Thank you everyone for your answer!

jremington:

The total voltage drop along a wire is easily calculated from the maximum current (I), the total length of the wire carrying the current (L) and the resistance per unit length (rho), which you can look up in the wire table.

Vdrop=I*L*rho

You decide on how much voltage drop can be tolerated, and then buy wire of at least the minimum diameter. From a 12V source, a drop of 1V means that 8% of the available power is wasted just heating up the wire.

How much is normally acceptable?

As I said, it will be running only occasionally. Most of the cable will be on the cold side of the Peltier (less than 1 m3 in volume). That means that, since the peltier will be cooling, it will only reduce the output of the Peltier. My primary objective is that the cable does not burn.

I guess I could calculate the exposed cable heat emission and work out how much it will heat the environmnent.

The other part of the cable will be very small and will be enclosed in an insulated box, together with the rest of the electronics.

jremington:

The total voltage drop along a wire is easily calculated from the maximum current (I), the total length of the wire carrying the current (L) and the resistance per unit length (rho), which you can look up in the wire table.

Vdrop=I*L*rho

You decide on how much voltage drop can be tolerated, and then buy wire of at least the minimum diameter. From a 12V source, a drop of 1V means that 8% of the available power is wasted just heating up the wire.

Excellent, I will check the maximum temperature tolerated by the cable, together with the mentioned calculation of the heat emitted.

jremington:

If this JDT-001 is the power adapter you plan to use, it won't work. 3A @ 12V = 36 Watts.

In the OP, you state that the devices draw a total of about 230 Watts, or around 19 Amperes from a nominal 12V source. I would suggest to use a 12V, 25A adapter instead.

You are right, it is wrong. It is 72 W max (24 V * 3 A). I guess that if the system draws full power, my arduino (also connected to this power source), will not get enough power. I guess the best idea would be either to limit the System output or as you mentioned, use a more powerful adapter.

JohnRob:

Voltage drop calculation is easy:

You need the resistance of the wire, look here wire chart

You can estimate the length, remember your length must include both the positive and negative wires.

Your current:

60W + 70W+ 95W= 225Watts

Current = watts / volts so 225 / 12 = 18.7 Amps.

Your current will be slightly larger ~ 22 amps because a car is not 12V but 13 - 14 volts (when running).

Voltage drop will be 22 a * wire resistance = ______

I think you might be able to get but with 2 mm wire but 2.5 mm is much safer.

John

The regulator can be set exactly to the required voltage (it is not a car but a plug adapter what supplies energy).

Intensity = 225 / 12 = 18.7 A

Taking the Gauge 12 wire from your table: Wire resistance = 5.21 Ohm/km = 0.00521 Ohm/m * 1 m = 0.00521 Ohm

Voltage drop = I * wire resistance = 18.7 * 0.00521 = 0.097 V --> 0,81 % Voltage drop

I think I could go for a lower cable gauge, I will let you know!