For a more general system I would do this differently. Maybe it's overkill in your case, but this isn't that hard. I've done something similar, although the power requirements were higher.
In my system I used CAN as a multi-drop bus to all the nodes. In the common configuration it uses differential signalling, which is a good idea for long distances where you could pick up common mode noise. I reserved one pair for the CAN lines. The other three pair were used for power, with a power and ground on each pair. That way the total common mode current of each twisted pair will still be close to zero. In my case I used 48 V because that is the maximum where you generally don't have to worry about safety issues much. There are plenty of microcontrollers with CAN built in, and the silicon deals with collisions and retries automatically.
In your case the power requirements are less, so 24 V might be a good choice for the power. Lots of transistors and buck regulator chips work up to 30 V. 28 V would be better if you want to push the most power and still use the cheapest power supplies at the ends, but I said 24 V because you're not pushing the limit and that's a very commonly available off the shelf power supply voltage.
At each node, put a small buck regulator. These are small and cheap nowadays. The MCP16301 can handle up to 30 V in, costs under $1, includes the switch, and comes in a nice and small SOT-23 package. Make sure to put a decent ceramic cap right at the input to the buck regulator though. The 24 V needs to be low impedance at high frequencies for the buck switcher to work. You probably want something like a 10 µF 30 V cap.
The advantage of this scheme is that you can tolerate a fairly wide power voltage at each node, but due to the higher voltage and therefore lower current, you won't have much drop. There will also be less heat at each node since the buck switcher will be more efficient than a linear regulator after you raise the power voltage enough to cover all the worst case conditions with headroom.
Another significant advantage is that there will be less ground offset between nodes, again due to the lower power current at the higher voltage.
Mechanical Generation of Power is basically how we first got electricity. It's a fairly common application, and there are over a billion of them running around outside. Automobiles have been using rotating shafts to power themselves since the very start.
The schematic for it is fairly simple. An Alternator going into a rectification stage which is then tapped into a battery for power storage, or directly to the rest of the circuit if the shaft can generate a fairly steady speed and voltage.
![enter link description here](https://sub.allaboutcircuits.com/images/05019.png)
Automotive alternators have been design in a way to couple the alternator to the shaft without directly mounting it around it, through the use of a belt or pully. As long as the shaft is fixed in place, and the alternator can be mounted along side, this works great (Safety first of course). There is a minimum RPM for stable generation, and output voltage will vary based on the RPM, so you have to work around that (Use a regulator circuit). The alternator's ideal rpm should match the Shaft's range of speeds.
A not typical setup for illustrative purposes:
![enter image description here](https://i.stack.imgur.com/RduUY.jpg)
Best Answer
As high as possible to minimize the cable cross section, since that will give you the lowest copper cross section. At some point the insulation will start to dominate unless the wires are physically separated. Let's suppose you use 350VDC.
Something rated to be safe at 350VDC. You have not divulged any environmental or regulatory limitations, so let's say it's just regular wire rated for 240VAC mains.
The current will be 1300W/(0.8 * 350) assuming 80% efficiency in your converter, or a bit under 5A (that's ignoring wire losses). The wire size calculation is another question and it depends a lot on the conditions. Note that the current might be higher if the loss in the wire turns out to be high, since you'll have to pump more in at one end to get it out the other. Let's suppose you use AWG20 wire with 200°C PTFE insulation. Resistance should be maybe 8 ohms (you will need to find the self-heating and do a calculation if you want it to be accurate). That means the voltage loss at 5A will be 40V. So the current now is more like 5.6A, not too far off.
Doesn't much matter for only 100m, but let's say DC because you get more power for a given wire size and peak voltage rating.
An Artesyn model UFE1300-5 should do it. 272.8mm x 140.0mm x 40.6mm, would be an estimate of the size. Weight is 2.5kg for the converter alone. Add maybe 1.2kg for the wire and you're up to 3.7 kg. 2.5~3kg is typical for 1300~1500W DC-DC converters.
The above is just a bunch of rough guesses without a bunch of important information to make a reliable and safe design, but from the information you gave, it does not appear to be too unreasonable a goal for an R&D project, but perhaps not quite off-the-shelf.