Intermediate neon switch

[333rocky333]

I think I see what you mean about supplying the lights from the switches,

Supply the lights from or in the same cable, as the indicators NOT the switches, there are no neutrals at the switches.
The indicators can off course be fitted next to the switches though.
The way you show on the drawing is 44m cu to last switch .38m back through indicators, and 38 m through the lights.
A whopping 120 metres

Not too keen to work it out exact tonight, but as a QUIDE only , from the drawing, your last light is approx 120 metres of cable,which is too much, due to volt drop and earth loop impedance .
I would start thinking of at least 1.5 mm with a 6amp mcb.

Regarding the Earth loop impedance, on site quide afaik says
1.0 at 6a may limit the cable to 38 metres
1.5 at 6a may limit the cable to 59 metres
This may vary with installation methods.

The amps you say 500watt so 2.17 amps

volt drop is I think 44mv/Amp/metre for 1.0mm and 29mv/Amp/metre for 1.5mm.

So
volt drop for 1.5mm is 29mv X 120met X 2.17 amps = 5664mv
divide by 1000 is 7.5 volts

And
volt drop for 1.0mm is 44mv X 120met X 2.17amps = 11.45 volts

Max volt drop afaik is 9.2 volts max, this is from the mcb to the last light.!

IMO, 1mm is pointless.

Think of the indicators as the lights and vise versa, you sound like you want to use twin and earth, therefore wire as 10 or so lights, as a radial starting at the first indicator as your drawing.

Forget the switch wires for now as you know how to wire them,
The lights and switches and indicator are close by each door i presume.

So run the 2core +e as a radial, from ind1 to light 1 to ind 2 ,to light 2, to ind 3 to light 3 etc, or similar.
This will reduce the run a lot , you need to aim for 60 metres max I think.

If poss redo your drawing the proposed wiring route including measurements.

 

[regsmyth]

333rocky333,
sorry for delay, I was stunned by your brilliance.
(Also trying to hold down the day job!)
I've redrawn the layout, which saves on cable, but I'm not sure which bits you include in your calcs. Methinks it might still be too long...

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[Space cat]

For a simple volts-drop calculation, include all the wire from the CU to the last light. That's a total of 108m as you've drawn it. Using 333rocky333's figures for voltage drop, that will give you 6.8V and 10.3V for the two cable sizes. On this basis, 1.5 sq mm will be fine.

A more precise calculation goes like this:

Voltage drop due to last lamp on its own = 1.7V
Second last lamp = 1.6V (because it's only 99m).
Third last lamp = 1.3V
First lamp = 0.9V

Add them all up to get the drop at the last lamp: 5.5V for 1.5 sq mm and 50% more for 1.0 sq mm - which is just a bit too much.

Apart from the first 10m from the CU, you have two earths in parallel and each is about half the total length. A rough calculation tells me that your earth impedance will be equivalent to that of 35m of cable. For best results you should tie the two earths together in each switch. I think your drawing shows this.