Ghiringhelli S.r.l.: la dinamica del movimento

SELECTING A WORM WHEEL SET
The input power P1 then the torque T2 shown in the table are nominal, calculated for an eight-hour workday without overloads, with a maximum of 10 operating cycles per hour, room temperature of 30°C and synthetic oil lubrification.
Forecast life Lh = 25,000 hours
To adjust the effective operating conditions, you must calculate the service factor, fs, to find the effective input power or the effective torque. In this way, you can calculate the worm gear power (P1e) as shown in Index 1 or the torque (T2e) on the rotating axis as shown in Index 2.
Therefore we have:
On the worm gear P1e = P1 x fs
On the wheel T2e = T2 x fs
The max. torque T2 max. shown in the table can be held for no more than 4-5 seconds.

Suggestion: any theoretic calculation cannot replace practical experience. Always contact us.

SERVICE FACTOR
The Service Factor, fs, takes into account the operating conditions into account.
It is determined with the following parameters :

  • f1: factor that depends on the type of machine activated and the type of motor used (table 1 and table 2)
  • f2: factor that takes in account the frequency of operating cycles (table 3)
  • f3: factor that is in function of the operations duration percent (ED in%) (table 4)
  • f4: factor that takes into account the room temperature where the worm wheel set operates (table 5)
  • f5: factor that depends on the type of lubricant: for mineral oil see table 6 - for synthetic oli fs = 1
  • f6: factor that depends on the housing's capacity to dissipate heat generated within the gearbox with a fan (table 7)

Therefore:

fs =

fsm = f1 x f2 x f5

fst = f3 x f4 x f5 x f6

Use the larger result of the two.

The fsm factor takes into accunt the mechanical and dynamic characteristics of the transmission, while the fst factor takes into account the type of lubricating oil used.

EXAMPLE 1
Worm wheel set to operate mixers, chemical industry, installed in a housing with a fan and synthetic oil lubricant operating 16 hours a day.

  • Motor power: P1 = 10kW - n1 = 1500 rpm
  • Reduction ratio: i = 21,5
  • Torque required T2: 1100 Nm
  • Max. torque T2 max = 3000 Nm
  • Operating cycles: 20 per hour
  • Cycle ED: 40%
  • Room temperature: 30ƒC
  
TABLE 2 Operating cycles per hour (factor f1)
Motor type Operating hours per day
Load type
U
M
F
Electric motor
Turbine
Hydraulic motor
Up to 2 (intermit) 0.9 1 1.25
From 2 to 10 1 1.25 1.5
From 10 to 24 1.25 1.5 1.75
Alternative motor
4/6 cylinders
Up to 2 (intermit) 1 1.25 1.5
From 2 to 10 1.25 1.5 1.75
From 10 to 24 1.5 1.75 2
Alternative motor
1/3 cylinders
Up to 2 (intermit) 1.25 1.5 1.75
From 2 to 10 1.5 1.75 2
From 10 to 24 1.75 2 2.25

TABLE 3 Starting frequency (factor f2)
Starts per hour Up to 10 From 10 to 60 From 60 to 150 Over 150
f2 1 1.1 1.2 1.3

TABLE 4 Cycles per hour (factor f3)
ED % 100 80 60 40 20
f3 1 0.95 0.87 0.75 0.58

TABLE 5 Room t° C (factor f4)
°C Up to 10° 20° 30° 40° 50°
f4 0.86 1 1.2 1.5 2

TABLE 6 Mineral oil (factor f5)
Size 80-100 125-160 180-250 280-400
f5 1.2 1.28 1.32 1.3

TABLE 7 Housing w/o fan volume
(factor f6)
Size 80-100 125-160 180-250 280-400
f6 1.3:2.5 14:3 15:4 1.6:5
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