i use a 2ohm and vape at around 4volt.. depends man
Best resistance on the head depends on where you like to vape. With a 3.0 ohm carto for example, you will only be able to push the wattage to about 7.7 (that may very well not be a problem, but bears mentioning). It will be easier on the battery...it's counterintuitive, but higher volts and higher resistance is more efficient and easier on battery life at any given wattage.
With a 2.0 ohm head, you will have a greater range with this particular device: up to about 11.5 watts, and you're not likely to want to go any higher, if even that high. However, at the same 7.7 watts on a 2.0 ohm cartomizer, you're going to be turning the device to about 3.9 volts and using a lot more battery juice (2.4 amp draw versus drawing 1.6 amps on the higher resistance head. Your batteries will last a lot less time at any given wattage.
Hopefully I didn't make that too confusing.![]()
Thanks for all of your advice and comments. The clearos I purchased were 2.4 ohms so I guess those should work quite well. I checked out the Vivi Novas. I may order one to try at home but it is just makes for too big of a "contraption" for me to feel comfortable with in public. To be honest, the 1000 twist with a clearo is pushing it for me.. I did look into the Vivi mini. Are they pretty much the same as the V2 Visions?
This assumption that higher voltages are easier for battery is wrong. Sorry. Yes, with higher voltage we need lower amperage to get the same wattage. But if there is a voltage increasing circuit between battery and the coil the electric current in battery will be not the same as current in battery. It will be proportionally higher. Because of energy conservation law of physics.
If you were right then using 3000 ohm atty and 4800 volts (this is possible to build a VV battery like this, why not?) we would have thousand times lower amperage with the same wattage. eGo battery would work a year after one charge. Make voltage ten times higher and eGo battery would drive an electric car.
That's a good point, and one that hadn't occurred to me. The way I stated it assumes the battery itself is delivering the voltage unaided by the boost circuit - not the case with the Twist. Any idea how to calculate the actual amp draw when you boost a nominally 3.7V battery up to, say, 4.5 volts at any given resistance?
ETA: I missed that you said "proportionally". Is it proportional? Would a 10% increase in output voltage increase the amp draw 10% at the same resistance? You have me genuinely curious now....
Hey @jimrug1 Here's something that can help you figure things out with your watts/volts/ohms and such. This calculator has a section where you can plug in your volts and your ohm and determine the watts you'll get. Most seem to like between 6 and 8 watts for the "sweet spot".
http://www.the12volt.com/ohm/ohmslawcalculators.asp#power
I guess the rule is simple, the wattage consumed by the coil (Ec*Ic) should be equal to the wattage produced by the battery (Eb*Ib) (well, minus some energy loss in the circuitry). So if battery voltage Eb is, say, 3.7v and the voltage on the coil is, say, 4.0v with the 2Ohm atty, then Ic=2A and Ib=Ec*Ib/Eb=4*2/3.7=2.16A.
<--- Nic Nic Nic... Pounding my head against the wall.. I forgot my original question... ;-)That's a good point, and one that hadn't occurred to me. The way I stated it assumes the battery itself is delivering the voltage unaided by the boost circuit - not the case with the Twist. Any idea how to calculate the actual amp draw when you boost a nominally 3.7V battery up to, say, 4.5 volts at any given resistance?
ETA: I missed that you said "proportionally". Is it proportional? Would a 10% increase in output voltage increase the amp draw 10% at the same resistance? You have me genuinely curious now....
<--- Nic Nic Nic... Pounding my head against the wall.. I forgot my original question... ;-)
Sorry, Jim...didn't mean to hijack your thread. No, you don't have to do any math to vape! You can look it up in the chart someone posted. I'm just trying to make sure that the advice I gave you (higher volts and ohms are going to result in a battery charge lasting longer) is true.