Calculate different gear units - the most important formulas for gear drives

Basic knowledge Transmission

Since it is of great importance in the design of gears that the gear wheels involved engage correctly and wear is minimised, various basic calculations must be carried out. Terms such as modulus, pitch diameter and tooth count play a major role here. In this article, we deal with the most important aspects of gear calculation and what to consider when calculating gear units.

Important parameters for gear calculation

Many parameters must be observed and dimensions determined in gear construction in order to adapt the gear geometry ideally to the requirements of later application.

The parameters for gear geometry are:

Axle spacing a
Gear ratio i
Modulus m
Pitching p
Tooth count z
Pitch diameter d_w
Root diameter d_f
Tip diameter d_a

The modulus of gear wheels

The modulus (plural: moduli) is a dimension used for gear calculation, which is specified in millimetres and standardised according to DIN 780.

The modulus measures the size of the teeth of gear wheels.

When designing gear wheel pairings, care must be taken to use only gear wheels with the same modulus. The modulus is calculated as follows:

m=\frac{ p }{ \pi } = \frac{ d_{w} }{ z } = \frac{ d_{a} }{ (z+2) }

3 different gear wheel diameters

When calculating the gear wheel, three important variables are relevant for the diameter.

The tip diameter d_a

The tip diameter d_a denotes the diameter that runs along the tooth tips of a gear wheel. It results from the pitch diameter and the tip height.

d_{a}= d_{w} + 2 \times m

d_{a}= m \times (z+2)

The root diameter d_f

The root diameter d_f denotes the diameter that runs along the tooth root of a gear wheel. It results from the pitch diameter and the root height.

d_{ f } = d - 2h_{f}

The pitch diameter d_w

The pitch diameter d_w describes an imaginary line that runs between the tip diameter and the root diameter. The pitch diameter is a firmly defined dimension of a gear wheel and can be used to determine the axial spacing.

d_{ w } = m \times z

Axial spacing of gear wheels in a gear unit

The axial spacing a defines the distance between the two centre points of two gear wheels and results from the pitch diameters of the two gear wheels (d_f,1, d_f,2).

a = \frac {d_{f,1} + d_{f,2}} {2}

a = \frac {z_{1} + z_{2}} {2} \times m

Tooth spacing of gear wheels in gear units

The tooth count z indicates how many individual teeth are on the mesh surface of the gear wheel. It is derived from the pitch diameter and the modulus.

z = \frac {d_{a} + 2m} { m }

Calculation of gear units

Combining two or more gear wheels together is the simplest form of a gear unit. The most important parameters of all gear unit types include the gear ratio and efficiency.

Gear ratio of gear units

One of the main characteristics of gear units is to achieve a conversion of the input speed (drive) into an output speed (output). This property is called the gear ratio and, depending on the dimensioning of the gear wheels, can be greater than the input speed (gear ratio) or less than the input speed (reduction ratio).

The gear ratio i can be expressed as the ratio of the drive speed n_an and the output speed n_ab.

i = \frac {n_{an} } { n_{ab} }

Alternatively, the gear ratio can be determined using the tooth count (z_an, z_ab) or the pitch diameter (d_an, d_ab).

i = \frac {z_{ab} } { z_{an} } = \frac {d_{ab} } { d_{an} }

In the case of multi-stage gear units, the gear ratios of the individual stages are multiplied by one another and ultimately result in an overall gear ratio of i_ges for stages 1, 2, n.

i_{ges} = \frac {n_{an, 1} } { n_{ab, 1} } \times \frac {n_{an, 2} } { n_{ab, 2} } \times \frac {n_{an, n} } { n_{ab, n} } = {i_{1} } \times { i_{2} } \times i_{n}

Calculate efficiency

The efficiency η of a gear unit is defined as the ratio of the usable power P_use and the supplied power P_pe. The difference between the usable and supplied power is primarily lost as thermal energy, which is caused by the friction between the materials of the gear unit components. The higher the sliding friction between the gear wheels, bearings and axles, the lower the efficiency of the gear unit.

\eta = \frac {P_{use} } { P_{pe} }

In the case of multi-stage gear units, the efficiency of the individual stages are multiplied by one another and ultimately result in an overall efficiency η_ges for stages 1, 2, n.

\eta_{ges} = \frac {P_{use, 1} } { P_{pe, 1} } \times \frac {P_{use, 2} } { P_{pe, 2} } \times \frac {P_{use, n} } { P_{pe, n} } = {\eta_{1} } \times { \eta_{2} } \times \eta_{n}

Simple calculation example for gear drives

A common scenario for the use of gear drives is a given distance between two shafts on which a force is to be transmitted in a given gear ratio.

The following calculation example - with practical values - is based on a simplified dimensioning. The objective is the calculation of the design parameters for the drive wheel and output wheel.

In practice, exact values are not realistic; therefore, the parameters are stated with a tolerance of 5%.
All length units are in millimetres [mm].
The calculation of gear drives depends on experience from practical applications. Follow the advice when designing the gear unit.
The force is usually transferred from the large gear wheel (drive wheel) to the smaller gear wheel (output wheel).
Index 1 belongs to the large drive wheel (e.g. d_w,1).
Index 2 belongs to the smaller output wheel (e.g. d_w,2).

The following is given:

The gear ratio i = 1.9 ... 2.1 - the desired transmission is 2
The axial spacing a = 33.25 mm ... 36.75 mm - the actual axial spacing is 35 mm.
The minimum number of teeth of the smaller gear wheel z_2,min = 11.
The constant for the tip clearance k = 1.25.

Advice: Always provide at least 11 teeth. Otherwise, wear occurs because the gear wheels do not engage with each other exactly.

The necessary design parameters are sought:

The actual axial spacing.
The diameters of the pitch, root and tip diameter.

First, the tooth count of the drive is calculated.

The specified tooth count z₂ of the output is used for this. Due to the tolerance, we use the lower limit of the transmission once and the upper limit once.

First the lower limit:

z_{1,min} = 11 \times 1.9

z_{1,min} = 20.9

Then the upper limit:

z_{1,max} = 11 \times 2.1

z_{1,max} = 23.1

The tooth count is always in whole numbers and is rounded up or down accordingly. Furthermore, we always select odd tooth counts.

Advice: A prime tooth count (prime number) is particularly advantageous - this improves the durability of the gear unit.

We therefore select the tooth count pairing z₁ = 23 und z₂ = 11.

We calculate the modulus from the tooth count and the axial spacing

For this purpose, we change the formula of the axial spacing and use the values for z₁ = 23 and z₂ = 11, as well as the actual axial spacing a = 35 mm:

m = \frac {2 \times 35 \mathrm{mm}}{23 + 11}

m = 2.06 \mathrm{mm}

We select the modulus with 2 mm.

The actual gear ratio and the axial spacing must be determined

Due to the rounding up or rounding down of the number of teeth z₁ = 23 and z₂ = 11 it must be ensured that the actual gear ratio and the axial spacing are still within the specified tolerances.

The actual gear ratio:

i_{tat} = \frac{23}{11}

i_{tat} = 2.09

Actual gear ratio is within tolerance. This calculation can be continued.

The actual axial spacing:

a = \frac {23 + 11} {2} \times 2 \mathrm{mm}

a = 34 \mathrm{mm}

The actual axial spacing is also within the tolerance.

Now the design parameters of the gear wheels can be calculated with the known formulas

The root diameters and the tip diameters depend on the pitch diameters. Therefore, the respective pitch diameters are calculated first.

The pitch diameter of the drive wheel with the modulus m=2 mm and z₁ = 23:

d_{ w,1 } = 2 \mathrm{mm} \times 23

d_{ w,1 } = 46 \mathrm{mm}

The pitch diameter of the output wheel with the modulus m=2 mm and z₂ = 11:

d_{ w,2 } = 2 \mathrm{mm} \times 11

d_{ w,2 }= 22 \mathrm{mm}

The root diameter of the drive wheel with the tip clearance k=1.25:

d_{ f,1 } = 46 \mathrm{mm} - 2 \times 1.25 \times 2 \mathrm{mm}

d_{ f,1 } = 41 \mathrm{mm}

The root diameter of the output wheel with the tip clearance k=1.25:

d_{ f ,2} = 22 \mathrm{mm} - 2 \times 1.25 \times 2 \mathrm{mm}

d_{ f ,2} = 17 \mathrm{mm}

The tip diameter of the drive wheel:

d_{a,1} = 2 \mathrm{mm} \times (23 + 2)

d_{a,1} = 50 \mathrm{mm}

The tip diameter of the output wheel:

d_{a,2} = 2 \mathrm{mm} \times (11 + 2)

d_{a,2} = 26 \mathrm{mm}

The fully designed gear wheels

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