Planes of projection: HP, VP and the XY line
We use two flat reference planes. HP (horizontal plane) is like the floor. VP (vertical plane) is like a wall standing on it. They meet in a straight line called the reference line XY.
Looking straight down on the object gives the top view (plan), drawn on HP. Looking straight at it from the front gives the front view (elevation), drawn on VP. Then HP is turned down into the paper so that both views sit on one sheet, with the top view below XY and the front view above XY.
Figure flat on HP (parallel to HP)
A thin plane figure has area but almost no thickness. When it lies parallel to HP:
- The top view is the true shape and true size.
- The front view is a straight line parallel to XY, as long as the figure is wide.
So the first job in every problem is to draw the true shape in the top view.
Figure tilted to HP: the two-stage method
Say the figure rests on one edge on HP, and that edge is parallel to VP. The figure is tilted by angle θ to HP.
- Stage 1: draw the figure flat. Top view = true shape, front view = a line.
- Stage 2: tilt the front-view line by θ about the resting edge. Then draw vertical lines up from the corners of the top view and horizontal lines across from the tilted front view. Where they meet are the corners of the new top view.
What changes? Only the depth (the size across the tilt). Top view depth = depth × cos θ. Front view height = depth × sin θ. Widths stay the same. The top view area = true area × cos θ.
When θ = 90° the top view is a line and the front view is the true shape.
Triangle, square, pentagon and hexagon
Draw the true shape from the side length a, then use the "depth" (the distance from the resting edge to the farthest point):
- Equilateral triangle: depth (altitude) = 0.866 a.
- Square: depth = a.
- Regular pentagon on a side: depth = 1.539 a. Draw it with a pentagon construction (five equal 72° turns).
- Regular hexagon with a side on HP: width across corners = 2a, depth across flats = 1.732 a.
The side resting on HP keeps its true length in the top view. If the figure rests on a corner instead, tilt it first about an edge and then turn it again; that needs a third step and is done at the next level.
Circle and semicircle
A circle of diameter d tilted to HP shows in the top view as an ellipse: the long axis stays d, the short axis becomes d cos θ. In the front view it becomes an ellipse of short axis d sin θ, and at 90° it is a line.
To draw it, mark 8 or 12 equal points on the circle (like the hours of a clock), tilt, project each point, and join them smoothly by a free curve.
A semicircle of diameter d is done the same way. Usually its diameter rests on HP, so the straight edge stays true length and the half-ellipse has depth (d/2) cos θ.
Key formulas and definitions
- Top view depth = true depth × cos θ
- Front view height = true depth × sin θ
- Widths (parallel to the resting edge) do not change
- Top view area = true area × cos θ
- Triangle altitude = 0.866 a; hexagon across flats = 1.732 a; pentagon depth = 1.539 a
Worked examples
1. A square of side 40 mm lies flat on HP. What are its views?
Top view: a square 40 × 40 mm (true shape). Front view: a straight line 40 mm long, parallel to XY.
2. The same square rests on an edge and is tilted at 30° to HP. Find the top view depth and front view height.
Depth = 40. Top view depth = 40 cos 30° = 34.64 mm, width 40 mm, so a rectangle 40 × 34.64. Front view height = 40 sin 30° = 20 mm.
3. An equilateral triangle of side 50 mm rests on a side on HP and is tilted at 60°. Find both views.
Altitude = 0.866 × 50 = 43.3 mm. Top view altitude = 43.3 cos 60° = 21.65 mm (base 50 mm still). Front view height = 43.3 sin 60° = 37.5 mm.
4. A regular hexagon of side 30 mm has a side on HP and is tilted at 45°. Find the top view depth.
Across flats = 1.732 × 30 = 51.96 mm. Top view depth = 51.96 cos 45° = 36.74 mm. The width across corners stays 60 mm.
5. A circle of diameter 50 mm touches HP at one point (on its edge) and is tilted at 60°. Describe its top view.
An ellipse. Long axis = 50 mm. Short axis = 50 cos 60° = 25 mm. The front view height is 50 sin 60° = 43.3 mm.
6. A square of area 1600 mm² is tilted at 60° to HP with an edge on HP. What is the area of its top view?
Area = 1600 cos 60° = 1600 × 0.5 = 800 mm². Check: side 40, so 40 × (40 × 0.5) = 40 × 20 = 800.
Common mistakes
- Changing the width as well. Only the depth (across the tilt) shrinks; widths along the resting edge do not.
- Mixing up cos and sin. The floor (top view) uses cos θ; the wall (front view) uses sin θ.
- Forgetting that the top view is the true shape only when the figure is parallel to HP.
- Joining the points of an ellipse with straight lines. Join them with a smooth curve.