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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
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| 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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