As long as the loads can be treated like a straight, linear resistance, this is a simple resistor ladder network. Start at the end: you have the last load and its two conductors in series, with that group in parallel with the next-to-last load.
Figure out that paralleled resistance, then place that value in series with the pair of resistances of the next-to-last conductor pair. Then you parallel that resistance with the third-to-last load, add that section's resistances, etc.
Once you reach the source point, you can calculate the current of the entire ladder, which gives you the voltage drop across the first (nearest source) pair of conductors. The voltage at the load end of this section is the first load's voltage.
You then use that same voltage to calculate the current through, and voltage lost along, the second section of conductors, giving you the second load's voltage and starting voltage for calculating the third run's current and its voltage drop, etc.
You will eventually arrive at the voltage across the last load, where you need to determine if that voltage is acceptable. If not, you need to upsize the conductors, starting with the run nearest the source, as it carries the entire circuit's current.
You'll probably end up cascading conductor sizes, such as #10 on the last section, #8 for the next-to-last, and so forth. If someone asked me to design this without doing the math, that's what I'd probably do, but it's better to do it by the numbers.