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: $\theta = \frac{r}{L} \times 360^\circ$ 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 $e'$, $f'(h')$, $g'$.

Figure: Step 98.1: Projections of square prism resting on HP with faces equally inclined to VP

Step 98.1 of the solution showing the initial projections of a square prism:

• Reference line labelled 'y' on both sides (XY line).
• Top view (below XY line): A square rotated at 45° to the XY line (corners at top, bottom, left, right). The corners are labelled: left corner 'a(e)', bottom corner 'b(f)', right corner 'c(g)', top corner 'd(h)'. The lower-right side is dimensioned with '□40'.
• Front view (above XY line): A rectangle representing the elevation of the prism, resting on the XY line. Base points are labelled 'e'' (left), 'f' (h')' (center vertical line), and 'g'' (right). Top corners are labelled 'a'' (left), 'b' (d')' (center), and 'c'' (right). Center vertical line connects 'b'(d')' to 'f'(h')'. Vertical height dimension on the left is labelled '70'.
• Projection lines connect the corresponding corners between top and front views.
• A small table next to the front view states 'B = 40' and 'H = 70'.

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 $c'$.

Figure: Section of a square prism - cutting plane representation (Step 98.2)

Step 98.2 of the projection and sectioning of a square prism:

• Reference line labeled y-y.
• Top View: A square of side 40 mm with all sides inclined at 45° to the reference line. Corners are labeled a(e) on the left, b(f) at the bottom, c(g) on the right, and d(h) at the top. Dimensioned with square symbol 40 (□40).
• Front View: A square prism of base side 40 mm and height 70 mm resting on HP. Base corners labeled e' (left), f'(h') (center vertical edge), and g' (right). Top corners labeled a' (left), b'(d') (center vertical edge), and c' (right). Overall height dimension is 70 mm. A horizontal line is located 15 mm below the top face.
• Section Plane (V-T): A cutting plane inclined at 55° to the horizontal passes through the prism, cutting edge a'-e' at point 1', axis/edge b'(d') at point 2', and the top face at point 3'. Section plane ends are thickened with perpendicular arrows marked 'V' at the bottom-left and 'T' at the top-right.
• Data Box: '98.2', 'B = 40', 'H = 70'.
• Text: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

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.

Figure: Sectional top view and front view of the cut square prism (Step 98.3)

Step 98.3 showing the completed sectional views of the square prism:

• Reference line labeled y-y.
• Front View: Retained lower portion of the prism is outlined with heavy dark lines passing through e'-f'-g'-c'-3'-2'-1'-e'. The cutting plane V-T is inclined at 55° to HP, cutting through 1', 2', and 3'. Height dimensions are 70 mm total and 15 mm from the top face.
• Sectional Top View: The cut section is shown with 45° hatching lines across the cut region between points 1 on edge a(e), 2 near d(h), 3 on the top face, and symmetric points 3_1 and 2_1 toward b(f). Corner c(g) remains unsectioned outside the hatched area. Edge length is dimensioned as □40.
• Data Box: '98.3', 'B = 40', 'H = 70'.
• Text: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

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.

Figure: Sectional views and true shape of section of a square prism cut by an inclined cutting plane

Technical drawing labeled 98.4 showing the sectional projections and true shape of a section of a vertical square prism:

• Inset box: '98.4', 'B = 40', 'H = 70', 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.
• Front View: A square prism of base side 40 mm and height 70 mm rests on HP on horizontal reference line y-y. Base corners are marked e', f'(h'), g'; top corners are a', b'(d'), c'. A cutting plane V-T is inclined at 55° to the horizontal, intersecting the left vertical edge at point 1', the central vertical edge at point 2', and the top base at point 3' (dimensioned 15 mm from the right edge c').
• Sectional Top View: The prism is oriented with base edges inclined at 45° to the VP, dimensioned with a square symbol '□40'. Corners are marked a(e)₁, b(f)₂, c(g), d(h). The portion cut by the plane is hatched between section points 1, 2, and 3.
• True Shape of Section: Projected perpendicular to the cutting plane V-T onto an auxiliary plane with reference line x₁-y₁. Projectors from points 1', 2', 3' yield vertices 1₁, 2₁, 3₁, etc., forming a hatched polygon representing the true shape of the section.

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 $o'$.

Figure: Projection of a pentagonal pyramid (Step 86.1)

Step 86.1: Orthographic projection of a regular pentagonal pyramid. Givens: Base side B = 30 mm, Height H = 60 mm. Front view above ground line x-y shows apex o' with height 60 mm, base on x-y line with points a'(e'), b'(d'), and c'. Top view below x-y shows a regular pentagon a-b-c-d-e with apex o at the center; edge a-e is oriented vertically with length 30 mm. Lines connect apex o to each base vertex (a, b, c, d, e). Scale: 1:1. All dimensions are in mm.

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 $o'c'$.

Figure: Section plane cutting pentagonal pyramid (Step 86.2)

Step 86.2: Cutting plane setup on the pentagonal pyramid. Givens: Base side B = 30 mm, Height H = 60 mm. The cutting plane V-T is inclined at 45° to the base/x-y line and passes through the axis at a height of 30 mm from the base. Slant edges are intersected at points 1', 2', and 3'. Front view and top view of the pyramid are identical in outline to Step 86.1. Scale: 1:1. All dimensions are in mm.

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.

Figure: Orthographic projections and sectional top view of a cut pentagonal pyramid (Step 86.3)

Step 86.3 of worked solution for Problem 86: Section of a regular pentagonal pyramid.

• Given box: Base side B = 30 mm, Axis height H = 60 mm.
• Reference line: xy.
• Front view:
  • Base rests on xy: points a'(e') at left, b'(d') in middle, c' at right.
  • Apex o' at height 60 mm above xy.
  • A section plane VT is inclined at 45° to the base/xy line, sloping upwards to the right.
  • It cuts slant edge o'-c' at point 3' at a height of 30 mm from the base.
  • It cuts slant edges o'-b' and o'-d' at point 2'.
  • It cuts slant edges o'-a' and o'-e' at point 1'.
  • Section plane line is indicated with heavy ends labelled V and T with viewing direction arrows.
• Top view:
  • Regular pentagon base a-b-c-d-e with vertical base edge a-e (length 30 mm) on the left, and apex o at the center.
  • Slant edges o-a, o-b, o-c, o-d, o-e.
  • Projectors from the front view determine section points in top view: 1 on o-e, 1₁ on o-a, 2 on o-d, 2₁ on o-b, and 3 on o-c.
  • The resulting sectional area 1-2-3-2₁-1₁ is hatched.
• Notes: SCALE 1:1, ALL DIMENSIONS ARE IN mm.

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 $3$.

Figure: True shape of section of the pentagonal pyramid (Step 86.4)

Step 86.4 of worked solution for Problem 86: Construction of the true shape of the section.

• Given box: Base side B = 30 mm, Axis height H = 60 mm.
• Answer box: W = 36 mm.
• Front view and top view are identical to Step 86.3, showing the cut pentagonal pyramid with section plane VT inclined at 45° to xy, cutting o'-c' at height 30 mm (point 3'), o'-b'(d') at 2', and o'-a'(e') at 1'.
• True shape construction:
  • Projectors are drawn perpendicular to the cutting plane line VT from points 1', 2', and 3' towards the upper-left.
  • Distances of section points from the center line in the top view are transferred along these projectors to plot points 1, 1₁, 2, 2₁, and 3.
  • The resulting polygon 1-2-3-2₁-1₁ is hatched, representing the true shape of the section.
  • Dimension W (width across true shape) is dimensioned with answer W = 36 mm.
• Notes: SCALE 1:1, ALL DIMENSIONS ARE IN mm.

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 $x$-$y$.

Figure: Projections of a hexagonal pyramid resting on HP

Orthographic projections of a hexagonal pyramid (Step 91.1):

• Top view (below xy reference line): A regular hexagon representing the base, oriented with edges f-e and b-c parallel to the horizontal reference line xy. Vertices are labelled counter-clockwise: a (leftmost), b, c, d (rightmost), e, f. Center apex is labelled o, connected to each base vertex with slant edge lines. One base edge length is dimensioned as 30 mm.
• Front view (above xy reference line): The base lies along the reference line xy, labelled with endpoints x and y. Base corner projections are labelled a', b' f', c' e', d'. Center axis line extends vertically up to apex o' at a height of 60 mm (dimensioned on the right as 60). Slant edges connect o' to a', b' f', c' e', and d'.
• Info box on the left shows: '91.1', 'B = 30', 'H = 60'.
• Scale notation at bottom: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

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 $o'd'$.

Figure: Step 91.2: Front view and top view of hexagonal pyramid showing cutting plane location

Step 91.2 of engineering drawing problem 91 for a regular hexagonal pyramid (Base B = 30 mm, Height H = 60 mm). Top view shows a regular hexagon with corners labelled a, b, c, d, e, f and apex o, oriented such that corners a and d lie on a horizontal line. Slant edges connect each corner to apex o. Base side length is dimensioned as 30 mm. Projectors lead up to the front view across the reference line x-y. The base rests on x-y with points a', b', f', c', e', d'. Apex is o' at height 60 mm above x-y. A section/cutting plane line V-T is inclined at 30° to the horizontal, intersecting the axis at a distance of 25 mm below the apex o'. The cutting plane cuts the slant edges at points labelled 1' (on o'-a'), 2' (on o'-b' and o'-f'), 3' (on o'-c' and o'-e'), and 4' (on o'-d'). Includes boxes: '91.2', 'B = 30, H = 60', and note 'SCALE 1:1, ALL DIMENSIONS ARE IN mm'.

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.

Figure: Step 91.3: Final sectional top view and front view of the truncated hexagonal pyramid

Step 91.3 of problem 91 showing the completion of the sectional top view and darkening of retained edges. The cutting points 1', 2', 3', 4' from the front view are projected downwards to the respective edges in the top view: 1 on oa, 2 on of, 2₁ on ob, 3 on oe, 3₁ on oc, and 4 on od. These points form a closed section polygon 1-2-3-4-3₁-2₁ in the top view which is hatched. The retained base edges (a-b-c-d-e-f) and remaining parts of the slant edges connecting the base to the cut section are drawn with thick solid lines. In the front view, the lower portion of the pyramid below the cutting plane (points a'-1'-2'-3'-4'-d') is drawn in thick solid lines. Includes boxes: '91.3', 'B = 30, H = 60', and note 'SCALE 1:1, ALL DIMENSIONS ARE IN mm'.

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.

Figure: Orthographic projection and true shape of section of a hexagonal pyramid

Step 91.4 of a section of solids problem for a hexagonal pyramid:

• Top View: A regular hexagon with corners a, b, c, d, e, f and apex o at the center. Base side length is dimensioned as 30 mm. Cut section points are marked: 1 on slant edge o-a, 2 and 2_1 on o-b and o-f, 3 and 3_1 on o-c and o-e, and 4 on o-d. The sectioned inner region 1-2-3-4-3_1-2_1 is hatched with thin lines.
• Front View: Base line lies on reference line x-y between points a' and d', with projector points b', f' and c', e' indicated. Apex o' is at height 60 mm above the base. Cutting plane line V-T is inclined at 30° to the horizontal/xy-line, passing through slant edges with cut points marked 1', 2', and 3', 4'. A vertical height of 25 mm is dimensioned from the apex o' down to the section plane at the right slant edge.
• True Shape: Projectors perpendicular to the cutting plane V-T extend upwards to an auxiliary reference line x1-y1 parallel to V-T. The true shape is plotted with vertices 1, 2, 3, 4, 3_1, 2_1 and hatched.
• Side box notes: '91.4', 'B = 30', 'H = 60', 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

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.

Figure: Step 87.1: Orthographic projections of a pentagonal pyramid

Step 87.1 showing the projections of a regular pentagonal pyramid.

• Given box: B = 30, H = 60.
• Top view (below XY line x-y): regular pentagon a-b-c-d-e with apex o at the center. Base edge e-a is vertical (perpendicular to XY) with dimension 30 mm.
• Front view (above XY line x-y): base lies on x-y line at points a'(e'), b'(d'), and c'. Apex o' is on the vertical centerline at height 60 mm from the base. Slant edges connect o' to base corners a'(e'), b'(d'), and c'.
• Dimension 60 mm indicated for the axis height.
• Note below: SCALE 1:1, ALL DIMENSIONS ARE IN 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 $o'c'$.

Figure: Step 87.2: Pentagonal pyramid cut by a section plane inclined at 60 degrees

Step 87.2 showing the pentagonal pyramid intersected by a cutting plane.

• Given box: B = 30, H = 60.
• Front view: cutting plane (trace marked with cutting plane indicators V and T) inclined at 60° to the horizontal/XY line. The cutting plane cuts the right slant edge o'-c' at a height of 15 mm above the base line x-y. The intersection points with the edges are marked as 1' on slant edge o'-a'(e'), 2' on o'-b'(d'), and 3' on o'-c'.
• Top view: regular pentagon a-b-c-d-e with apex o; side length e-a dimensioned 30 mm.
• Dimensions indicated: total height 60 mm, cutting height 15 mm, section plane angle 60°, base edge 30 mm.
• Note below: SCALE 1:1, ALL DIMENSIONS ARE IN mm.

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.

Figure: Section of a pentagonal pyramid - Step 87.3

Projection of a regular pentagonal pyramid cut by an auxiliary inclined plane:

• GIVEN: Base edge $B = 30\text{ mm}$, Pyramid height $H = 60\text{ mm}$.
• Front View: Apex $o'$ at height $60\text{ mm}$ above the $xy$ line. Base on $xy$ line represented by points $a'(e')$, $b'(d')$, and $c'$.
• Section Plane (Cutting Plane VT): Inclined at $60^\circ$ to the $xy$ line, passing through a point on the right slant edge $o'c'$ at a vertical height of $15\text{ mm}$ above the base. Arrows labeled $V$ and $T$ mark the trace ends.
• Cutting plane cuts slant edges at points labeled $3'$ (on $o'c'$), $2'$ (on $o'b'$ and $o'd'$), and $1'$ (on $o'a'$ and $o'e'$ or base).
• Top View: Pentagonal base with corners $a, b, c, d, e$ and center/apex $o$, with vertical side $e-a = 30\text{ mm}$ on the left. The sectional area is hatched, with vertices labeled $1, 2, 3, 2_1, 1_1$.
• Scale: 1:1. All dimensions are in mm.

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 $3$.

Figure: True shape of section of a pentagonal pyramid - Step 87.4

Complete projection and true shape of the section of a pentagonal pyramid:

• GIVEN: Base edge $B = 30\text{ mm}$, Height $H = 60\text{ mm}$.
• Front view and top view are identical to Step 87.3.
• An auxiliary view (true shape of the section) is projected perpendicularly from the cutting plane $V-T$.
• The true shape is a hatched polygon with vertices labeled $1, 2, 3, 2_1, 1_1$.
• The overall width/dimension of the true shape is indicated as $W$.
• ANSWER: $W = 39\text{ mm}$.
• Scale: 1:1. All dimensions are in mm.

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.

Figure: Projections of a cylinder resting on its base (Step 114.1)

Step 114.1 shows the initial orthographic projections of a cylinder resting on the HP on its base:

• Top View: A circle with diameter 40 mm (dimensioned as Ø40), divided into 8 equal divisions with lines intersecting at the center point 'o'. The division points along the circumference are labeled clockwise: a(i) at 9 o'clock, h(p) at 10:30, g(o) at 12 o'clock, f(n) at 1:30, e(m) at 3 o'clock, d(l) at 4:30, c(k) at 6 o'clock, and b(j) at 7:30.
• Reference line: Horizontal line labeled 'x' on the left and 'y' on the right.
• Front View: A rectangle of width 40 mm and height 60 mm (dimensioned as 60 on the left). The bottom base points on xy are labeled i', j'(p'), k'(o'), l'(n'), m'. The top base points are labeled a', b'(h'), c'(g'), d'(f'), e'. Vertical projection lines link each point in the top view to the corresponding generator in the front view.
• Inset box to the upper-left of the front view: 'D = 40', 'H = 60'.

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.

Figure: Projection of cylinder cut by an inclined section plane (Step 114.2)

Step 114.2 of a cylinder cut by an inclined section plane.

• Given box: D = 40 mm, H = 60 mm; SCALE 1:1; ALL DIMENSIONS ARE IN mm.
• Top View: A circle of diameter Ø40 divided into 8 equal sectors with generator points labelled along the perimeter: a(i), b(j), c(k), d(l), e(m), f(n), g(o), h(p), and center o.
• Front View: A cylinder of height 60 mm and base width 40 mm standing on the horizontal reference line x-y.
  • Base points: i', j'(p'), k'(o'), l'(n'), m'.
  • Top face points: a', b'(h'), c'(g'), d'(f'), e'.
  • Generators are drawn vertically connecting top and bottom points.
  • An inclined cutting plane V-T cuts across the cylinder, inclined at 30° to the central vertical axis (k'(o')).
  • The cutting plane intersects the central axis at point 3', located at a height of 30 mm above the base x-y line.
  • Intersection points along the generators are marked as: 1' (on generator a'-i'), 2' (on generators b'-j' and h'-p'), 3' (on the axis / generators c'-k' and g'-o'), 4' (on generators d'-l' and f'-n'), and 5' (on generator e'-m').

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.

Figure: Sectional top view and front view of the cut cylinder (Step 114.3)

Step 114.3 showing the completed sectional views of the cut cylinder.

• Given box: D = 40 mm, H = 60 mm; SCALE 1:1; ALL DIMENSIONS ARE IN mm.
• Front View: The retained lower portion of the cylinder is outlined in bold lines: base i'-m', left vertical edge i'-1', inclined cut face line 1'-2'-3'-4'-5', and right vertical edge 5'-m'. The cut-away upper portion is indicated with thin construction lines. The cutting plane line V-T remains inclined at 30° to the vertical axis, passing through the axis at height 30 mm above the base.
• Sectional Top View: The circle of diameter Ø40 (with generators a(i) through h(p)) is entirely covered with 45° hatching lines, representing the sectional top view of the cut 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.

Figure: Projections and true shape of section of a cylinder cut by an inclined plane (step 114.4)

Engineering drawing showing orthographic projections and true sectional shape for a cylinder (step 114.4). Dimensions given in a box: D = 40, H = 60, Scale 1:1, all dimensions in mm. Top view shows a circle with diameter Ø40 divided into 12 equal sectors with generator points labeled a(i), b(j), c(k), d(l), e(m), f(n), g(o), h(p) around the perimeter and center o. Front view is drawn above the reference line xy, showing a cylinder of diameter 40 mm and height 60 mm, with base points i', j'(p'), k'(o'), l'(n'), m' and top points a', b'(h'), c'(g'), d'(f'), e'. A cutting plane line V-T is inclined at 30° to the horizontal/xy line, passing through the axis at a height of 30 mm from the base. The cutting plane intersects generators at points 1', 2', 3', 4', 5'. Auxiliary projection lines perpendicular to the cutting plane project points to an auxiliary datum line x1-y1, constructing the true shape of the section as a hatched ellipse.

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.

Figure: Step 115.1: Projections of a cylinder resting on HP

Step 115.1 of cylinder projection:

• Parameters: D = 40 mm, H = 60 mm.
• Reference line labelled x-y.
• Top View: A circle of diameter $\Phi 40$ divided into 8 equal sectors with center 'o'. The perimeter points are labelled counterclockwise: a(i) at 9 o'clock, b(j) at ~7:30, c(k) at 6 o'clock, d(l) at ~4:30, e(m) at 3 o'clock, f(n) at ~1:30, g(o) at 12 o'clock, h(p) at ~10:30.
• Front View: A vertical rectangle of width 40 mm and height 60 mm sitting on reference line x-y.
  • Base points along x-y: i', j'(p'), k'(o'), l'(n'), m'.
  • Top face points: a', b'(h'), c'(g'), d'(f'), e'.
  • Centerline shown passing through k'(o') - c'(g').
  • Height dimensioned as 60 on the left.
• Note: SCALE 1:1, ALL DIMENSIONS ARE IN mm.

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 $7'$.

Figure: Step 115.2: Cylinder cut by an inclined section plane

Step 115.2 of cylinder section:

• Parameters: D = 40 mm, H = 60 mm.
• Reference line labelled x-y.
• Top View: Same as in step 115.1, circle of $\Phi 40$ divided into 8 equal sectors.
• Front View: Vertical cylinder of height 60 mm.
• A cutting/section plane V-T inclined at 50° to the horizontal/x-y line cuts across the cylinder.
  • The cutting plane is indicated by thick end lines marked 'V' and 'T'.
  • Intersection points with generators are labelled 1', 2', 3', 4'.
  • A height dimension of 45 mm from the x-y line to the cutting point on the generator d'(f') - l'(n') is shown on the right.
  • Angle of the cutting plane with the axis or horizontal: 50°.
• Note: SCALE 1:1, ALL DIMENSIONS ARE IN mm.

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.

Figure: Orthographic projections of a truncated cylinder (Step 115.3)

Front view (elevation) and top view (plan) of a cut vertical cylinder. In the top view, a circle of diameter 40 mm is divided into 8 equal parts labeled a(i), b(j), c(k), d(l), e(m), f(n), g(o), h(p) with center o. The rightmost segment between f(n) and d(l) is hatched. In the front view, the cylinder has base width 40 mm and height 60 mm. Projectors from the top view map the base points i', j'(p'), k'(o'), l'(n'), m' along the horizontal axis x-y, and top points a', b'(h'), c'(g'), d'(f'), e'. A cutting plane line V-T is inclined at 50° to the horizontal, intersecting the cylinder generators at points 1', 2', 3', and 4', where point 4' is at a vertical height of 45 mm from the base. A small vertical cut is made on the right between 4' and the top face at d'(f').

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.

Figure: Auxiliary view showing true shape of section of a truncated cylinder (Step 115.4)

Orthographic projections and true sectional view of the truncated cylinder. In addition to the top view (circle of diameter 40 mm) and front view (cylinder of diameter 40 mm, height 60 mm, cut by plane V-T inclined at 50° up to height 45 mm), an auxiliary reference line x1-y1 is drawn parallel to the cutting plane line V-T. Perpendicular projectors from cut points 1', 2', 3', 4' extend across x1-y1 to construct the true shape of the section, which is a hatched parabolic/elliptical shape truncated by a straight chord corresponding to the right cut face.

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.

Figure: Projections of a cone resting on HP (Step 117.1)

Orthographic projections of a cone resting on HP before sectioning. Top view is a circle with diameter Ø50 mm divided into 8 equal sectors with points labeled a, b, c, d, e, f, g, h and center o. Vertical projection lines extend up to reference line x-y. Front view is a triangle with base points a', b'(h'), c'(g'), d'(f'), e' on the xy line, apex o' at height 60 mm, with generators drawn from o' to the base points. Beside the front view is a label box '117.1' with 'D = 50', 'H = 60'. Bottom note specifies 'SCALE 1:1, ALL DIMENSIONS ARE IN mm'.

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.

Figure: Step 117.2: Cone with section plane inclined at 50 degrees

Orthographic projections of a cone (step 117.2). GIVEN box contains D = 50, H = 60. Front View (above reference line x-y): An isosceles triangle with apex o' and height 60 mm, base diameter 50 mm represented by base points a', b'(h'), c'(g'), d'(f'), e'. A cutting plane line inclined at 50° to the x-y line passes through point 1' (coincident with a') and cuts the generators at points 2', 3', 4', and 5'. Viewing direction arrows T (from top-right, perpendicular to the section plane) and V (from bottom-left along the plane) are indicated. Top View (below x-y line): A full circle of diameter Ø50 mm with center o, divided into 8 equal sectors with points labeled a, b, c, d, e, f, g, h. Thin projectors connect corresponding points between views. Notes: SCALE 1:1, ALL DIMENSIONS ARE IN mm.

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 $o$-$g$.

Figure: Step 117.3: Sectional top view and front view of truncated cone

Orthographic projections showing sectioned cone (step 117.3). GIVEN box contains D = 50, H = 60. Front View: Lower retained portion of the cone is drawn with thick continuous lines, bounded by base line, generator to the right, and the cutting plane line 1'-2'-3'-4'-5' inclined at 50° to the horizontal. Vertical projectors are projected down from the section points 1', 2', 3', 4', 5' to intersect corresponding generators and circle boundaries in the top view. Top View: Shows the sectional top view with a hatched region bounded by points 1, 2, 3, 4, 5, 4_1, 3_1, 2_1. Outer circle diameter Ø50 mm with base points a, b, c, d, e, f, g, h. Retained outline of the cone is highlighted. Notes: SCALE 1:1, ALL DIMENSIONS ARE IN mm.

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.

Figure: Orthographic projections and true shape of section of a cone cut by an inclined section plane

Projection drawing of a cut cone:

• Front View: A triangle representing the cone resting on the reference line $xy$. Base points are $a', b'(h'), c'(g'), d'(f'), e'$, and apex is $o'$. Cone base diameter is 50 mm, axis height is 60 mm. A cutting plane $V-T$ cuts through the cone parallel to the leftmost generator $o'a'$ at an angle of $50^\circ$, with section points labelled $1', 2', 3', 4', 5'$.
• Top View: A circle representing the base with 8 generator divisions labelled $a, b, c, d, e, f, g, h$ and center $o$, diameter $\varnothing 50$. The cut section is shown hatched between points $1, 2, 3, 4, 5, 4_1, 3_1, 2_1$.
• True Shape: An auxiliary view projected perpendicular to the cutting plane $V-T$ shows the true shape of the section (a parabola), hatched, with numbered points $1, 2, 3, 4, 5, 4_1, 3_1, 2_1$.
• Labels & Notes: Box marked '117.4', with 'D = 50', 'H = 60', 'SCALE 1:1', and 'ALL DIMENSIONS ARE IN mm'.

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.

Figure: Orthographic projections of a cone - Step 122.1

Step 122.1 showing the initial orthographic projections of a vertical cone. A specification box indicates 'D = 50', 'H = 60'. Front view above reference line xy shows an isosceles triangle with base on the xy line, apex labeled o', base diameter 50 mm, and vertical height 60 mm. Top view below the xy line shows a full circle of diameter Ø50 mm with center labeled o, aligned with the front view's vertical centerline. Additional annotations: 'SCALE 1:1' and 'ALL DIMENSIONS ARE IN mm'.

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.

Figure: Section of a cone - Step 122.2

Step 122.2 showing the introduction of a section plane parallel to an extreme generator of the cone. In the front view, a cutting plane is drawn parallel to the left generator at an offset distance of 10 mm, marked with perpendicular offset dimension lines '10' and section plane trace markers labeled 'T' and 'V'. The cutting plane intersects the base (xy line), and a vertical projector is projected down into the top view to cut the Ø50 mm circular base, showing a section chord with an indicated dimension of 41 mm. Boxed text: 'D = 50', 'H = 60'. Additional labels: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

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.

Figure: Step 122.3: Section of a cone with cutting plane parallel to generator

Engineering drawing step 122.3 showing the front view and top view of a cone cut by an inclined section plane. Givens box: D = 50, H = 60. Front View: Cone with vertex o' and base a'b' resting on the reference line xy. Height is 60 mm, base diameter is 50 mm. A cutting plane VT is inclined and drawn parallel to the extreme generator o'a' at a perpendicular distance of 10 mm. The section plane cuts the base at point 1' and the surface at points 2', 3', 4', and apex/limit 5'. Top View: A circle of diameter Ø50 with centre o. The cut surface is projected down, showing a straight cut along the base of length 41 mm between points 1 and 1_1, and a curved section passing through points 2, 3, 4, 5, 4_1, 3_1, 2_1. The sectional area is hatched. Notes: SCALE 1:1, ALL DIMENSIONS ARE IN mm.

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.

Figure: Step 122.4: True shape of the section of a cone

Engineering drawing step 122.4 showing the determination of the true shape of the parabolic section of a cone. Givens box: D = 50, H = 60. Front and Top Views: Cone of base diameter 50 mm and height 60 mm with cutting plane VT offset by 10 mm from the generator o'a'. Section points 1', 2', 3', 4', 5' on the cutting plane. True Shape Construction: Projectors are drawn perpendicular to the cutting plane VT from points 1', 2', 3', 4', 5'. Distances from the axis in the top view are transferred along these projectors to plot the true shape with points 1, 2, 3, 4, 5, 4_1, 3_1, 2_1, 1_1 (base width is 41 mm). The resulting parabolic profile is fully hatched. Notes: SCALE 1:1, ALL DIMENSIONS ARE IN mm.

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.

Figure: Projections of a cone, step 118.1

Step 118.1: Orthographic projection of a cone resting on HP with base on XY line.

• Given box: D = 50, H = 60.
• Front view: Isosceles triangle with apex o' and base points on reference line xy: a', b'(h'), c'(g'), d'(f'), e'. Height of apex o' from xy is 60 mm. Generators connect o' to base points.
• Top view: Circle of diameter Ø50 centered at o, divided into 8 equal sectors with points labeled counter-clockwise: a (left, 9 o'clock), b, c (bottom, 6 o'clock), d, e (right, 3 o'clock), f, g (top, 12 o'clock), h. Thin projection lines link points between views.
• Note: SCALE 1:1, ALL DIMENSIONS ARE IN mm.

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 $od$.

Figure: Projections of a cone with cutting plane line, step 118.2

Step 118.2: Cutting plane introduced in top view.

• Given box: D = 50, H = 60.
• Front view and top view are identical to step 118.1 (base diameter Ø50, height 60 mm).
• In the top view, a section plane line H-T is drawn parallel to the horizontal centerline a-e at a distance of 10 mm below it. Arrows at ends labeled H and T point downwards.
• Intersections of the cutting plane with the cone features are numbered 1 through 5 from left to right: 1 on the outer circumference, 2 on generator o-b, 3 on generator o-c, 4 on generator o-d, and 5 on the outer circumference.
• Note: SCALE 1:1, ALL DIMENSIONS ARE IN mm.

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.

Figure: Orthographic projections of a cone cut by a section plane, showing a hyperbolic cross-section.

Step 118.3 of a worked problem showing the section of a cone producing a hyperbola.

• Top-left box: 'D = 50', 'H = 60', with text 'HYPERBOLA'.
• Front View (above line x-y): Triangular elevation of the cone with apex o' and base a'-e' of diameter 50 mm, height 60 mm. The sectional area is hatched with boundary points 1, 2, 3, 4, 5 forming a hyperbolic profile. Base points labelled a', 1, b'(h'), c'(g'), d'(f'), 5, e'. Apex labelled o'.
• Top View (below line x-y): Circular plan with center o and diameter Ø50, generator lines radiating to points a, h, g, f, e. A horizontal cutting line H-T is positioned 10 mm offset from the center/axis, intersecting the base circle at points 1, 2, 3, 4, 5.
• Projectors connect top view points 1, 2, 3, 4, 5 to front view points 1, 2, 3, 4, 5.
• Annotations: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

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.

Figure: Initial orthographic projections of a pentagonal pyramid resting on HP

Step 88.1 showing the initial projections of a pentagonal pyramid:

• Left text box: 'B = 30', 'H = 60'.
• Reference line labelled x on the left and y on the right.
• Top view (below xy line): A pentagon abcde oriented such that the vertical edge e-a is perpendicular to the xy line (length dimensioned as 30 mm). Apex o is at the center, connected to all vertices a, b, c, d, e by slant edge lines.
• Front view (above xy line): Triangular elevation with apex o' at height 60 mm above the base on the xy line. Base points are a'(e') at the left, b'(d') along the middle, and c' at the right. Solid lines connect o' to a'(e'), b'(d'), and c'. Vertical axis line runs from o' down through the base.
• Vertical projection lines connect each vertex between top and front views.
• Notes at bottom: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN 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 $oc$.

Figure: Step 88.2 of section of a pentagonal pyramid

Projection of a pentagonal pyramid cut by an auxiliary inclined section plane. Box at upper left indicates: '88.2', 'B = 30', 'H = 60'. Front view shows apex o' with height 60 mm above reference line x-y; base points on x-y are a'(e'), b'(d'), and c'. Top view shows a regular pentagon base a-b-c-d-e with center/apex o, base side length 30 mm (dimensioned along edge a-b). A section plane H-T is inclined at 30° to the vertical projector and passes at a perpendicular offset of 10 mm from the apex o. It intersects edge e-a at point 1, slant edge o-a at point 2, slant edge o-b at point 3, and edge b-c at point 4. Note at bottom right: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

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 $1'2'3'4'$.

Figure: Step 88.3 showing the sectional front view of the pentagonal pyramid

Completed sectional front view and top view of the pentagonal pyramid. Box at upper left indicates: '88.3', 'B = 30', 'H = 60'. Top view shows the pentagonal base a-b-c-d-e with base side 30 mm, apex o, and cutting plane line H-T inclined at 30° at a distance of 10 mm from o, intersecting at points 1, 2, 3, and 4. Projectors transfer these points to the front view: point 1 to base at 1', point 2 to slant edge o'-a' at 2', point 3 to slant edge o'-b' at 3', and point 4 to base at 4'. The resulting cut surface 1'-2'-3'-4' in the front view is hatched. The retained lower portion of the solid is shown in firm outlines. Height of pyramid is dimensioned as 60 mm. Reference line is labeled x-y. Note at bottom right: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

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.

Figure: Orthographic projections and true shape of section of a hexagonal pyramid

Projection drawing showing the front view, top view, and true shape of a section of a hexagonal pyramid.

• Front view shows pyramid apex o', base points a'(e'), b'(d'), c', height 60 mm, with cut section hatched between points 1', 2', 3', 4'. Base width/side B = 30, Height H = 60.
• Top view (plan) shows hexagonal base with vertices a, b, c, d, e, apex o, side length 30 mm. A section plane line cuts through edges at points 1, 2, 3, 4, inclined at 30° to the horizontal/reference, with offset distance 10 mm indicated to the axis. Cutting plane is indicated with section arrows labelled H and T.
• An auxiliary reference line x1-y1 is drawn parallel to the cutting plane line.
• The true shape of the section is projected perpendicular to x1-y1 and hatched, with vertices labelled 1, 2, 3, 4, showing width dimension W = 34.
• Inset box: '88.4', 'B = 30', 'H = 60'.
• Inset box: 'ANSWER W = 34'.
• Note: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

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.

Figure: Step 92.1: Projections of a hexagonal pyramid

Orthographic projections of a hexagonal pyramid standing on the HP with axis perpendicular to the HP. Front view shows apex o' at height H = 60 mm above reference line xy. Base points on xy are marked a', b' f', c' e', d'. Top view shows a regular hexagon with base edge B = 30 mm (corners labelled a, b, c, d, e, f, and center/apex o). Slant edges connect each corner to apex o. Dimension 60 mm indicates the axis height. Scale 1:1, all dimensions are in 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 $oc$.

Figure: Step 92.2: Section plane in top view of hexagonal pyramid

Projections of the hexagonal pyramid showing a cutting plane H-T in the top view. The section plane is inclined at 20° to the horizontal/reference direction and is located at a perpendicular distance of 15 mm from the axis o. The cutting plane cuts the edges of the base and slant faces at points labelled 1, 2, 3, 4. The base dimension is 30 mm and height is 60 mm. Scale 1:1, all dimensions are in mm.

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.

Figure: Orthographic projections of a truncated hexagonal pyramid (step 92.3)

Step 92.3 of the projection and section of a hexagonal pyramid.

• Parameters in box: B = 30, H = 60. SCALE 1:1, ALL DIMENSIONS ARE IN mm.
• Front View: A hexagonal pyramid standing on the HP on reference line x-y. Apex is labelled o', base vertices are a', b', f', c', e', d'. Height is dimensioned as 60 mm. A cutting plane appears inclined, with cut points 1', 2', 3', 4' on generators/edges, and the sectioned portion hatched.
• Top View: A regular hexagon with base edge dimensioned as 30 mm, with vertices a, b, c, d, e, f, and center/apex o. An inclined section plane H-T is shown inclined at 20° to the horizontal/edge direction, at a distance of 15 mm from center o. Cut points 1, 2, 3, 4 are marked where the cutting plane intersects the edges/base.

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.

Figure: Projections and true shape of section of a hexagonal pyramid (step 92.4)

Step 92.4 of the section of a hexagonal pyramid showing the true shape of the section.

• Parameters in box: B = 30, H = 60. SCALE 1:1, ALL DIMENSIONS ARE IN mm.
• Front View: Triangular outline of hexagonal pyramid with apex o' and base on x-y line (a', 1', b', f', c', e', 4', d'). Section cut points 1', 2', 3', 4' connected and hatched, height 60 mm.
• Top View: Regular hexagon with corners a, b, c, d, e, f, center o. Cutting plane H-T inclined at 20°, offset by 15 mm from apex o, cutting points 1, 2, 3, 4.
• Auxiliary Plane View: An auxiliary reference line x1-y1 parallel to the cutting plane H-T (or perpendicular projectors from cut points 1, 2, 3, 4). The true shape of the section is drawn with vertices 1, 2, 3, 4, hatched, using the heights/distances from the front view.

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.

Figure: Step 89.1: Orthographic projections of the pentagonal pyramid before sectioning

Step 89.1 shows the front and top projections of a pentagonal pyramid.

• Given box: 'B = 30', 'H = 60'.
• Reference line: Horizontal line labeled 'x' on the left and 'y' on the right.
• Top View (below xy line): A regular pentagon labeled with vertices a, b, c, d, e, and apex at center o. Edge 'e-d' is horizontal (parallel to the xy line). Base edge 'a-b' is dimensioned as 30. Slant edges connect apex o to each base corner: o-a, o-b, o-c, o-d, o-e.
• Front View (above xy line): The base lies on xy, with points labeled a', e', b', d', c' from left to right, projected vertically from the top view. Apex o' is located at a height of 60 mm above xy along the vertical center axis. Slant edges are drawn from o' to each base point (o'a', o'e', o'b', o'd', o'c').
• Height dimension: 60 mm between base on xy and apex o'.
• Footer notes: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN 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 $3$.

Figure: Step 89.2: Projections of a pentagonal pyramid with a section plane in the top view

Worked step labelled 89.2. GIVEN box:

• B = 30
• H = 60

Projections:

• Reference line labelled x on the left and y on the right.
• Top view (below xy line): A regular pentagon base a-b-c-d-e with center apex o. One side is dimensioned as 30. A cutting plane line H-T is inclined at 45° to the base edge and passes at a perpendicular distance of 10 mm from the apex o. The cutting plane intersects the edges at points 1, 2, 3, and 4.
• Front view (above xy line): Triangle with apex o' and base on the xy line with corners a', e', b', d', c'. The vertical height of the pyramid from xy to o' is dimensioned as 60. Thin projection lines link the top view vertices to the front view base points.
• Caption: SCALE 1:1, ALL DIMENSIONS ARE IN mm.

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 $3'$.

Figure: Step 89.3: Sectional front view of the pentagonal pyramid with hatching

Worked step labelled 89.3. GIVEN box:

• B = 30
• H = 60

Projections:

• Reference line x - y.
• Top view (below xy): Pentagonal pyramid a-b-c-d-e with apex o and cutting plane line H-T inclined at 45° and offset 10 mm from center o, cutting at points 1, 2, 3, 4. An arc from o with radius to c rotates to horizontal point c1.
• Front view (above xy): The cutting plane points are projected upward: point 1 to 1' on the base line, point 2 to 2' on slant edge o'a', point 3 to 3' on slant edge o'b' (height dimensioned as 31 from the xy line to point 3'), and point 4 to 4' near the base. The sectioned face bounded by 1'-2'-3' is hatched with diagonal lines. The retained portion of the pyramid is drawn with thick dark outlines, while cut-away edges are faint.
• Caption: SCALE 1:1, ALL DIMENSIONS ARE IN mm.

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.

Figure: Section of a pentagonal pyramid and true shape of section

Worked solution drawing labelled 89.4 for a pentagonal pyramid (base side B = 30 mm, height H = 60 mm, scale 1:1, all dimensions in mm).

• Top view: Regular pentagon a-b-c-d-e with center apex o, base edge dimensioned as 30 mm. A vertical cutting plane line T-T is inclined at 45° to the reference line xy, at an offset distance of 10 mm perpendicular to the axis o. It intersects edges at points 1, 2, 3, 4.
• Front view: Shows the triangular outline on ground line x-y. The section plane projects vertically to cut the edges at points 1', 2', 3', 4'. The sectioned upper portion between 1'-2'-3'-4' and the base is hatched with section lines. Heights are dimensioned (vertical dimension 31 mm shown on the right).
• True shape of section: Projected perpendicular to cutting plane line T-T onto reference line x1-y1, forming the hatched true shape quadrilateral 1-2-3-4.

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.

Figure: Orthographic projections of a cone (Step 119.1)

Step 119.1 of the solution. Given parameters: D = 50 mm, H = 60 mm. The drawing shows the initial front view (elevation) and top view (plan) of a vertical cone:

• Top view (below xy line): A circle of diameter Ø50 mm with center o, divided into 8 equal parts by four diameters with perimeter points labeled clockwise starting from the left: a, h, g, f, e, d, c, b (a-e is horizontal, g-c is vertical).
• Front view (above xy line): An isosceles triangle with apex o' and base lying on the reference line xy. Height of apex o' above xy is 60 mm. The base points are projected directly upward from the top view and labeled: a' (leftmost), b'(h'), c'(g') (axis center line), d'(f'), and e' (rightmost). Generators connect each base point to the apex o'.
• Notes: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN 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.

Figure: Section plane cutting the top view of the cone (Step 119.2)

Step 119.2 of the solution. Given parameters: D = 50 mm, H = 60 mm.

• Front view: Same as step 119.1, triangular elevation of the cone with apex o', height 60 mm, base on xy with points a', b'(h'), c'(g'), d'(f'), e'.
• Top view: The circular top view (Ø50 mm, 8 divisions: a, b, c, d, e, f, g, h) intersected by a cutting plane line H-T.
• Cutting plane: Inclined at 45° to the vertical (parallel to the 45° diameter line h-o-d), offset at a perpendicular distance of 10 mm from the center o towards the bottom. Arrows labeled 'H' and 'T' at the ends of the cutting plane line point downward/inward perpendicular to the plane.
• Intersection points along the cutting plane line are numbered:
  • 1: intersection with the base circle between a and h
  • 2: intersection with generator o-a
  • 3: intersection with generator o-b
  • 4: intersection with generator o-c
  • 5: intersection with the base circle between c and d.
• Notes: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

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.

Figure: Orthographic projection step 119.3 of a cone cut by an inclined section plane

Step 119.3 showing orthographic projection of a cone cut by a section plane. Boxed label '119.3'. Given parameters: D = 50, H = 60. Note: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'. Front view shows cone of apex o', base points a', b'(h'), c'(g'), d'(f'), e' on the horizontal reference line x-y. Height of cone is 60 mm. The sectional front view shows a hatched curve resulting from the cut. Top view shows the circular base of diameter Ø50 with center o and generators radiating to base points. A cutting plane line labeled H-T is inclined at 45° to the x-y axis, offset by 10 mm from the base perimeter/centerline, cutting the cone generators at points 1, 2, 3, 4, 5. Projector lines project these cut points upwards to the corresponding generators in the front view to construct the hatched section.

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.

Figure: Orthographic projection step 119.4 showing true shape of the hyperbolic section

Step 119.4 completing the sectional projection and constructing the true shape of the section. Boxed label '119.4'. Given parameters: D = 50, H = 60. Text notes: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm', 'HYPERBOLA'. Contains the front view and top view from step 119.3, along with an auxiliary projection plane x₁-y₁ drawn parallel to the cutting plane line H-T (inclined at 45° with a 10 mm offset). Perpendicular projection lines are drawn from cut points 1, 2, 3, 4, 5 on line H-T across reference line x₁-y₁. Distances from the front view are transferred to plot points 1, 2, 3, 4, 5, defining a smooth, hatched curve representing the true shape of the section, labeled 'HYPERBOLA'.

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 $x'y'$.

Figure: Step 93.1: Projections and initial side view of a hexagonal pyramid

Engineering drawing showing step 93.1 of the projection of a hexagonal pyramid:

• A horizontal reference line labeled x-y and a vertical reference line labeled x'-y' intersecting at the axis origin.
• Box at the top-left of the drawing indicates: B = 30, H = 60.
• Top View (Plan): A regular hexagon with base side 30 mm, oriented such that vertical edges (a-f and c-d) are perpendicular to the XY line. Vertices are labeled counterclockwise: a (bottom-left), b (bottom-center), c (bottom-right), d (top-right), e (top-center), f (top-left), with center apex labeled o. Internal generators connect each base corner to o.
• Front View (Elevation): The base rests directly on the xy line with points a'(f'), b'(e'), and c'(d'). The apex o' is at a height of 60 mm along the central axis projector from o. Triangular faces are outlined by joining o' to the base corners.
• Side View (Right Profile View): Horizontal projection lines from corners of the top view extend rightward to the vertical axis x'-y', then arc 90 degrees upward to the xy line using circular transfer arcs, yielding points e'', f'' d'', a'' c'', and b''. Apex o'' is projected horizontally from o' at height 60 mm. Generators connect o'' to each base point.
• Dimension 30 mm shown on the base edge (between f and a).
• Dimension 60 mm shown for the height from xy to o'.
• Bottom note: SCALE 1:1, ALL DIMENSIONS ARE IN mm.

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.

Figure: Step 93.2: Orthographic projections and side view construction for a cut hexagonal pyramid

Step 93.2 of the solution for problem 93. Box at top-left shows given dimensions: base side B = 30 mm, height H = 60 mm. Bottom-right text: 'SCALE 1:1, ALL DIMENSIONS ARE IN mm'.

• Reference axes: Horizontal ground line x-y and vertical reference line x'-y' intersecting at an origin.
• Front view (above x-y): Triangular elevation of the regular hexagonal pyramid with apex o' at height 60 mm (dimensioned on the left). Base vertices on x-y are a'(f'), b'(e'), and c'(d'). An axis/edge line runs from o' to b'(e').
• Top view (below x-y): Regular hexagon with apex o at the center and vertices e (top), d (top-right), c (bottom-right), b (bottom), a (bottom-left), f (top-left). Hexagon width dimensioned as 30 on the left. A vertical cutting plane is located 15 mm to the left of the central axis (dimensioned 15 at bottom), cutting edges at point 1 (on f-e), point 2 (on f-o), point 3 (on a-o), and point 4 (on a-b). Arrows labeled 'T' at top and 'V' at bottom point rightward indicating the viewing direction for the sectional side view.
• Side view construction (right of x'-y'): Apex o'' is at height 60 mm, projected horizontally from o'. Horizontal projectors from top view vertices (e, d, c, b, a, f) extend to the x'-y' axis and are rotated 90 degrees via quarter-circle arcs centered at the axis origin onto the x-y line, establishing base points e'', f'' d'', a'' c'', and b''. Slant edges connect o'' to each base point. Projection lines from the cutting points also begin extending rightward.

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.

Figure: Step 93.3: Sectional side view and final views of the cut hexagonal pyramid

Step 93.3 of the solution for problem 93. Box at top-left shows given dimensions: base side B = 30 mm, height H = 60 mm. Bottom-right text: 'SCALE 1:1, ALL DIMENSIONS ARE IN mm'.

• Top view (below x-y): Regular hexagon with vertices a, b, c, d, e, f and center o, cut by a vertical section plane at 15 mm from the axis, with cutting points 1, 2, 3, 4.
• Front view (above x-y): Elevation of the pyramid with apex o' and base a'(f'), b'(e'), c'(d'). Points 3', 2' are marked along the cutting plane line, projected horizontally from the side view.
• Sectional side view (right of x'-y'):
  • The cutting plane points are projected via quarter-circle arcs and vertical lines into the side view:
    • Points 1 and 4 on the base project to 1'' and 4'' on the x-y base line.
    • Points 2 and 3 on the slant edges project to points 2'' and 3'' at their corresponding heights on the slant edges.
  • The cross-sectional surface 1''-2''-3''-4'' is hatched with evenly spaced diagonal section lines at 45 degrees.
  • A horizontal projector links height 2'', 3'' back to 3', 2' in the front view.
  • The visible outline of the cut solid is outlined in thick solid lines: from e'' to o'', o'' to b'', along the base to 4'', up through the section boundary 4''-3''-2''-1'', and along the base to e''.

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 $x'y'$.

Figure: Step 120.1: Initial projection showing front view, top view, and right side view setup of a cone resting on its base on HP

Orthographic projections of a cone:

• Reference lines: horizontal XY line labelled 'x' on the left and 'y' on the right; vertical axis labelled 'y'' at the top and 'x'' at the bottom.
• Top view (below XY line, left of vertical axis): A circle of diameter 50 mm (labelled 'Ø50') centered at 'o', divided into 8 equal sectors with circumferential points labelled counterclockwise from left: a, b, c, d, e, f, g, h.
• Front view (above XY line, left of vertical axis): Triangle with base on the XY line and apex at 'o''. Base points correspond to top-view projections: a', b'(h'), c'(g'), d'(f'), e'. Cone height is dimensioned as 60 mm.
• Side view construction (to the right of vertical axis): Horizontal projection lines from top view points g, f, e, d, c extend rightward to the vertical axis, then curve via quarter-circle arcs to the XY line, meeting base points labelled g'', f''h'', a''e'', b''d'', c''. Above the XY line, generators extend to the side-view apex o''.
• Annotation box: '120.1' with 'D = 50', 'H = 60'.
• Scale note at bottom right: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

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.

Figure: Orthographic projection step 120.2 showing construction lines for a sectioned cone

Step 120.2 of orthographic projection of a cone cut by a section plane parallel to VP. Given parameters in a box on the left: D = 50, H = 60. The layout shows:

• Reference axes: Horizontal axis x-y and vertical reference axis y'-x' separating front view and side view.
• Top view: A circle of diameter Ø50 with center 'o', divided into 8 equal parts labeled a, b, c, d, e, f, g, h. A vertical cutting plane T-T cuts the circle along chord h-b (length 40 mm). Intermediate points 1, 2, 3, 4, 5 are marked along the section line.
• Front view: Cone elevation with base on x-y line of points a', b'(h'), c'(g'), d'(f'), e', and apex o' at height 60 mm. Light construction lines connect base points to apex o'. Cutting plane T-T is projected upwards as a dashed line.
• Side view (profile view): Shows apex o'' at height 60 mm. Projection arcs centered on the origin transfer horizontal positions from the top view to the profile baseline, locating points g'', f''h'', a''e'', b''d'', c'', which connect to apex o''.
• Annotations: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

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.

Figure: Orthographic projection step 120.3 showing the completed views and hatched section of the cone

Step 120.3 of orthographic projection of the cut cone. Given box: D = 50, H = 60.

• Top view: Shows the retained portion of the circular base (Ø50) with the circular segment beyond chord h-b removed. The cut section line b-h has length 40 mm and contains points 1, 2, 3, 4, 5.
• Front view: The remaining truncated cone is drawn with dark outlines, showing the vertical cut surface rising from base point b'(h') to the outer generator, and base line from b'(h') to e'.
• Side view (left profile view): Shows the true shape of the vertical section surface as a hatched hyperbolic area with a base width of 40 mm (centered at a''e'' and spanning to f''h'' and b''d''). The outline of the cone is also darkened.
• Annotations: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

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 $o'$.

Figure: Initial orthographic projections of a square pyramid

Projection step labeled 96.1 showing the initial setup of a square pyramid of base side 40 mm and height 60 mm. The reference line xy separates the views. In the top view below xy, square base abcd is drawn with horizontal sides ab and cd, vertical sides ad and bc, and diagonals intersecting at center o. A dimension of square 40 is indicated below edge ab. Projectors extend upward to the xy line: a and d project to a'(d'), b and c project to b'(c'). The apex o projects along a centerline to o' at a height of 60 mm above xy. The front view forms an isosceles triangle o'-a'(d')-b'(c'). A text box next to the axis indicates 'B = 40' and 'H = 60'.

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 $\theta$.

Figure: Orthographic projections of a cut square pyramid (Step 96.2)

Step 96.2 of problem 96. Shows the front view and top view of a truncated square pyramid.

• Given box: B = 40, H = 60, SCALE 1:1, ALL DIMENSIONS ARE IN mm.
• Front view: Apex o', base a'(d') to b'(c') of width 40 mm resting on the XY line. Axis height is 60 mm. A cutting plane line cuts the pyramid through point 1' on slant edge o'-a'(d') and point 2' on slant edge o'-b'(c') at an inclination angle θ = 31° to the horizontal. Viewing arrows marked 'T' (perpendicular to cutting plane) and 'V' (from the right).
• Top view: Base square a-b-c-d of side length 40 mm, center apex o. Section area is hatched with vertices 1, 2, 2_1, 1_1. The vertical distance between 1 and 1_1 is 20 mm, and between 2 and 2_1 is 30 mm.
• Answer box: θ = 31°.

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.

Figure: Auxiliary view showing the true shape of the section of a square pyramid (Step 96.3)

Step 96.3 of problem 96. Demonstrates the construction of the true shape of the section.

• Base projections (front and top views) identical to Step 96.2.
• Auxiliary projection: An auxiliary reference line x1-y1 is drawn parallel to the cutting plane line 1'-2'. Projectors are drawn perpendicular to x1-y1 from cutting points 1' and 2' to construct the true shape of the cut section with hatched quadrilateral boundary 1-1_1-2_1-2. Width is marked as W.
• Given box: B = 40, H = 60, SCALE 1:1, ALL DIMENSIONS ARE IN mm.
• Answer box: θ = 31°, W = 29 mm.

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.

Figure: Step 100.1: Initial front and top views of the square prism resting on HP

Step 100.1 showing the initial orthographic projections of a square prism:

• Top View: A square resting with its vertical faces equally inclined at 45° to the VP (diamond orientation). The vertices are labelled counter-clockwise: left corner a(e), bottom corner b(f), right corner c(g), top corner d(h). The edge length is dimensioned as '□40'.
• Front View: Projected above the horizontal reference line (marked 'y' on both ends). The base lies on the reference line with vertices e' on the left, f'(h') in the centre, and g' on the right. The top face has vertices a' on the left, b'(d') in the centre, and c' on the right. A central vertical axis/edge connects f'(h') to b'(d'). The total height is dimensioned as 70 mm on the left.
• Step label box at upper-right: '100.1', with 'B = 40' and 'H = 70'.
• Bottom text: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN 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 $\theta$.

Figure: Step 100.2: Orthographic projections of a truncated square prism

Front view and top view of a truncated square prism standing on HP. The square base has side 40 mm, height H = 70 mm. In the top view, the square base is oriented with diagonals parallel and perpendicular to the reference line y-y; corners are labelled a(e) at left, b(f) at bottom, c(g) at right, and d(h) at top. The side is dimensioned as square 40. A section plane cuts through the prism, with the hatched cut section bounded by points 1, 2, 2_1, and 1_1. The vertical distance between points 1 and 1_1 is 20 mm, while the total width across the corner is 40 mm. In the front view, the prism is drawn on ground line y-y with height 70 mm. Points at top: a', b'(d'), c'; points at base: e', f'(h'), g'. The cut section is shown by inclined line 1'-2'. Arrow V points downwards from above b'(d') towards 2', and arrow T points upwards. A box at the upper right indicates '100.2' with givens B = 40, H = 70. Note at bottom right: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

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.

Figure: Step 100.3: True shape of section of the truncated square prism using an auxiliary plane

Auxiliary projection to determine the true shape of the section of the truncated square prism. The front view and top view are reproduced from step 100.2 with base B = 40 mm and height H = 70 mm. The cutting plane in the front view is inclined at angle θ to the horizontal ground line. An auxiliary reference line x_1-y_1 is drawn parallel to the cutting plane line 1'-2'. Projectors are drawn perpendicular to x_1-y_1 from points 1' and 2' to obtain the true shape of the cut section, labelled with vertices 1, 2, 2_1, 1_1 and hatched. The true length of the section is dimensioned as W. A box labelled '100.3' shows 'B = 40', 'H = 70'. An 'ANSWER' box indicates 'θ = 82°', 'W = 71mm'. Note at bottom right: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

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.

Figure: Step 103.1: Initial top view and base front view of the square prism

Initial orthographic projections (step 103.1) of a square prism with base edge 40 mm. Reference line xy is horizontal. The top view below xy is a square rotated 45 degrees, with vertices labeled clockwise: top vertex d(h) lying on the vertical centerline, right vertex c(g), bottom vertex b(f), and left vertex a(e). Dimension of square side is marked as □40. Vertical projectors extend upwards from a(e), d(h), and c(g) to the xy line, marking base points e', f'(h'), and g' on the xy line. A small box above the drawing is labeled '103.1' and 'B = 40'. Notes at the bottom read 'SCALE 1:1' and 'ALL DIMENSIONS ARE IN mm'.

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.

Figure: Step 103.2 of orthographic projection and true shape construction

Projection drawing showing the top view (below xy reference line) of a cut hexagonal prism or shape with vertices labelled a(e), b(f), c(g), d(h). The left section between vertical line of length 30 and d(h)-b(f) is hatched. Above xy are projected points e', f'(h'), g'. To the right is an auxiliary hexagonal view with side dimension 40 and total length L = 75, with horizontal projection alignment lines to the top view. Given parameters: B = 40, Scale 1:1, all dimensions in 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.

Figure: Step 103.3 finding height H = 71 mm using arc of radius L = R75

Continuation of step 103.2 showing the construction of the front view height H. An arc with radius L = R75 swings from point b(f) up to the projector from e', determining the overall height H. Beside it is the answer box 'ANSWER: H=71mm'. The top view shows the hatched hexagonal shape with side length 30, vertices a(e), b(f), c(g), d(h), and the right-hand view with side 40 and overall length L = 75.

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.

Figure: Section of solid orthographic projections and true shape of section labeled 103.4

Technical drawing for problem 103.4 showing sections of a solid:

• Information boxes: 'B = 40', 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm', and ANSWER box: 'H = 71mm', 'θ = 69°'.
• Front View (above reference line x-y): Vertical elevation of total height H, intersected by a cutting plane VT inclined at angle θ to the horizontal. The cutting plane passes through points marked 1', 2', 3', with a vertical dimension of 30 mm shown. Corners of the base and solid are labeled e', f'(h'), g'.
• Top View (below reference line x-y): Plan view with vertices labeled a(e), b(f), c(g), d(h). The cut section inside is hatched and connects points labeled 1, 2, 3, 4 with a side dimension of 40.
• True Shape of Section: Projected perpendicular to the cutting plane VT onto an auxiliary plane line x1-y1. The resulting section is a hatched symmetric polygon with vertices labeled 1, 2, 3.
• Additional view to the right: A six-sided symmetrical figure with side length 40 and total length dimension L = 75.

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.

Figure: Orthographic projections of a square prism / cube standing vertically

Step 106.1 shows the initial projections of a vertical square prism (or cube) with a side of base B = 40 mm. Top View: A square inclined at 45° to the XY line with vertices labeled a(e) on the left, d(h) at the top, c(g) on the right, and b(f) at the bottom. One edge length is dimensioned as '□40'. Front View: Located above the XY line, drawn as a rectangle resting on the reference line xy. The base vertices on xy are labeled e', f'(h'), and g'. The top face vertices are labeled a', b'(d'), and c'. Vertical projection lines link corresponding points between the top and front views. Labels and annotations: 'B = 40' in a rectangle above; 'SCALE 1:1' and 'ALL DIMENSIONS ARE IN mm' at the bottom right.

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 $b'(d')-f'(h')$.

Figure: Section of a square prism cut by an inclined cutting plane

Step 106.2 shows the prism cut by an inclined section plane T-V. Front View: An inclined cutting plane represented by a center line with thick ends and arrows marked 'T' (upper left) and 'V' (lower right) cuts through the front view. The cutting plane intersects the leftmost vertical edge a'-e' at point 1', the central vertical axis/edges at point 2', and the rightmost vertical edge c'-g' at point 3'. Top View: The entire square cross-section is hatched with section lines inclined at 45°. Points corresponding to the cutting plane are marked 1 at vertex a(e), 2 at vertex d(h), 3 at vertex c(g), and 2₁ at vertex b(f). Labels and annotations: '106.2' and 'B = 40' at the top; 'SCALE 1:1' and 'ALL DIMENSIONS ARE IN mm' at the bottom.

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.

Figure: Orthographic projections and true shape of a section of a cube

Worked solution 106.3 for a cube section. Top view shows a square of side 40 mm resting on a corner with edges inclined at 45 degrees to the reference line xy, vertices labeled a(e), b(f), c(g), d(h). The front view shows the elevation of the cube bounded by a', b'(d'), c' on top and e', f'(h'), g' on base line xy. A cutting plane VT passes through the cube at an angle theta to xy, cutting edges at points labeled 1', 2', 3'. The sectional top view shows hatching across the cut area between points 1, 2, 3. Projectors at angle perpendicular to VT project onto an auxiliary ground line x1-y1 to draw the true shape of the section, which is a hatched rectangle of width W = 72 mm labeled with vertices 1, 2, 3, 2_1. Answer box specifies: W = 72mm, theta = 35°, B = 40, SCALE 1:1, ALL DIMENSIONS ARE IN mm.

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.

Figure: Orthographic projections of a square prism with its base inclined

Step 109.1 of a worked problem showing orthographic projections of a square prism:

• Label box at top: '109.1', 'B = 40'.
• Front view (above reference line x-y): A rectangle representing the front view of the prism standing on the HP. Bottom base line on x-y has points e', f'(h'), and g'. Top face has points a', b'(d'), and c'. An internal vertical line represents the front-most edge connecting f'(h') and b'(d').
• Top view (below reference line x-y): A square rotated at 45 degrees (diamond shape) with corners labelled a(e) on the left, d(h) at the top, c(g) on the right, and b(f) at the bottom. Side dimension marked as '□40'.
• Projection lines connect corresponding points between front view and top view: a(e) to e' and a', d(h) and b(f) to f'(h') and b'(d'), and c(g) to g' and c'.
• Text at bottom right: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

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 $f'(h')-g'$.

Figure: Section of a square prism cut by an inclined section plane

Step 109.2 of the worked problem showing the prism cut by an inclined cutting plane:

• Label box at right: '109.2', 'B = 40'.
• Front view: An inclined cutting plane V-T slopes downward to the right, intersecting the top face a'-b'(d') at point 1', the central vertical edge b'(d')-f'(h') at point 2', and the base line f'(h')-g' at point 3' on the x-y reference line. The portion of the prism to the left of the cutting plane remains in thick visible lines.
• Top view: Projectors from cutting points 1', 2', and 3' drop down to the top view to define the sectioned area:
  • Point 1' projects to 1 on top edge a-d and 1₁ on bottom edge a-b.
  • Point 2' projects to 2 on corner d(h) and 2₁ on corner b(f).
  • Point 3' projects to 3 on edge d-c and 3₁ on edge b-c.
• The resulting section is a six-sided polygon with vertices 1, 2, 3, 3₁, 2₁, 1₁, shown filled with 45-degree hatching lines.
• Retained visible edges of the prism in the top view are drawn with solid lines, and the cut side is bounded by the section.
• Base side dimension is marked as '□40'.
• Text at bottom right: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

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.

Figure: Section of a cube showing sectional views and hexagonal true shape

Orthographic projection and true shape of a section of a cube:

• Reference line labelled x - y.
• Front view: cube elevation with top corners a', b'(d'), c' and base corners e', f'(h'), g'. An inclined cutting plane T-V passes through, intersecting edges at points 1', 2', 3'. The plane is inclined at angle θ = 55° to the vertical.
• Top view: cube positioned with diagonal horizontal, corners labelled a(e), b(f), c(g), d(h), with side dimension marked □40. The cut section is shown hatched with cutting points 1, 2, 3 and corresponding symmetric points 1₁, 2₁, 3₁.
• True shape view: projected along reference line x₁-y₁ parallel to cutting plane line T-V. Shows a hatched hexagonal section of width W = 28 mm.
• Text boxes beside the drawing:
  • 'ANSWER: W = 28mm, θ = 55°'
  • Label '109.3'
  • Box 'B = 40'
  • 'SCALE 1:1'
  • 'ALL DIMENSIONS ARE IN mm'

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.

Figure: Projection of a square prism, step 107.1

Step 107.1: Orthographic projections of a square prism with base side 40 mm. Front view above xy-line shows a rectangle: top corners labelled a', b' (d'), c'; bottom corners on xy-line labelled e', f' (h'), g'. Top view below xy-line is a diamond-oriented square (sides inclined at 45° to xy) labelled at vertices: top corner d(h), left corner a(e), right corner c(g), and bottom corner b(f). Dimension of the square side is marked as □40. Scale is 1:1. Box at top reads 'B = 40'. Note at bottom: 'ALL DIMENSIONS ARE IN mm'.

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 $2'$.

Figure: Section of a square prism, step 107.2

Step 107.2: Square prism cut by an inclined section plane VT. In the front view, the cutting plane line passes through point 1' near b'(d') down to point 2' near g', marked with cutting plane indicators labelled 'T' at the top and 'V' at the bottom right. The remaining portion of the prism is outlined in dark lines: a'-b'-2'-g'-e'-a', with line b'(d')-f'(h') retained. In the top view, the sectioned surface corresponds to vertices 1, 2, 1₁, forming a hatched triangular region across the right half of the square between d(h), c(g), and b(f). Vertices labelled: d(h) with point 1, c(g) with point 2, b(f) with point 1₁, and a(e). Side dimension marked as □40. Box at top reads 'B = 40'. Note at bottom: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

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.

Figure: Orthographic projections and true shape of cut prism (Step 107.3)

Projection of a vertical square prism (base side 40 mm, marked as square 40) cut by an inclined section plane:

• Horizontal reference line marked x-y.
• Front view: Base points e', f'(h'), g'; top points a', b'(d'), c'. A cutting plane passes through point 1' on the vertical edge b'(d') down to point 2' at the bottom right corner g', making an inclination angle θ with the horizontal. Viewing direction arrows marked T and V.
• Top view: Square base resting on HP with edges inclined at 45°, corners labelled a(e) at left, b(f) at bottom, c(g) at right, and d(h) at top. The cut region between section points 1 (along edges d-h and b-f) and 2 (at corner c) is cross-hatched.
• True shape of section: Drawn on an auxiliary plane x1-y1 parallel to the cutting plane line 1'-2'. Perpendicular projection lines project points 1' and 2' to form a hatched triangular true shape with vertices 1_1, 1_1, and 2, with width marked W.
• Data/result boxes:
  • ANSWER: W = 57mm, θ = 55°
  • Label: 107.3
  • B = 40
  • SCALE 1:1, ALL DIMENSIONS ARE IN mm

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 $o'$.

Figure: Orthographic projections of a triangular pyramid (Step 110.1)

Projections of a triangular pyramid:

• Horizontal reference line labelled 'x' on the left and 'y' on the right.
• Front view (above xy): Base lies on the xy line with corners labelled a', c' at the left end and b' at the right end. Apex o' is located centrally above the base, connected to a', c' and b'. A dashed axis line extends vertically down from o'.
• Top view (below xy): Triangular base with vertical edge ac on the left (length 50 mm) and corner b to the right. Apex o is centrally located and connected to corners a, b, and c with solid lines.
• Labels and annotations: Label box '110.1', given box 'B = 50', note 'SCALE 1:1', and 'ALL DIMENSIONS ARE IN mm'.

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 $ob$.

Figure: Section of triangular pyramid by an inclined cutting plane (Step 110.2)

Section of the triangular pyramid:

• Front view (above xy): Cutting plane line marked with arrows 'T' and 'V' cuts through the pyramid at an angle of 35° to the horizontal. It intersects the left edges at point 1' and the right edge at point 2'. The lower retained portion is drawn with thick continuous lines; the removed upper portion with apex o' is drawn in thin lines.
• Top view (below xy): Shows the true/apparent section created by the cutting plane, represented by a hatched triangle with vertices 1, 2, and 1₁. The vertical distance between points 1 and 1₁ along the left side is dimensioned as 30 mm. Total base edge length ac is 50 mm.
• Labels and annotations: Label box '110.2', given box 'B = 50', note 'SCALE 1:1', and 'ALL DIMENSIONS ARE IN mm'.

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

Figure: Orthographic projections and true shape of section for a tetrahedron cut by an inclined plane (Step 110.3)

Step 110.3 showing the section of a tetrahedron. Front view above ground line x-y with base points a', c' and b', and apex o'. A cutting plane with trace T-V passes through the tetrahedron at points 1' and 2', making an angle θ with the horizontal. In top view below x-y, triangle abc has edge ac of length 50 mm oriented vertically, with an internal dimension of 30 mm, apex o, and hatched section area with vertices 1, 2, 1_1. Projectors from the cutting plane 1'-2' perpendicular to an auxiliary reference line x1-y1 (with dimension 35 mm) construct the true shape of the cut section with points 1, 1_1, 2 hatched. Beside the drawing are an ANSWER box stating 'θ = 17°', a label '110.3', a box 'B = 50', and notes 'SCALE 1:1' and 'ALL DIMENSIONS ARE IN mm'.

Answer: Inclination $\theta = 35°$.