BE Civil Engineering (IOE, TU) Engineering Drawing II (IOE, CE 451) Question Paper 2080 Nepal

Given data

• Base diameter $D = 60\text{ mm}$ $\Rightarrow$ base radius $r = 30\text{ mm}$
• Axis height $h = 80\text{ mm}$
• Section plane: $\perp$ VP, inclined $45^{\circ}$ to HP, cutting the axis at $50\text{ mm}$ above base.

Step 1 — Slant height of the full cone

$$L = \sqrt{r^{2}+h^{2}} = \sqrt{30^{2}+80^{2}} = \sqrt{900+6400} = \sqrt{7300} = 85.44\text{ mm}$$

Step 2 — Sector angle of the full development

The lateral surface of a cone develops into a sector of radius $L$ and arc length equal to the base circumference $2\pi r$.

$$\theta = \frac{r}{L}\times 360^{\circ} = \frac{30}{85.44}\times 360^{\circ} = 126.4^{\circ}$$

Step 3 — Divide the base circle

Divide the base circle into 12 equal parts, numbered $1,2,\dots,12$. Each division subtends $\theta/12 = 126.4^{\circ}/12 = 10.53^{\circ}$ on the development. Generators $o\text{-}1$ through $o\text{-}12$ are drawn from apex $o$ in both the front view and the development; in the development each generator has true length $L = 85.44\text{ mm}$.

Step 4 — Locate the cut points (true lengths along generators)

In the front view the section line is a straight line at $45^{\circ}$ passing through the axis point $50\text{ mm}$ up. Where this line crosses each generator gives the cut point. Because generators in the front view are foreshortened (except the contour generators $o\text{-}1$ and $o\text{-}7$), the cut points must be transferred horizontally to a contour generator to read true distances $o\text{-}P_i$ from the apex.

The section plane has equation, taking apex $o$ as origin with axis pointing down: at axis the cut is $80-50 = 30\text{ mm}$ below apex. A $45^{\circ}$ plane through that axis point. For a generator at radial offset $x$ (measured in the VP-projected plane), the cut height below apex is $30 \pm x$ depending on the side. The true distance from apex along each generator is obtained graphically; representative computed true lengths $o\text{-}P_i$ (rounded) are:

Each true length is $o\text{-}P_i = (\text{cut height below apex})\times \dfrac{L}{h} = (\text{height})\times\dfrac{85.44}{80} = (\text{height})\times1.068$.

Step 5 — Construct the development

1. With apex $O$ as centre and radius $L = 85.44\text{ mm}$, draw an arc.
2. Step off 12 chord divisions to mark generators $O\text{-}1', O\text{-}2', \dots$ spanning the total sector $126.4^{\circ}$.
3. Along each developed generator $O\text{-}i'$, mark $P_i'$ at the corresponding true length from Step 4.
4. Join $P_1' P_2' \dots P_{12}' P_1'$ with a smooth curve. This curve is the true shape of the cut on the lateral surface.

        O (apex)
       /|\  \  \
      / | \  \  \   radius L = 85.44
     /  |  \  \  \
   P7'  P5' P3' P1'   <- cut curve (smooth)
   1'  2' 3' ... 12'   <- base arc (12 divisions, total 126.4 deg)

Result

The development is a sector of radius $85.44\text{ mm}$ and included angle $126.4^{\circ}$, with the cut shown as a smooth curve joining points whose true distances from the apex range from $12.8\text{ mm}$ (generator 7) to $64.1\text{ mm}$ (generator 1).

Given data

• Vertical cylinder: diameter $D = 70\text{ mm}$, radius $R = 35\text{ mm}$, axis vertical on HP.
• Horizontal cylinder: diameter $d = 50\text{ mm}$, radius $\rho = 25\text{ mm}$, axis horizontal, $\parallel$ VP, intersecting the vertical axis at $55\text{ mm}$ above HP.

Since the smaller cylinder ($50\text{ mm}$) penetrates the larger ($70\text{ mm}$) completely, two curves of intersection are produced (entry and exit).

Step 1 — Set up the views

• Front view: vertical cylinder appears as a rectangle of width $70\text{ mm}$; horizontal cylinder as a rectangle of width $50\text{ mm}$ centred at $55\text{ mm}$ above the XY line.
• Top view: vertical cylinder is a circle of $\varnothing70$; horizontal cylinder is a rectangle.
• Side view: the horizontal cylinder appears as a circle of $\varnothing50$ — this circle is divided into 12 equal generators.

Step 2 — Generator (cutting-plane) method

Divide the end circle of the horizontal cylinder (seen in the side view) into 12 equal parts: $1,2,\dots,12$. Each point traces a generator running parallel to the horizontal axis. Each generator is at a horizontal offset $y_i$ from the common axis:

$$y_i = \rho\cos\phi_i,\qquad z_i = \rho\sin\phi_i,\qquad \phi_i = 30^{\circ}\,(i-1)$$

where $z_i$ is the vertical offset of the generator above/below the axis ($55\text{ mm}$ level).

Step 3 — Find where each generator meets the vertical cylinder

A generator at vertical offset $z_i$ lies at height $55 + z_i$. In the top view it is a horizontal line at distance $y_i$ from the centre; it meets the $\varnothing70$ circle at

$$x_i = \sqrt{R^{2}-y_i^{2}} = \sqrt{35^{2}-y_i^{2}}.$$

Tabulating ($\rho = 25$, $R = 35$):

(Points 8–12 mirror 6–2 below the axis: heights $55 - z_i$.)

Step 4 — Plot the intersection curve (front view)

For each generator, the intersection point in the front view has:

• vertical coordinate = height $55 \pm z_i$,
• horizontal coordinate = $\pm x_i$ measured from the vertical axis (front view shows the projection, so the meeting point projects onto the contour at horizontal distance read from the top view).

Join the plotted points $P_1,P_2,\dots$ with a smooth curve. Two symmetric curves appear — one on the front (entry) and one on the back (exit); in the front view they may coincide or appear as two bowed curves crossing near the top and bottom of the penetration.

 Front view (vertical cyl. width 70):
  ____________________
 |        ____        |
 |      /      \      |   <- upper intersection curve (peaks at z=+25, h=80)
 |---- (  horiz  ) ---|   axis at h = 55
 |      \ ____ /      |   <- lower intersection curve (dips to z=-25, h=30)
 |____________________|

Result

The intersection appears as two smooth, symmetrical curves bulging toward the top and bottom of the horizontal cylinder. The curve reaches its extreme heights of $80\text{ mm}$ (top) and $30\text{ mm}$ (bottom) at the generators where $y_i = 0$, and meets the contour of the vertical cylinder at the midheight $55\text{ mm}$ where $y_i = \pm25\text{ mm}$.

Given data

• Internal size: $7.0\text{ m}\times5.0\text{ m}$; external wall $t = 230\text{ mm}$; partition $115\text{ mm}$.
• Floor-to-ceiling height $H = 3.0\text{ m}$; plinth $= 0.45\text{ m}$.
• Door D1 $= 1.0\times2.1\text{ m}$; Windows W1, W2 $= 1.2\times1.5\text{ m}$, sill $= 0.9\text{ m}$.

Step 1 — Overall (plinth) dimensions

External length $= 7.0 + 2(0.23) = 7.46\text{ m}$ External width $= 5.0 + 2(0.23) = 5.46\text{ m}$ Plinth area $= 7.46\times5.46 = 40.73\text{ m}^2$.

Step 2 — Carpet (usable floor) area

The $7.0\text{ m}$ internal length is split into two rooms by a $115\text{ mm}$ partition. If the partition is placed to give Room A $= 4.0\text{ m}$ and Room B the remainder:

• Room A clear length $= 4.0\text{ m}$; width $5.0\text{ m}$ $\Rightarrow 20.0\text{ m}^2$
• Room B clear length $= 7.0 - 4.0 - 0.115 = 2.885\text{ m}$; width $5.0\text{ m}$ $\Rightarrow 14.43\text{ m}^2$

Carpet area $= 20.0 + 14.43 = 34.43\text{ m}^2$.

(Built-up/internal area within external walls $= 7.0\times5.0 = 35.0\text{ m}^2$ less the partition footprint $0.115\times5.0 = 0.575\text{ m}^2$ gives the same $34.43\text{ m}^2$.)

Step 3 — Plan (drawn to scale, e.g. 1:100)

The plan is a horizontal section taken at about $1.0\text{ m}$ above floor, looking down. Show:

• External walls $230\text{ mm}$ thick (hatched as masonry: 45° cross-hatch or brick convention).
• Internal partition $115\text{ mm}$.
• Door opening D1 in an external wall with the door leaf shown swung open at $90°$ (quarter-circle arc indicates swing).
• Two window openings W1, W2 shown by two thin parallel lines across the wall thickness (the lintel/sill convention).
• Dimension lines: overall, room-to-room, openings (chained dimensioning). Mark wall centre-lines.
• North arrow, room names, finished floor levels (FFL $\pm0.000$).

 Plan (schematic):
  <------------- 7.46 -------------->
  +================================+
  |        |                       |
  | Room A |        Room B         | 5.46
  | 4.0x5.0|     2.885x5.0         |
  |   [W1] | [D1]          [W2]    |
  +========+=======================+
           ^ 115 partition

Step 4 — Sectional elevation (vertical section, scale 1:100)

Assume a cutting plane through the door and one window. Show:

• Foundation & footing below ground, plinth $0.45\text{ m}$ above ground (DPC line at plinth top).
• Floor-to-ceiling clear height $3.0\text{ m}$.
• Door: head at $2.1\text{ m}$ above FFL; lintel over the door.
• Window: sill at $0.9\text{ m}$, head at $0.9 + 1.5 = 2.4\text{ m}$ above FFL; lintel over window.
• Roof slab/parapet above ceiling; level marks at FFL, sill, lintel, ceiling, roof.
• Cut masonry hatched; show DPC, plinth protection, ground line.

 Sectional elevation:
  ====== roof slab ======
  |                     | ceiling +3.000
  |   [---window---]    | head +2.400
  |   [          ]      |
  |---[ ]----[door]-----| sill +0.900 / door head +2.100
  |   [ ]    [    ]     |
  ==FFL +0.000==========
   plinth 0.45  | DPC
  --ground------+-------
   [ footing ]

Step 5 — Conventions used

• Hatching: masonry walls cross-hatched at 45°; concrete shown with dotted-triangle/aggregate symbol; earth with diagonal broken lines.
• Door swing: quarter-circle arc; window: sill/lintel double lines.
• Lines: thick continuous for cut edges, thin continuous for projection/dimension lines, chain-dotted for centre-lines.
• Levels: given as $\pm0.000$ FFL with reduced levels.
• Scale: 1:100 typical for plans/elevations, 1:50 for details.

Result

Plinth area $= 40.73\text{ m}^2$; Carpet area $= 34.43\text{ m}^2$. The plan is a horizontal section at $\sim1\text{ m}$ with walls hatched and openings/doors shown by convention; the sectional elevation shows the $0.45\text{ m}$ plinth, $3.0\text{ m}$ clear height, door head at $2.1\text{ m}$, and window from sill $0.9\text{ m}$ to head $2.4\text{ m}$.

Given data

• Block: $4\text{ m}\times3\text{ m}\times3\text{ m}$.
• Faces at $30^{\circ}$ and $60^{\circ}$ to PP (the two sets of horizontal edges).
• Station point (SP) distance from PP $= 7\text{ m}$.
• Horizon height (eye level) $= 1.5\text{ m}$ above ground line (GL).
• One vertical edge touches the PP $\Rightarrow$ that edge is drawn in true height.

Step 1 — Principle of two-point perspective

In angular perspective, the two sets of horizontal edges are inclined to the PP, so each set converges to its own vanishing point (VP) on the horizon line (HL). Vertical edges remain vertical (no vertical VP). Each VP is found by drawing, from the SP in plan, a line parallel to the corresponding set of edges; where it meets the PP, project down to the HL.

Step 2 — Locate the vanishing points (plan construction)

Let the SP be at perpendicular distance $7\text{ m}$ from PP. From SP draw two lines parallel to the two edge directions ($30^{\circ}$ and $60^{\circ}$ to PP). Their intersections with the PP give the piercing points, projected vertically down to the HL as $VP_L$ and $VP_R$.

The horizontal distance of each VP from the point directly below SP (the centre of vision, CV) is:

$$x_{VP} = \frac{\text{(SP distance)}}{\tan(\text{angle of that face to PP})}$$

For the face at $30^{\circ}$ (its edges make $30^{\circ}$ with PP):

$$x_1 = \frac{7}{\tan 30^{\circ}} = \frac{7}{0.5774} = 12.12\text{ m}$$

For the face at $60^{\circ}$:

$$x_2 = \frac{7}{\tan 60^{\circ}} = \frac{7}{1.7321} = 4.04\text{ m}$$

Check: the two VP-to-CV distances satisfy $x_1 x_2 = 12.12\times4.04 = 48.96 \approx 7^2 = 49$ (since the two edge sets are mutually perpendicular, $x_1 x_2 = (\text{SP dist})^2$). ✓

The two VPs lie on opposite sides of CV, so the total span between them is

$$x_1 + x_2 = 12.12 + 4.04 = 16.16\text{ m (measured along HL at PP scale)}.$$

Step 3 — Heights

• Ground line (GL) drawn first; horizon line (HL) at $1.5\text{ m}$ above GL.
• The vertical edge touching the PP is measured in true height $= 3\text{ m}$ from the GL upward.
• All other vertical edges are drawn by projecting their top/bottom along the lines to the two VPs (heights diminish with distance).

Step 4 — Drawing the block

1. Draw GL, then HL $1.5\text{ m}$ above it. Mark $VP_L$ at $12.12\text{ m}$ to the left of CV and $VP_R$ at $4.04\text{ m}$ to the right (or vice versa).
2. Erect the true vertical edge ($3\text{ m}$) where the block touches the PP.
3. From its top and bottom draw lines to both $VP_L$ and $VP_R$.
4. From the plan, project the visual rays from SP to the far corners; where they pierce the PP, drop verticals to cut the VP-lines — these fix the foreshortened edges of the $4\text{ m}$ and $3\text{ m}$ horizontal sides.
5. Complete the receding faces; far vertical edges are bounded by the converging VP lines.

        true ht 3 m
          |\
   VP_L   | \        VP_R
  --------*--*--------*---  HL (1.5 m up)
         /|   \      /
        / |    \    /
  -----/--+-----\--/----- GL
      (block in 2-pt perspective)

Result

$VP_R$ lies $4.04\text{ m}$ from the centre of vision (on the $60^{\circ}$ side) and $VP_L$ lies $12.12\text{ m}$ from it (on the $30^{\circ}$ side), both on the horizon $1.5\text{ m}$ above the ground line; the vertical edge on the PP is drawn in true height $3\text{ m}$, and all horizontal edges converge to these two vanishing points.

Given data

• Square prism: base side $40\text{ mm}$, height $80\text{ mm}$, vertical, faces equally inclined to VP (i.e. edges at $45^{\circ}$ to VP).
• Square pyramid: base side $40\text{ mm}$, axis $80\text{ mm}$, axis horizontal, meeting prism axis at $40\text{ mm}$ above HP.

Step 1 — Method of intersection (edge / line method)

Both solids have plane faces, so the intersection is a series of straight line segments. Use the line (edge) method: locate the points where the edges of one solid pierce the faces of the other, then join those piercing points face-by-face.

Key edges to track:

• The 4 lateral edges of the pyramid (apex-to-base-corners) running horizontally toward the prism.
• The 4 vertical faces of the prism that the pyramid enters and exits.

Step 2 — Set up views

• Front view: prism is a rectangle $40\text{ mm}$ wide (since faces equally inclined, the projected width is $40\sqrt2/...$ — actually the contour width is $40\sqrt2 \approx 56.6\text{ mm}$ across the diagonal seen edge-on; the two front faces appear as a rectangle). The pyramid appears as a triangle of base $40\text{ mm}$ and apex projecting $80\text{ mm}$ horizontally, centred at $40\text{ mm}$ height.
• Top view: prism is a square shown as a diamond (corners toward/away from VP); pyramid is a triangle pointing into the prism.
• Side view: pyramid base is a $40\times40$ square (divided to give its edge points).

Step 3 — Locate piercing points

The pyramid's four lateral edges and its base diagonals are projected; where each pyramid edge crosses a prism face (read in top view, where the prism faces appear as lines), mark the piercing point, then project up to the front view onto the corresponding prism face line. Because the prism faces are flat and the pyramid edges are straight, joining consecutive piercing points across each prism face gives straight segments. The complete intersection is a closed broken (polygonal) line entering one pair of prism faces and exiting the other.

The pyramid axis at $40\text{ mm}$ height is the prism mid-height; by symmetry the intersection is symmetric about that horizontal line and about the vertical axis.

Step 4 — Development of the prism

1. Develop the prism lateral surface as a rectangle: width $= 4\times40 = 160\text{ mm}$ (perimeter), height $= 80\text{ mm}$.
2. Mark the four vertical fold lines at $40, 80, 120\text{ mm}$ dividing it into four faces (each $40\text{ mm}$ wide).
3. Transfer the heights of the piercing points from the front view onto the corresponding face of the development (measure each point's height above HP and its horizontal position within its face).
4. Join the transferred points to outline the hole (entry and exit openings) cut by the pyramid. The hole appears twice — once on the entry pair of faces, once on the exit pair — each as a closed polygon symmetric about the $40\text{ mm}$ height line.

 Development of prism (160 x 80):
  +-----+-----+-----+-----+
  |     |     |     |     | 80
  |  __ |     |  __ |     |
  | /  \|     | /  \|     | <- hole outline (entry & exit)
  | \__/|     | \__/|     | mid-height 40
  |     |     |     |     |
  +-----+-----+-----+-----+ 0
    F1    F2    F3    F4

Step 5 — Conventions

• The cut/hole edges on the development are drawn as continuous thick lines.
• Fold lines as thin chain or thin continuous lines.
• Since the pyramid penetrates fully, two hole-openings are shown on opposite faces.

Result

The intersection is a closed polygonal (straight-segment) line found by the edge/piercing-point method; the prism develops into a $160\text{ mm}\times80\text{ mm}$ rectangle on which the pyramid's penetration is shown as two symmetric polygonal holes centred at the $40\text{ mm}$ mid-height.

The four methods of development

Key note

• Developable surfaces (prism, cylinder, cone, pyramid) develop exactly; the first three methods give true developments.
• Non-developable (sphere, ellipsoid) surfaces cannot be developed exactly — only the approximate method (zone/lune) is used.

Result

Parallel-line (prisms/cylinders), Radial-line (pyramids/cones), Triangulation (transition pieces/oblique cones), and Approximate (spheres/domes).

Definition

CAD (Computer-Aided Design/Drafting) is the use of computer hardware and software to create, modify, analyse and document engineering drawings and designs, replacing manual drafting on a board.

Four advantages over manual drafting

1. Accuracy & precision — coordinates and dimensions are exact (to many decimal places).
2. Speed & easy editing — drawings are copied, mirrored, arrayed and revised quickly without redrawing.
3. Storage & reuse — files are stored digitally, shared, and reused (blocks, templates, layers).
4. Scalability & output — easy zoom, scaling and plotting at any scale; consistent line/text standards.

(Other valid points: 3D modelling, automatic dimensioning, integration with analysis/BIM.)

Four basic AutoCAD commands

(Other acceptable: COPY, MOVE, MIRROR, ARRAY, FILLET, EXTEND, ERASE.)

Result

CAD = computer-aided drafting; advantages: accuracy, speed/editing, storage/reuse, scalability; commands: LINE, CIRCLE, TRIM, OFFSET with the functions above.

Detailing of a simply supported singly-reinforced beam (span 4 m)

Components

• Main (tension) bars: placed at the bottom (sagging moment is maximum at midspan, tension at bottom). e.g. $3$–$\varnothing16\text{ mm}$ bars.
• Anchor / hanger bars: $2$–$\varnothing12\text{ mm}$ at the top to hold stirrups and resist any minor negative moment/handling stresses.
• Stirrups (shear reinforcement): $\varnothing8\text{ mm}$ two-legged vertical stirrups; closer spacing near supports, wider at midspan.
• Clear cover: typically $25\text{ mm}$ (nominal) to protect bars from corrosion and provide fire/bond protection.
• Anchorage: bottom bars carry a development length $L_d$ and are bent up (hook / $90°$ bend) into the support to develop full bond.

Sketch (longitudinal section)

   <------------ 4 m clear span ------------>
   ___________________________________________
  |  o o o (top hanger 2-Ø12)                  |
  | ||  |   |    |     |    |   |  || stirrups  |  Ø8 @ closer near support,
  |                                            |     wider at midspan
  |  ===  ===  ===  (bottom main 3-Ø16) ===    |
  |__hook_____________________________hook_____|
  [supp]                                  [supp]

Cross-section at midspan

   +-----------+
   | o   o     |  2-Ø12 top
   | [stirrup] |  Ø8 @ spacing
   | o  o  o   |  3-Ø16 bottom (main)
   +-----------+
   |<-- cover 25 mm all round -->|

Why closer stirrup spacing near supports

In a simply supported beam under uniform load, the shear force is maximum at the supports and zero at midspan (it varies linearly). Diagonal tension (shear cracks) is therefore most severe near the supports, so stirrups are spaced closer there to resist the higher shear, and may be spaced wider toward midspan where shear is small.

Result

Bottom main bars (tension), top hanger bars, two-legged vertical stirrups closely spaced near supports (max shear) and wider at midspan, $25\text{ mm}$ clear cover, and anchorage (development length + hooks) into the supports.

Given data

• Floor-to-floor height $h = 3.0\text{ m} = 3000\text{ mm}$.
• Riser $R = 150\text{ mm}$, Tread (going of one step) $T = 250\text{ mm}$.
• Dog-legged stair $\Rightarrow$ two equal flights with a half-landing.

Step 1 — Number of risers

$$N_R = \frac{h}{R} = \frac{3000}{150} = 20 \text{ risers}$$

Step 2 — Number of treads

Number of treads $= N_R - 1$ per continuous flight; for a dog-legged stair split into two flights of 10 risers each:

$$\text{Treads per flight} = 10 - 1 = 9$$

Total treads (both flights) $= 2\times9 = 18$. (One landing replaces the missing tread between flights.)

Step 3 — Going (horizontal run) of each flight

Each flight has 10 risers $\Rightarrow 9$ treads:

$$\text{Going per flight} = 9\times T = 9\times250 = 2250\text{ mm} = 2.25\text{ m}$$

Step 4 — Comfort check

$$2R + T = 2(150) + 250 = 300 + 250 = 550\text{ mm}$$

This falls at the lower bound of the acceptable range $550$–$700\text{ mm}$, so the stair is acceptable (comfortable, on the steeper-going/generous-tread side).

Also check the pitch angle:

$$\tan\alpha = \frac{R}{T} = \frac{150}{250} = 0.60 \Rightarrow \alpha = 30.96^{\circ}$$

This is within the comfortable range ($25^{\circ}$–$35^{\circ}$). ✓

Result

20 risers (10 per flight), 18 treads total (9 per flight), going of each flight $= 2.25\text{ m}$; $2R+T = 550\text{ mm}$ and pitch $\approx31^{\circ}$ — the stair is acceptable/comfortable.

Types of perspective

Definitions

• Picture Plane (PP): the transparent vertical plane between the observer and the object on which the perspective image is formed (the "glass screen").
• Station Point (SP): the position of the observer's eye from which the object is viewed.
• Horizon Line (HL): the horizontal line on the PP at the level of the observer's eye; all VPs of horizontal lines lie on it.
• Vanishing Point (VP): the point on the PP (usually on the HL) to which a set of parallel receding lines appears to converge.

Result

One/two/three-point perspectives use 1/2/3 vanishing points respectively, depending on how many sets of parallel edges are inclined to the PP; PP = image plane, SP = eye position, HL = eye-level horizontal, VP = convergence point of parallel lines.

The two general methods

1. Line (generator / edge) method

The surface of one solid is represented by a series of lines (generators or edges). The point where each such line pierces the surface of the other solid is located in the views, and these piercing points are joined to form the line of intersection.

• Best when one solid has straight generators (cylinder, prism, cone, pyramid).

2. Cutting-plane (auxiliary section) method

A series of imaginary cutting planes is passed through both solids. Each plane cuts a simple section (line, circle or polygon) on each solid; the intersection of these two sections gives points on the required curve. Planes are chosen so the sections are easy to draw (e.g. planes containing the apex of a cone give straight generators; planes parallel to a cylinder axis give rectangles).

• Best when sections by suitable planes are simple — especially useful for cone + cylinder or sphere intersections.

Preferred method for a cylinder penetrating a cone

The cutting-plane method is preferable. If the cutting planes are chosen to pass through the apex of the cone, each plane cuts the cone along two straight generators and cuts the cylinder along straight lines — both simple sections — making the piercing points easy and accurate to plot. (Horizontal cutting planes are an alternative, giving circles on the cone and lines on the cylinder.)

Why curves result

When at least one solid has a curved (single- or double-curved) surface, consecutive piercing points do not lie on a straight line; their locus is a smooth curve. Only when both solids are bounded by plane faces (prism + prism, prism + pyramid) does the intersection become a series of straight line segments.

Result

Two methods: the line/generator method and the cutting-plane method; for a cylinder penetrating a cone, the cutting-plane method (planes through the cone apex) is preferred because it yields simple straight-line sections; the intersection is a curve because at least one surface is curved.