﻿ Replace Gear Segments with Four-bar Mechanism

# Replace Gear Segments with a Mechanism

## Replace Gear Segments with a Four-bar Mechanism

It is often the case in machine design, that a motion needs to be transmitted between shafts for small angles of rotation, say an oscillation of 30º. The shafts may rotate in the same and the opposite direction to give 'positive' and 'negative' gear ratios.

Frequently, Gear Segments are used. As an alternative, you can use a four-bar mechanism to replace the gear segments.

This topic gives the calculations to find the near optimum four-bar mechanism that can replace the gear segments.

At the design position, the gear ratio is EXACT. The accuracy is within 0.2% over 30º of input shaft rotation.

### Design Methods

There are three design methods:

 • Method 1: Perpendiculars
 • Method 2: Inflection Circle and Cubic of Stationary Circle
 • Method 3: Collineation Axis

With all three methods, you have some design choices to make:

 • Methods 1 & 2 :  One Design Variable
 • Method 3 : Two Design Variables

Methods 1 & 3: use only the Part-Editor.

Method 2: One or two simple calculations.

Preamble

 Kinematically, the required gearing motion - one shaft rotates at a speed relative to a different shaft - is obtained with two pitch cylinders rolling on each other without slipping. Where these two cylinders touch, the corresponding points on each cylinder have identical velocities. It is the same with gears. Such a Point, I, is called the Instantaneous Centre of Rotation for the relative motion of the two cylinders. or gears. Linkages and Instantaneous Centres In the Image, A0B0 is the Frame, B0B and A0A are the input and output links, AB is the coupler that transfers the motion from the input to the output link. The lines drawn to I are construction lines. We can find the ratio of the links in the four-bar linkage so that links A0A and B0B move with a relative velocity equal to the Gear Ratio of the gear segments we want to replace. The input and output links will rotate with the relative speeds at the 'design position', and also for a number of degrees to each side of the design position. Equation : locate Point 'I' Every four-bar linkages has an Instantaneous Centre, I. The first step in all three methods is to find the Instantaneous Centre, I, that gives the Gear Ratio at the design position. Find length a to locate Point I. a = GD/(1-G)....Equation 1 where G = Gear Ratio, D = distance between the Point Ao and Bo G = Input Speed / Output Speed. You should choose the Input to be the faster gear. Usually the shorter Link. G is positive when the gears segments rotate in the same direction, and negative when in the opposite direction. G should be between –1 and +1. Swap the input and output shafts if G lies outside this range. D is a scaling factor. For convenience, it is easy to make the length A0B0 = 100mm Note If G = 0, then B.B0 become zero length. If G = 1, then you get a parallelogram.

#### Method 1: Perpendiculars

 This method to find the lengths of all the links uses Equation 1 and then construction lines. Positive Gear Ratio Find Point I from Equation 1 Draw a line from Point I, making an angle, α, with the frame, A0B0. This line becomes the Coupler, AB Draw lines from B0 and A0 so they intersects the coupler at B and A, and are perpendicular to the Coupler. In the image, I have drawn the two perpendiculars in the mechanism at B and A Negative Gear Ratio Find Point I from Equation 1. The Point I is between the two fixed centres. Draw a line from Point I, making an angle, α, with the frame, A0B0. This line becomes the Coupler, AB Draw lines from B0 and A0 so they intersects the coupler at B and A, and are perpendicular to the Coupler. In the image, I have drawn the two perpendiculars in the mechanism at B and A

#### Method 2: Inflection Circle and Cubic of Stationary Curvature

From the ratios and the radii of the gear segments. we calculate the diameters of the Inflection Circle and Cubic of Stationary Curvature. These are two important conceptual functions found with Instantaneous Kinematics. Even though the functions are very simple, I find it best to use Excel®. This way, I can make quick changes to the gear ratios and their radii. In the future, we will add the equations and include them as a design tool in MechDesigner.

We can use the Constraint Based Sketch Editor to position the Inflection Circle and the Cubic of Stationary Curvature (sometimes called the Circle Point Curve, or Circles of Stationary Curvature) in a sketch. We also calculate the Centre-of-Curvature for the design position (on the Centre Point Curve). It is then possible to add Points and Lines to represent the Pin-Joints and Parts for the optimum four-bar mechanisms.

We transfer the Points and Lines to Pin-Joints and Parts to give the Four-bar Mechanism.

We use a graph to plot the output speed as a function of the input speed. This shows how near the ratio is over a reasonable range of input and output shaft angles.

#### Opposite Rotations - External Gear Segment - Negative Gear Ratio

In this example, the gears have a gear ratio: -2 : 1

They operate over a range of 30º of the smaller gear.

The small Gear has a radius of 120mm and the large gear a radius of 240mm.

This gives a centre distance of 360mm

There are three Bodies, or Parts:

 • Gear Segment 1
 • Gear Segment 2
 • Frame = 3

First, we replace the Gear Segments with a schematic mechanism.

The schematic has four Parts, designated as A0ABB0

The mechanisms transmits the motion from axis B0, to axis A0.

We want to find the length of Parts so that Part B0B rotates two times faster than A0A for the range of 30º of the smaller gear.

The first step in the synthesis is to fix Gear Segment 1, A0A.

Then we can imagine that we will roll the small gear segment around the outside of the larger gear segment.

This helps with the application of instantaneous kinematics.

Inflection Circle

The First Step is to find the Diameter of the Inflection Circle. - I.C. The inflection circle is the locus of points on the moving body that happen to move in a straight-line.

 • I is the Pole, or the Instantaneous Centre
 • I.K is the diameter of the Inflection Circle.

The Euler-Savary Equation gives the diameter of the Inflection Circle.

Euler Savary Equation:

1/D = 1/R2 - 1/R1

Note that the radius of the small gear is negative

1/240 - 1/-120 = 1/D

D = -80mm = diameter I.K

R2 = Radius of small gear

R1 = Radius of fixed gear

D = Diameter of Inflection Circle

The negative result identifies the diameter is to the left of the Instantaneous Centre, I.

Cubic of Stationary Curvature

Then we need to calculate the Cubic of Stationary Curvature - C.S.C

The polar coordinate equation of the C.S.C is found with the following equation.

1/ M.sinψ -1/ N.cosψ - 1/r =0

This is not usually a circle. However, 1/N = 0 in this case.

Therefore, M.sinψ = r . An equation for a circle in polar coordinates.

The angle ψ is measured from the tangent between the Gears.

1/M = 1/3 (1/R2 - 1/D)

1/M = 1/3 (1/-120 - 1/80)

M = 144mm

M = Diameter of Cubic of Stationary Curvature

Choose a Point B on the C.S.C circle.

B is fairly arbitrary. However, we choose it so it lies at an angle of 30º above the centre line, or 60º from the Pole Tangent.

This is a reasonable choice to give a reasonable transmission between the Parts.

When we choose B, we calculate the length of the Part AB.

Point A is the Centre of the Curvature at the design position.

We use a different form of the Euler-Savary Equation.

A.B = I.B2/J.B

A.B = 124.72/55.4 = 280.6mm

J.B = I.B - J.I

J.I = D sin60º = 69.3

J.I = -80sin60 =

I.B = M sin60º = 124.7

J.B = 55.4

To find the lengths of the Links B.Bo and A.Ao use the Cosine Rule

B.Bo = 63.5

A.Ao = 130.77

The input shaft, B.Bo, rotates 30º.

This means the angular velocity of the input Part is 30º.

This is the Graph of the Angular Velocity of the driven Part, A.Ao.

You can see it is exactly 15º at the 'Sweet Spot' - the position where the Parts are exactly as we have designed with the Inflection Circle and the Cubic of Stationary Curvature

The error, or deviation, over ±15º is no more than -0.065º/sec. About 0.4% error.

If you do not use this technique, you should expect about 1.5% error.

[The Cubic of Stationary Curvature are those points on the moving body that happen to move with a curvature that is momentarily at a minimum].

#### Co-Rotations - Internal Gear Segment - Positive Gear Ratio

In this example, the gears have a gear ratio: 3 : 1

Notice the Gear Ratio is 'Positive'. This means that both shafts rotate in the same direction.

The small gear rotates over an angle of 30º

There are three Bodies:

 • Gear Segment 1 - Big internal Gear
 • Gear Segment 2 - Small External Gear
 • Frame = 3

We can use a schematic mechanism to replace the Gear Segments.

First, we replace the Gear Segments with a schematic mechanism.

The schematic has four Parts, designated as AoABBo

The mechanisms transmits the motion from axis Bo, to axis Ao.

We want to find the length of Parts so that Part BoB rotates three times faster than BoB for the range of 30º of the smaller gear.

The first step in the synthesis is to fix Gear Segment 1, AoA.

Then we can roll the small gear segment around the inside of the larger gear segment.

This helps with the application of instantaneous kinematics.

Inflection Circle

The First Step is to find the Diameter of the Inflection Circle. - I.C. The inflection circle is the locus of points on the moving body that happen to move in a straight-line.

 • Point I is the Pole and
 • I.D indicates the diameter of the Inflection Circle. [ I.C.]

The diameter of the Inflection Circle can be found from the Euler-Savary Equation:

Euler Savary Equation

1/D = 1/R2 - 1/R1

1/240 - 1/80 = 1/D

D = 120mm

R2 = radius of small gear

R1 = radius of fixed gear

D = Diameter of Inflection Circle

The Positive value indicates it is to the right.

Cubic of Stationary Curvature

Then we need to calculate the Cubic of Stationary Curvature* - C.S.C

The general equation to find the C.S.C is found with the following equation.

 • 1/ M sin ψ -1/ N cos ψ - 1/r =0

This is not usually a circle. However, 1/N = 0 in this case.

 • M.sin ψ = r .

An equation for a circle in polar coordinates.

The angle ψ is measured from the tangent between the Gears.

1/M = 1/3 (1/R2 - 1/D); 1/M = 1/3 (-1/80 - 1/120)

M = -144mm

M = Diameter of Cubic of Stationary Curvature

Choose a Point B on the C.S.C circle. We can choose B so it lies at an angle of 15º above the Pole Tangent, to put the link Bo.B above the centre line.

This is simply a reasonable choice to give a reasonable transmission between the Parts.

When we choose B, we can calculate the length of the Part AB. AB must be on a ray through 'I', the Instantaneous Centre of the Gears.

We can use a different form of the Euler-Savary Equation.

A.B = I.B2/J.B

A.B = 37.272/6.21 = 223.62mm

J.B = I.B - J.I

J.I = D.sin15º = -31.06

I.B = M.sin15º = 37.27

J.B = 6.21

To find the lengths of Links B.Bo and A.Ao use the Basic Trigonometry, such as the Sine and Cosine Rule

B.Bo = 79.4

A.Ao = 339.82

*The Cubic of Stationary Curvature are those points on the moving body that happen to be move with a curvature that is momentarily at a minimum.

This is the linkage that you can use to replace the gear segments.

The relative speed ratio is 3 : 1

The Angular Velocity of the input Part, B.Bo, is 30º.

The total rotation of the input shaft is 30º

This is a graph of the Angular Velocity of the output Part, A.Ao.

You can see it is exactly 10º at the Sweet Spot - the position where the Parts are exactly as we have designed with the Inflection Circle and the Cubic of Stationary Curvature

The distortion from 10º over the complete 30º rotation of the input shaft is no more than -0.015º/sec. About 0.15% error.

If you do not use this technique, you should expect about 3% error.

It is actually not easy to find a mechanism without this technique, at least for me.

#### Method 3: Use the Collineation Axis.

In this method, we use the instantaneous centre, Point I, with Angle ,α, as a free choice. This ensures that in the design position the velocity ratio provided by the four-bar mechanism is exactly right (equal to G).

It is also possible to make the design position one where the velocity ratio is reaching a maximum or a minimum. Such a condition is achieved when the Collineation axis is perpendicular to the Coupler.

We use the Constraint Based Sketch Editor to construct the mechanism.

All we need to do is calculate position I for the ratio of the gear segments you want to replace.

Use the Constraint-Based Sketch-Editor to construct examples of Positive and Negative Gear Ratios

The Construction

The Collineation Axis is the Line I.H, with I the Instantaneous Centre again.

A condition of maximum or minimum velocity ratio is achieved when the Collineation Axis is perpendicular Coupler.

H is the Instantaneous Centre of rotation of the Coupler relative to the Frame.

At the design position, shown in the image, the coupler is moving as though it had a fixed centre of rotation at H.

The distance I.H is called p. In this design, p can be chosen arbitrarily.

All that is necessary is to draw a line through I at right angles to the ray at angle α, and then mark of H on this second line at a distance p from I. H is then joined to the two fixed centres, Ao and Bo, and so define the moving joints where the two lines cross the 'Alpha' line, at A and B.

Positive Gear Ratio

 1 Find I - from the Gear Ratio, G, you want, and the length of D (AoBo).

Use a = GD/(1-G)

 2 Draw a Ray at angle α. from the Point I relative to the frame line.
 3 Draw a perpendicular from Point I a length you decide, p, to give Point H
 4 Draw rays from the point H to Bo and Ao.

Where these rays cross the Ray from I along the angle α, are the moving centres B and A.

Negative Gear Ratio

 1 Find I - from the Gear Ratio, G you want, and the length of D, (AoBo).

Use a = GD/(1-G)

 2 Draw a Ray at angle α. from the Point I relative to the frame line.
 3 Draw a perpendicular from Point I a length you decide, p, to give Point H
 4 Draw rays from the point H to Bo and Ao.

Where these rays cross the Ray from I along the angle α, are the moving centres B and A.

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