Special Curves

Given: Major axis = 70 mm, Minor axis = 40 mm
Semi-major axis $a = 35$ mm, Semi-minor axis $b = 20$ mm

Steps of Construction:

1. Draw two concentric circles with centre $O$, one with radius $a = 35$ mm (outer) and one with radius $b = 20$ mm (inner).
2. Draw the major axis $AA'$ (horizontal, 70 mm) and minor axis $BB'$ (vertical, 40 mm) through $O$.
3. Divide both circles into 12 equal parts (every 30°) by drawing radial lines through $O$. Number the division points 1, 2, 3, … 12 on both circles.
4. From each point on the outer circle, draw a line parallel to the minor axis (i.e., vertically downward/upward).
5. From the corresponding point on the inner circle, draw a line parallel to the major axis (i.e., horizontally).
6. The intersection of these two lines gives a point on the ellipse.
7. Repeat for all 12 divisions to get 12 points on the ellipse.
8. Join all the points with a smooth freehand curve (French curve) to complete the ellipse.

Result: The required ellipse with major axis 70 mm and minor axis 40 mm is constructed.

Given: Semi-major axis $a = 40$ mm, Semi-minor axis $b = 25$ mm
Major axis $AA' = 80$ mm, Minor axis $BB' = 50$ mm

Steps of Construction:

1. Draw the major axis $AA' = 80$ mm (horizontal) and bisect it to get centre $O$. Draw the minor axis $BB' = 50$ mm (vertical) through $O$.
2. Locate the foci: The distance of each focus from centre is $$OF = \sqrt{a^2 - b^2} = \sqrt{40^2 - 25^2} = \sqrt{1600 - 625} = \sqrt{975} \approx 31.2 \text{ mm}$$ Mark foci $F_1$ and $F_2$ on the major axis at 31.2 mm on either side of $O$.
3. Divide the major axis $AA'$ into a number of equal parts (say 8–10 parts). Mark points $1, 2, 3, \ldots$ between $A$ and $A'$.
4. With $F_1$ as centre and radius $= A$-to-point-1 (i.e., $A1$), draw arcs above and below the major axis.
5. With $F_2$ as centre and radius $= A'$-to-point-1 (i.e., $A'1$), draw arcs to intersect the previous arcs. The intersections are points on the ellipse. (Note: $A1 + A'1 = AA' = 2a = 80$ mm always.)
6. Repeat for all division points to get sufficient points on the ellipse.
7. Join all points with a smooth curve using a French curve.

Result: The required ellipse with semi-major axis 40 mm and semi-minor axis 25 mm is drawn.

Given: Major axis $= 60$ mm, Minor axis $= 40$ mm
Semi-major axis $a = 30$ mm, Semi-minor axis $b = 20$ mm

Steps of Construction (Rectangle / Oblong Method):

1. Draw the major axis $AA' = 60$ mm (horizontal) and minor axis $BB' = 40$ mm (vertical), intersecting at centre $O$.
2. Construct a rectangle $PQRS$ with length $= 60$ mm and width $= 40$ mm, with $AA'$ and $BB'$ as its centre lines.
3. Divide $OA$ (half of major axis) into any number of equal parts, say 4. Label them $1, 2, 3$ (from $O$ towards $A$). Similarly divide $PA$ (the top half of the rectangle side) into the same 4 equal parts, labelling $1', 2', 3'$ from $P$ towards $A$.
4. From $B$ (top of minor axis), draw lines to points $1, 2, 3$ on $OA$.
5. From $O$, draw lines to points $1', 2', 3'$ on $PA$.
6. The intersections of corresponding lines (line $B$-to-$1$ with line $O$-to-$1'$, etc.) give points on the ellipse.
7. Repeat the same construction for the other three quadrants by symmetry.
8. Join all points with a smooth curve.

Result: The required ellipse with major axis 60 mm and minor axis 40 mm is constructed.

Given: Width of parabola $= 60$ mm, Depth $= 25$ mm

Steps of Construction (Rectangle / Tangent Method):

1. Draw a rectangle $ABCD$ with width $AB = 60$ mm and height (depth) $AD = BC = 25$ mm.
2. Mark the midpoint $M$ of $AB$ (the base). The vertex $V$ of the parabola will be at $M$. The apex of the parabola is at the midpoint of $DC$, call it $E$.
3. Divide $AE$ (half the top edge) into any number of equal parts, say 4. Label them $1, 2, 3$ from $A$ towards $E$.
4. Divide $AD$ (the left vertical side) into the same 4 equal parts. Label them $1', 2', 3'$ from $A$ downward towards $D$ (i.e., towards $M$).
5. From $E$ (apex), draw lines to points $1', 2', 3'$ on $AD$.
6. From $M$ (vertex at base), draw vertical lines through points $1, 2, 3$ on $AE$ (i.e., lines parallel to $AD$).
7. The intersections of corresponding lines give points on the parabola.
8. Repeat symmetrically on the right half ($EB$ and $BC$).
9. Join all points with a smooth curve passing through $E$ at the top and $M$ at the base.

Result: The required parabola with width 60 mm and depth 25 mm is drawn.

Given: Distance between directrix and focus $= 20$ mm

Concept: For a parabola, the vertex $V$ lies midway between the directrix and the focus. So $VF = 10$ mm and the distance from vertex to directrix $= 10$ mm.

Steps of Construction (Focus-Directrix Method):

1. Draw the directrix $DD'$ as a vertical line.
2. Mark the focus $F$ at a horizontal distance of 20 mm from the directrix.
3. Mark the vertex $V$ midway between the directrix and $F$, i.e., at 10 mm from each.
4. Draw the axis of the parabola as a horizontal line through $F$ and $V$.
5. On the axis, mark points $1, 2, 3, \ldots$ at convenient distances from $V$ (e.g., 10, 20, 30 mm).
6. Through each point, draw lines perpendicular to the axis.
7. With $F$ as centre and radius equal to the distance of each point from the directrix (i.e., $10 +$ distance of point from $V$), draw arcs to cut the corresponding perpendicular lines. These intersections are points on the parabola. (By definition: distance from focus $=$ distance from directrix.)
8. Obtain sufficient points on both sides of the axis.
9. Join all points with a smooth curve.

Result: The required parabola with focus-directrix distance 20 mm is drawn.

Given: Radius of circle $r = 15$ mm, so diameter $= 30$ mm, Circumference $= 2\pi r = 2 \times \pi \times 15 \approx 94.25$ mm

Concept: An involute of a circle is the curve traced by the free end of a thread unwound from the circle, kept taut.

Steps of Construction:

1. Draw the given circle with centre $O$ and radius $= 15$ mm.
2. Divide the circle into 12 equal parts. Label the points $P_0, P_1, P_2, \ldots, P_{12}$ (with $P_0 = P_{12}$).
3. Draw tangents to the circle at each of these 12 points.
4. The length of the thread unwound after $n$ divisions $= n \times \dfrac{\text{Circumference}}{12} = n \times \dfrac{94.25}{12} \approx n \times 7.85$ mm.
5. At point $P_1$, mark a length of $1 \times 7.85 = 7.85$ mm along the tangent at $P_1$ to get point $Q_1$.
6. At point $P_2$, mark a length of $2 \times 7.85 = 15.7$ mm along the tangent at $P_2$ to get point $Q_2$.
7. Continue similarly up to $P_{12}$, where the length $= 12 \times 7.85 = 94.25$ mm, giving $Q_{12}$.
8. Join $P_0, Q_1, Q_2, \ldots, Q_{12}$ with a smooth curve.

Result: The required involute of the circle of radius 15 mm is drawn.

Given: Radius of rolling circle $r = 20$ mm, Diameter $= 40$ mm
Circumference $= \pi d = \pi \times 40 \approx 125.66$ mm

Concept: A cycloid is the curve traced by a point on the circumference of a circle rolling along a straight line without slipping.

Steps of Construction:

1. Draw the directing (base) line $AB$ of length equal to the circumference $= 125.66$ mm.
2. Divide $AB$ into 12 equal parts. Label them $0, 1, 2, \ldots, 12$. Each part $= \dfrac{125.66}{12} \approx 10.47$ mm.
3. Draw the rolling circle of radius 20 mm with centre $C_0$ above point $0$ on $AB$. Mark the tracing point $P_0$ at the bottom of the circle (touching $AB$).
4. Draw a line $C_0C_{12}$ parallel to $AB$ at height $= r = 20$ mm (locus of centre of rolling circle).
5. Divide $C_0C_{12}$ into 12 equal parts to get centres $C_1, C_2, \ldots, C_{12}$.
6. With each centre $C_n$ and radius $= 20$ mm, draw arcs of the rolling circle.
7. Divide the rolling circle into 12 equal parts. The tracing point moves through $\dfrac{360°}{12} = 30°$ for each division.
8. For position $n$: with centre $C_n$, the tracing point $P_n$ is located by drawing a circle of radius 20 mm and marking the point at angle $n \times 30°$ from the bottom (measured in the direction of rotation).
9. Plot all 12 points $P_1, P_2, \ldots, P_{12}$.
10. Join $P_0, P_1, P_2, \ldots, P_{12}$ with a smooth curve.

Result: The required cycloid for a circle of radius 20 mm is drawn.

Given: Diameter of cylinder $= 40$ mm, Pitch $= 36$ mm

Concept: A helix is the curve traced by a point moving uniformly along the surface of a cylinder while the cylinder rotates uniformly. The pitch is the axial distance covered in one complete revolution.

Steps of Construction:

1. Draw the front view of the cylinder: a rectangle of width $= 40$ mm and height $=$ pitch $= 36$ mm.
2. Draw the top view: a circle of diameter $= 40$ mm.
3. Divide the top-view circle into 12 equal parts. Label them $0, 1, 2, \ldots, 12$.
4. Divide the pitch height (36 mm) in the front view into the same 12 equal parts. Each part $= \dfrac{36}{12} = 3$ mm. Label horizontal lines $0, 1, 2, \ldots, 12$ from bottom to top.
5. Project the division points of the top-view circle vertically upward into the front view to get vertical lines.
6. The intersection of the vertical projector from division point $n$ (top view) with horizontal line $n$ (front view) gives a point on the helix.
7. Plot all 12 such points.
8. Join the points with a smooth curve. The visible portion is drawn as a continuous curve and the hidden portion as a dashed curve.

Result: The required helix with diameter 40 mm and pitch 36 mm is drawn.

Given: Diameter of circle $\phi = 40$ mm, so radius $r = 20$ mm

Concept: A sine curve is obtained by plotting the projection of a point moving on a circle against the angle of rotation. The height of the point at angle $\theta$ is $r\sin\theta$.

Steps of Construction:

1. Draw the given circle of diameter 40 mm with centre $O$. Mark the horizontal diameter as the reference.
2. Divide the circle into 12 equal parts (every 30°). Label the points $0°, 30°, 60°, 90°, \ldots, 360°$.
3. Draw the x-axis (horizontal line) to the right of the circle. Choose a suitable scale for the x-axis, e.g., let $1° = 1$ mm, so the full x-axis length $= 360$ mm (or use a compressed scale such as $30° = 10$ mm, giving total length $= 120$ mm).
4. Divide the x-axis into 12 equal parts corresponding to $0°, 30°, 60°, \ldots, 360°$.
5. From each division point on the circle, draw horizontal projectors to the right.
6. At each x-axis division point $n$, draw a vertical line. The intersection of this vertical with the horizontal projector from the corresponding circle point gives a point on the sine curve.
7. Plot all 12 points.
8. Join the points with a smooth S-shaped curve passing through $0°$, reaching maximum $+20$ mm at $90°$, returning to $0$ at $180°$, reaching minimum $-20$ mm at $270°$, and returning to $0$ at $360°$.

Result: The required sine curve for a circle of diameter 40 mm is drawn.

Concept: The shape of the cross-section (conic section) of a cone depends on the angle at which the cutting plane meets the cone.

(a) To get a circular surface:
$$\boxed{\text{Cut parallel to base}}$$
When the cutting plane is parallel to the base of the cone, the section is a circle.

(b) To obtain an elliptical surface:
$$\boxed{\text{Cut at an angle to the height}}$$
When the cutting plane is inclined to the axis (height) of the cone but is not parallel to any slant side, the section is an ellipse.

(c) To get a triangular surface:
$$\boxed{\text{Cut along the height}}$$
When the cutting plane passes through the apex and along the height (axial section), the section is a triangle.

(d) To obtain a parabola:
$$\boxed{\text{Cut parallel to the slant height}}$$
When the cutting plane is parallel to one of the slant sides (generators) of the cone, the section is a parabola.

(i) Circular machine parts:
1. Shafts and axles (circular cross-section)
2. Gears (spur gears — circular pitch circle)
3. Flanges and washers

(ii) Elliptical machine parts:
1. Elliptical leaf springs (used in automobiles)
2. Elliptical cam profiles
3. Elliptical manhole covers (used in pressure vessels)

(iii) Parabolic machine parts / engineering applications:
1. Parabolic reflectors (in headlights and searchlights)
2. Parabolic dish antennas
3. Parabolic arches in bridges and structural members

(i) Involute:
- Gear tooth profile: The most important application. The flanks (sides) of gear teeth are cut in the involute profile. This ensures smooth, uniform transmission of motion and constant velocity ratio between meshing gears.
- Also used in the design of scroll compressors and certain cams.

(ii) Cycloid:
- Gear tooth design (cycloidal gears): Cycloidal curves are used for the tooth profile of clock gears and precision instrument gears.
- Used in the design of cycloidal speed reducers.
- The cycloid is the path of least time (brachistochrone) — important in theoretical mechanics.

(iii) Helix:
- Screw threads: The thread on a bolt or nut follows a helical path — the most common engineering application.
- Springs (coil springs): The wire of a helical spring follows a helix.
- Worm gears, drill bits, augers, and conveyor screws all use helical geometry.

(iv) Sine-curve (Sinusoidal wave):
- Electrical Engineering: Alternating current (AC) voltage and current are sinusoidal in nature. Sine curves are used to represent AC waveforms.
- Electronics / Signal Processing: Radio waves, sound waves, and carrier waves are sinusoidal.
- Mechanical vibrations: Simple harmonic motion (SHM) of springs and pendulums is represented by a sine curve.
- Cam design: Certain cams are designed to produce sinusoidal follower motion for smooth operation.

Answer: The point $C$ is the Centre of the ellipse.

*Justification:* The centre of an ellipse is the midpoint of both the major axis and the minor axis, i.e., the point equidistant from both vertices and both ends of the minor axis.

Answer: Length $A$-$A'$ is the Major Axis of the ellipse.

*Justification:* The major axis is the longest diameter of the ellipse, passing through both foci and the centre.

Answer: Length $B$-$B'$ is the Minor Axis of the ellipse.

*Justification:* The minor axis is the shortest diameter of the ellipse, perpendicular to the major axis and passing through the centre.

Answer: Length $CA = CA'$ is called the Semi-Major Axis of the ellipse.

*Justification:* It is half the length of the major axis, measured from the centre $C$ to either vertex $A$ or $A'$.

Answer: Length $CB = CB'$ is called the Semi-Minor Axis of the ellipse.

*Justification:* It is half the length of the minor axis, measured from the centre $C$ to either end $B$ or $B'$ of the minor axis.

Answer: The points $F$ and $F'$ are each known as a Focus of the ellipse (together called Foci).

*Justification:* The foci are two fixed points on the major axis such that for any point $P$ on the ellipse, the sum of distances $PF + PF' =$ constant $=$ Major Axis length $= 2a$.