#### Universal joints with friction bearing, Type EG

The table shows the transferable output N and/or torques M of universal joints DIN 808, type EG (single friction bearing) in relation

The values are only applicable to a constant speed of rotation, constant load and an operating inclination angle of max. 10Â°. They are not applicable to universal joints in Stainless Steel.

For larger inclination angles ÃŸ a nominal output N increased by the correction coefficient k and/or a nominal torque M has to be selected (see example below).

Conversion formular:

Torque M [Nm] = 9550 N[kW]/n [min-^{1}]

Output N [kW] = M [Nm] x n [min-^{1}]

1 kW = 1.36 PS / 1 PS = 0.736 kW

Output to be transferred N = 0.65 kW

R.p.m. n = 230 min-^{1}

Angle of inclination ÃŸ = 10Â°

Correction coefficient k = 1

Indicative output Nâ€˜= Nominal output N

Intersection point P results from 0.65 kW and 230 min-^{1} (which corresponds to a torque of 27 Nm).

The next size up universal joint corresponding to point P is the model with a diameter d_{1} = 25.

**Example 2**

Torque to be transferred M = 27 Nm

R.p.m. n = 230 min-^{1}

Angle of inclination ÃŸ = 30Â°

Correction coefficient k = 2.25

Indicative torque = 2.25 x 27 Nm = 60 Nm

Intersection point P_{1} results from 61 Nm and 230 min-^{1} (which is equivalent to an indicative output N = 1.47 kW).

The next size up universal joint corresponding to P_{1} is the model with a diameter d_{1} = 36.

#### Universal joints with needle bearing, Type EW

The table shows the transferable output N and/or torques M of universal joints DIN 808, type EW (single needle bearing) in relation

The values are only applicable to a constant speed of rotation, constant load and an operating inclination angle of max. 10Â°.

For larger inclination angles ÃŸ a nominal output N increased by the correction coefficient k and/or a nominal torque M has to be selected (see example below).

Conversion formular:

Torque M [Nm] = 9550 N[kW]/n [min-^{1}]

Output N [kW] = M [Nm] x n [min-^{1}]

1 kW = 1.36 PS / 1 PS = 0.736 kW

**Example 1**

Torque to be transferred N = 5.5 kW

R.p.m. n = 2300 min-^{1}

Angle of inclination ÃŸ = 10Â°

Correction coefficient k = 1

Indicative output Nâ€˜= Nominal output N

Intersection point P results from 5.5 kW and 2300 min-^{1} (which corresponds to a torque of 23 Nm).

The next size up universal joint corresponding to point P is the model with a diameter d_{1} = 28.

**Example 2**

Torque to be transferred M = 23 Nm

R.p.m. n = 2300 min-^{1}

Angle of inclination ÃŸ = 18Â°

Correction coefficient k = 1.43

Indicative torque = 1.43 x 23 Nm = 33 Nm

Intersection point P_{1} results from 33 Nm and 2300 min-^{1} (which is equivalent to an indicative output N = 7.9 kW).

The next size up universal joint corresponding to P_{1} is the model with a diameter d_{1} = 32.