S1 EEE Engineering Graphics · Isometric Projection

Isometric Projections of Solids and Combinations of Solids

This notebook teaches how to construct isometric projections and isometric views of three-dimensional engineering solids from their two-dimensional orthographic views. Learners will master the box method, the four-center ellipse method, the true-radius rule for spheres, and the coaxial offset method for stacking multiple solids and real-world mechanical parts.

Before you start: Basic 2D geometric constructions (squares, regular pentagons, regular hexagons, and circles); Reading and drawing first-angle orthographic projections (front view and top view across an x-y reference line); Using standard drafting instruments (compass, set squares, and protractor).

Your path through this module

Isometric Axes, Planes, and the Isometric Scale: By tilting a 3D object so that all three coordinate axes make equal angles with the drawing plane, its three main axes appear at 30 degrees left, 30 degrees right, and vertical, with all parallel edges foreshortened to about 81.6% of their true length.
The Box and Coordinate Offset Method for Polyhedra: Because non-isometric lines do not stay at their true angles or lengths, we never measure them directly; instead, we enclose the shape in a rectangular box and transfer the corners by measuring along the box's isometric sides.
Circles, Cylinders, Cones, and Hemispheres (Four-Center Method): A circle enclosed in a square becomes an ellipse inside an isometric rhombus, which can be drawn smoothly with four compass arcs centered at the obtuse corners and the intersections of lines to the opposite midpoints.
Coaxial Combinations of Solids and the Sphere Rule: To stack solids coaxially, you locate the upper solid's base or center on the lower solid's top face using isometric offsets; if the upper solid is a sphere, its center rises by the isometric radius while its outline remains a circle of true radius.
Practice: Prisms, Pyramids, and Polyhedral Combinations: Apply the bounding-box and offset method to orient single prisms and stack two or three flat-faced polyhedra along a common vertical axis.
Practice: Combinations with Cylinders, Cones, and Holes: Combine the box method with four-center ellipses to place cones, cylindrical holes, and pyramids accurately onto flat polygonal or hemispherical surfaces.
Practice: Spheres Resting on Blocks, Cylinders, and Spheres: Place spheres onto square, cylindrical, hexagonal, and spherical bases by stepping up along the vertical axis by the sphere's isometric radius and drawing a circle with its true radius.
Practice: Frustums, Multi-Solid Stacks, and Engineering Applications: Synthesize two-base frustum constructions, four-center ellipses, and the sphere rule to draw complex multi-solid stacks and real-world mechanical shapes.

Isometric Axes, Planes, and the Isometric Scale

You already know how to represent 3D objects using separate 2D front and top orthographic views; now we begin learning how to combine those views into a single 3D pictorial drawing.

Imagine holding a cardboard shipping box and tilting it toward your eyes until you see its top, front-left, and front-right faces equally well. When the three mutually perpendicular corners of the box make equal angles with your flat drawing sheet, their projected lines are called the isometric axes. On your paper, if you draw a horizontal reference line, the three isometric axes meet at a single point: one goes up and left at 30°, one goes up and right at 30°, and the third goes straight up vertically at 90°. This means the three axes are spaced equally at 120° from one another on the flat page.

Once these axes are set up, any edge of an object that runs parallel to an isometric axis is called an isometric line, and any flat surface formed by two isometric axes is an isometric plane. Edges that are tilted and do not run parallel to any of the three isometric axes are called non-isometric lines.

Because the box is tilted away from your sheet, its edges appear slightly shorter than they really are, just as a pencil looks shorter when you point it toward your eye. This visual shrinking is called foreshortening. For isometric lines, the ratio of the apparent isometric length to the actual true length is a constant value given by $\frac{\cos 30^\circ}{\cos 45^\circ} = \sqrt{\frac{2}{3}} \approx 0.8165$ . In plain words, every isometric line shrinks to about 81.6% of its real size, while non-isometric lines shrink by different amounts and cannot be measured directly along their tilted paths.

Engineers use two slightly different styles depending on whether they apply this shrinkage, and you must always read the problem carefully to see which one is requested:

Feature	Isometric Projection	Isometric View (or Isometric Drawing)
Lengths used along axes	Foreshortened isometric lengths (81.6% of true size)	True lengths (100% of true size)
Why we use it	Mathematically exact optical representation	Faster to draw directly with a standard ruler

Worked example

Let us construct a graphical isometric scale to convert a true length of 40 mm into an isometric length without using a calculator, and then draw the isometric projection of a 40 mm cube.

Draw a horizontal baseline starting at point $O$ . From $O$ , draw a line at 45° to represent the true-length scale, and a lower line at 30° to represent the isometric-length scale.
Using your standard ruler, mark 40 mm along the 45° line from $O$ to a point $A$ . We use 45° because the trigonometry of a 45° and 30° right triangle naturally produces the ratio $\frac{\cos 30^\circ}{\cos 45^\circ}$ .
From point $A$ , drop a vertical line straight down (90° to the horizontal baseline) until it hits the 30° line at point $B$ . The distance $OB$ is the exact isometric length of 40 mm.
Pick a new starting corner $P$ on your sheet and draw a horizontal line through it. From $P$ , draw three isometric axes: 30° to the left, 30° to the right, and 90° straight up.
Set your compass to the isometric length $OB$ and step off this distance from $P$ along all three axes to locate the three adjacent corners of the cube. Complete the visible faces by drawing lines parallel to the isometric axes from these points.

Remember

Always draw isometric axes at 30° left, 30° right, and 90° vertical relative to a horizontal baseline.
Measure lengths only along isometric lines (lines parallel to the three isometric axes).
Use foreshortened lengths ( $\approx 81.6\%$ via the 45°/30° scale) for an isometric projection, and true ruler lengths ( $100\%$ ) for an isometric view.

The Box and Coordinate Offset Method for Polyhedra

Now that you can draw lines along the three isometric axes, you need a reliable way to draw inclined edges—like the sides of hexagons, pentagons, and pyramids—whose angles and lengths change in isometric views. This lesson teaches you how to trap any flat-faced solid (called a polyhedron) inside a simple rectangular box so you only ever have to measure along the isometric axes.

Imagine shipping an oddly shaped hexagonal nut by packing it snugly inside a standard rectangular cardboard box. In drafting, any line not parallel to the three isometric axes is a non-isometric line. Because non-isometric lines do not foreshorten equally and their angles distort, you can never measure their lengths or angles directly on an isometric drawing. Instead, you use the box method: draw a light rectangle, labeled $1$ - $2$ - $3$ - $4$ , tightly around the polygon in your 2D orthographic view, and then draw that rectangle in isometric as a parallelogram with $30°$ sides.

To put the polygon inside your isometric box, you use coordinate offsets, which are the distances from the box corners ( $1, 2, 3, 4$ ) to the polygon's corners (vertices) measured strictly along the box edges. Take these distances from the 2D view with your compass (using isometric lengths for an isometric projection, or true lengths for an isometric view) and step them off along the $30°$ edges of the isometric parallelogram. Connect the marked points with straight lines, and the angled edges appear automatically.

Once the base is drawn, you build the 3D solid by lifting it along the third axis. A prism has two identical parallel bases; to draw it, raise vertical lines of equal height from every base vertex and connect their tops. A pyramid tapers to a single point called the apex; find the center of the base by crossing the box diagonals, raise a vertical axis line to the required height to locate the apex, and join the apex to each base corner. For a frustum (a pyramid with its top sliced off flat), enclose the whole solid in a 3D box, draw the bottom base on the floor, and use coordinate offsets on the box's top face to plot the smaller top base before connecting the sloping edges. Finally, apply visibility rules: darken all outer boundary lines and front-facing edges, and leave rear edges that pass behind the solid out (or dashed if required).

Worked example

Draw the isometric view of a regular hexagonal prism with base side $20\text{ mm}$ and height $50\text{ mm}$ standing upright.

Draw the 2D box: Draw a regular hexagon of side $20\text{ mm}$ and enclose it snugly in a rectangle $1$ - $2$ - $3$ - $4$ . This turns angled sides into simple horizontal and vertical offsets along the rectangle.
Draw the isometric base box: From corner $1$ , draw edges $1$ - $2$ and $1$ - $4$ at $30°$ left and right, copying the rectangle's true width and depth. Complete the parallelogram $1$ - $2$ - $3$ - $4$ so you have an isometric frame.
Transfer offsets: With your compass, transfer the distance from each box corner to the six hexagon vertices $A$ through $F$ onto the parallelogram edges, then join $A$ - $B$ - $C$ - $D$ - $E$ - $F$ lightly. This locates the non-isometric edges without measuring angles.
Extrude the height: From each vertex $A$ through $F$ , draw a vertical line $50\text{ mm}$ up and connect the top points to form the parallel top hexagon.
Set visibility: Darken the entire top hexagon, the front bottom edges, and the four visible vertical edges, leaving the hidden back edges light.

Remember

Never measure the length or angle of a non-isometric line directly; always locate its two endpoints using offsets along the isometric box.
For pyramids, find the base center using box diagonals, step off the axis height along the third isometric axis to find the apex, and join the apex to the base corners.
Darken the top face, outer boundary, and front edges last; omit or dash the rear edges hidden behind the solid.

Circles, Cylinders, Cones, and Hemispheres (Four-Center Method)

You can now enclose and draw any flat-faced polygon using a rectangular box. In this lesson, you will apply that same enclosing-box idea to circles so you can draw curved solids like cylinders, cones, and hemispheres.

If you tilt a round coin in your hand, its circular edge looks like an oval. In an isometric drawing, any circle parallel to a coordinate plane projects as an ellipse, a smooth, symmetrical oval. To draw it, we first enclose the true circle in a square touching it at four midpoints ( $a, b, c, d$ ). When drawn along isometric axes, this square becomes a 30° rhombus (a tilted diamond with four equal sides) whose corners we label $1, 2, 3, 4$ .

Drawing a freehand ellipse is clumsy, so we use the four-center method, a compass technique that approximates the ellipse using four smooth circular arcs. Look at your isometric rhombus: two opposite corners are obtuse corners (wider than 90°), and two are acute (narrower than 90°). Draw straight lines from both obtuse corners to the midpoints ( $a, b, c, d$ ) of their opposite sides. These four lines cross each other inside the rhombus at two new points. Your four compass centers are the two obtuse corners (for the two long, shallow arcs) and the two inner intersection points (for the two tight end arcs).

Once you can draw one isometric circle, you can build curved 3D solids. A generator is an imaginary straight line on a curved surface that shapes its outer boundary, like the straight side silhouette of a tin can or a party hat. For a cylinder or a frustum (a cone with its pointed top sliced off flat), draw the bottom and top ellipses, then join their outer edges with straight lines that are common tangents—lines that smoothly kiss both ellipses without cutting inside them. For a cone, draw the base ellipse, Step up vertically from the rhombus center to the apex, and draw two tangent generators from the apex to the ellipse.

A hemisphere is an exact half-ball with one flat circular face and one curved dome. Because a ball looks equally wide from every viewing angle, the curved dome contour is drawn as a simple semicircular arc touching the extreme left and right edges of the ellipse. Whether the flat face rests on the ground or faces the sky, first draw the flat circular face as an isometric ellipse (using isometric length for an isometric projection), and then sweep a single compass arc from the ellipse's center to form the round dome.

Worked example

Let us draw the isometric view of a vertical cylinder of diameter 40 mm and height 60 mm.

Draw the base rhombus: From a front corner $1$ , draw two 40 mm lines at 30° left and 30° right, then complete the top-view rhombus $1\text{-}2\text{-}3\text{-}4$ so that corners $1$ and $3$ are obtuse. Mark the side midpoints $a, b, c, d$ .
Find the four centers and draw the base: Join obtuse corner $1$ to opposite midpoints $b$ and $c$ , and obtuse corner $3$ to $a$ and $d$ . Where these lines cross, mark centers $C_1$ and $C_2$ . Use $1$ and $3$ as centers to swing the two large arcs, and $C_1$ and $C_2$ to swing the two small end arcs.
Lift the top face and connect: Shift the four centers ( $1, 3, C_1, C_2$ ) and the four midpoints vertically upward by 60 mm to draw the identical top ellipse. Finally, connect the extreme left and right curves of both ellipses with two vertical tangent lines, and erase the hidden back half of the bottom ellipse.

Remember

Always start the four-center construction from the two obtuse corners of the isometric rhombus, never the acute ones.
Connect the outer boundaries of cylinders, cones, and frustums using straight lines tangent to the ellipses, not by joining the rhombus corners.
For a hemisphere, construct the flat circular face as a four-center ellipse first, then draw the curved dome as a circular arc.

Coaxial Combinations of Solids and the Sphere Rule

Now that you can draw any single polyhedron or curved solid using bounding boxes and four-center ellipses, we can combine them into real engineering parts. This lesson shows you how to stack one solid centered on top of another, including the special case of a sphere.

Think of stacking a small square gift box squarely in the middle of a larger one. When two solids share the same central vertical line, we call them coaxial. To stack two square-based solids coaxially, the center of the upper solid's base must sit directly on the center of the lower solid's top face. If the lower top face has side $B_1$ and the upper base has side $B_2$ , you can step inward from the lower edges by the symmetric offset distance $d = (B_1 - B_2) / 2$ along both 30° isometric axes. On your drawing, this frames a smaller isometric rhombus right in the middle of the larger one.

What if the top object is a ball, like a finial on a fence post? A sphere is a 3D solid where every surface point is at an equal distance—its radius—from the center. When a sphere rests on a flat face, it touches at a single point called the point of contact. Because you look at a sphere along tilted sightlines, the vertical line from the contact point to the sphere's center is foreshortened, just like any vertical edge. However, no matter what angle you view a ball from, its silhouette is always a full-sized circle. This gives us the Sphere Rule: in an isometric projection, raise the sphere's center above the contact point by the isometric radius ( $R_{\text{iso}} \approx 0.816 R$ ), then set your compass to the true radius ( $R$ ) to draw the outer circle. Because $R$ is larger than $R_{\text{iso}}$ , the circle's bottom edge will dip below the contact point on paper—this is normal and shows the front belly of the sphere curving toward you.

Finally, whenever an upper solid sits on a lower one, it blocks part of the view. This inter-solid visibility means you must keep construction lines light and darken only the visible front edges, leaving the rear parts of the lower top face erased where they pass behind the upper solid.

Worked example

Draw the isometric projection of a sphere of radius 25 mm resting centrally on the top face of a square slab of side 60 mm and height 20 mm.

Convert slab dimensions to isometric lengths: Multiply 60 mm and 20 mm by 0.816 to get $B_{\text{iso}} = 49$ mm and $H_{\text{iso}} = 16.3$ mm. We use isometric lengths because this is a true isometric projection.
Draw the lower slab: Construct a 30° isometric box with a 49 mm $\times$ 49 mm base and 16.3 mm vertical edges.
Find the point of contact: Lightly draw the two diagonals of the slab's top rhombus. Their intersection $P$ is the exact center where the sphere touches.
Locate the sphere's center: From $P$ , draw a vertical line straight up of length $R_{\text{iso}} = 25 \times 0.816 = 20.4$ mm to mark the center $C$ . The vertical radius foreshortens like any isometric axis.
Draw the sphere and check visibility: With center $C$ and your compass set to the true radius $R = 25$ mm, draw a full circle. Leave the back edges of the slab's top face light wherever they fall inside this circle.

Remember

To center a base of width $B_2$ inside a face of width $B_1$ , measure inward by $d = (B_1 - B_2) / 2$ along the 30° isometric axes.
The Sphere Rule: Step up from the contact point by the isometric radius ( $R_{\text{iso}}$ ) to find the center, then draw a circle of true radius ( $R$ ).
Always draw the lower solid's top face lightly first, since the upper solid will hide part of its rear edges.

Practice: Prisms, Pyramids, and Polyhedral Combinations

Having learned the box method and coaxial offset rules in theory, we start our practice with purely flat-faced solids where every vertex can be plotted directly by coordinates. You will now apply the bounding-box and offset method to orient single prisms and stack two or three flat-faced polyhedra along a common vertical axis.

The idea

Flat-faced solids (polyhedra) have straight edges that meet at sharp vertices. Even when a prism is tilted onto a side face or several solids are stacked on top of one another, you never need curves: you only need to locate corner points and join them with straight lines.

When a prism rests on a rectangular side face with its axis horizontal, its polygonal end faces are vertical. Because the polygon's slanted edges are non-isometric lines, you enclose the polygon in a 2D rectangle in the orthographic view and draw the corresponding 3D isometric box oriented vertically. You then transfer each polygon corner to the vertical box face using coordinate offsets before projecting the prism's length along the horizontal isometric axis.

When two or three polyhedra share a coaxial vertical axis, you build them sequentially from bottom to top. First, complete the bottom slab and locate the top face of its enclosing box. To center an upper solid of base width $B_2$ inside a lower box face of width $B_1$ , calculate the symmetric offset $(B_1 - B_2) / 2$ along both horizontal isometric directions. Stepping inward by these isometric offsets gives a smaller grid on the lower solid's top face, which serves as the exact footprint for raising the next solid.

Worked example

Given: A square slab of base edge 60 mm and height 20 mm is surmounted coaxially by a cube of side 40 mm. Find: Draw the isometric projection of the combination.

Construct an isometric scale converting the true lengths 20 mm, 40 mm, and 60 mm into foreshortened isometric lengths, because an isometric projection requires all parallel dimensions to be scaled by $\cos 30^\circ / \cos 45^\circ$ .
Draw the orthographic front and top views of the coaxial pair to visualize how the 40 mm cube sits inside the 60 mm square slab, leaving a uniform border of $(60 - 40) / 2 = 10$ mm on all four sides.

Draw the bottom isometric rhombus $ABCD$ of side ISO 60 mm with edges inclined at 30° to the horizontal to represent the base of the square slab.

Raise vertical edges of height ISO 20 mm from corners $A, B, C, D$ and join their tops to complete the lower square slab.

On the top face of the slab, step inward by an isometric offset of ISO 10 mm from each side to draw parallel grid lines that locate the centered base $PQRS$ of the 40 mm cube.

Erect vertical lines of height ISO 40 mm from $P, Q, R, S$ to complete the upper cube, darkening all visible front and top edges while omitting hidden rear lines.

Answer: Isometric projection of the 40 mm cube centered with a 10 mm offset on the 60 mm by 20 mm square slab.

Practice: Combinations with Cylinders, Cones, and Holes

Now that you can stack flat-faced blocks and pyramids, we introduce the four-center ellipse method into stacked combinations involving cylinders, cones, through-holes, and hemispherical bases. You will combine the box method with four-center ellipses to place cones, cylindrical holes, and pyramids accurately onto flat polygonal or hemispherical surfaces.

The idea

Whenever a combination includes a circular edge—whether it is the base of a cone, the rim of a cylinder, a drilled through-hole (a cylindrical cut passing completely through a solid), or the flat top of a hemisphere—that circle lies in an isometric plane and projects as an ellipse. Because you cannot align two ellipses directly by their curves, you must always enclose each circle in a square bounding box first. In isometric projection, these squares become concentric rhombuses (rhombuses that share the same center and axes) on the contact plane.

The key rule is to align the bounding boxes using isometric offsets before constructing any curves. To place a cone or drill a hole on a lower solid's top face, first find the center of that face, or step inward from the outer box edges by the isometric offset $(B_1 - B_2)/2$ . Once the inner rhombus is positioned, apply the four-center method inside it to draw the ellipse. For a cone, project the apex vertically from the rhombus center and draw tangent generators (straight lines from the apex that just touch the outer edges of the base ellipse).

When the lower solid is an upward-facing hemisphere, always start at its flat circular top face. Construct the top ellipse inside its isometric rhombus first, then sweep the lower spherical outline using the true radius (the exact unscaled radius $R = D/2$ ) centered at the ellipse's center, or tangent to the ellipse's major axis. Any upper solid—even when both solids share the exact same diameter—then sits directly on that flat top rhombus.

Worked example

Given: A cone of base diameter 50 mm and height 70 mm rests coaxially on the flat circular face of a hemisphere of diameter 50 mm. Find: Draw the isometric projection of the combined toy.

Draw the orthographic front and top views of the 50 mm cone resting on the 50 mm hemisphere, enclosing the circular top view in a square $1234$ of side 50 mm.
Construct an isometric rhombus $1234$ of side 50 mm with edges at 30° to the horizontal to represent the shared circular contact plane.
Use the four-center method inside rhombus $1234$ to draw the isometric ellipse for the shared 50 mm circular face.
Draw the lower curved outline of the hemisphere as a circular arc tangent to the outermost corners of the ellipse.
From the center of rhombus $1234$ , erect a vertical axis of isometric height 70 mm to locate the cone's apex.
Draw two straight generators from the apex tangent to the outer left and right arcs of the ellipse, and darken only the visible front arc of the shared base.

Answer: Isometric projection of a 50 mm diameter, 70 mm high cone coaxially mounted on a 50 mm diameter hemisphere.

Practice: Spheres Resting on Blocks, Cylinders, and Spheres

You already know how to construct flat-faced blocks and four-center cylinders, and how to locate centers on their top surfaces. Now we isolate and practice the special true-radius rule to place spheres onto square, cylindrical, hexagonal, and spherical bases.

The idea

When a sphere is tilted into an isometric position, its outer contour remains a circle of its full, unscaled true radius ( $R$ ), because a sphere looks the same from every viewing direction. However, the center of the sphere sits on a vertical isometric axis that foreshortens just like any other isometric line.

This creates a two-step rule for placing any sphere in an isometric projection. First, find the contact point on the supporting surface—such as the diagonal intersection of a square or the center of a top ellipse—and step vertically upward by the sphere's isometric radius ( $R_{\text{iso}} = 0.816 R$ ). Second, place your compass at that elevated center and draw a circle using the true radius ( $R$ ). (Even when a problem asks for an "isometric view" where the base block is drawn with true lengths, a sphere's center still steps up by its isometric radius while its outline is drawn with the true radius.)

When a second sphere rests coaxially on top of a lower sphere of radius $R_1$ , their point of tangency lies along the vertical axis between their centers. To find the upper sphere's center, step vertically upward from the lower sphere's center by the sum of both isometric radii ( $R_{1,\text{iso}} + R_{2,\text{iso}}$ ), then draw both circles with their respective true radii $R_1$ and $R_2$ . Finally, darken only the base edges that lie outside the overhanging circular outline of the sphere.

Worked example

Problem 205: Sphere resting on a square block Given: A square block of base side 70 mm and height 20 mm, with a sphere of diameter 60 mm resting centrally on its top face. Find: The isometric view of the combination.

Draw the orthographic front and top views of the 70 mm square block of height 20 mm with the 60 mm diameter sphere centered on top to establish the vertical axis and contact point.
Construct the isometric base rhombus $ABCD$ with sides of 70 mm at 30° to the horizontal, then project verticals of 20 mm from each corner to form the top square face of the block.
Draw the diagonals on the top face of the square block to intersect at the central contact point $O_1$ , because the sphere touches the flat top face right at its geometric center.
From the contact point $O_1$ , step vertically upward by the isometric radius of the sphere ( $R_{\text{iso}}$ for $R = 30$ mm) to locate the sphere's isometric center $O_2$ , since vertical distances along the axis are foreshortened.
With center $O_2$ and the sphere's true radius of 30 mm, draw a full circle for the sphere's outline, and darken the visible front edges of the square block outside the circle.

Answer: The isometric view is completed with a 70 mm by 20 mm square block and a 30 mm true-radius circle centered at $R_{\text{iso}}$ above the top face center.

Practice: Frustums, Multi-Solid Stacks, and Engineering Applications

Having mastered polyhedra, curved solids, and spheres on simple slabs, you are ready to combine all these techniques to draw tapered frustums, three-part curved assemblies, and practical engineering objects like rivets and bowls. In this section, you will synthesize two-base frustum constructions, four-center ellipses, and the sphere rule to draw complex multi-solid stacks and real-world mechanical shapes.

The idea

A frustum is a pyramid or cone sliced parallel to its base, leaving two unequal parallel faces. Because the sloping lateral edges or generators are non-isometric lines, you cannot measure their lengths or angles directly. Instead, you must use a two-plane box construction: build the isometric bounding box for the bottom base, locate its center, step vertically up by the frustum's isometric height, and construct a second, smaller isometric bounding box centered at that upper point. Connecting corresponding corners of the two polygons (or drawing common tangents to the two four-center ellipses for a cone) automatically gives the correct sloping edges.

When stacking a curved solid onto a frustum or cylinder, the orientation and size of the upper solid dictate how you locate its center. If a sphere or a bowl-like hemisphere rests on its curved surface, its point of contact is at the center of the lower solid's top face; you must step vertically upward by the isometric radius to find the sphere's center, then draw its outer contour using the true radius. Conversely, if a hemisphere rests flat on its circular base—or overhangs a narrower cylindrical shank like an engineering rivet head—its flat circular face lies directly in the top plane of the lower solid. You construct its four-center ellipse right on that plane (stepping outward if it overhangs) and sweep the dome tangentially using its true radius.

Worked example

Problem Q170: Sphere on a Frustum of a Square Pyramid

Given: A frustum of a square pyramid (bottom base 50 mm, top base 30 mm, height 40 mm) surmounted centrally by a sphere of diameter 30 mm. Find: The isometric projection of the combination.

Draw the front and top orthographic views of the square frustum and sphere using true dimensions to establish all coordinate offsets.
Construct an isometric rhombus $abcd$ of isometric side 50 mm with edges at 30° to the horizontal because the bottom square base lies on the horizontal plane.
Locate the base center by intersecting the diagonals of $abcd$ , then project vertically upward by the isometric height of 40 mm to establish the top face's center.
Around this upper center, construct a parallel isometric rhombus $efgh$ of isometric side 30 mm, which represents the smaller top square face at the correct elevation.
Join the visible corresponding corners ( $a$ to $e$ , $b$ to $f$ , and $c$ to $g$ ) with straight lines to form the non-isometric sloping edges of the frustum.
From the center of the top face $efgh$ , measure vertically upward by the sphere's isometric radius of 15 mm to locate the isometric center of the sphere.
With the compass set to the sphere's true radius of 15 mm, draw a full circle centered at the sphere's isometric center, because a sphere's contour always projects as its true radius.
Darken all visible edges of the frustum and erase the back edges of the top face that are hidden behind the opaque sphere.

Answer: The completed isometric projection shows the square frustum (isometric bases 50 mm and 30 mm, isometric height 40 mm) topped by a sphere of true radius 15 mm centered at an isometric height of 15 mm above the top face.