Isometric Projection

Given: Equilateral triangle, base side = 50 mm, plane in V.P. (Vertical Plane).

Concept: When a plane figure lies in the V.P., its isometric projection is drawn on the isometric front face. All lines parallel to the isometric axes are foreshortened by the isometric scale factor $\approx 0.816$ (i.e., multiply true length by 0.816). In isometric drawing (full-size method), true lengths are used directly.

Step-by-step procedure:

Step 1 – Draw the reference axes.
Draw the isometric axes: one vertical axis and two axes at 30° to the horizontal (left and right). Since the triangle is in the V.P., it will appear on the vertical isometric plane (the front face).

Step 2 – Enclose the triangle in a rectangle.
Enclose the equilateral triangle of side 50 mm in a rectangle.
- Height of equilateral triangle $= \dfrac{\sqrt{3}}{2} \times 50 = \dfrac{1.732}{2} \times 50 \approx 43.3$ mm.
- Rectangle dimensions: width = 50 mm, height = 43.3 mm.

Step 3 – Draw the isometric rectangle on the V.P. face.
On the isometric drawing, the V.P. face uses the vertical axis and the 30° right (or left) axis.
- Mark point $A$ at the bottom-left corner.
- Along the 30° axis, mark $B$ at 50 mm from $A$ (base of rectangle).
- From $A$ and $B$, draw vertical lines of height 43.3 mm to get points $D$ and $C$ respectively.
- $ABCD$ is the isometric rectangle enclosing the triangle.

Step 4 – Locate key points of the triangle inside the rectangle.
- The base of the triangle coincides with $AB$.
- The apex $P$ is at the midpoint of $DC$, i.e., midpoint of the top side of the rectangle.
- Mark $M$ = midpoint of $AB$; the apex $P$ is directly above $M$ at height 43.3 mm.

Step 5 – Draw the triangle.
Join:
- $A$ to $B$ (base, along 30° axis).
- $A$ to $P$ (left slant side — draw as a straight line between the two points).
- $B$ to $P$ (right slant side — draw as a straight line between the two points).

Step 6 – Isometric scale (if isometric projection, not drawing).
Multiply all dimensions by 0.816:
- Base $= 50 \times 0.816 = 40.8$ mm.
- Height $= 43.3 \times 0.816 = 35.3$ mm.
Redraw with these scaled dimensions following the same procedure.

Result: The isometric projection of the equilateral triangle (base 50 mm) in V.P. is obtained — it appears as a triangle on the isometric front face with the base along the 30° axis and the apex located vertically above the midpoint of the base.

Given: Square lamina, side = 55 mm, plane in H.P. (Horizontal Plane).

Concept: When a plane figure lies in the H.P., its isometric projection is drawn on the top (horizontal) isometric face. The two isometric axes used are the two axes at 30° to the horizontal (left and right). Vertical measurements do not apply here.

Step-by-step procedure:

Step 1 – Draw the isometric axes.
Draw the three isometric axes from a common origin $O$: one vertical, one at 30° to the left, and one at 30° to the right.

Step 2 – Draw the isometric parallelogram (rhombus) for the square.
The square in H.P. appears as a rhombus on the isometric top face.
- From origin $O$, mark point $A$ along the right 30° axis at 55 mm.
- From origin $O$, mark point $B$ along the left 30° axis at 55 mm.
- From $A$, draw a line parallel to $OB$ for 55 mm to get point $C$.
- From $B$, draw a line parallel to $OA$ for 55 mm to get point $C$ (same point).
- $OACB$ (or $OABC$) forms the isometric rhombus representing the square.

Step 3 – Check angles.
In the isometric rhombus, the angles are 60° and 120° (since the two 30° axes are at 60° to each other, the parallelogram has included angles of 60° and 120°).

Step 4 – Isometric scale (for true isometric projection).
Multiply side by 0.816:
$$\text{Isometric side} = 55 \times 0.816 = 44.9 \text{ mm} \approx 45 \text{ mm}$$
Redraw the rhombus with side 44.9 mm along both 30° axes.

Step 5 – Complete the figure.
Darken the four sides of the rhombus $OACB$. Label the corners.

Result: The isometric projection of the square lamina (side 55 mm) in H.P. is a rhombus with sides $\approx 44.9$ mm drawn on the horizontal isometric face, with sides along the two 30° isometric axes.

Given: Regular pentagon, base side = 35 mm, plane in V.P.

Concept: For irregular plane figures (pentagon), enclose the figure in a rectangle, locate key points by their coordinates (offsets), transfer those coordinates onto the isometric face using the isometric axes, and join the points.

Step-by-step procedure:

Step 1 – Draw the true shape and find coordinates.
Draw the regular pentagon $ABCDE$ with base $AB = 35$ mm in true shape (orthographic view).
- Interior angle of regular pentagon $= 108°$.
- Draw $AB$ as the base (horizontal, 35 mm).
- Locate all five vertices. Using geometry:
- Let $A = (0, 0)$, $B = (35, 0)$.
- $C = (35 + 35\cos 72°,\; 35\sin 72°) \approx (45.8,\; 33.3)$ mm.
- $E = (-35\cos 72°,\; 35\sin 72°) \approx (-10.8,\; 33.3)$ mm.
- $D$ = apex at midpoint of $CE$ horizontally $= (17.5,\; 35 + 35\sin 72° \times \tan 54°)$.
- More practically: height of pentagon $= \dfrac{35}{2}(1 + \sqrt{5})\sin 54° \approx 35 \times 1.539 \approx 53.9$ mm (total height from base to apex).
- Enclose in a rectangle of width = 35 + 2(35 cos 72°) $\approx$ 46.6 mm and height $\approx$ 53.9 mm.
- Mark all five vertices with their $x$ (horizontal) and $y$ (vertical) offsets from the bottom-left corner of the enclosing rectangle.

Step 2 – Draw the isometric enclosing rectangle on the V.P. face.
- On the isometric drawing, the V.P. face uses the 30° axis (horizontal direction) and the vertical axis.
- Draw the isometric rectangle: width $\approx 46.6$ mm along the 30° axis, height $\approx 53.9$ mm along the vertical axis.
- (For isometric projection, multiply all dimensions by 0.816 before drawing.)

Step 3 – Transfer all five vertices.
For each vertex, measure its $x$-offset along the 30° axis and its $y$-offset along the vertical axis from the reference corner, and mark the point on the isometric rectangle.

Step 4 – Join the vertices.
Join $A \to B \to C \to D \to E \to A$ with straight lines in sequence.

Step 5 – Isometric scale.
All offsets used in Step 3 should be multiplied by $0.816$ for true isometric projection (or used as-is for isometric drawing).

Result: The isometric projection of the regular pentagon (side 35 mm) in V.P. is obtained as an irregular five-sided figure on the isometric front face.