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.

Given: Regular hexagon, base side = 30 mm, plane in H.P.

Concept: Enclose the hexagon in a rectangle, find key point coordinates, and transfer them onto the isometric top face using the two 30° axes.

Step-by-step procedure:

Step 1 – Draw the true shape and find the enclosing rectangle.
Draw the regular hexagon with side 30 mm in true shape (flat, in H.P.).
- Place the hexagon with one pair of sides horizontal (flat orientation).
- Width of hexagon $= 2 \times 30 = 60$ mm (distance between two parallel vertical sides).
- Height of hexagon $= \sqrt{3} \times 30 = 51.96 \approx 52$ mm (distance between two parallel horizontal sides).
- Enclosing rectangle: $60$ mm wide $\times$ $52$ mm tall (in true shape).
- Label the six vertices $A, B, C, D, E, F$ and note their offsets from the bottom-left corner of the rectangle:
- $A = (15, 0)$, $B = (45, 0)$, $C = (60, 26)$, $D = (45, 52)$, $E = (15, 52)$, $F = (0, 26)$.

Step 2 – Draw the isometric enclosing rectangle on the H.P. face.
- On the isometric drawing, the H.P. face uses the right 30° axis (for width = 60 mm) and the left 30° axis (for depth = 52 mm).
- From origin $O$:
- Mark $P$ at 60 mm along the right 30° axis.
- Mark $Q$ at 52 mm along the left 30° axis.
- Complete the parallelogram $OPRQ$ (where $R = P + Q$ vectorially).

Step 3 – Transfer all six vertices onto the isometric parallelogram.
For each vertex, its $x$-offset is measured along the right 30° axis and its $y$-offset along the left 30° axis:
- $A$: 15 mm along right axis, 0 along left axis.
- $B$: 45 mm along right axis, 0 along left axis.
- $C$: 60 mm along right axis, 26 mm along left axis.
- $D$: 45 mm along right axis, 52 mm along left axis.
- $E$: 15 mm along right axis, 52 mm along left axis.
- $F$: 0 along right axis, 26 mm along left axis.

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

Step 5 – Isometric scale.
For true isometric projection, multiply all dimensions by $0.816$:
- Side $= 30 \times 0.816 = 24.5$ mm; all offsets scaled similarly.

Result: The isometric projection of the regular hexagon (side 30 mm) in H.P. is obtained as a six-sided figure on the isometric top face, appearing as a flattened hexagon due to the 30° projection.

Given: Circle, diameter = 50 mm (radius = 25 mm), plane in H.P.

Concept: A circle in isometric projection appears as an ellipse. It is drawn using the four-centre method (approximate method), which constructs the ellipse using four circular arcs.

Step-by-step procedure:

Step 1 – Determine the isometric square.
The circle is enclosed in a square of side = diameter = 50 mm.
In isometric projection (using isometric scale): side $= 50 \times 0.816 = 40.8$ mm $\approx 41$ mm.
*(For isometric drawing, use 50 mm directly.)*

Step 2 – Draw the isometric square (rhombus) on the H.P. face.
- From origin $O$, draw two lines at 30° (right and left) each of length 50 mm (or 41 mm for projection).
- Complete the rhombus $ABCD$ where $AB$ and $CD$ are along the right 30° axis and $AD$, $BC$ are along the left 30° axis.
- The rhombus has angles of 60° and 120°.

Step 3 – Find the four centres for the four-centre ellipse method.
- Mark the midpoints of all four sides of the rhombus: $M_1$ (mid of $AB$), $M_2$ (mid of $BC$), $M_3$ (mid of $CD$), $M_4$ (mid of $AD$).
- The two obtuse-angle corners of the rhombus (120° corners) are centres $O_1$ and $O_2$.
- Draw perpendiculars from $O_1$ to sides $AB$ and $AD$; they meet at $M_1$ and $M_4$.
- Draw perpendiculars from $O_2$ to sides $BC$ and $CD$; they meet at $M_2$ and $M_3$.
- The intersections of lines $O_1 M_1$ extended and $O_2 M_2$ extended give centres $O_3$ and $O_4$ (the other two centres, located near the acute-angle corners).

Step 4 – Draw the four arcs.
- Arc 1: Centre $O_1$, radius $= O_1 M_1 = O_1 M_4$, draw arc from $M_4$ to $M_1$ (large arc, near obtuse corner).
- Arc 2: Centre $O_2$, radius $= O_2 M_2 = O_2 M_3$, draw arc from $M_2$ to $M_3$ (large arc, near other obtuse corner).
- Arc 3: Centre $O_3$, radius $= O_3 M_1 = O_3 M_2$, draw small arc from $M_1$ to $M_2$.
- Arc 4: Centre $O_4$, radius $= O_4 M_3 = O_4 M_4$, draw small arc from $M_3$ to $M_4$.

Step 5 – Complete the ellipse.
The four arcs join smoothly at $M_1, M_2, M_3, M_4$ to form the approximate ellipse representing the circle in isometric projection.

Result: The isometric projection of the circle (dia 50 mm) in H.P. is an ellipse drawn by the four-centre method, inscribed in the isometric rhombus on the horizontal face.

Given: Semi-circle, radius = 30 mm (diameter = 60 mm), plane in H.P.

Concept: The semi-circle is half of a full circle. First, construct the full isometric ellipse of the circle (diameter 60 mm) using the four-centre method on the H.P. face, then retain only the half that represents the semi-circle.

Step-by-step procedure:

Step 1 – Determine dimensions.
- Diameter of full circle $= 2 \times 30 = 60$ mm.
- Isometric scale factor $= 0.816$.
- Isometric diameter $= 60 \times 0.816 = 48.96 \approx 49$ mm (for isometric projection).
*(Use 60 mm for isometric drawing.)*

Step 2 – Draw the enclosing isometric rectangle for the semi-circle.
- The semi-circle fits in a rectangle of width = 60 mm and height = 30 mm (radius).
- On the H.P. isometric face, draw the isometric rectangle:
- From origin $O$, mark 60 mm along the right 30° axis to point $B$.
- From $O$, mark 30 mm along the left 30° axis to point $D$.
- From $B$, mark 30 mm parallel to $OD$ to point $C$.
- $OBCD$ is the isometric rectangle (parallelogram) enclosing the semi-circle.

Step 3 – Construct the full isometric rhombus (for reference).
Extend the rectangle to a full rhombus of side 60 mm to use the four-centre method:
- From $O$, mark 60 mm along the left 30° axis to point $A$.
- Complete the full rhombus $OABC'$ (where $C'$ is the fourth corner of the full rhombus).

Step 4 – Apply the four-centre method to the full ellipse.
Follow the same four-centre method as in Question 5 (with diameter 60 mm) to locate all four centres $O_1, O_2, O_3, O_4$ and midpoints $M_1, M_2, M_3, M_4$.

Step 5 – Draw only the half ellipse (semi-circle portion).
- The diameter line $OB$ (60 mm along the right 30° axis) divides the full ellipse into two halves.
- Retain only the upper half (or the half corresponding to the semi-circle in the original orientation) of the four-centre ellipse.
- This consists of:
- The large arc (centre $O_1$ or $O_2$) on the far side of the diameter.
- The two small arcs ($O_3$ and $O_4$) at the ends of the diameter.
- Draw the straight diameter line $OB$ (60 mm along 30° axis, scaled to 49 mm for projection).

Step 6 – Complete the figure.
- The semi-ellipse (half ellipse) plus the straight diameter line $OB$ together form the isometric projection of the semi-circle.
- Darken the semi-ellipse and the diameter line.

Result: The isometric projection of the semi-circle (radius 30 mm) in H.P. is a half-ellipse on the horizontal isometric face, with the straight diameter edge along the 30° isometric axis and the curved portion drawn using the relevant arcs of the four-centre method.