New Variable Voltage device from smoktech?
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Finally, a question I can answer. That's simple - neither! LMAO
OK, I'll think about that as it applies to my questions. I'll trust you that some edification lies within.
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The team that got the free lunch?
I'm changing my answer; I'm on fuzzi's team! lol
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Lastly: Gotta give proper props to this joke of fuzzi's that I hope no one missed:
"I suspect you will find that with a vmax in hand....resistance is futile". Nice one.
And if you haven't done so yet, please stop what you're doing and watch five minutes of one of the most incredibly dark, funny, and disturbingly insightful scenes in cinematic history:
http://www.youtube.com/watch?v=59YKlP--PhU
"Dr. Strangelove addresses the War Room", from the amazing film "Dr. Strangelove, or How I Learned To Stop Worrying and Love the Bomb", from the great Stanley Kubrick, and starring the late, great Peter Sellers and George C. Scott.
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No, no... remember I said "you get to volunteer". If you're building pyramids in the Egyptian desert, in the summer, you probably have no choice. Both teams get dinner when it's done. But if you are on the 1K lb team u might never get diner. What I'm trying to get across is the EFFORT required. Moving 1K lb is gonna bust ur balls... it's a huge effort. Moving 500 lbs is not easy but doable. Delivering 4A from some batteries might be a very 'straining' effort. 2A might not be easy but it's a 'eck of a lot less 'strain'.
The problem with these analogies is that the voltage at the battery effectively doesn't change.
volt = Joules/Coulomb
ampere = Coulomb/Second
watt = Joules/Second
wattage = voltage * amperage
Assume an ideal voltage converter that takes constant input voltage (Vi) and variable input current (Ii) to produce output Voltage (Vo) and output current (Io). The principle of conservation of energy holds that the amount of Joules coming out of the battery must be equal to the amount being output by the converter. Vi * Ii = Vo * Io
If:
Vi = 3.7
Vo = 5.0
Io = 1.8
Solve for Ii:
Ii = (Vo * Io)/Vi
= (5.0 * 1.8) / 3.7
= 2.43 amps
I'm happy to be proven wrong, but I don't see how it's possible.
volt = Joules/Coulomb
ampere = Coulomb/Second
watt = Joules/Second
wattage = voltage * amperage
Assume an ideal voltage converter that takes constant input voltage (Vi) and variable input current (Ii) to produce output Voltage (Vo) and output current (Io). The principle of conservation of energy holds that the amount of Joules coming out of the battery must be equal to the amount being output by the converter. Vi * Ii = Vo * Io
If:
Vi = 3.7
Vo = 5.0
Io = 1.8
Solve for Ii:
Ii = (Vo * Io)/Vi
= (5.0 * 1.8) / 3.7
= 2.43 amps
I'm happy to be proven wrong, but I don't see how it's possible.
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Been an interesting exchange of opinions, thanks for the contributions thus far. I'm running late for my next class, atty debridging 101, but before departing I'd like to thank John for his generous acknowledgement of my irreversibly warped attempts humour, together with that Strangelove clip. Love that scene, even more than the cowboy bomb ride.
Just a few more vapes and pages of Asimov before going down for the count tonight
Just a few more vapes and pages of Asimov before going down for the count tonight
Folks as you take the time to analyze EVERY tiny aspect of this very simple built yet wonderful mod (clearing throat...JOHN!) Jesus man...if you can't get the answer your looking for out of all those posts and analogies Im gonna have to draw you a diagram..sheeesh !!! lol lol lo llllllooollll
Please folks just try to keep in mind that as Smoktech was designing this device they were merely listening to ALL the b!..tch!..n and moaning how no mods would fire a DUAL COIL or a high resistant carto (which SMOKTECH MAKES) without the protection kickin in or a E1 code or feeling like vaping air. They designed this mod around DUAL coils and 3.0, 3.2, 4.0, 4.5, 5.0, 5.5 ohm single coils!!!!! NOT LR ANYTHING. ..... Not 2.0 ohm like Phil's review (which was good and bad in many points IMO) not 1.7's, not 1.5's (unless its a DC) . DC's and LR atty's/cartos are 2 entirely different situations,although having the same ending ohm load 1 has 2 coils that are 3 ohms each or higher. If you pay close attention to the top of the VMAX you'll notice the tapered top matches seamlessly to a 3.5mL DCT when you put 1 on. Heres you sign !!!! My avatar was your 1st hint. Take care everyone. And yes the both of my Provari's V1 & V2 have a loose wiggly button.
Please folks just try to keep in mind that as Smoktech was designing this device they were merely listening to ALL the b!..tch!..n and moaning how no mods would fire a DUAL COIL or a high resistant carto (which SMOKTECH MAKES) without the protection kickin in or a E1 code or feeling like vaping air. They designed this mod around DUAL coils and 3.0, 3.2, 4.0, 4.5, 5.0, 5.5 ohm single coils!!!!! NOT LR ANYTHING. ..... Not 2.0 ohm like Phil's review (which was good and bad in many points IMO) not 1.7's, not 1.5's (unless its a DC) . DC's and LR atty's/cartos are 2 entirely different situations,although having the same ending ohm load 1 has 2 coils that are 3 ohms each or higher. If you pay close attention to the top of the VMAX you'll notice the tapered top matches seamlessly to a 3.5mL DCT when you put 1 on. Heres you sign !!!! My avatar was your 1st hint. Take care everyone. And yes the both of my Provari's V1 & V2 have a loose wiggly button.
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I don't think anyone's saying the button doesn't wiggle, but even on my brother's well-used almost year old ProVari the button barely makes an audible noise.
I have no doubt that the VMax is a great match with dual-coils. You can only push so many watts through a coil before things get too hot, adding another coil and running them both at a high wattage can only mean insane vapor.
I'm keeping my eye on the VMax because I anticipate a need for a high current device and having VV would be quite nice. I'm a little spoiled by the ProVari's accurate voltage, I guess, but it definitely can't do 20W and the VMax can.
And this, from a guy who doesn't know how to use quotes. LMAO!!!
You are a loose wiggly button. Still ...... at you, you frickin' backstabber! lol
OK, loose ends (although I still feel, even after VAP's tongue-lashing, that there are still some electronic depths to plumb):
"The VMax needs to be recalibrated": back to my Corvette analogy; as long as my car goes like a bat outta hell, and there's no speed limit, I really don't care if the speedometer's off.
"The button rattles a little": beats the livin' crap out of one that sticks. Maybe Smoktech just wanted to make damn sure that there wouldn't be any "sticking button" issues like on the e-Powers to some extent, and on the Woo to a large extent. Mission accomplished.
"Should have an LCD, not an LED, display: wouldn't that have to be backlit, requiring battery power and adding some complexity? (Not sure if LCD's consume less energy than LED's - VAP? (you jerk?).
"It uses stacked batteries" - and I love it for it. When used with IMR batteries, I consider that an absolute non-issue.
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I think we're addressing two different aspects of this...
The voltage of a battery changes. You can measure 4.2v at the battery w/o load but that drops as amperage increases. It also drops as you drain it. Even if it is the same energy expended from this chemical conversion.
If a 1A load is put on a battery I will get x mA out of it until it drops to a per-determined voltage. If a .5A load is put on it the total mA I get out of it will be more and over a longer period of time. As far as I know this is true of any type of Li-Ion and I think NiCad also. See HERE for actual measurements.
The original thing I was addressing was "Why is it easier on the batteries to draw lower amperage". What I failed to address was that part of the indirect question was that there was a device here being used to adjust the load on the batteries and I think that is your point.
In this case the device is the load not the atty. The device can be adjusted to draw 4A or 3A or 2A from the batteries. All aspects of the battery chemistry will be 'better' at the 2A draw and the battery voltage will be higher and remain higher for a longer period of time. Which I thought was the original question... maybe not.
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Yeah, I was addressing the claim that running your VV device at a higher voltage/resistance for the same output wattage will give you better battery life because the load on the battery would be less. It won't and it's a common misconception that drives me crazy every time I see it.
edit: I should say... it probably will give slightly better life to run at the top of the VMax's range because typically the converter is more efficient closer to the input voltage. Buck converters can be very efficient so it may not make a noticeable difference. As a general rule, the output wattage of an efficient VV device should be the primary determining factor of your battery life.
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First, a point of order. I request that we leave voltage drop out of it, and for purposes of simplicity assume that the voltage remains exactly the same as current is delivered to the coil(s).
Second, my original question involves completely understand at the molecular level why it is easier on the battery to provide the same amount of [heat, when all is said and done] to a coil at lower amps than at higher amps, when it ultimately has to deliver the same total amount of energy. In my defense, I invoke the First Law of Thermodynamics.
I'll keep it short for once, cuz I think that about sums it up.
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OK, I was mistaken. One more thing. The 2 x 500 lb. stone blocks thing isn't working for me because we don't have time to make two trips. The Pharoah (the coils) wants his stones delivered when he wants them, not when I have the time to bring them to him. Time cannot be discounted.
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Ahhhh, just caught it. The wattage out of the batteries is NOT the same.
OK then the voltage out of the batteries is a constant (not really but OK for now). You can only vary the amperage which in turn will vary the wattage out of the batteries. It still not this simple but for now this'll do.
Next we can calculate how many beans the dinosaur ate before he died so we can calculate the quality of the gas we extracted from the earth to know how much energy we can get from it when we convert it to mechanical energy run the turbines to generate AC to feed to our house that ran our charger that charged our batteries, OK?
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OK, please continue to help me out.
If a hypothetical single coil is getting 8 watts, how can these two sets of batteries, Set A and Set B, not each be putting out 8 watts worth of heat /electrons /mojo /call it what you will:
Set A: 2 x 3.7 volts (nominal) delivering 8 watts to a single coil at 6 volts
Set B: 2 x 3.7 volts (nominal) delivering 8 watts to a single coil at 3 volts.
How does this square with "The wattage out of the batteries is NOT the same".
"Watt's" up with that? When is a watt not a watt? If a coil is drawing/using/getting 8 watts, they have to come from the battery or batteries. (?)
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Plus, what adami just said:
And VAP, get prepared to bite me. LOL
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how would a tripple coil carto do on the vmax? they are really high resistance right? Also i have OLed screen on my vision tox and in the sunlight i can't see what it says at all, this bright red LED should be really easy to see when your outside.
You do not control the voltage coming out of the battery. If you have fully charge stacked 18350s you have about 8.4v as the input to your device. Since we've decided to call that a constant the only thing that changes is the amperage (and therefore wattage drawn) from the batteries. All Li-ion and NiCad can produce more total mA from a low drain than from a high drain.
When you change the device voltage and amperage that is the demand on the regulator output. What changes on the input side is the amperage drawn from the batteries (still at 8.4v).
BUT I totally misunderstood the question earlier... I thought it was purely about load on a battery not how it relates to a VV regulator hitting those batteries. So while my measurements earlier showing more mA total from same battery at a lower load it probably doesn't apply to what you're saying here. If your device had 100% efficiency at all settings then X watts out requires x watts input from the battery. It's just that to the battery side is all just an amperage at 8.4v. So in this case there is no benefit at x watts on the device of how you get it from the battery. Sorry, I didn't get what was being ask the first time.
Lower amps from battery = better for battery. But that requires lower wattage at the device.
When you change the device voltage and amperage that is the demand on the regulator output. What changes on the input side is the amperage drawn from the batteries (still at 8.4v).
BUT I totally misunderstood the question earlier... I thought it was purely about load on a battery not how it relates to a VV regulator hitting those batteries. So while my measurements earlier showing more mA total from same battery at a lower load it probably doesn't apply to what you're saying here. If your device had 100% efficiency at all settings then X watts out requires x watts input from the battery. It's just that to the battery side is all just an amperage at 8.4v. So in this case there is no benefit at x watts on the device of how you get it from the battery. Sorry, I didn't get what was being ask the first time.
Lower amps from battery = better for battery. But that requires lower wattage at the device.
I agree. I totally misunderstood what the original question was really about.
I honestly dont recommend using a triple coil on ANY device EVER!!!! Their just battery killers and juice hoggers. They serve no purpose thats positive. "IMO" of course.
So, if I run 4000 amps at 10 volts over a 10 mile power line, I'll get the same power at the other end as I would if I ran 4000 volts at 10 amps?
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