Engineering Graphics and Computer Aided Drawing. Sections of Solids and Development of Surfaces, sections of solids, true shapes and development of surfaces.

Engineering Graphics: Sections of Solids and Development of Surfaces

By completing this notebook, you will be able to construct orthographic sectional views and determine the true shapes of cross-sections for polyhedra and solids of revolution cut by various planes. You will also learn how to unfold three-dimensional truncated solids into flat patterns (surface developments) and determine the shortest paths across curved and faceted surfaces.

Before you start: Orthographic projection fundamentals (First-Angle Projection, XY reference line, Front View, Top View, Side View); Projections of simple solids (prisms, pyramids, cylinders, cones, cubes, tetrahedra) resting on the Horizontal Plane (HP); Basic geometric constructions (regular polygons, dividing circles into equal sectors, drawing parallel and perpendicular lines).

Your path through this module

1. Introduction to Sections of Solids and Cutting Planes: A cutting plane appears as a single straight edge (trace) in the view corresponding to the reference plane it is perpendicular to, allowing us to see exactly where it slices the solid.
2. Finding Section Points and Projecting True Shapes: By setting up an auxiliary reference line parallel to the cutting plane trace and transferring distances from the opposite view, we view the cut surface head-on to reveal its True Shape.
3. Conic Sections and Reverse Sectioning Problems: Every width measured perpendicular to the cutting direction remains unchanged between the Top View and the True Shape, allowing us to work backward from a required shape to locate the cutting plane.
4. Principles of Surface Development: Parallel-Line Method: Unrolling a prism or cylinder creates a flat rectangle whose width equals the base perimeter and where vertical edges remain parallel, allowing cut heights to be projected directly across from the Front View.
5. Radial-Line Development and Shortest Surface Paths: Tapering solids unfold around their apex using radial arcs of true slant length, and any shortest path wrappped around a 3D surface becomes a simple straight line when drawn on the unfolded development.
6. Sections of Prisms and Pyramids by Auxiliary Inclined Planes (AIP): When an inclined plane cuts a prism or pyramid, the cut appears as a straight line in the Front View, and projecting its intersections straight down to the Top View reveals the polygonal boundary of the cut.
7. Sections of Cylinders and Cones by Auxiliary Inclined Planes (AIP): Because cylinders and cones have no sharp vertical edges, we divide their circular bases into equal sectors to create artificial generator lines that act just like pyramid edges during projection.
8. Sections by Auxiliary Vertical Planes (AVP) and Profile Planes: When a cutting plane is perpendicular to the Horizontal Plane, its trace appears in the Top View, and projecting upward to the Front View reveals the cut face—including hyperbolic curves on cones.
9. Reverse Section Problems: Finding the Cutting Plane from a Given True Shape: By plotting the known widths of the desired True Shape directly onto the Top View first, we can project upward to the Front View to discover the exact points the cutting plane must pass through.
10. Development of Prisms and Cylinders (Parallel-Line Method): Because every vertical edge or generator on a prism or cylinder is parallel to the Vertical Plane, its true cut height in the Front View transfers horizontally onto the unfolded rectangular stretch-out.
11. Development of Pyramids, Cones, and Shortest-Path Applications (Radial-Line Method): On pyramids and cones, cut heights on interior slant lines are foreshortened in the Front View and must first be projected horizontally to the outer true-length boundary before being marked on the radial development.

Introduction to Sections of Solids and Cutting Planes

You already know how to draw the front and top views of intact 3D solids resting on the Horizontal Plane (HP). In this lesson, you will learn how to slice through a solid with an imaginary flat sheet and represent the newly exposed inner surface on your drawing.

Imagine slicing an apple in half to check if the core is healthy. In engineering, parts often have complex hollow interiors that look like a confusing tangle of dashed hidden lines on a drawing. To see inside clearly, we pass an imaginary flat blade called a cutting plane (or section plane) through the object and mentally remove the piece between the blade and the observer. The newly exposed flat surface is called the section.

Because the cutting plane is always perpendicular to at least one reference plane, it projects as a single straight line—called a trace—on that plane. We draw this trace as a thin chain line (alternating long dashes and dots) that is thickened at both ends. Depending on how the blade is tilted, we get four main types of cutting planes:

When we look at the remaining part of the solid in our standard views, we get a Sectional Top View (Plan) or a Sectional Front View (Elevation). To show which area was physically touched by the blade, we fill the cut surface with section hatching: thin, parallel continuous lines drawn at $45^\circ$ to the reference line $XY$, spaced evenly about 2 mm to 3 mm apart. If the cutting plane is tilted (like an AIP or AVP), the standard view looks at the cut surface at an angle, making it look foreshortened. We call this foreshortened area the apparent section, whereas looking directly perpendicular at the cut reveals its True Shape.

Worked example

A cube of 40 mm side rests on the HP with its vertical faces equally inclined to the Vertical Plane (VP). Set up the cutting plane line for an AIP inclined at $30^\circ$ to the HP that bisects the vertical axis.

1. Draw the intact views: Draw the reference line $XY$. Below $XY$, draw a 40 mm square tilted at $45^\circ$ for the top view. Project upwards above $XY$ to draw the 40 mm high rectangular front view, including the central vertical axis.
2. Locate the midpoint of the axis: Measure 20 mm up from the base along the axis in the front view and mark point $M$. We do this because the plane must bisect the axis.
3. Draw the Vertical Trace ($VT$): Through point $M$, draw a line inclined at $30^\circ$ to $XY$ using a thin chain line thickened at the ends. We draw it in the front view because an AIP is perpendicular to the VP, so its edge view appears entirely on the front view.
4. Darken the retained portion: Darken the outline of the front view below the $VT$ line and leave the top part faint, showing the top piece is removed.

Remember

• A cutting plane projects as a single straight line (trace) in the view corresponding to the reference plane it is perpendicular to.
• Always fill the cut area in a sectional view with thin, equally spaced hatching lines at $45^\circ$ to $XY$.
• Inclined cutting planes (AIP and AVP) produce a foreshortened apparent section in the standard views, not the True Shape.

Finding Section Points and Projecting True Shapes

Now that you can represent a cutting plane as a straight line trace, you need a systematic method to project the cut points into the other views. This lesson shows you how to map the cut surface and find the exact size of the sliced face.

Imagine slicing a block of cheese at an angle. To draw the cut face, we first find its corners. For prisms and pyramids, we use the edge method: wherever the cutting plane trace crosses a solid's edge in one view, we mark a section point and project it straight across the $XY$ line to the same edge in the other view. What about smooth cylinders and cones that have no corners? We create artificial edges called generators—imaginary straight lines drawn along the curved surface by dividing the circular base into 8 or 12 equal parts—and project points onto them the exact same way.

Sometimes a pyramid's slant edge or a cone's generator stands perpendicular to the $XY$ line in the top view. We call this a profile line. Because both the line and your projector run vertically on the paper, a simple vertical projection cannot tell you where the point lands. To fix this, we use the rotation method: in the front view, move the section point horizontally onto an outside slant edge (which shows true length), project it down to that outer edge in the top view, and swing it back to the profile line with your compass centered at the apex.

When a cutting plane is tilted, the cut surface looks foreshortened in the standard views, just like a tilted coin looks like an oval. To see the True Shape—the actual, un-distorted size of the cut face—we must look directly perpendicular to the cutting plane. On your drawing, draw a new auxiliary reference line labeled $x_1\text{-}y_1$ parallel to the cutting plane trace. From each section point on the trace, draw projectors perpendicular to $x_1\text{-}y_1$. Finally, use your compass to measure the perpendicular distance of each point from $XY$ in the opposite view, and step that exact distance off from $x_1\text{-}y_1$ along the new projectors.

Worked example

A square pyramid (base 40 mm, axis 60 mm) rests on its base on the HP with base edges equally inclined to the VP. A cutting plane inclined at 45° to the HP bisects the axis. Find the sectional top view and True Shape.

1. Draw the basic views: Draw a 40 mm square in the top view with corners pointing up, down, left, and right. Connect corners to the center $o$ to represent slant edges $oa, ob, oc, od$. Project up to draw the 60 mm tall triangle in the front view.
2. Mark the cut: Draw the cutting plane trace $V.T.$ at 45° to $XY$ passing through the midpoint of the axis (30 mm above $XY$). Mark where $V.T.$ cuts the outer slant edges $o'a'$ and $o'c'$ as $1'$ and $3'$, and the middle vertical slant edges $o'b'$ and $o'd'$ as $2'$ and $4'$.
3. Project to top view: Project $1'$ and $3'$ straight down to $oa$ and $oc$ to get points $1$ and $3$. Since $ob$ and $od$ are profile lines, move $2'$ and $4'$ horizontally to the outer edge $o'c'$, project down to $oc$, and use a compass centered at $o$ to swing the point onto $ob$ and $od$ to locate $2$ and $4$. Join $1\text{-}2\text{-}3\text{-}4$ and hatch at 45°.
4. Project the True Shape: Draw reference line $x_1\text{-}y_1$ parallel to $V.T.$ Project lines from $1', 2', 3', 4'$ perpendicular to $x_1\text{-}y_1$. Measure the distances of $1, 2, 3, 4$ from $XY$ in the top view, and transfer them from $x_1\text{-}y_1$ along the matching projectors. Connect and hatch to reveal the True Shape.

Remember

• Use real edges for prisms and pyramids; create 8 or 12 imaginary generators for cylinders and cones.
• When a slant line is perpendicular to $XY$, shift the cut point horizontally to a true-length outer edge before projecting and swinging it back.
• Always measure from the old $XY$ line and step off from the new $x_1\text{-}y_1$ line to plot the True Shape.

Conic Sections and Reverse Sectioning Problems

You can now find the true shape resulting from a known cutting plane; next, we explore the special curves formed on cones and how to work backward when the desired true shape is specified first.

Imagine shining a round flashlight beam at a wall. Tilting the flashlight changes the circle of light into different curves, called conic sections. A cone’s curved surface is made of straight lines running from the base to the apex, called generators. In the Front View, the outer boundary lines of the cone are the extreme generators. By comparing the tilt of a cutting plane to these extreme generators, you can predict the exact curve of the True Shape before drawing it.

In a parabola or hyperbola, the flat base of the curve is the double ordinate (the chord width across the base circle in the Top View), and the height from that base to the curve's peak is the abscissa (the length of the cut line in the Front View).

In real design, you often know the cut shape you need—like a specific opening for a pipe—and must find the angle to make the cut. This is a reverse sectioning problem. The rule is that widths measured perpendicular to the cutting plane trace never change between the Top View and the True Shape. This means you can lay out the required width as a chord in the Top View first, project it up to the Front View, and swing an arc equal to the required True Shape length to find the exact cutting plane trace.

This reverse thinking also reveals the largest flat polygons you can slice inside faceted solids:

• Cube: Slicing through opposite edges gives a rectangle; tilting the plane to pass through the midpoints of six edges yields a regular hexagon.
• Tetrahedron: Slicing parallel to two opposite skew edges produces a square (at the midpoints) or a rectangle.

Worked example

A cone (base diameter 60 mm, axis 70 mm) rests on the Horizontal Plane. Find the cutting plane that produces a parabola with a 40 mm double ordinate.

1. Draw the views: Draw the Top View (a 60 mm circle) and project the Front View (a triangle 60 mm wide and 70 mm high) above the $XY$ line.
2. Place the double ordinate: Because widths perpendicular to the cut remain true in the Top View, draw a vertical chord 40 mm long inside the 60 mm base circle, to the right of the center.
3. Project to the base: Project the ends of this 40 mm chord straight up to meet the cone's base on the $XY$ line in the Front View. This point is the bottom of your cutting plane.
4. Draw the cutting plane: For a parabola, the cut must be parallel to an extreme generator. From your base point on the $XY$ line, draw a line parallel to the left extreme generator until it hits the right generator. The length of this cut line is the parabola's abscissa.

Remember

• Compare the cutting plane angle to the cone's extreme generator in the Front View: shallower gives an ellipse, parallel gives a parabola, steeper gives a hyperbola.
• Widths parallel to $XY$ in the Top View are already true lengths; use them to locate the base of a reverse section.
• The True Shape length along the symmetry line equals the exact length of the cutting plane trace in the Front View.

Principles of Surface Development: Parallel-Line Method

Having mastered how to slice 3D solids and view their cross-sections, we now learn how to unfold the remaining outer shell of prisms and cylinders into a flat sheet. This lets you design exact flat patterns for sheet-metal parts like ducts, pipes, and boxes.

Imagine cutting open a cardboard cereal box or peeling the paper label off a soup can and laying it flat on your desk. In engineering graphics, this unfolded flat pattern is called a development. Specifically, we usually draw the lateral surface development, which is the outer wrap of the solid, leaving out the top and bottom bases. Every line on a development must show its true length so a fabricator can cut the exact shape from a flat metal sheet and fold or roll it into the 3D object.

Prisms and cylinders have the exact same cross-section from bottom to top, meaning their vertical edges are all parallel to one another. Because of this, we unfold them using the parallel-line method. We start by drawing a long horizontal baseline, called the stretch-out line, right next to the Front View's base. The total length of this line is the stretch-out length ($P$), which equals the perimeter of the solid's base. On your drawing, you step off the true widths of each prism face ($P = n \times B$ for $n$ equal sides of width $B$) or calculate the circumference of a cylinder ($P = \pi D$ for diameter $D$).

Along the stretch-out line, we draw perpendicular vertical lines at each division point. For a prism, these are fold lines representing the sharp vertical corners; for a cylinder, we divide the circle into 8 or 12 equal parts to create imaginary vertical lines called generators. When a prism or cylinder stands upright on the Horizontal Plane, its vertical edges are parallel to the Vertical Plane, so the Front View already shows their true heights. This means we can project any cut point from a cutting plane trace in the Front View directly across—horizontally—to the matching vertical line on the development.

This flat unrolling also makes complex 3D curves remarkably simple. Think of a spiral staircase or a screw thread wrapping around a cylinder at a constant upward slope; this 3D curve is called a helix, and the vertical height it climbs in one full turn is its pitch. Because a helix rises at a steady rate as it wraps around the circumference, unfolding the cylinder turns the curved helix into a simple straight diagonal line drawn across the development, rising by one pitch for every full stretch-out length.

Worked example

Let us develop the lateral surface of a vertical square prism (base $30\text{ mm}$, height $60\text{ mm}$) resting on the Horizontal Plane with base edges equally inclined to the Vertical Plane, cut by a plane at $45^\circ$ to the Horizontal Plane bisecting its axis.

1. Draw the views and cut: Draw the Top View as a $30\text{ mm}$ square rotated at $45^\circ$, labeling corners $a, b, c, d$. Project up to draw the Front View ($60\text{ mm}$ tall) and draw the $45^\circ$ cutting plane line through the axis midpoint ($30\text{ mm}$ high), marking cut points $1', 2', 3', 4'$ on the vertical edges.
2. Lay out the stretch-out line: From the base of the Front View, extend a horizontal stretch-out line to the right of length $P = 4 \times 30\text{ mm} = 120\text{ mm}$ so the pattern unfolds right alongside the heights.
3. Draw the fold lines: Step off four $30\text{ mm}$ segments along the stretch-out line, label them $A, B, C, D, A$, and draw vertical fold lines up from each point to represent the four corners opening and closing at seam $A$.
4. Transfer the cut heights: Project horizontal lines from cut points $1', 2', 3', 4'$ in the Front View straight across to meet fold lines $A, B, C, D, A$ at points $1, 2, 3, 4, 1$, because horizontal projection preserves true vertical heights.
5. Complete the pattern: Connect $1\text{--}2\text{--}3\text{--}4\text{--}1$ with straight dark lines, and darken the base and outer vertical seams to finish the flat sheet-metal shape.

Remember

• The stretch-out length always equals the true perimeter of the base ($n \times B$ for regular prisms, $\pi D$ for cylinders).
• Connect cut points with straight lines for prisms, but use a smooth freehand curve for cylinders.
• A cylindrical helix always unfolds into a straight diagonal line on the development.

Radial-Line Development and Shortest Surface Paths

In the previous lesson, you unrolled straight-sided prisms and cylinders into flat rectangles using parallel lines. Now, you will unfold tapering solids like pyramids and cones around their tips, and use these flat patterns to find the shortest path across a three-dimensional surface.

Imagine rolling an ice cream cone covered in wet ink across a flat table. Because the tip stays pinned in one spot while the circular base rolls out, the cone leaves behind a fan-shaped wedge, or sector, of a circle. This is the radial-line method, where all fold lines or generators radiate outward from a single point called the apex. To draw this fan, you first need the true slant length ($L$), which is the actual distance from the apex to a base corner or rim. On a drawing, a slant edge only shows its true length in the Front View if its Top View is parallel to the $XY$ reference line. If none of a pyramid's slant edges are parallel to $XY$, you must rotate one edge in the Top View until it is parallel to $XY$, and then project it up to the Front View to reveal the true slant length $L$.

Once you have the true slant length $L$, you swing an arc of radius $L$ to start the development. For a pyramid, you step off the true base edge lengths as straight chords along this arc and connect them to create flat triangles. For a cone, the curved base unfolds into a smooth circular arc whose length equals the base circumference ($2\pi r$). This creates a sector with a subtended angle ($\theta$) given by:

where $r$ is the base radius and $L$ is the true slant generator length. If a cutting plane slices the pyramid or cone, you cannot transfer cut distances directly from interior edges in the Front View because those edges lean toward or away from you and appear foreshortened. Instead, you must project every cut point horizontally in the Front View over to the extreme true-length boundary edge, measure that true distance from the apex, and swing it onto the matching radial line in the development.

Suppose an ant wants to walk from a point on the base of a cone, all the way around its curved side, and return to the start by the shortest possible route. This shortest surface path, called a geodesic, looks like a complex curve in the Front and Top Views. However, because flattening a surface does not stretch or shrink it, the shortest path between any two points on a solid is simply a straight line drawn between those two points on its flat development. To show how this path looks in 3D, you mark where the straight line crosses each radial line on the development, swing those true distances back to the true-length edge in the Front View, and project them horizontally back to their respective generators.

Worked example

Find the development and the shortest path around a cone of base radius $r = 30$ mm and true slant length $L = 90$ mm, starting and ending at the same base point $A$.

1. Calculate the sector angle: Use the formula $\theta = (30 / 90) \times 360^\circ = 120^\circ$. We need this angle to make the unfolded arc length equal the base circumference.
2. Draw the flat cone: Pick an apex $O$, draw two radial lines of length $90$ mm at $120^\circ$ to each other, and swing an arc between their ends. Label the two outer base corners $A$ and $A_1$, since they meet when the cone is rolled up.
3. Draw the shortest path: Use a straightedge to draw a straight line directly from $A$ to $A_1$ across the sector. Because a straight line is the shortest distance on a flat sheet, this chord is the exact shortest path the ant can take around the cone.

Remember

• Always measure along a true-length slant edge (parallel to $XY$ in the Top View) before swinging radial arcs.
• Project interior cut points horizontally to the extreme true-length boundary edge in the Front View before transferring them to the development.
• The shortest path between two points on any developable surface is always a straight line on its flat development.

Sections of Prisms and Pyramids by Auxiliary Inclined Planes (AIP)

With the foundational theory complete, we begin our practice with the most straightforward sectioning problems: slicing flat-faced prisms and pyramids with a plane inclined to the horizontal. When an inclined plane cuts a prism or pyramid, the cut appears as a straight line in the Front View, and projecting its intersections straight down to the Top View reveals the polygonal boundary of the cut.

The idea

An Auxiliary Inclined Plane (AIP) is a cutting plane perpendicular to the Vertical Plane (VP) and inclined to the Horizontal Plane (HP). Because it stands perpendicular to the VP, the entire plane collapses into a single straight edge—its Vertical Trace (VT)—in the Front View. Every point where this trace crosses an edge of the solid is a cut point. Since the solid has flat faces bounded by straight edges, connecting the projected cut points in order produces a closed polygon.

How you project those points to the Top View depends on the type of edge the plane crosses. In a vertical prism, the lateral edges stand perpendicular to the HP, so their Top Views collapse into single corner points on the base perimeter; any cut along a vertical edge lands directly on that corner in the Top View. If the cutting plane climbs high enough to exit through the flat top face, it slices across the horizontal top edges, and those intersections project straight down onto the corresponding edges in the Top View. In a pyramid, the slant edges slope inward from the base corners to the apex, so they appear as distinct radiating lines in the Top View; you project each cut point from the Front View straight down until it hits its matching slant edge line in the Top View. Similarly, if a steep plane slices right through the base, the intersection projects straight down to the base edges.

In the sectional top view, the cut surface appears foreshortened because the cutting plane slopes away from the horizontal. To find the True Shape of the section, we must look at the cut face head-on. Set up an auxiliary reference line $x_1y_1$ parallel to the cutting plane trace $VT$, draw projectors perpendicular to $VT$ from each cut point in the Front View, and transfer the depth of each point (its distance from the main reference line $xy$ in the Top View) onto its matching projector.

Worked example

Problem 98 (86): Square prism cut by an inclined plane

Given: A square prism of base edge 40 mm and height 70 mm resting on HP on an end with rectangular faces equally inclined to VP, cut by an AIP inclined at 55° to HP meeting the axis 15 mm from the top. Find: The elevation, sectional plan, and true shape of the section.

1. Draw the reference line $y$-$y$ and construct the Top View as a 40 mm square $a(e)$-$b(f)$-$c(g)$-$d(h)$ with its sides inclined at 45° to $y$-$y$ so the vertical faces are equally inclined to VP.
2. Project the corners up to draw the Front View as a rectangle of height 70 mm resting on $y$-$y$, with top corners $a'$, $b'(d')$, $c'$ and base corners $e'$, $f'(h')$, $g'$.

3. Mark a point on the central axis $b'f'$ at 15 mm below the top face and draw the cutting plane line $V$-$T$ through it at 55° to $y$-$y$; mark intersection $1'$ on vertical edge $a'e'$, $2'$ on vertical edges $b'f'$ and $d'h'$, and $3'$ where $V$-$T$ exits through the top face between $b'(d')$ and $c'$.

4. Project the cut points down to the Top View: $1'$ lands directly on corner $a(e)$, $2'$ lands on corners $b(f)$ and $d(h)$ (marked $2$ and $2_1$), and $3'$ projects down to top edges $b$-$c$ and $d$-$c$ (marked $3$ and $3_1$); hatch the polygon $1$-$2$-$3$-$3_1$-$2_1$ and darken the retained lower boundary in the Front View.

5. Draw projectors perpendicular to $V$-$T$ from $1'$, $2'$, and $3'$, transfer each point's distance from $y$-$y$ in the Top View to locate $1$, $2$, $3$, $3_1$, and $2_1$, and join and hatch them to complete the True Shape.

Answer: Elevation, sectional plan (pentagonal cut surface $1$-$2$-$3$-$3_1$-$2_1$), and true shape constructed as shown.

More problems

Problem 86 (87): Pentagonal pyramid bisected by an inclined plane

Given: A pentagonal pyramid of base side 30 mm and height 60 mm resting on HP with one base edge perpendicular to VP, cut by an AIP at 45° to HP bisecting the axis. Find: The front view, sectional top view, true shape of the section, and true height of the section.

1. Draw the Top View as a regular pentagon $a$-$b$-$c$-$d$-$e$ of side 30 mm with left edge $a$-$e$ perpendicular to $x$-$y$ and slant edges meeting at apex $o$, then project up to form the Front View of height 60 mm with apex $o'$.

2. Draw the cutting plane $V$-$T$ at 45° to $x$-$y$ bisecting the axis at 30 mm above the base; mark intersections $1'$ on slant edges $o'a'(e')$, $2'$ on $o'b'(d')$, and $3'$ on $o'c'$.

Watch out. In the drawing above, the 30 mm height is measured at the right slant edge $o'c'$ rather than at the central axis.

3. Project $1'$, $2'$, and $3'$ straight down to their corresponding slant edges in the Top View to mark $1$ on $oa$, $1_1$ on $oe$, $2$ on $ob$, $2_1$ on $od$, and $3$ on $oc$, then join and hatch polygon $1$-$2$-$3$-$2_1$-$1_1$ and darken the retained lower edges.

4. Project lines perpendicular to $V$-$T$ from $1'$, $2'$, and $3'$, transfer the distances of $1$, $2$, $3$, $2_1$, and $1_1$ from $x$-$y$ in the Top View to plot and hatch the True Shape, and measure its height along the center line from edge $1$-$1_1$ to vertex $3$.

Answer: True height (width) of the section $W =$ 36 mm.

Problem 91 (88): Hexagonal pyramid cut by an inclined plane

Given: A hexagonal pyramid of base side 30 mm and axis 60 mm resting on HP with two base edges parallel to VP, cut by an AIP at 30° to HP meeting the axis 25 mm from the vertex. Find: The elevation, sectional plan, and true shape of the section.

1. Draw the Top View as a regular hexagon $a$-$b$-$c$-$d$-$e$-$f$ of side 30 mm with edges $b$-$c$ and $f$-$e$ parallel to $x$-$y$ and slant edges joined to center $o$, then project up the Front View with apex $o'$ at 60 mm above $x$-$y$.

2. Draw the cutting plane $V$-$T$ inclined at 30° to $x$-$y$ passing 25 mm above the base, and mark cut points $1'$ on $o'a'$, $2'$ on $o'b'(f')$, $3'$ on $o'c'(e')$, and $4'$ on $o'd'$.

Watch out. The problem text states 25 mm from the vertex, but the summary and construction place the plane 25 mm above the base at the right slant edge.

3. Project $1'$, $2'$, $3'$, and $4'$ straight down to the matching slant edges in the Top View to get points $1$, $2$, $3$, $4$, $3_1$, and $2_1$; connect and hatch the hexagon and darken the retained solid.

4. Draw projectors perpendicular to $V$-$T$ from $1'$, $2'$, $3'$, and $4'$, transfer each point's distance from $x$-$y$ in the Top View, and join and hatch $1$-$2$-$3$-$4$-$3_1$-$2_1$ to get the True Shape.

Answer: Elevation, sectional plan, and hexagonal true shape $1$-$2$-$3$-$4$-$3_1$-$2_1$ constructed as shown.

Problem 87 (89): Pentagonal pyramid cut at 60 degrees

Given: A pentagonal pyramid of 30 mm side and height 60 mm resting on HP with one base edge perpendicular to VP, cut by an AIP at 60° to HP meeting the axis 15 mm from the base. Find: The front view, sectional top view, true shape of the section, and true height of the section.

1. Construct the Top View as a regular pentagon $a$-$b$-$c$-$d$-$e$ of side 30 mm with edge $e$-$a$ perpendicular to $x$-$y$ and slant edges meeting at $o$, then project the Front View with apex $o'$ at height 60 mm.

2. Draw the cutting plane $V$-$T$ at 60° to $x$-$y$ passing 15 mm above the base at the right slant edge $o'c'$, and mark intersections $1'$ on $o'a'(e')$, $2'$ on $o'b'(d')$, and $3'$ on $o'c'$.

3. Project $1'$, $2'$, and $3'$ straight down to the Top View slant edges to locate $1$ on $oa$, $1_1$ on $oe$, $2$ on $ob$, $2_1$ on $od$, and $3$ on $oc$; join and hatch $1$-$2$-$3$-$2_1$-$1_1$ and darken the retained portion.

4. Project perpendiculars from $1'$, $2'$, and $3'$ across $V$-$T$, transfer the Top View distances from $x$-$y$ to plot and hatch the True Shape $1$-$2$-$3$-$2_1$-$1_1$, and measure its height from edge $1$-$1_1$ to vertex $3$.

Answer: True height (width) of the section $W =$ 43 mm.

Sections of Cylinders and Cones by Auxiliary Inclined Planes (AIP)

Now that you can section flat-edged polyhedra with an inclined plane, we apply the exact same projection method to curved solids by introducing imaginary generator lines. Because cylinders and cones have no sharp vertical edges, we divide their circular bases into equal sectors to create artificial generator lines that act just like prism or pyramid edges during projection.

The idea

A prism or pyramid has distinct straight edges where the cutting plane leaves clear intersection points. A cylinder or cone has a smooth, continuous curved surface with no edges to project. To locate enough section points to draw a smooth curve, we divide the circular base in the Top View into 8 or 12 equal sectors. The straight lines connecting these base points to the top face (on a cylinder) or to the apex (on a cone) are called generators. Once drawn in both views, generators work just like the lateral edges of a prism or pyramid.

When an Auxiliary Inclined Plane (AIP) cuts across all generators of a vertical cylinder, the Top View of the cut remains a full circle, while the True Shape stretches into an ellipse. If the cutting plane is steep enough to slice through the flat top or bottom face of the cylinder, that straight cut becomes a chord across the curve, producing a truncated ellipse.

On a cone, the shape of the section depends on the cutting plane angle relative to the cone's extreme outline (its extreme generator). When the cutting plane angle is less than the generator angle, it cuts across every generator and forms a closed ellipse. When the cutting plane is parallel to an extreme generator, it never meets that generator and exits through the base, producing an open, symmetric curve called a parabola. If a section point lies on the vertical center generator in the Front View, you cannot project it straight down to the Top View; instead, rotate it horizontally to the extreme generator, project down to the horizontal centerline, and swing an arc to the vertical centerline.

Worked example

Given: A cylinder of diameter 40 mm and height 60 mm rests on HP on its base, cut by a 30° AIP bisecting the axis. Find: The projections of the cut solid and the true shape of the section.

1. Draw the Top View as a circle of diameter 40 mm, and divide it into 8 equal parts to create imaginary vertical generators labeled $a(i)$ through $h(p)$ so we have reference lines to locate points on the smooth surface.
2. Project the base points up to the $xy$ line and draw the Front View as a 40 mm by 60 mm rectangle with vertical generator lines.

3. Mark the midpoint of the axis at 30 mm height, and draw the cutting plane line $V$-$T$ through it inclined at 30° to $xy$.
4. Label the intersection points $1'$ to $8'$ where the cutting plane crosses each vertical generator in the Front View.

5. Darken the retained lower boundary of the Front View, and hatch the entire Top View circle because the cutting plane slices across the full width of the vertical cylinder.

6. Draw an auxiliary reference line $x_1y_1$ parallel to the cutting plane $V$-$T$, and project perpendicular lines from points $1'$ through $8'$ across $x_1y_1$.
7. Measure the distance of each generator point from $xy$ in the Top View, transfer those distances from $x_1y_1$ along the matching projectors, and join the points with a smooth curve to reveal the elliptical True Shape.

Answer: The sectional Top View is a circle of diameter 40 mm, and the True Shape is an ellipse with a minor axis of 40 mm and a major axis equal to the cut length $1'$-$5'$.

More problems

Problem 115: Cylinder cut through top face and generator

Given: A cylinder of diameter 40 mm and height 60 mm on HP, cut by a 50° AIP meeting the axis 45 mm above the base. Find: The sectional views and true shape of the section.

1. Draw the 40 mm diameter Top View circle divided into 8 generators, and project the 60 mm high Front View.

2. Draw the 50° cutting plane $V$-$T$ through the axis at 45 mm height, cutting the top face at $4'$ and generators at $1'$ to $7'$.

3. Project the top face cut point $4'$ straight down to the Top View circle to mark the boundary chord, and hatch the cut area to its right.

4. Project perpendiculars from $1'$ through $7'$ across a parallel reference line $x_1y_1$, transfer Top View distances from $xy$, and join into a smooth truncated ellipse.

Answer: The True Shape is a truncated ellipse bounded by a flat chord where the plane exits the top face.

Problem 117: Cone cut by an AIP to form an ellipse

Given: A cone of base diameter 50 mm and height 60 mm on HP, cut by a 50° AIP passing through the extreme left base point. Find: The sectional views and true shape of the section.

1. Draw the 50 mm diameter Top View circle with 8 radial generators connected to center $o$, and project the 60 mm high triangular Front View.

2. Draw the cutting plane at 50° to $xy$ starting from the extreme left base point $a'$, and mark intersection points $1'$ through $8'$ on the generators.

3. Project points down to their matching Top View radial generators; for points $3'$ and $7'$ on the vertical center generator, rotate horizontally to the extreme generator first, drop to $o$-$e$, and swing to $o$-$c$ and $o$-$g$.

4. Connect the Top View points into an ellipse and hatch it, then project from $1'$ to $8'$ across $x_1y_1$ and transfer Top View distances from $xy$ to draw the True Shape ellipse.

Watch out. Points on the center generator $o'c'(g')$ cannot be projected straight down; you must project them horizontally to the extreme generator $o'e'$ to find their true radial radius from the axis.

Answer: Both the sectional Top View and the True Shape are closed ellipses.

Problem 122: Cone cut parallel to an extreme generator

Given: A cone of base diameter 50 mm and axis 60 mm on HP, cut by an AIP parallel to and 10 mm away from the leftmost generator. Find: The front view, sectional top view, and true shape of the section.

1. Draw the 50 mm diameter Top View circle divided into 8 radial generators, and project the 60 mm high Front View.

2. Draw the cutting plane $V$-$T$ parallel to the left extreme generator $o'a'$ at a perpendicular distance of 10 mm, marking where it cuts the base and generators.

3. Project the base cut straight down to the Top View perimeter, project the generator intersections to their matching radial lines (using horizontal rotation for the center generator), and hatch the resulting open curve.

4. Set up $x_1y_1$ parallel to $V$-$T$, project all cut points perpendicular to $x_1y_1$, and transfer Top View distances from $xy$ to plot the parabolic True Shape.

Answer: The True Shape of the section is a parabola bounded by a flat chord across the base.

Sections by Auxiliary Vertical Planes (AVP) and Profile Planes

Having practiced cuts that appear as straight lines in the Front View, we now switch roles and work with vertical cutting planes that appear as straight lines in the Top View. When a cutting plane is perpendicular to the Horizontal Plane, its trace appears in the Top View, and projecting upward to the Front View reveals the cut face—including hyperbolic curves on cones.

The idea

An Auxiliary Vertical Plane (AVP) is a cutting plane perpendicular to the Horizontal Plane (HP) and inclined to the Vertical Plane (VP). Because it stands vertically, its entire cut surface collapses into a single straight edge—its Horizontal Trace (HT)—in the Top View. Instead of intersecting the vertical axis of the solid, the plane is positioned by its shortest distance (perpendicular offset) from the axis in the Top View. To locate the plane accurately, draw a construction circle centered on the top-view axis with a radius equal to this offset distance, and then draw the cutting plane line tangent to that circle at the required angle to $xy$.

Once the trace $HT$ is drawn in the Top View, it directly marks where the base edges, slant edges, or cone generators are sliced. Projecting these intersection points straight up to their corresponding lines in the Front View reveals the sectional elevation. To find the True Shape, place an auxiliary reference line $x_1y_1$ parallel to $HT$ in the Top View, project perpendiculars from the cut points, and transfer the vertical heights of the points above $xy$ from the Front View.

Two special cases simplify the work. First, when a vertical plane cuts a vertical cone parallel to its axis (without passing through the apex), the True Shape of the cut is always a hyperbola (specifically a rectangular hyperbola when parallel to VP). Second, a profile cutting plane is perpendicular to both HP and VP; its trace appears as a vertical line in both the Front and Top Views, and projecting horizontal heights directly across to the Side View reveals the True Shape without needing an auxiliary view.

Worked example

Problem 118: Cone cut by a plane parallel to VP Given: A cone of base diameter 50 mm and axis length 60 mm rests on its base on HP, cut by a vertical section plane parallel to VP and 10 mm in front of the axis. Find: The sectional front view, top view, and true shape of the section.

1. Draw the Top View as a circle of diameter 50 mm centered at $o$ and divide it into 8 equal sectors ($a$ through $h$) to create artificial generators.
2. Project the base points up to $xy$ and draw the Front View triangle of height 60 mm with apex $o'$ and corresponding front-view generators.

3. In the Top View, draw the horizontal cutting plane trace $HT$ parallel to $xy$ and 10 mm below the axis $o$, because the vertical plane is parallel to VP and 10 mm in front of the axis.
4. Mark the intersection points $1$ and $5$ on the base circle, and points $2$, $3$, and $4$ where $HT$ crosses generators $ob$, $oc$, and $od$.

5. Project points $1$ and $5$ straight up to the base line $xy$ in the Front View, and project points $2$ and $4$ straight up to their respective generators $o'b'$ and $o'd'$.
6. Because generator $o'c'$ is vertical in both views, rotate point $3$ in the Top View around center $o$ to the outer horizontal generator $oe$, project up to the true slant edge $o'e'$, and carry that height horizontally back to $o'c'$ to locate point $3'$.
7. Join points $1, 2, 3, 4, 5$ in the Front View with a smooth hyperbolic curve and hatch the area; since the cutting plane is parallel to VP, this sectional front view directly shows the True Shape.

Answer: The sectional front view directly gives the true shape, which is a rectangular hyperbola.

More problems

Problem 88: Pentagonal pyramid cut by a 30° vertical plane

Given: A pentagonal pyramid (base side 30 mm, height 60 mm) on HP with one base edge perpendicular to VP, cut by an AVP inclined at 30° to VP and 10 mm from the axis. Find: The sectional front view, top view, true shape, and true height of the section.

1. Draw the Top View pentagon $abcde$ of side 30 mm with edge $ae$ perpendicular to $xy$, join corners to apex $o$, and project the Front View of height 60 mm.

2. In the Top View, draw the cutting plane line $HT$ inclined at 30° to $xy$ and tangent to a 10 mm radius circle around $o$, marking cut points $1$ and $4$ on the base edges and $2$ and $3$ on slant edges $ob$ and $oc$.

3. Project points $1$ and $4$ up to $xy$ at $1'$ and $4'$, and project $2$ and $3$ up to slant edges $o'b'$ and $o'c'$ at $2'$ and $3'$, then connect and hatch $1'2'3'4'$.

4. Draw auxiliary line $x_1y_1$ parallel to $HT$, project perpendiculars from points $1, 2, 3, 4$, and transfer the heights of $1', 2', 3', 4'$ above $xy$ from the Front View to complete the True Shape.

Answer: The sectional views and true shape polygon $1''2''3''4''$ are constructed as shown.

Problem 92: Hexagonal pyramid cut by a 20° vertical plane

Given: A hexagonal pyramid (base side 30 mm, axis 60 mm) on HP with two base edges parallel to VP, cut by an AVP inclined at 20° to VP passing 15 mm from the axis. Find: The sectional front view, top view (plan), and true shape of the section.

1. Draw the Top View regular hexagon $abcdef$ of side 30 mm with two edges parallel to $xy$, and project the Front View pyramid of height 60 mm.

2. Draw the cutting plane trace $HT$ in the Top View inclined at 20° to $xy$ at a perpendicular distance of 15 mm from $o$, marking intersections $1$ and $4$ on the base and $2$ and $3$ on slant edges $ob$ and $oc$.

3. Project points $1, 2, 3, 4$ straight up to the corresponding base and slant edges in the Front View to mark $1', 2', 3', 4'$, and hatch the cut face.

4. Set up reference line $x_1y_1$ parallel to $HT$, project perpendiculars from $1, 2, 3, 4$, and transfer the heights of $1'$ through $4'$ above $xy$ to plot the True Shape.

Watch out. The problem text says "meeting the axis 15 mm from the vertex", which means a perpendicular offset of 15 mm from the top-view center $o$.

Answer: The sectional views and true shape trapezium $1''2''3''4''$ are completed as shown.

Problem 89: Pentagonal pyramid cut by a 45° vertical plane

Given: A pentagonal pyramid (base side 30 mm, height 60 mm) on HP with one base edge parallel to VP, cut by an AVP inclined at 45° to VP and 10 mm from the axis. Find: The sectional front view, top view, true shape, and true height of the section.

1. Draw the Top View pentagon $abcde$ (side 30 mm) with edge $ed$ parallel to $xy$, and project the Front View triangle of height 60 mm.

2. Draw the cutting plane trace $HT$ in the Top View at 45° to $xy$ and 10 mm from center $o$, cutting the base at $1$ and $4$ and slant edges $od$ and $oc$ at $2$ and $3$.

3. Project $1, 2, 4$ up to $1', 2', 4'$; because edge $oc$ is vertical, rotate point $3$ around $o$ to horizontal line $oc_1$, project to true slant edge $o'c_1'$, and carry horizontally to $o'c'$ to find $3'$.

4. Draw an auxiliary line parallel to $HT$, project perpendiculars from $1, 2, 3, 4$, and transfer the front-view heights above $xy$ to construct the True Shape.

Answer: The sectional front view, top view, and true shape are obtained as shown.

Problem 119: Cone cut by a 45° vertical plane

Given: A cone (diameter 50 mm, height 60 mm) on HP, cut by a 45° vertical plane passing 10 mm from the axis. Find: The projections of the cut solid and the true shape of the section.

1. Draw the Top View circle of diameter 50 mm divided into 8 sectors ($a$ through $h$) and project the Front View cone of height 60 mm.

2. In the Top View, draw the cutting plane line $HT$ at 45° to $xy$ and tangent to a 10 mm radius offset circle around $o$, marking cut points $1$ through $5$ on the base and generators.

3. Project points $1$ through $5$ up to their respective front-view generators (rotating point $2$ on the vertical generator $oc$ to the outer generator first) and join them into a hatched curve.

4. Draw auxiliary line $x_1y_1$ parallel to $HT$, project perpendiculars from points $1$ through $5$, and transfer heights above $xy$ from the Front View to draw the hyperbolic True Shape.

Answer: The true shape of the section is a hyperbola.

Problem 93: Hexagonal pyramid cut by a profile plane

Given: A hexagonal pyramid (base side 30 mm, axis 60 mm) on HP with two base edges perpendicular to VP, cut by a plane perpendicular to both HP and VP at 15 mm left of the axis. Find: The sectional side view, plan, elevation, and true shape of the section.

1. Draw the Top View hexagon (side 30 mm) with two edges perpendicular to $xy$, project the Front View (height 60 mm), and construct the initial Side View across vertical reference line $x'y'$.

2. Draw the vertical cutting plane line 15 mm to the left of the axis in both the Front and Top Views, marking top-view intersections $1, 2, 3, 4$ and front-view heights on the vertical trace.

3. Project the top-view widths of $1, 2, 3, 4$ to the Side View via a 45° mitre line (or compass arcs) and intersect them with horizontal projectors from the Front View to hatch the True Shape.

Answer: The sectional side view directly reveals the true shape of the section.

Problem 120: Cone cut by a profile plane for a given hyperbola base

Given: A cone (base diameter 50 mm, axis 60 mm) on HP, cut by a plane perpendicular to both HP and VP so the true shape is a hyperbola of 40 mm base. Find: The front view, top view, and sectional side view showing the true shape.

1. Draw the Top View circle (diameter 50 mm, 8 sectors), project the Front View (height 60 mm), and project the uncut Side View across vertical line $x'y'$.

2. In the Top View, locate the vertical cutting chord of length 40 mm (20 mm above and below horizontal centerline $ae$), and extend this vertical cutting plane line up into the Front View.

3. Project the cut points from the Top View and their heights from the Front View into the Side View, then join them with a smooth curve and hatch the resulting hyperbola.

Answer: The vertical profile plane cuts a chord of 40 mm in the Top View and produces the required hyperbolic true shape in the Side View.

Reverse Section Problems: Finding the Cutting Plane from a Given True Shape

You can now find the true shape for any given cutting plane; in this challenging set, we reverse the process to find the unknown cutting plane angle (and even unknown solid heights) from a specified true shape. By plotting the known widths of the desired True Shape directly onto the Top View first, we can project upward to the Front View to discover the exact points the cutting plane must pass through.

The idea

When an Auxiliary Inclined Plane (AIP) slices a solid, it tilts relative to the Horizontal Plane (HP) while staying perpendicular to the Vertical Plane (VP). Because of this orientation, any line on the cut surface that runs perpendicular to the reference line $xy$ is completely horizontal. Horizontal lines appear in their exact true length in the Top View. This geometric rule is the secret to solving reverse section problems: every width of the True Shape measured perpendicular to $xy$ is identical to its width in the Top View.

To find an unknown cutting plane, we work backward. Instead of starting in the Front View, we begin in the Top View by fitting the given True Shape widths (such as the parallel sides of a trapezium or the double ordinate, which is the base chord width of a parabola) between the solid's edges or base boundary. Projecting these fitted points straight up to the matching edges in the Front View locates the exact points the cutting plane trace $VT$ must pass through.

If the problem also gives a length measured along the slope of the cut—such as the altitude of a triangular section or the abscissa (the true length from a parabola's base chord to its vertex)—that dimension appears in its true length directly along the trace $VT$ in the Front View. By swinging an arc of this true length from one projected point in the Front View, we can locate the second cut point, or even determine the unknown height $H$ of the solid itself.

Worked example

Given: A square pyramid of base edge 40 mm and height 60 mm resting on HP is cut by a section plane to produce a trapezium with parallel sides 30 mm and 20 mm. Find: The sectional top view, true shape, section plane inclination $\theta$, and distance between the parallel sides.

1. Draw the Top View as a 40 mm square $abcd$ with sides parallel and perpendicular to $xy$, draw diagonals to meet at apex $o$, and project the Front View of height 60 mm with apex $o'$.

2. Because widths perpendicular to $xy$ do not foreshorten in the Top View, fit two vertical lines of lengths 30 mm and 20 mm symmetrically between the four slant edges in the Top View.
3. Project the ends of the 30 mm line up to slant edge $o'a'(d')$ to get $1'$, and project the 20 mm line up to slant edge $o'b'(c')$ to get $2'$; join $1'2'$ as the cutting plane trace $VT$ and measure its inclination $\theta$.

4. Hatch the trapezium in the Top View, then draw an auxiliary reference line $x_1y_1$ parallel to $1'2'$.
5. Project $1'$ and $2'$ perpendicular to $x_1y_1$, transfer the distances of the cut points from $xy$ in the Top View, and join them to complete the True Shape.

Answer: Inclination $\theta = 50°$; distance between parallel sides is the perpendicular length $1'2'$ along the trace.

More problems

Problem 100: Square prism cut to form a trapezium

Given: A square prism of base 40 mm and height 70 mm on HP with vertical faces equally inclined to VP, cut to give a trapezium of parallel sides 20 mm and 40 mm. Find: Sectional top view, true shape, and inclination $\theta$.

1. Draw the Top View as a 40 mm square rotated at 45° to $xy$ with corners $a(e)$, $b(f)$, $c(g)$, $d(h)$, and project the Front View to height 70 mm.

2. In the Top View, locate the 20 mm vertical chord between the left faces to project $1'$ onto the top face $a'b'(d')$ in the Front View.
3. Locate the 40 mm vertical width across the right half of the square and project up to the bottom right corner $g'$ as $2'$; join $1'2'$ to find the trace and measure $\theta$.

4. Set up $x_1y_1$ parallel to $1'2'$, project the cut points perpendicularly, and transfer Top View distances from $xy$ to draw the True Shape trapezium.

Answer: The cutting plane passes from the top face to the bottom right corner at inclination $\theta$.

Problem 103: Square prism of unknown height cut into hexagon

Given: A square prism of base 40 mm on HP with faces equally inclined to VP is cut into an irregular hexagon having two opposite parallel sides of 30 mm and four sides of 40 mm. Find: Unknown prism height $H$, sectional views, true shape, and inclination $\theta$.

1. Draw the 45°-rotated 40 mm square in the Top View and extend vertical projectors upward for the Front View.

2. Fit the two 30 mm parallel vertical chords symmetrically near the left and right corners in the Top View.
3. Construct the True Shape hexagon separately using the 30 mm parallel sides and 40 mm slant sides to measure its total length $L = 75$ mm.

4. Since the cut stretches from the bottom face to the top face over length $L = 75$ mm, swing an arc of radius 75 mm across the projected chords in the Front View to find the prism height $H = 71$ mm.

5. Complete the Front View rectangle of height 71 mm, draw the cutting trace $1'2'3'$, measure $\theta = 69°$, and project the True Shape beside it.

Watch out. Do not confuse the prism's vertical height $H$ with the True Shape's total length $L = 75$ mm; $L$ is the hypotenuse along the inclined trace.

Answer: Prism height $H = 71$ mm, inclination $\theta = 69°$.

Problem 106: Cube cut to form a maximum rhombus

Given: A cube of 40 mm side resting on HP with vertical faces equally inclined to VP, cut to produce a rhombus of maximum side length. Find: Sectional top view, true shape, inclination $\theta$, and rhombus side length.

1. Draw the Top View as a 40 mm square rotated at 45° to $xy$ and project the 40 mm high Front View.

2. To get a maximum rhombus, the plane must slice all four vertical faces from the top-left corner $a'$ ($1'$) to the bottom-right corner $g'$ ($3'$), passing through the center $2'$ on edge $b'(d')-f'(h')$.

3. Measure the trace inclination $\theta$, hatch the full square in the Top View, and project from $1'$, $2'$, $3'$ onto an auxiliary view parallel to $VT$ to draw the True Shape rhombus and measure its side.

Answer: The cutting plane connects opposite elevation corners $a'$ and $g'$ at inclination $\theta$, yielding the maximum rhombus.

Problem 109: Cube cut to form a regular hexagon

Given: A cube of 40 mm side resting on HP is cut by a section plane so the true shape is a regular hexagon. Find: Sectional top view, true shape, inclination $\theta$, and hexagon side length.

1. Draw the 45°-rotated 40 mm square in the Top View and project the 40 mm high cube in the Front View.

2. A plane forms a regular hexagon only when it slices through the midpoints of six edges of the cube; mark $1'$ at the midpoint of $a'-b'(d')$, $2'$ at the midpoint of $b'(d')-f'(h')$, and $3'$ at the midpoint of $f'(h')-g'$.

3. Draw the trace $VT$ through $1'-2'-3'$, project the six midpoints down to hatch the Top View, and project perpendicular to $VT$ to construct the regular hexagon and measure its side.

Watch out. The printed figure labels the angle $\theta = 55°$ against a vertical line, which is actually the complement; the true inclination of the trace to the horizontal $xy$ line is $55°$.

Answer: Inclination $\theta = 55°$; hexagon side length equals half the face diagonal ($28.3$ mm).

Problem 107: Cube cut to form a maximum equilateral triangle

Given: A cube of 40 mm side is cut by a section plane to produce an equilateral triangle of maximum size. Find: Sectional top view, true shape, inclination $\theta$, and triangle side length.

1. Draw the 45°-rotated 40 mm square in the Top View and project the 40 mm high Front View.

2. To get the largest equilateral triangle, the plane must pass through three non-adjacent corners of the cube; draw the trace $VT$ from the top central corner $b'(d')$ as $1'$ down to the bottom right corner $g'$ as $2'$.

3. Hatch the right half of the square in the Top View, measure $\theta$ in the Front View, and project $1'$ and $2'$ perpendicular to $VT$ to draw the True Shape equilateral triangle.

Answer: The cutting plane connects $b'(d')$ to $g'$ at inclination $\theta$, producing an equilateral triangle whose side equals the face diagonal.

Problem 110: Tetrahedron cut to form a given triangle

Given: A tetrahedron of 50 mm edge resting on HP is cut by an AIP to give a triangle of base 30 mm and altitude 35 mm. Find: Projections of the cut solid, true shape, and inclination $\theta$.

1. Draw the equilateral triangle $abc$ of side 50 mm in the Top View with edge $ac$ vertical, locate centroid $o$, and project the Front View using true slant edge 50 mm to find apex $o'$.

2. Fit a vertical line of 30 mm between edges $ab$ and $bc$ in the Top View, and project it up to base $a'(c')-b'$ or slant edge $o'a'(c')$ to locate $1'$.
3. From $1'$, swing an arc of radius equal to the altitude 35 mm to cut the right slant edge $o'b'$ at $2'$; join $1'2'$ to get trace $VT$ at $\theta = 35°$ and project $2'$ down to $ob$.

4. Project $1'$ and $2'$ perpendicular to $VT$ and transfer widths from the Top View to draw the True Shape triangle.

Answer: Inclination $\theta = 35°$.

Problem 111: Tetrahedron cut to form a maximum square

Given: A tetrahedron of 50 mm edge resting on HP is cut by an AIP to produce a square of maximum size. Find: Projections of the cut solid, true shape, and inclination $\theta$.

1. Construct the Top View and Front View of the 50 mm tetrahedron with base edge $ac$ perpendicular to $xy$.

2. A section of a regular tetrahedron is a maximum square of side 25 mm (half the 50 mm edge) when cut through the midpoints of four edges; locate the vertical chord of 25 mm in the Top View and project up to find $1'$ and $2'$ at the slant edge midpoints.

3. Draw trace $VT$ through $1'2'$, measure inclination $\theta$, and project perpendicular to $VT$ to draw the 25 mm True Shape square.

Answer: The cutting plane passes through the midpoints of the slant edges at inclination $\theta$, forming a 25 mm square.

Problem 121: Cone cut to form a parabola

Given: A cone of base diameter 50 mm and height 60 mm on HP is cut by an AIP to give a parabola of double ordinate 40 mm. Find: Projections of the cut solid and true shape.

1. Draw the 50 mm diameter circle divided into sectors in the Top View, and project the 60 mm high cone in the Front View.

2. Fit the 40 mm double ordinate as a vertical chord in the Top View circle and project it up to the Front View base line $xy$ as point $1'$.
3. Because a parabolic section on a cone requires the cutting plane to be parallel to an extreme generator, draw trace $VT$ from $1'$ parallel to the right-hand generator $o'b'$.

4. Mark the intermediate generator intersections $2', 3', 4', 5'$ along $VT$ and project them down to the Top View generators (using the horizontal circle arc method for the central generator) to hatch the parabola.

5. Project $1'$ through $5'$ perpendicular to $VT$ and transfer the chord widths from the Top View to plot the smooth True Shape parabola.

Answer: The cutting plane starts at the 40 mm base chord and runs parallel to the extreme generator.

Problem 123: Cone of unknown height cut into parabola

Given: A cone of base diameter 50 mm on HP is cut by an AIP to form a parabola of double ordinate 40 mm and abscissa 50 mm. Find: Unknown height of the cone, sectional views, and true shape.

1. Draw the 50 mm base circle in the Top View, fit the 40 mm vertical chord (double ordinate) on the left side, and project it up to the $xy$ line to locate $1'$.

2. For a parabola of abscissa 50 mm, the trace $1'5'$ has length 50 mm and is parallel to the opposite extreme generator; by similar triangles, the extreme generator length is also 50 mm, giving a vertical cone height of 57 mm (or 60 mm in the given box).

3. Draw the cone Front View, mark generator intersections $1'$ through $5'$ along trace $VT$, and project them down to complete and hatch the sectional Top View.

4. Project $1'$ through $5'$ perpendicular to $VT$ and transfer Top View distances from $xy$ to draw the True Shape parabola.

Watch out. The drawing's given box prints $H = 60$ mm while the elevation dimension marks 57 mm; moreover, a cone with a 50 mm base and 50 mm slant edge actually has a vertical height of $\sqrt{50^2 - 25^2} = 43.3$ mm.

Answer: Cone height is dimensioned as 57 mm (with $H = 60$ mm in the data box).

Development of Prisms and Cylinders (Parallel-Line Method)

You already know how to slice prisms and cylinders with inclined cutting planes and project their sectional views. Moving from cross-sections to surface fabrication, we now apply the Parallel-Line Method to unfold truncated prisms and cylindrical pipes into flat patterns. Because every vertical edge or generator on a prism or cylinder is parallel to the Vertical Plane, its true cut height in the Front View transfers horizontally onto the unfolded rectangular stretch-out.

The idea

When a sheet-metal worker makes a duct or pipe, they start with a flat sheet, cut a specific pattern, and fold or roll it into a 3D shape. Finding that flat pattern is called surface development. For a right prism or cylinder standing on the Horizontal Plane (HP), every vertical edge or generator is parallel to the Vertical Plane (VP). Therefore, the Front View shows the exact, true length of every vertical line on the surface.

To unfold the solid, we draw a stretch-out line right next to the Front View, aligned with the base. Its total length equals the perimeter of the base ($n \times B$ for an $n$-sided prism of base side $B$, or $\pi \times D$ for a cylinder of diameter $D$). We divide this stretch-out into segments matching the faces of the prism or the 8 equal sectors of the cylinder, and draw vertical fold lines or generators at each division. Since heights do not change when the surface unrolls horizontally, we simply project each cut point horizontally from the Front View to its corresponding vertical line on the development.

Connecting these projected points gives the upper boundary of the flat pattern: straight segments for faceted prisms, and a smooth sine-like curve for cylinders. In compound pipe cuts—such as a $45^\circ$ mitre cut used to join two pipes at a right angle—both the top and bottom ends are sliced, so we project both cut profiles across. Finally, because the shortest distance between two points on a flat sheet is a straight line, any shortest path or helix (a curve rising at a constant rate around a cylinder, Advancing by one pitch per full turn) is drawn as a straight diagonal on the development first, and its intersection points are projected back to the Front View.

Worked example

Problem 126: Square prism cut by an inclined plane Given: A square prism of base side 40 mm and height 70 mm rests on HP with a base edge parallel to VP, cut by an AIP at 45° to HP bisecting the axis. Find: The sectional top view and the lateral surface development of the lower portion.

1. Draw a 40 mm square below the $xy$ line with edges parallel to $xy$ for the Top View, and project the 70 mm high rectangular Front View above $xy$ because the vertical edges are parallel to VP.
2. Mark the axis midpoint at 35 mm height in the Front View and draw the cutting plane line $V\text{-}T$ at 45° to $xy$, marking cut points $1'$ on the left edges $a'(d')$ and $2'$ on the right edges $b'(c')$.

3. Darken the retained lower portion in the Front View and hatch the entire square in the Top View at 45° because the cutting plane slices across the full width of the prism.

4. Draw a horizontal stretch-out line to the right of the Front View of length $4 \times 40 = 160$ mm, divide it into four 40 mm faces ($E\text{-}F\text{-}G\text{-}H\text{-}E$), and erect vertical fold lines.
5. Project cut heights $1'$ and $2'$ horizontally from the Front View to the corresponding vertical fold lines on the development, and connect them with straight dark lines to complete the pattern.

Answer: Stretch-out length = 160 mm; cut heights on vertical edges steps from 15 mm on $A$ and $D$ up to 55 mm on $B$ and $C$.

More problems

Problem 128: Truncated pentagonal prism

Given: A pentagonal prism ($B = 30$ mm, $H = 70$ mm) rests on HP with a base edge parallel to VP, cut by a 45° AIP bisecting the axis. Find: The sectional top view and lateral surface development.

1. Draw a regular pentagon of side 30 mm below $xy$ with back edge $ed$ parallel to $xy$, and project the 70 mm high Front View.
2. Draw the 45° cutting plane through the axis midpoint (35 mm height), marking intersections $1', 2', 3', 4', 5'$ on vertical edges $a', e', b', d', c'$.

3. Darken the lower prism in the Front View and hatch the entire pentagon in the Top View to represent the cut surface.

4. Draw a stretch-out line of length $5 \times 30 = 150$ mm divided into five 30 mm strips, project $1'$ through $5'$ horizontally to their respective fold lines, and join them with straight segments.

Answer: Stretch-out length = 150 mm divided into five 30 mm faces.

Problem 129: Truncated hexagonal prism

Given: A hexagonal prism ($B = 30$ mm, $H = 70$ mm) on HP with two base edges perpendicular to VP, cut by a 30° AIP bisecting the axis. Find: The sectional top view and lateral surface development.

1. Construct a 30 mm regular hexagon in the Top View with two vertical sides perpendicular to $xy$, and project the 70 mm tall Front View.
2. Draw the cutting plane at 30° to $xy$ through the axis midpoint (35 mm height), labeling cut points $1', 2', 3'$ on the projected vertical edges.

3. Outline the retained lower portion in the Front View and hatch the complete hexagonal Top View.

4. Lay out a stretch-out line of $6 \times 30 = 180$ mm with six 30 mm faces, project heights $1', 2', 3'$ horizontally across, and connect the points with straight lines.

Answer: Stretch-out length = 180 mm divided into six 30 mm faces.

Problem 131: Shortest path of an ant on a hexagonal prism

Given: A hexagonal prism ($B = 30$ mm, $H = 70$ mm) on HP; an ant travels from a bottom corner to the diametrically opposite top corner. Find: The lateral surface development and the ant's path in the Front View.

1. Draw the Top View hexagon of side 30 mm with two edges parallel to $xy$, and project the 70 mm high Front View.

2. Construct the $180 \text{ mm} \times 70 \text{ mm}$ rectangular development alongside the Front View, divided into six 30 mm vertical strips labeled $A$ through $A$.
3. Draw a straight line from bottom corner $A$ to top corner $D$ (three faces over, 90 mm horizontally) because a flat straight line gives the shortest surface path.
4. Mark points $1', 2', 3', 4'$ where this diagonal crosses fold lines $A, B, C, D$, project them horizontally back to edges $a', b', c', d'$ in the Front View, and connect them.