No you can't.
No, you can't.
it should work as long as the batteries are connected in series + to - what you would need to check is if the charger for the cart charges 24v or 8v if it is 8v you would need to get a 24v charger for it it would be possible to charge the batteries with a 12v charger but it would need to be hooked to each battery separately
12V, 8V, 5V, 3.3V, and 2V.
Is it 4v-7+8v+4-5 ? If so then you're answer is 12v-8
4v-7+8v+4-5 - add the like terms =12v-8 12v and 8 cannot be subtracted because they are not like terms. Like terms have the same variable.
Why would you want to do that? With the (4) 12V Deep Cycle's your run-time will be MUCH less.Club Car golf cart's for 2009 just switched back to the (6) 8V batt system for their new Precedent models. Too many voltage problems with the (4) - 12V's, that's why Club Car went back to the (6) 8V's.Optimum 48V set up is (8) 6V's for long run time and easy 12V accessory. With the (6) 8V's, you just get a 48V to 12V REDUCER to run whatever accessories you like. Can be found on eBay for $60.
4v-7+8v+4-5 - add the like terms =12v-8 12v and 8 cannot be subtracted because they are not like terms. Like terms have the same variable.
Yes, but you will need the complete wiring harness and E.C.U (from the doner car)for an easier swap.
The 8v was 115BHP.
Without knowing what you have for a regulator, I can't say this will help. I did one last spring for a friend on his old antique boat. It is a 6v system which we changed to 8v (positive ground at that). This regulator is on a generator, not alternator, probably same as you, basically a small metal box with a relay inside. The premis of this regulator is the spring tension is just so that at about 7.5 to 8 volts, it would trip and disconnect the generator. Simply adjust the tension so it trips at a higher voltage. You will need a very accurate volt meter, preferably digital. Also remember your target isn't actually 8 volts, 10 is probably close enough. 8.8 volts is static charge, I think 9.2 is operating charge so 9.6 or so would be peak voltage under charge (a 12V battery will show 14.4 sitting and about 15 while under charge).. This worked great for us, doesn't show signs of boiling the battery but does keep it at peak.. Greg
If that's v3 + 8v, the answer is v(v2 + 8)
56v2 + 121v + 63 = 56v2 + 49v + 72v + 63 = 7v*(8v + 7) + 9*(8v + 7) = (8v + 7)*(7v + 9)