[Solved] Choosing Loudspeaker Cable Diameter: 1mm2 vs 2.5mm2 vs 4mm2 for 3-Meter, 150W & 500W Peak, 8 Ohm

For some time I have been looking for information on the choice of cable diameter for loudspeakers. E.g. for a 3 meter long cable. I know that 2.5mm2 is enough for me and 4mm2 is for sure.
But I still don't know why such a diameter and not e.g. 1mm2?

I am looking for proof in calculation form that the current that flows to the loudspeaker through a cable with a diameter of e.g. 2.5mm2 will definitely be enough to drive it 100%.
Nowhere did I come across the choice of diameter in a computational way. Probably because there is no "regular" current in the cable

The only thing I noticed was some data from the manufacturer of the columns that could be used for calculations. And yes:
150W column and 500W peak, 8 Ohm
efficiency 97db / 2.83V

I do not know exactly what this voltage means, but I assumed that the loudspeaker, in its largest element, will consume 500W of power at 2.83V. Apart from the fact that I do not have such a large amplifier, I would like to calculate how much ampere will flow in the cable and astronomical 177 A.
Sometimes I count voltage drops for direct current, but I see that in this way I won't get to the diameter selection.

Is there a formula or calculator to calculate the actual cable diameter needed?

P = U * I
or
P = I ^ 2 * R
or
P = U ^ 2 / R

(strict, please do not rebel, that I accept resistance)

The tension from the performance specification has nothing to do with it, you have taken it unnecessarily

There is no philosophy in this, only Ohm's law.
To read.
https://loudspeaker.pl/dlugosc-przewodu-do-glosnikow/

Read also about the power and efficiency of the speakers. The power is - how much the loudspeaker is able to absorb, and the effectiveness - how loud it will sound at a given delivered power. And not as you wrote.

I know that efficiency is loudness, but I thought that the voltage value given by the manufacturer at efficiency is the voltage value at the output of the amplifier at maximum loudspeaker power.

Well, the point is for the cable to supply the required power to the speaker. We have power but we don't have tension. How to count the current in amps? And what can be the acceptable voltage drop in the speakers to make them sound 100%?

Interesting article, but they write only that with speaker power over 100W and a cable distance of up to 24 meters it will be 2.5mm2. I already know that but how they calculated it. Also, if 2.5mm2 is enough for 20 meters, why 1.5mm2 is not enough for 3 meters?

Sorry for the inquisitiveness

adikas wrote:

We have power but we don't have tension. How to count the current in amps?

Ohm's law bows in a belt ... You have power and resistance - the latter applies to resistance and the loudspeaker and the cable itself.
For various reasons, it is assumed that the cable power drop should not exceed 10% of the column power. Of course, it can be less, but it is already wasteful - a larger diameter cable is more expensive. The greater resistance of the cable means that the amplifier devotes too much power to unnecessary heating of the cable. There will always be losses in the cable, but here you should center the pros and cons that result from too much reduction of these losses.

adikas wrote:

if 2.5mm2 is enough for 20 meters, why 1.5mm2 is not enough for 3 meters?

And doesn't a longer cable have more resistance?
The easiest way is to use a ready table instead of counting:

All right thanks guys. Cool tables and everything. I know what cable to buy, but I don't understand
You know, how I count voltage drops at direct current, that's clear to me.
For example, the voltage drop on a 2.5mm2 cable over a distance of 1m at a current of 1A is about 9.6mV (also taken from the table). Knowing that the device consumes e.g. 1A and the cable is 100m long, the voltage will drop by almost 2V and okay it is acceptable, or turn the 2V on the power supply.
But with these loudspeakers, I can't get to China.
Thanks for the help of colleagues

You can count (if you necessarily want to count) for direct current. It doesn't matter in this case.
I do not know if you count the length of both wires or the cable as such? Because the resistance on the cable occurs on both wires so it should be added up.
And as for "overclocking" the amplifier can only give back as much power as results from its supply voltage (minus voltage drops on transistor structures, emitter resistors, etc.), so it is not so light. It may turn out that you "unscrew" as much as you can and still the column will not get full power.
The amplifier is not a power supply - the power supply voltage is what it is ...

I gave 2x2.5