), inner ∅20 mm (top face).
- Top View: trapezium with base 40 mm, top 20 mm, height 50 mm, bottom edge on XY.

Step 2 — Rotate the top view so that the axis makes 30° to XY:
- Redraw the Top View with the axis at 30° to XY (towards right).
- Divide the circles into 12 parts; project each point to get the new front view.

Drawing Procedure:
1. Draw XY. Complete Step 1 views.
2. Redraw the top view: rotate the trapezium so that its axis makes 30° to XY.
3. Project all points from the rotated top view vertically up to XY.
4. Draw horizontal lines from the Step 1 front view points to intersect with the projectors from Step 3.
5. Join the intersection points with smooth curves to get the new front view (elliptical outlines).

Result: Top View = tilted trapezium at 30° to XY; Front View = frustum outline with elliptical ends.

Given:
- Square duct (frustum of square pyramid): bottom side = 90 mm, top side = 60 mm, length (axis) = 110 mm
- Axis parallel to H.P. and inclined at 60° to V.P.
- Thickness negligible.

Two-step method (axis parallel to H.P., inclined to V.P.):

Step 1 — Assume axis ⊥ to V.P.:
- Front View: two concentric squares — outer 90 mm × 90 mm (base), inner 60 mm × 60 mm (top face), both centred.
- Top View: trapezium (side view of frustum): base 90 mm, top 60 mm, height 110 mm, bottom edge on XY.

Step 2 — Rotate the top view so that the axis makes 60° to XY:
- Redraw the Top View with the axis at 60° to XY.
- Project the new Front View from the rotated top view.

Drawing Procedure:
1. Draw XY. Complete Step 1 views.
2. Redraw the top view: rotate the trapezium so that its axis makes 60° to XY.
3. Project all corners of the rotated top view vertically up to XY.
4. Draw horizontal lines from the Step 1 front view corners to intersect with the projectors from Step 3.
5. Join the intersection points to get the new front view (foreshortened frustum).

Result: Top View = tilted trapezium at 60° to XY; Front View = foreshortened square frustum outline.

Given:
- Square prism: base edge = 30 mm, axis = 55 mm
- Resting on base on H.P.
- Axis inclined at 30° to V.P.

Two-step method (axis ⊥ to H.P., inclined to V.P.):

Step 1 — Assume one side of base parallel to V.P.:
- Top View: square (30 mm sides) with one side parallel to XY.
- Front View: rectangle 30 mm × 55 mm with base on XY.

Step 2 — Rotate the top view so that the axis makes 30° to XY:
- Redraw the Top View with the square rotated so that the axis (perpendicular to base) makes 30° to XY...
- Actually, since the axis is ⊥ to H.P. (resting on base), the axis is vertical. "Axis inclined at 30° to V.P." means the axis (when viewed from top) makes 30° to V.P., i.e., one side of the base makes 30° to XY.
- Redraw the Top View with one side of the square at 30° to XY.
- Project the new Front View.

Drawing Procedure:
1. Draw XY. Complete Step 1 views.
2. Redraw the top view: rotate the square so that one side makes 30° to XY.
3. Project all corners of the rotated top view vertically up to XY.
4. Draw horizontal lines from the Step 1 front view corners to intersect with the projectors from Step 3.
5. Join the intersection points to get the new front view.

Result: Top View = square with one side at 30° to XY; Front View = foreshortened prism outline, 55 mm tall.

Given:
- Hexagonal prism: base edge = 20 mm, axis = 50 mm
- Resting on base on H.P.
- Axis parallel to H.P. and inclined at 40° to V.P.

Two-step method (axis parallel to H.P., inclined to V.P.):

Step 1 — Assume axis ⊥ to V.P.:
- Front View: regular hexagon (20 mm sides) with one side on XY (resting on H.P.).
- Top View: rectangle 50 mm × (height of hexagon) with bottom edge on XY.

Step 2 — Rotate the top view so that the axis makes 40° to XY:
- Redraw the Top View with the rectangle rotated so that its length (axis) makes 40° to XY.
- Project the new Front View.

Drawing Procedure:
1. Draw XY. Complete Step 1 views.
2. Redraw the top view: rotate the rectangle so that its length makes 40° to XY, keeping the base edge on XY.
3. Project all corners of the rotated top view vertically up to XY.
4. Draw horizontal lines from the Step 1 front view corners to intersect with the projectors from Step 3.
5. Join the intersection points to get the new front view (foreshortened hexagonal prism).

Result: Top View = rectangle at 40° to XY; Front View = foreshortened hexagonal prism outline.

Given:
- Triangular prism: base edge = 50 mm, axis = 60 mm
- Resting on base on H.P.
- Axis inclined at 45° to V.P.

Two-step method (axis ⊥ to H.P., inclined to V.P.):

Step 1 — Assume one side of base parallel to V.P.:
- Top View: equilateral triangle (50 mm sides) with one side parallel to XY.
- Front View: rectangle 50 mm × 60 mm with base on XY.

Step 2 — Rotate the top view so that the axis makes 45° to XY:
- Since the axis is ⊥ to H.P. and the prism rests on its base, "axis inclined at 45° to V.P." means one side of the triangular base makes a certain angle to XY.
- Rotate the Top View so that the axis direction (perpendicular to the base edge) makes 45° to XY.
- Project the new Front View.

Drawing Procedure:
1. Draw XY. Complete Step 1 views.
2. Redraw the top view: rotate the equilateral triangle so that the axis (perpendicular from base to apex) makes 45° to XY.
3. Project all corners of the rotated top view vertically up to XY.
4. Draw horizontal lines from the Step 1 front view corners to intersect with the projectors from Step 3.
5. Join the intersection points to get the new front view.

Result: Top View = equilateral triangle rotated at 45°; Front View = foreshortened triangular prism, 60 mm tall.

Given:
- Hexagonal pyramid: base edge = 25 mm, axis = 60 mm
- Resting on base on H.P.
- Axis inclined to V.P. at 30°.

Two-step method (axis ⊥ to H.P., inclined to V.P.):

Step 1 — Assume one side of base parallel to V.P.:
- Top View: regular hexagon (25 mm sides) with one side parallel to XY, and apex at the centroid.
- Front View: triangle (isosceles) with base 50 mm (across flats) and height 60 mm, base on XY.

Step 2 — Rotate the top view so that the axis makes 30° to XY:
- Rotate the Top View (hexagon with apex) so that the axis direction makes 30° to XY.
- Project the new Front View.

Drawing Procedure:
1. Draw XY. Complete Step 1 views.
2. Redraw the top view: rotate the hexagon so that the axis (from centroid to apex) makes 30° to XY.
3. Project all corners (6 base corners + apex) of the rotated top view vertically up to XY.
4. Draw horizontal lines from the Step 1 front view corners to intersect with the projectors from Step 3.
5. Join the intersection points to get the new front view (foreshortened hexagonal pyramid).

Result: Top View = hexagon with apex, axis at 30° to XY; Front View = foreshortened hexagonal pyramid outline, 60 mm tall.