Universal BUT Special

High-Performance

Joints&Shafts

CRITERIA for Joints selection.

The table shows the Maximum allowable torque (expressed in Kgm) calculated on the basis with an angle of 10° and with continuous use. If the angle is more than 10°, the indicated values will be reduced in accordance with the torque factors shown below.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Example: criteria for selection of joint after taking into account the power to be transmitted,

the speed and the angle of inclination.

 

Example:

 

- power N 3 CV

- speed 2000 RPM.

- angle a 20°

 

To calculate the corresponding torque moment, follow the example:

 

                                                                  716,2 x N        716,2 x 3

                                                         Mt =                   =                   = 1,074Kgm

                                                                        n                 2000

 

 

The torque to be transmitted is 1074 Kgm but, since the joint angle is 20°, the user must select a joint with larger dimension and torque in order carry the needed torque on it.

 

Since the torque factor for 20° is 0,75 (as indicated on the table), divide Mt by F. Follow the example.

 

                                                                             MT       1,074

                                                                                    =            MT 1,432Kgm

                                                                              F          0,75

 

 

The appropriate joint should have a torque capability of 1432 Kgm or higher.

Per you reference, selected the proper Joint from the table of joint with needle bearings (, in the example, the V Series becasue of its speed). Model, in the example, is 105V.

 

Pay attention that 1 Kgm corresponds 9,80665Nm.

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