What wire do I need to power 10m of SK9822 LED strip

Question

I have 10 meters of SK9822 LEDs (60 LED/m) which I want to lead around my room. Because there's some furniture in the way, I'll need to cut this in a few pieces and create segments joint with a piece of wire.

These LEDs are 5V and draw the maximum of 60mA each. I know the maximum amount of power these LEDs will need is 36A (60 * 60 * 10) so I bought a transformer that provides 5V, 40A of power.

I read that powering the whole 10 meter strip only from one end wouldn't work, so I'll put the power source somewhere in the middle with branch A: 400cm of LEDs (plus 20cm of joints) on one side and branch B: 600cm of LEDs (plus about 270cm of joints) on the other side. Now B is probably still a bit too long for powering from one end, so my idea is to supply power to the distant end of it by leading wire from the source, along the LEDs and thus powering the long side from both ends. This would all go from the same supply (40A transformer).

I'm quite new to this but doing as much as I can to catch up with all the electrician knowledge. A couple of things I'm not sure about though:

1. Will the current get naturally distributed wherever is needed? Especially when connecting the strip on both ends, will it draw roughly a half of the total required current on each side?
2. Do I need a special wire for the joints and for the power extension? Some wires have maximum current rating on them (like 3 or 6A), so if the answer to the first question is yes and my calculation is correct (6m of LEDs will draw the maximum of 18A), will I need a wire that can handle 9A of current for the power extention (and possibly for the joints as well)?
3. Also, what about the wire from the power source to the middle of the strip, should I use separate wires to send power to each branch?

Answer 1

Look closely at the PCB traces on the LED strips and ask yourself "can I realistically expect this to carry 36A?"

Mind you, Electrical Code says a 36A load is too big even for 10AWG wire. Go to a home improvement store and handle some 10AWG wire for yourself. See what I mean? It's pretty big stuff.

Much worse, even though your distance isn't far, voltage drop is rather ouchey when you're starting with only 12V. So run your numbers through a voltage drop calculator just to make sure you're not dropping too much.

Feeder everywhere

Look at any trolley-bus system in Dayton, San Francisco, Philly, Vancouver, Toronto, Boston, or any former Eastern Bloc country. You can do it on Google Earth, Hayes St. in San Francisco is a good example. They have two #2/0 trolley wires feeding the bus. But up on the electric poles is also a pair of Very Fat Wires, and every 3-4 poles a fat jumper wire from a trolley wire to one of the fat wires. They are almost 2" diameter and solid aluminum. They are feeder. The 2/0 trolley wires simply can't carry enough current to supply a whole line full of trolley buses, so the feeder takes most of the line's current. Any bus is supplied via the trolley wire from both directions down to the nearest jumper to the feeder.

So you do the same thing. You won't have more than 30A in either direction, so 10AWG feeder will probably suffice. You run positive and negative (traditional colors red and black) parallel to the strips the entire length. And you tie it to the LED strips proper everywhere you possibly can, such as all those breaks that you need anyway.

You could even feed each strip from both ends if you want to, in which case the feeder is paralleled to the internal bus in the strips. That is fine, like the trolleybus, because it is DC. Do not parallel AC wires.