Volt/Watt/Amp???
- Thread starter ckn71nm
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You're absolutely correct, my wording was sloppy and wrong. Here, I'll show you what I mean to avoid miscommunication:
Assume 4v:
Resistance Power Current
.5 32 8
.4 40 10 .1 decrease, 8w increase, 2a increase
.3 53 13.3 .1 decrease, 13w increase, 3.3a increase
.2 80 20 .1 decrease, 27w increase, 6.7a increase
.1 160 40 .1 decrease, 80w increase, 20a increase
Plot a graph of resistance against current, or resistance against power and it's not a straight line, which means it's not a linear relationship. It looks a whole lot like he following graph (just an example)
In theory (all else equal) it doesn't matter what your resistance or current level is, only the power level will correlate directly, and in all circumstances, to the amount of vapor production.
But it's pretty much impossible to make "all else equal" in these circumstances. If you achieve the same power with different resistance, that means you are either changing wire thickness, or wire length, which is going to throw off your surface area to power ratios.
So technically speaking your experiment is flawed, but it's as close as you're going to get. juice consumption should be similar, especially if using commercially available equipment instead of home built coils since there would be less variables to take into account. But it's a very very unscientific comparison with many variables that could effect juice consumption to the point of totally skewing the results, especially when dealing with levels of comparison that are so close to each other to begin with.
Well, it's not any more difficult to calculate amperage than wattage. In fact it's exactly as simple when it comes to doing the calculation. APVs already measure coil resistance, and already measure/set voltage, so it's just as easy for the chip to do the calculation for amperage as it is wattage.
The simple reason variable current devices don't exist (in the sense of being able to set a certain amperage as all variable devices on the market are technically also variable current devices) is because it's completely pointless. Setting current to determine your heat or vapor production simply just doesn't work that way.
If you had a variable current device, and put a 1.5 ohm atty on it, and adjusted the current, guess what the PV would actually be doing? Adjusting voltage. Which all our PVs already do. It's just a dead-end path, that's why there isn't one made.
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Resistance vs current is 100% straight line. Decrease your R, and I increases linearly for a given input voltage. If you vary voltage AND RESISTANCE...you are dealing with a 3d plot.
V = IR is a linear equationSent from my Provari Mini using Tapatalk
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Correct, this will be interesting. If you increase resistance and keep current the same, power will increase and you will get a very different experience.
Take 1ohm @ 3 amp. Simple power calc is I^2 * R. 9 watts.
Take 2 ohm @ 3 amp. I^2 * R = 9*2 = 18 watts.
I'm guessing you will see an enormous difference!!!
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In the above two examples you would need to set your PV to 3v and 6v to get equal current in the two scenarios. You will not only notice a difference, you will be blown away. I guarantee it
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It's just a measurement. Just like length, doesn't matter if you measure something in mm, inches, feet, whatever, its still the same length.
Same thing with watts, amps, volts.... When one changes, so does the other.
With the same resistance Atty, you up the volts, your upping the watts...
No need to make VA mod, you want noobs setting it to 8amps with a dinky little 30g coil? People can't even grasp the concept of variable voltage and wattage, let alone adding another thing to confuse them.
Same thing with watts, amps, volts.... When one changes, so does the other.
With the same resistance Atty, you up the volts, your upping the watts...
No need to make VA mod, you want noobs setting it to 8amps with a dinky little 30g coil? People can't even grasp the concept of variable voltage and wattage, let alone adding another thing to confuse them.
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Because the current is a byproduct of the power being injected into the circuit. Using a hydro-electric dam as an example it's like saying the speed of the water is what causes the cogs to turn and therefore generate the sough after result. In actuality the amount of water (voltage) and the slope it travels on (resistance) are causing the speed(current) to be determined, which will result in the sough after power or in rba's heat(ie wattage). Honestly I can't think of a more simplistic and straightforward way to illustrate this. If you keep sticking to your position I'm going to have to assume this is just some strange form of trolling your attempting, nouveau-trolling anyone? =] keep on vaping you cloud-taming cowboys. ( I kind of like this sign off, I may make it my new mainstay, what do you guys think corny or do you like it?)
Amps/current is not used to measure heat because the same amount of heat will use different amounts of amps/current based on the voltage!!!
Voltage is the input...the potential difference..... You know, like your house wiring is 240/120 volts. That power comes IN from the transformer on the pole. It's not measured in freaking watts.
You can have plenty of voltage thru a circuit, without any power/watts. You will not have any watts without a resistance in that circuit, nor will you have any amps.
Now watts, watts is the outcome of voltage put thru a resistance......, is your 800watt microwave measured in amps? Is your electric stove measured in amps? Is a lightbulb measured in amps? No, it's watts. A higher WATT lightbulb puts out more heat and more light than a lower WATT light bulb. Its not measured in amps because an 80w lightbulb made for a 120v circuit will use more amps than an 80w bulb made for 240v circuit. Yet it will be just as bright, and put out just as much heat.
It's basic middle school science....
And this is why non qualified homeowners should do any sort of wiring in their house, because they have no clue about the basic understanding of electricity.
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another reason why and a primary reason is because unlike the adjustment of voltage or wattage which is incremental and orderly ie 3.1,3.2,3.3'etc amperage would adjust In big leaps and in a simply unpredictable manner off the top of an average persons mind, they'd have to have an ohms law calculator at hand constantly to see where they'd be adjusting to and if it was safe to even adjust to that level given the power cell they were incorporating in the circuit at that time. Simply put, variable current is neither practical or effective, it would be an ... backwards way of going about modulating a PV. Endofstory
Resistance vs current is 100% straight line. Decrease your R, and I increases linearly for a given input voltage. If you vary voltage AND RESISTANCE...you are dealing with a 3d plot.
V = IR is a linear equationSent from my Provari Mini using Tapatalk
Here are three graphs, the first is resistance x power at 4v, the second is resistance x current at 4v. Neither are a straight line, and it's not because my graphs are sloppy. Last is both together.
I'm not sure what else I can say/do?
He SHOULD get a very different experience, but he is claiming he sees similar performance when current matches, in other words his basic premise is that power levels aren't responsible for vapor, but current is.
Your two examples should, and would, perform very differently. But with the OP's theory they would perform same.
OP,
I think the reason is that your build comparisons aren't taking into account all the variables. If you say performance is similar when you build to a similar amperage, that's not because the amperage itself is determining the vapor, but because you just happened to have some builds where the variables meet up to give a similar vape.
I'm not trying to sound rude, but there is no need to theorize here, we already know wattage is responsible for heat. These concepts aren't new or unique to vaping, they are very basic scientific principals that are well understood.
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The problem with trying to set the current can be illustrated by looking at 2 amp though the coil.
2 amp though a 2 ohm coil means you have 4 volts.
Ok so that's a typical realistic scenario which is why I chose 2 amps;
4 volt, 2 ohms, current = 2amps.
Not let's put in a ridiculous resistance: 0.1 ohms.
Ok so we want to get 2 amps though 0.1 ohms so we need 0.2 volts!!
Heck if it were that easy to get a good vape, we wouldn't need all the fancy gizmos; it's easy find low resistance wire (obviously not kanthal or nichome) and easy to get low voltages.
2 amp though a 2 ohm coil means you have 4 volts.
Ok so that's a typical realistic scenario which is why I chose 2 amps;
4 volt, 2 ohms, current = 2amps.
Not let's put in a ridiculous resistance: 0.1 ohms.
Ok so we want to get 2 amps though 0.1 ohms so we need 0.2 volts!!
Heck if it were that easy to get a good vape, we wouldn't need all the fancy gizmos; it's easy find low resistance wire (obviously not kanthal or nichome) and easy to get low voltages.
Not quite an accurate comparison because with mm, feet, inches, etc they are all measuring the same exact thing, which is length.
With watts, volts, amps, etc there are not measuring the same thing, there is no overall thing (like length) they are measuring.
A better comparison would be like inches, liters, and pounds and a bucket of water.
If you have a large bucket of water, it is a certain number of inches tall, contains a certain number of liters of water, and weighs a certain number of pounds. All of those are measuring different things, but they are all still related and depend on each other.
If the inches go down, it's shorter, holds less water, so both liters and pounds go down. If inches go down horizontally it's thinner, same thing happens.
If you If pounds go down, liters go down, and inches go down. Obviously buckets of water don't automatically scale themselves to a given variable so the comparison kinda breaks down now
If your goal was to figure out how much water that bucket carries, obviously liters would be the best way (like wattage)
Using current to measure vapor would be like using inches to measure how much water is in the bucket. You can't. Because if the water bucket is "4 inches tall" that doesn't tell you how wide it is, so you have no idea how much water is in there. Best to stick to liters (watts)
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Amps are a byproduct.
Why is this thread still alive?!?!
Please stop with the equations, as you are feeding the OP.
I'm not sure if he's trolling, or just stupid.
THERE ARE ADJUSTABLE AMP MODS!
/close thread
Why is this thread still alive?!?!
Please stop with the equations, as you are feeding the OP.
I'm not sure if he's trolling, or just stupid.
THERE ARE ADJUSTABLE AMP MODS!
/close thread
Here are three graphs, the first is resistance x power at 4v, the second is resistance x current at 4v. Neither are a straight line, and it's not because my graphs are sloppy. Last is both together.
View attachment 302650 View attachment 302653 View attachment 302656
I'm not sure what else I can say/do?
.
I stand corrected. If you solve for current or wattage, it's inversely proportional.
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Not quite an accurate comparison because with mm, feet, inches, etc they are all measuring the same exact thing, which is length.
With watts, volts, amps, etc there are not measuring the same thing, there is no overall thing (like length) they are measuring.
A better comparison would be like inches, liters, and pounds and a bucket of water.
If you have a large bucket of water, it is a certain number of inches tall, contains a certain number of liters of water, and weighs a certain number of pounds. All of those are measuring different things, but they are all still related and depend on each other.
If the inches go down, it's shorter, holds less water, so both liters and pounds go down. If inches go down horizontally it's thinner, same thing happens.
If you If pounds go down, liters go down, and inches go down. Obviously buckets of water don't automatically scale themselves to a given variable so the comparison kinda breaks down now
If your goal was to figure out how much water that bucket carries, obviously liters would be the best way (like wattage)
Using current to measure vapor would be like using inches to measure how much water is in the bucket. You can't. Because if the water bucket is "4 inches tall" that doesn't tell you how wide it is, so you have no idea how much water is in there. Best to stick to liters (watts)
You are correct, it wasn't the best way to explain it, but other than that paragraph I wrote, everything else explains it perfectly.
Sorry if I was nitpicky about comparisons, the amperage thing was getting to me again, the thread's issue being that amperage could be used in a fashion similar to wattage (like mm similar to inches) which it can't; I know you weren't supporting that idea, I was just trying to make it clear again to any silent observers of this thread that current can't be used to judge vapor or performance. I wasn't trying to pester ya
Honestly I drew the first graph for myself to figure out the relationship in the first place; I was having trouble figuring in my head if it was a linear function or not lol.
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....that current can't be used to judge vapor or performance....
LOL - that was established and scientifically proven on the first page of this thread....the rest is just convincing some cloud chasers of how Ohm's Law can't be broken....even if "oh wow man, I run 29.5 amps on my quad coil with a 30 amp battery, and the current is boss"
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