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OVERHEAD CONTACT WIRES

Wire cross-sections

Ordinary electrical wire is circular in cross-section but this is difficult to suspend from overhead fittings in such a way that the contact device of the tramcar, pressing on the underside, does not foul the attachment points.

On the earliest tramways, the suspension 'ears' were soldered to the top and sides of the circular wire.

Soldered ears

 

Where soldering was undesirable, 'mechanical ears' were used but a circular wire restricted their area of grip. Wires of non-circular cross-section were developed so that mechanical attachment could be made to the upper side without being fouled by the contact device running on the lower side.

A mechanical ear on figure-of-eight wire

 

Nowadays grooved wire is popular and the need for visually obtrusive overhead insulators is avoided by suspending the clamping system from an insulating synthetic rope.

 More on older fittings

 Appearance of overhead work

 Royal Fine Art Commission circular on tramway design

 

 

Electrical properties

The electrical resistance of overhead contact wire can be found from the following tables:

(The wire gauges follow historic tramway practice although present day manufacturers will often quote them in metric units.)

PROPERTIES OF HARD-DRAWN COPPER TROLLEY WIRE

 Gauge

 Diameter

Inch

 Sectional area

 Resistance at 15.5°C

 Weight

 Breaking stress

Sq. Inch

Sq. mm

 Ohms Per Km

 Ohms Per mile

Kg/Km

Lbs./mile

KN

Lbs.

 0000000  S.W.G.

 0.500

 0.1964

 126.7

 0.1363

 0.2193

 1131

 4005

 40.80

 9150

 000000    S.W.G.

 0.464

 0.1691

 109.1

 0.1582

 0.2546

 972

 3441

 37.14

 8330

 0000  B&S    

0.460

 0.1661

 107.2

 0.1613

 0.2596

 955

 3381

 37.32

 8370

 00000      S.W.G.

 0.432

 0.1466

 94.6

 0.1825

 0.2937

 843

 2983

 32.95

 7390

 000   B&S    

 0.410

 0.1320

 85.2

 0.1986

 0.3196

 759

 2687

 30.28

 6790

0000        S.W.G.

 0.400

 0.1257

 81.1

 0.2129

 0.3426

 722

 2557

 28.76

 6450

000         S.W.G.

 0.372

 0.1087

 70.1

 0.2461

 0.3961

 625

 2212

 24.97

 5600

 00    B&S    

 0.365

 0.1046

 67.5

 0.2506

 0.4033

 601

 2129

 24.03

 5390

00          S.W.G.

 0.348

 0.0951

 61.6

 0.2813

 0.4527

 547

 1935

 21.85

 4900

 0     B&S    

 0.325

 0.0829

 53.5

 0.3219

 0.5180

 478

 1692

 19.04

 4270

0            S.W.G.

 0.324

 0.0824

 53.2

 0.3245

0.5222 

 474

 1678

 18.95

 4250

 
 
PROPERTIES OF SILICIUM BRONZE TROLLEY WIRE

 Gauge

 Diameter

Inch

 Sectional area

 Resistance at 15.5°C

 Weight

 Breaking stress

Sq. Inch

Sq. mm

 Ohms Per Km

 Ohms Per mile

Kg/Km

Lbs./mile

KN

Lbs.

 0000000  S.W.G.

 0.500

 0.1964

 126.7

 0.1908

 0.3071

1131

 4005

62.16

13940

 000000    S.W.G.

 0.464

 0.1691

 109.1

 0.2220

 0.3574

972

 3441

54.08

12130

 00000      S.W.G.

 0.432

 0.1466

 94.6

 0.2562

 0.4123

 843

 2983

47.58

10670

 000   B&S    

 0.410

 0.1320

 85.2

 0.2844

 0.4577

 759

 2687

43.43

9740

0000        S.W.G.

 0.400

 0.1257

 81.1

 0.2989

 0.4811

 722

 2557

41.87

9390

000         S.W.G.

 0.372

 0.1087

 70.1

 0.3455

 0.5561

 625

 2212

36.34

8150

 00    B&S    

 0.365

 0.1046

 67.5

 0.3589

 0.5777

601

 2129

35.49

7960

00          S.W.G.

 0.348

 0.0951

 61.6

 0.3949

 0.6356

 547

 1935

32.28

7240

0            S.W.G.

 0.324

 0.0824

 53.2

0.4555

0.7330

 474

 1678

28.40

6370

Some sample calculations

Supposing a tramcar is at one end of a 1 Km length of 000 S.W.G. overhead copper wire, the sub station is supplying 750 volts to the other end. Calculate the voltage available to the tram when:

 

1)  The tram is switched off

From the above tables, the resistance of 1 Km of 000 S.W.G. copper wire is 0.2461 ohms
The current drawn through the wire = 0 Amps

Ohm's Law states that the voltage dropped equals  current  x  resistance

The voltage dropped in the wire in this case =  0  x  0.2461  = 0 volts
so the voltage available to the tram =  750 - 0  = 750 volts

 

2)  The tram accelerates hard and momentarily draws 200 Amps

Now the current through the wire is 200 Amps and its resistance is 0.2461 ohms

Ohms law says the voltage drop will be  200  x  0.2461  =  approx 50 volts

So the voltage available to the tram will be 750 - 50  = 700 volts

 

3)   The tram brakes and regenerates 100 amps

The voltage difference along the wire will be 100  x  0.2461  = approx 25 volts
but the current is now flowing from the tram to the sub-station end of the wire, the opposite direction from before. The voltage difference is now the other way around, it is a voltage gain.

The voltage at the tram will now be 750  +  25  = 775 volts

This is not magic, getting more than we started with. It shows that the tram is returning to the system, some of the energy it took when it accelerated



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