R=pL/A!!!

A 70 AMP wire@77F or 25C as is the normal test temperature, has a volt drop that is less obviously than a smaller wire, but is way more costly and not needed at all under any circumstances for the application, why not go the whole hog and get starter cable from the QE2 ?
The above formula is what you need to know to work out volt drop as well as temperature/length of wire, and of course the wire INSULATION.
If you were top use as an example I gave 28/030 that has a continuous carrying capacity of 28AMPS, pvc insulation, 25AMPS thinwall ie HARDGRADE PVC.
Assuming a length of 20 feet, 10 foot supply 10 foot return, you would have a RESISTANCE of 0.0536 OHMS, which as you know would be a VOLT DROP of 0.6968.
This is at the LOAD that the BOSCH 044 PUMP places on the circuit, by it's rated current consumption(AMPS) at 12VOLTS.
Of course this is an estimation of the length of the wire, if it is less then the RESISTANCE/VOLT DROP will be correspondingly lower still.
If you want to read CRC HANDBOOK of CHEMISTRY and PHYSICS 57th edition, 1976-1977 CRC press, p. F167-168, you would find lots of lovely useful info there.
Now bearing in mind the RATED VOLTAGE of the BOSCH PUMP 044 is 12 VOLTS, and most electrical systems will operate upwards of 12.8 VOLTS, and most nearer to 13.4/13.6VOLTS, one can easily see how 28/030 is more than capable, and a lot cheaper.
Without knowing the LOAD/TEMPERATURE/WIRE INSULATION TYPE/LENGTH etc etc etc etc it can only ever be a guess.
Some guess, and go over board to be safe, spending money unnecessarily on gauge of wire way too large, that is up to them, I prefer not to spend where not necessary, and do it the easier way, using above.
tabetha