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.