ORIGINAL: anci3nt
ORIGINAL: kingrustler
ORIGINAL: piranha661
I am running a Fusion speed control and a 19x1 Fantom. If you get a 6 cell pack you will get a longer runtime. The more voltage of the pack the shorter the run time.
no way dude you will have same run time out of a 6 cell 3000mah battery pack as you will a 7 cell 3000mah battery pack just with more punch because you have more volts which means more rpms out of the motor not less run time .look at it this way fill two glasses up one with 7.2 oz and one with 8.4 oz drain them both at the same flow rate the 8.4 will take a bit longer to drain but will drain faster at first due to the extra 1.2 oz of volume . . get it
flow rate =amps
volume = volts
flow rate +volume =pressure
amps +volts = watts
watts and pressure are the same . and watts turn motors more watts more power but just about the same amount of run time .that is about the best way i can describ it . to you i may be a little off but not much and the idea behind what i say is right even if i worded it wrong .
i don't think that is entirely correct. i think this may be a better analogy:
flow_rate*time = amp hour (this is like volume, or the amount of water in your analogy)
flow_rate = amps
pressure = volts (this is like the ability to 'force' current)
flow_rate*pressure = watts (this is like instantaneous power)
flow_rate*pressure*duration = joules (energy, or the amount of power you can apply in some duration of time)
so yes, more watts seen at the motor is more power, but i'm unsure about how that relates to a 6-cell to 7-cell runtime (I asked the same question in another thread). the explaination that was posted was because the voltage has increased, it has increased the current flow (assuming the same load). i'm unsure about that because while the added cell does increase the potential, it also brings added capacitance. now if the increase in voltage increases the current, and the draw (over the full discharge of the battery) outpaces any capacitance the extra cell added, then i could see that as being true.
the problem with your statement is that since the added cell is wired in series it does not add capacity but rather voltage.