# Construction of Cycloid – How to draw a Cycloid – Engineering Drawing

## Cycloids – Construction of Cycloid

Cycloidal curves are generated by a fixed point on the circumference of a circle, which rolls without slipping along a fixed straight line or a circle.

The rolling circle is called generating circle and the fixed straight line or circle is called directing line or directing circle.

**Definitions :-**

** Cycloid :** It is a curve generated by a point on the circumference of a circle, which rolls without slipping along a straight line.

** Trochoid :** It is a curve generated by a point fixed to a circle, within or outside its circumference, as the circle rolls without slipping along a straight line.

** Epicycloid :** It is a curve generated by a point on the circumference of a circle, which rolls without slipping along another circle outside it.

** Hypocycloid :** It is a curve generated by a point on the circumference of a circle, which rolls without slipping along another circle inside it.

** Epitrochoid :** It is a curve generated by a point fixed to a circle, within or outside its circumference, rolling on the outside of another circle.

** Hypotrochoid :** It is a curve generated by a point fixed to a circle, within or outside its circumference, rolling on the inside of another circle.

Further, classification of trochoid, epitrochoid and hypotrochoid is done on the position of a point fixed to a circle.

If this points is fixed outside the circle, then the curve is called as superior Trochoid, superior epitrochoid and superior hypotrochoid.

And if this point is fixed inside the circle, then the the curve is called as Inferior Trochoid,Inferior epitrochoid and Inferior hypotrochoid. Thus, there are the following nine different types of cycloidal curves :-

**Cycloid and Trochoid**

- Cycloid
- Inferior Trochoid
- Superior Trochoid

**Epicycloid and Epitrochoid**

- Epicycloid
- Inferior Epitrochoid
- Superior Epitrochoid

**Hypocycloid and Hypotrochoid**

- Hypocycloid
- Inferior Hypotrochoid
- Superior Hypotrochoid

## Cycloid and Trochoid – Construction of Cycloid

### 1. Cycloid

** Example:** A circle of 40 mm diameter rolls along a straight line without slipping. Draw the curve traced out by a point P on the circumference, foe one complete revolution of the circle. Name the curve. Draw a tangent to the curve at a point on it 27 mm from the line.

**Procedure :-**

- Draw a circle of diameter 40 mm and mark its centre as C.

- Mark point P on the circumference of the circle as the initial position of a point P.

- Divide the circle into 12 equal parts and mark on the circle 1′, 2′,…., 12′ in opposite direction to rotation.

- Draw a line PA tangent to the circle as the directing line and of length equal to the circumference of the circle = 𝛑D = 22/7 * 40 = 125.7 mm.

- Draw horizontal lines or parallel lines to the directing line through 1′, 2′,…12′.

- Divide line PA into 12 equal parts and mark them by 1, 2,….12.

- From C draw parallel line to directing line and mark on it C1, C2, …C12 corresponding to 1, 2, …. 12 of the directing line.

- Now, when the circle has rolled without slip by 1/12th of a revolution, the centre point C must have moved to C1 and the point P must have moved and achieved the height of point 1′ of the circle.

- Accordingly, with C1 as the centre and radius equal to 40/2 = 20 mm draw an arc to cut the horizontal line through 1′ of the circle at P1.

- Similarly, get arc-line intersection points P2, P3, … P12 by drawing arcs with C2, C3, … C12 as centres and radius equal to 40/2 = 20 mm to intersect with lines through 2′, 3′,…. 12′ of circle respectively.

- Join P, P1, P2,….., P12 by means of a smooth curve to get cycloid.

- Mark point R on cycloid which is lying on a line 27 mm from the directing line.

- Now with R as the centre and radius equal to 40/2 = 20 mm, draw an arc to cut centre line C1-C12 at some point B. Through B, draw a line BS perpendicular to the directing line PA and cutting it at S.

- S is the point of contact and B is the position of the centre of the generating circle when the generating point P is at R. Draw a line through R and S. The line is the required normal. Through R draw a line MN at right angles to RS. MN is the tangent to the cycloid.

** Example: **A wheel of 56 mm diameter rolls downwards on a vertical wall for half revolution and then on the horizontal floor for half revolution. Draw the locus of a point P on the circumference of the wheel, the initial position of which is the contact point with the wall. Name the curve.

### 2. Inferior Trochoid

** Example**: A circle of 50 mm diameter rolls along a straight line without slipping. Draw the curve traced by point Q located 15 mm inside the circle. Take initial position of point Q at the bottom on vertical centre line of circle. Name the curve traced.

**Procedure :-**

- Draw a circle of diameter 50 mm and mark C as its centre.
- Divide the circle into 12 equal parts and mark on the circle 1′, 2′,.. 12′ in the opposite direction to rotation.
- Draw line PA tangent to the circle as the directing line and of length equal to the circumference of the circle = 𝛑D = 22/7 * 50 = 157.14 mm.
- Draw horizontal lines or parallel lines to directing line through 1′, 2′, … 12′.
- Divide the line PA into 12 equal parts and mark on it C1, C2, …. C12 corresponding to 1, 2, …. 12 of the directing line.
- Do the same construction as done for cycloid and obtain points P, P1,… P12 as shown in the figure.
- Join CP and mark Q on it by taking PQ = 15 mm. Take Q inside.
- Similarly, find points Q1, Q2, …. Q12 on lines C1P1, C2P2, Q12 in sequence by a smooth curve to fet
**inferior trochoid.**

### 3. Superior Trochoid

** Example:** A circle of 50 mm diameter rolls along a straight line without slipping. Draw the curves traced by point R located 15 mm outside the circle. Take initial position of point R at the bottom on vertical centre line of circle. Name the curves traced.

**Procedure :-**

- Draw two concentric circles of diameters 50 mm and 80 mm.
- Draw tangential directing line at bottom most points to the circle of diameter 50 mm. Take length of directing line equal to circumference 𝛑D = 22/7 * 50 = 157.14 mm.
- From centre C draw path line of centre parallel to directing line and also of length equal to 157.14 mm.
- Divide this centre line into 12 equal parts and mark them by C, C1, C2, … C12. These points are going to be the centres of rolling circle during its rotation by 1/12th of a revolution and multiples of it.
- Divide the circle of diameter 80 mm into 12 equal parts and mark on it points 1′, 2′, ….12′ in opposite direction to that of rotation.
- Draw parallel lines to the directing line from 0, 1, 2,… 12.
- Now with C, C1, C2 … C12 as centres and radius equal to r=40 mm, draw arcs to cut lines through 0,1, 2, …12 at points Q,Q1,Q2, … Q12 respectively.
- Join points Q, Q1, Q2, … Q12 in sequence by a smooth curve to get superior trochoid.

## Epicycloid and Epitrochoid – Construction of Cycloid

### 1. Epicycloid

** Example:** Draw an epicycloid with rolling circle diameter 50 mm and directing circle diameter 150 mm. Draw tangent and normal at a point on the curve 110 mm from the centre of directing circle.

**Procedure :-**

- When rolling circle rolls by one revolution, it will advance on the directing circle at distance equal to 𝛑D = 22/7 * 50 = 157.1 mm.
- An angle subtended for directing circle of radius R is calculated as follows :-
- 𝛑D = R.𝚹
- Therefore, 𝚹 = 𝛑D/ R = 𝛑*50/(150/2) = 2𝛑/3 = 120°

- Draw an arc of radius R = 150/2 = 75 mm and subtending an angle of 120° at centre O.
- Draw rolling circle of radius r = 50/2 = 25 mm and touching outside the directing circle of radius R = 75 mm.
- Consider touching point as the starting position of point P.
- Divide rolling circle into 12 equal parts and mark them as 1′,2′,…12′ in opposite direction to that of rotation.
- Now with O as the centre and radii equal to 01′, 02′, ….0-12′ draw arcs of sufficient length.
- Draw arc with O as centre and OC as the radius to get path line of the centre.
- Divide angle of 120° into 12 equal parts and draw radial lines through O, as shown in figure, to intersect with the centre line at C, C1, C2, …. C12.
- With C,C1,C2,…. C12 as centres and radius equal to r = 25 mm draw arcs intersecting with arc lines through 1′, 2′, ….. 12′ at points P, P1, P2,… P12 respectively as shown in figure.
- Join P, P1, P2, ….. P12 in sequence by a smooth curve to get epicycloid.
- Now to draw tangent and normal to the curve at any point M (which is obtained on a curve 110 mm from the centre of the directing circle), draw arc with M as the centre and radius equal to r = 25 mm to intersect with the centre line at q. Draw QN line passing through O. Join NM which will be normal to the curve at point M and now draw right angle to it at the point M to get tangent ST.

** Example:** A circle of 50 mm diameter rolls outside and along another fixed circle of 150 mm diameter. Draw the locus of the point P on the circumference of the rolling circle for one complete revolution of the rolling circle, if initially point P is I05 mm from the centre of the directing circle. Name the curve. Draw a normal and a tangent to the curve at a point 110 mm from the centre of the directing circle.

### 2. Inferior Epitrochoid

** Example:** A wheel of 50 mm diameter rolls on outside on another circle of 150 mm diameter. Draw and name the curve traced out by point Q 15 mm from the centre on the wheel along a straight line passing through the centre of the wheel.

**Procedure:-**

- Do the same construct ion as done for epicycloid and get points C,C1,C2 … C12 .

- Mark initial position of Q by taking CQ = 15 mm

- Draw a circle with radius as CQ and divide it into twelve equal parts.

- Draw parallel arcs o the directing circle through these twelve points.

- With C1 as centre and radius equal r1 = C1Q1= 15 mm, mark a point Q1 on the parallel arc through first point.

- Similarly obtain Q2, Ql3 etc.

- Join Q, Q1 • Q2…Q12 in sequence by a smooth curve to get inferior epitrochoid.

### 3.Superior Epitrochoid

** Example:** A wheel of 50 mm diameter rolls on outside on another circle of 150 mm diameter. Draw and name the curve traced out by point R 35 mm from the centre on the wheel along a straight line passing through the centre of the wheel.

**Procedure:-**

- Do the same construction as done for epicycloid and get points C, C1 , C2 , .., C12.

- Mark initial position of Q by taking CQ = 35 mm.

- Follow the further procedure explained in Example 28 to get points QI, Q2,.. Q12.

- Join Q, QI, Q2, .. Q12 in proper sequence by a smooth curve to get Superior Epitrochoid.

## Hypocycloid and Hypotrochoid – Construction of Cycloid

### 1. Hypocycloid

** Example:** Draw a hypocycloid rolling circle 50 mm diameter and directing circle 150 mm diameter. Draw a tangent and normal to it at a point 40 mm from the centre of the directing circle.

**Procedure:-**

- When rolling circle rolls by one revolution, it will advance on the directing circle at a distance equal to πd = π x 50 = 157.1 mm

- An angle θ subtended for directing circle of radius R is calculated as follows:-

πd = R θ

Therefore, θ = πd/R = π*50/75 = 2 π/3 = 120°

- Draw an arc of radius R = 75 mm and subtending an angle of 120° at centre 0.

- Draw rolling circle of radius r = 50/2 = 25 mm and touching Inside the directing circle of radius R= 75 mm.

- Consider touching point as 1he starting position of point P.

- Divide rolling circle into 12 equal parts and mark them as I ‘. 2′, … 12’ in opposite direction to that or rotation.

- Now with 0 as the centre and radii equal 10 0-1′, 0-2′, .., 0-12′ draw arcs of sufficient length .

- Draw arc with 0 as centre and OC as the radius to get path line of the centre.

- Divide angle of 120° into 12 equal parts and draw radial lines through 0. as shown m the figure. to intersect with the centre line at C, C” C1• .. C12.

- With C, C” C2, .. C11as centers and radius equal to r = 25 mm draw arcs intersecting with arc lines through 1′.2’… 12′ at points P 1, P2… P 12 respectively as shown in the figure.

- Joint P, P1, P2… P12 in sequence by a smooth curve to get Hypocycloid.

- Now to draw tangent and normal to the curve at any point M (which is obtained on a curve 40 mm from the centre of the directing circle). Draw arc with M as the centre and radius equal to r = 25 mm to intersect with the centre line at Q. Draw NQ line passing through 0.Join NM which will be normal to the curve and now draw right angle to it at the point M to get tangent TS.

### 2. Inferior Hypotrochoid

** Example:** A circle of 50 mm, diameter rolls on another circle of 150 mm diameter. Draw and name the curve traced out by point Q lying on a straight line through the centre C of the rolling circle and 20 mm from it, when it rolls inside the other circle.

**Procedure:-**

- Do the same construction as done for hypocycloid and get points C, Cl, C2,.. C12.

- Mark initial position of Q by taking CQ = 20 mm.

- Draw a circle with radius as CQ and divide it into twelve equal parts.

- Draw parallel arcs to directing circle through these twelve points.

- With C 1 as centre and radius equal to r 1 = C1 Q1 = 20 mm mark a point Q1 on the parallel arc through first point.

- Similarly, obtain Q2, Q3, etc.

- Join Q, QI, Q2, … Q12 in sequence by a smooth curve to get inferior hypotrochoid.

### 3. Superior Hypotrochoid

** Example:** A circle of 50 mm diameter rolls on another circle of 150 mm diameter. Draw and name the curve traced out by point R lying on a straight line through the centre C of the rolling circle and 35 mm from it, when it rolls inside the other circle.

**Procedure:-**

- Do the same construction as done for hypocycloid and get points C, C1, C2, .. C12.

- Mark initial position of Q by taking CQ = 35 mm.

- Follow the further procedure explained in Example 31 to get points Q1, Q2, ..,Q1 2 .

- Join Q, QI’ Q2 , .., Q12 in sequence by a smooth curve to get a Superior Hypotrochoid .

## Special case of Hypocycloid – Construction of Cycloid

** Example:** Show by means of a drawing that when the diameter of the directing circle is twice that of the generating circle, the hypocycloid is a straight line. Consider the diameter of the generating circle equal to 150 mm.

**Procedure:-**

- Diameter of rolling circle (generating circle) = (1/2 ) x Diameter of directing circle = (150/2) = 75 mm.

- When rolling circle rolls by one revolution, it will advance on the directing circle a distance equal to πd = π x 75 mm.

- An angle θ subtended for directing circle of radius R is calculated as follows: d = π R x θ

Therefore, θ = πd/R = π*75/(150/2) = πc = 180°

- Draw an arc of radius R = 15O/2 = 75 mm and subtending an angle of 180° at centre O.

- Follow further, the procedure same as explained in the Example 30 & obtain the prints P, P 1,P2,.. P12.

- Observe that after joining P, P 1,P2,…,P 12 in sequence, we get a hypocycloid as a straight line and hence this is considered as the special case of hypocycloid .

- Also observe that when the diameter of the rolling circle is half the diameter of the directing circle, the hypocycloid is a straight line and is a diameter of the directing circle.