Output torques of gearboxes are generally given as table in the catalogues.Torque of a gearbox output shaft rnay be calculated as follows:
M2= Output torque(Nm)
N = Motor power (kW)
h = Gearbox efficiency
n2 = Gearbox output speed



EXAMPLE: 1 A1.5 kW electric motor is dkectiy coupled to a gearbox, output speed is 30 r.p.m and gearbox efficiency is 94% from the equation (1)



EXAMPLE: 2
G=700 kg load are to be carried on a conveyor belt, that is inclined ato=20° from the horizontal, to a lorry. Conveyor speed is v=0.4 m/sec.What is the required power of geared motor? Belt conveyor efficiency is 11^=0.85 and gearbox efficiency is T1r=0.94



Equation (2) is used for the solution of the problem. It is important to mention that belt conveyor system is working on ball bearings. So that we can use 1/8 value in the equation. If the load is carried without using bearings, that is sliding under friction, real fnction coefficient must be used instead of 1/8 value. Power requirement for frictionless conditions, never exceeds G value according to angle.When the values are put in their places in the equation (2)



EXAMPLE: 3
The conveyor is at horizontal position. (In this case cc=0) From the Equation (2) Standard motor power of 0.55 kW should be taken.



EXAMPLE: 4
If the load carried at a vertical position(oc=90). From the Equation (2)

To be able to calculate gearbox output speed, head pulley diameter of conveyor or lift must be known. In this case, gearbox output speed is calculated by the following equation:



V= Conveyor belt or rope speed (m/sec)
dt= Head pulley diameter (mm)
For example: Conveyor belt speed is 0.4 m/sec Head pulley diameter is 200 mm Gearbox output speed is.

(Service factor = Operation Coefficient-ki) This value of reductors, depends on the endurance time to all technical and characteristic features of the machine it would agitate. Machines generally show 3 types of characteristics in respect to loading.

1. LIGHT LOAD (U)

2. MEDIUM LOAD (M)

3. HEAVY LOAD (H)

Even though moments, which are being produced in three separate machine and work in three different loading ways, are equal to each other, a reductor with a bigger operation coefficient is used in the machine that works slower. Daily working hour, in respect to working cogwheel and tr ansmission elements being subject to fatigue of materials, affects negatively in case of excessive working hours. Given the fact that every machine is subject to the highest load during the start, it is considered dangerous when it comes to start-stop condition. This decreases at the following workings. We have shown how operation coefficient works with an example in our catalogue: First, we determine the characteristic of the machine from Table: 1 in line with its working field. If our machine is an elec tr ic engine agitated CHAIN BUCKET EKSKAVATOR then its load status is HEAVY (H). Since the machine will work 24 hours its minimum operation coefficient from Table 2 is ki = 2.

Engine power 5,5 kW

Explanations

1. 1. Gearbox (If ordered gearbox is a single housing type, information represented by symbols belongs to this type or gearbox.)
   
   
2. Output side gearbox. (Information represented by symbols belongs to this type of gearbox.)
   
   
3. Motor size. (Please refer to motor catalogue for detail.)
   
   
4. S : Short
   M : Medium
   L : Long
   
   
5. Pole number express motor speed.
   2=3000 d/d 4=1500 d/d
   6=1000 d/d 8=750d/d