# Power Transmission Technical Information

Below is power transmission information that can be used

## SI Units ## Common units related to Power Transmission ## Common multiples and sub-multiples ## Common conversion factors

The factors given below are typical of those used in Power
Transmission engineering :-
Metric to Imperial are on the left and the converse on the right

### Length

millimetres (mm) x 0.0394 = inches (ins)                             inches x 25.4 = mm
metres (m) x 39.37 = inches                                  inches x 0.0254 = metres (m)
metres x 3.281 = feet (ft)                                                      feet x 0.305 = metres
kilometres x 0.6213 = miles                                          miles x 1.61 = kilometres

### Force

Newtons x 0.225 = pound force (lbf)          lbf x 4.45 = Newtons (N)
kilogram force (kgf) x 2.205 = lbf                lbf x 0.454 = kgf
kgf x 9.81 = Newtons
N x 0.102 = kgf

### Torque

Newton metre (Nm) x 0.735 = pounds feet (lbf.ft)
Newton metre (Nm) x 8.85 = pounds inches (lbf.ins)
kilogram force metre (kgf.m) x 9.81 = Newton metre

### Power

kilowatt (kW) x 1.34 = horse power (hp) hp x 0.746 = kW
Note – the French Cheval-Vapeur (CV) and the German
Pferdestarke (PS) are virtually the same value of horse power.
To be exact, hp x 0.98 = CV or PS

### Inertia

kg.m² x 23.73 = lb.ft²

### Temperature

°C =  5/9 x (°F – 32)  °F = 9/5 x °C + 32

### Pi (π)

Ratios between the circumference and diameter of a circle is π
diameter x π = circumference
π = 3.1416
so diameter x 3.1416 = circumference .

## Common formulae useful in Power Transmission

### Torque, Power and Speed

Power (kW) = Torque (Nm) x rev/min
9550

Torque (Nm) = Power (kW) x 9550
rev/min

## ‘V’ - Drive shaft/bearing loads

The following simple formulae give a good indication as to the static and dynamic loads imposed on shafts/bearings by ‘V’- Belts It is useful formulae being based on the actual setting force used to tension the drive

Ts = static tension
Tc = centrifugal tension
Td = dynamic tension
Ts = 16 x 2 x P x B = N
Tc = M x S² x 2 x B = N
Td = Ts – Tc = N

where :-
16 = a constant
2 = the tight and slack sides of the belt
P = 80% of the higher tensioning force figure (1.3 x column) – kgf

B = the number of belts on the drive
M = belt mass per unit length – kilogram per metre – kg/m

S = belt speed in metres per second (m/s)
S = d x n            m/s
19100

whereby :-
d = small pulley pitch diameter – mm
n = rotational speed of small pulley – rev/min

### Example

Calculate the dynamic tension from the following drive. 90kW 1440 rev/min direct start electric motor to a Belt Conveyor running at 400 rev/min for 12 hours/day carrying copper ore and absorbing 81 kW. Motor shaft is 75 mm, conveyor shaft 105 mm. 1200 mm drive centres.
The drive chosen is :-

Motor Pulley: 280 x 5 SPB with a 3535 / 75 mm bore taper bush

Conveyor pulley: 1000 x 5 SPB with a 4545 / 105 mm bore taper bush

Belts: 5 off SPB 4500 Wedge Belts giving 1191 mm drive centres

Calculating the Dynamic Tension

Ts = 16 x 2 x P x B = 16 x 2 x (8.2 x 9.81 x 0.80) x 5 = 10297 N
Tc = M x S² x 2 x B
whereby :-
M = 0.19 kg/m
S = d x n =   280 x 1440 = 21.11 m/s
19100        19100
Tc = 0.19 x 21.11² x 2 x 5 = 847 N
Td = Ts – Tc
Td = 10297 – 847
= 9450 N

## The calculation of weights

Weight is mass measured vertically and simple empirical formulae can be used to calculate the weight of round and rectangular objects:
Round objects
diameter (mm) squared x length (m) x factor = weight (kgf)
factor for mild steel = 0.00617
factor for stainless steel = 0.00636
factor for cast iron = 0.00598

### Example

calculate the weight of a 25mm diameter rod of mild steel with a length
500 mm.
25² x 0.5 x 0.00617 = 1.928 kgf
Rectangular objects
depth (mm) x height (mm) x length (m) x factor = weight (kgf)
factor for mild steel = 0.00785
factor for stainless steel = 0.00809
factor for cast iron = 0.00761

### Example

calculate the weight of a rectangular mild steel bar 25 mm x 35 mm with
length 600 mm
25 x 35x 0.6 x 0.00785 = 4.121 kgf