20 gauge wire has a resistance of about \$ 33.31m\Omega / m \$. For a 20 meter cable, it's effectively a resistor \$33.31\frac{\Omega}{m} \cdot 20m \approx 1 \Omega \$. Your circuit is this:
You want 30A through the motor, and since this is a series circuit, that means 30A through R1 and R2, also. The voltage drop over a resistor is given by Ohm's law: voltage equals current times resistance:
\$ V_{R1} = 30A \cdot 1\Omega = 30V \$
So, you lose 30V in R1, and another 30V in R2, so you will need 60V plus whatever voltage is required at the motor to get 30A. Worse, power equals the product of voltage and current. We know the voltage across R1 is 30V, and current is 30A, so power lost in the wire is:
\$ P_{R1} = 30A \cdot 30 V = 900 W \$
For comparison, a typical electric heater is around 1000W. You have two wires, so your total losses are 1800W. It's quite likely you will trip a circuit breaker if you can magically prevent the wires from bursting into flames, and we haven't even powered the motor yet.
There are two obvious solutions:
- make the wires shorter
- make the wires fatter
If you can't do either of those, there's a less obvious solution:
- keep the power the same by raising the voltage and reducing the current
This is the solution the electric company uses to avoid huge losses in power transmission. If you combine the two previous equations, you can see that the power through a fixed resistance can be calculated through the current alone:
\$ P = I^2 R \$
12V at 30A is 360W. 360V at 1A is also 360W, and in theory, capable of producing the same mechanical power at your motor. But, the losses in your wires will be much less, by minimizing the \$I^2\$ term above.
To do this, look for a motor that operates at a higher voltage, or put a mechanism for converting the voltage near the motor.
Drastic edit of my answer, as I didn't fully understand the LED strips.
I would think the best option is very wide, tinned copper ground strap. They have these with very flexibly braid, used for grounding hinged interfaces. You should be able to solder individual strip power wires to them and just have to keep them isolated vertically. Tying the battery at the center of the strip would minimize the voltage difference among all strands.
Best Answer
You can use a higher frequency than the standard 50 or 60 Hz to reduce the volume and weight of the transformers (for example the aircraft industry use 400 Hz as the standard in airplanes).
Personally I'll don't go beyond a few kHz to avoid all kind of problems but it's just a rough estimation (and you might need a higher frequency than that to pass 2 kW with a 2 kg transformer), do the math to be sure. Also, if the frequency is relatively high for the wire thickness, think about using litz wire for the transformers windings, you lose on the volume but you win on the weight.
To drive the first transformer you can use a H-bridge (or even just a half-bridge with a center tapped primary on the transformer) and a crude astable since the efficiency isn't a concern. You can do better by approximating a sinus with multiple taps on the primary but it gets far more complicated.
On the other side, after the second transformer, just be careful about the switching speed of the diodes if you want to rectify the current (classic rectifying diodes might not be fast enough), use schottky diodes if you want to be sure to have no speed problem ever.