Why wattage, what we care about is current!

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GoBlue88

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Edited to add supr quickly demonstrated my argument wrong. Still feel free to read it though (there's a puppy at the end).


Long/ranty/mathy, but come along for the ride if you've got a few.

TLDR at the end

First the math:

(1) V=IR (Voltage = Current X Resistance)

(2) Wattage = VI (Voltage X Current)

So as a result

(3) W=I^2R

(4) I = Square root(W/R)


So a couple of points:

The current is the description of the energy delivered to vaporize the ejuice, and thus the vapor production/vapor heat.

Wattage is a description of the electrical work performed by the battery. It is not the energy delivered by the device. That's the current.

Take equation (3). You run a VW device at a high wattage , but a very high resistance. That's could actually be a really low vapor production/ cool vape, because your current would be small. So your doing tons of electrical work (using battery life), but not producing much current, and therefore not producing much vapor, and relatively cool vapor.

Wattage is often used as a stand in for "coolness" of vapor or the vapor production, but that really isn't necessarily true. It's only true for a constant resistance.

As a way of describing your battery capability wattage is useful, as a way of describing your vaping experience, it has to be referenced with resistance, so when people talk about wattage they are really referencing current without knowing it (or at least without explicitly referencing it)

Why does this matter:

Lets say I'm running an RDA at 1.3 ohms and 14 watts and enjoying the vapor production/throat hit/temperature (in fact, I am). What I should be doing is identifying the current associated with those specs using equation (4). I should then be using the current I discovered works for me, and dropping the resistance significantly down towards the bottom of what my mod can handle. Then, using equation (3), I can input my current which I've determined I like, the lowest resistance (or near lowest, because as coils heat up their resistance drops) that my mod can handle, and get the wattage I want to set my VW mod to. It will be significantly lower than the 14 watts described above, because, I dropped the resistance as low as I safely could. Now I'm running my desired current at a lower wattage, so I'm producing the vapor that I want while draining my battery less, because as I mentioned above wattage just tells you how fast your're draining your battery (the work being performed). You're gaining nothing by running at higher resistance than the lowest resistance (or near lowest) your mod is capable of on a variable wattage device other than draining your battery faster.

A weakness in this analysis is that dropping the resistance would require a coil build that would vary from my current build (thinner coil inner diameter, larger gauge, less coils), but for the purposes of the post I'm assuming that people aren't intensely attached to the particulars of their coil build like gauge, inner diameter, and number of coils.


TLDR: Wattage tells us how much were sapping our battery. Current tells us how much energy we're delivering to our wicked coils, thus producing vapor. Why don't we as vapors talk more about the current we're vaping at? It's the energy delivered to our ejuice, and thus the vapor produced/vapor temp.

If you made it this far, here's a puppy for your troubles. Thoughts?


3f5f837a50a5b899bcb74707bb693ad8.jpg
 
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suprtrkr

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I'm hardly an expert in these matters, but I do not think so. A 10 amp current at 4.2 volts makes 42 watts. A 10 amp current at 10 volts make 100. I think you would find a significant difference in the vape between those two points. While you are correct "watts" as applied to the battery is inclusive of things not directly associated with the production of vapor, losses on the board for example, the mod doesn't measure those. It measures the actual watts applied to the coil using (usually) the V^2/R formula. That is to say, it measures the atty resistance and calculates the voltage to apply to the coil to make the set watts. The voltage it considers, and measures, it the output voltage of the board. Under these circumstances, current is a limiting factor by what can be safely drawn from the batteries, but it doesn't affect the quality of the vape. A 1V, 10A current makes 10W, just as does a 10v, 1A current. This is the measure properly applied to the atty.
 

GoBlue88

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A 1V, 10A current makes 10W, just as does a 10v, 1A current. This is the measure properly applied to the atty.

This is my point exactly supr. You just described a situation in which wattages matched but 10X the energy (in amperage) is being delivered to the coils in one of the scenarios. That's a massive increase in vapor production/vapor temp at the exact same wattage. But if you changed resistance and matched amperage, the same energy would be delivered to the coils and the same vapor production/vapor temp would occur. Again, this isn't factoring in coil gauge, number of coils etc. but, what you said hammers home the value of amperage over wattage in determining what our vapor production/temp will be like.
 

GoBlue88

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PV companies are thinking Brilliant!! Now we just reprogram our chips to display current and resell them a mod they already have!!

Assuming the OP's thoughts are true. I wouldn't know lol. I simply think of voltage or wattage as "juice" I am delivering to the coil. Juice as in power

Wattage is power (in a physics sense), voltage isn't power, but all other variables held constant increasing voltage will increase wattage, and as a result increase power.
 

suprtrkr

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I'm sorry, my friend, but we disagree. "Amperage" does not measure energy. It measures current. "Energy," or more properly "power," or better still "work" in Newtonian physics, is measured by Watts. The example of mine you quote delivers the exact same amount of power to the coil, it just does it using different parameters. Your example of maintaining current while changing atomizer resistance isn't even possible, unless volts change as well. Ohm's Law describes the relationship between V, I and R; it says nothing about absolute valuation. If you change the atomizer resistance the current changes right along with it; resistance limits current, holding voltage constant.
 

suprtrkr

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Wattage is power (in a physics sense), voltage isn't power, but all other variables held constant increasing voltage will increase wattage, and as a result increase power.
That's true. The problem is you can't hold everything else constant. That's what Ohm's Law says. If you increase voltage, and hold resistance the same, amps change.
 
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dhood

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I'm too tired to work out the math in my head right now. But I can tell you that I have two Kayfuns. One has a 1.8 ohm coil in it and the other has a 1.1. One is a 28 gauge coil and the other 26. I find that both are perfect at 14.5 watts and that both tend to start tasting burned over 15 watts. that amperage is different, but it tastes the same to me at the same wattage. I'll do the math tomorrow.
 

GoBlue88

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That's true. The problem is you can't hold everything else constant. That's what Ohm's Law says. If you increase voltage, and hold resistance the same, amps change.

Sure, which would increase wattage. Voltage increase would increase increase current, so both variables in the W=V*I equation would increase, increasing wattage.
 

suprtrkr

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I'm too tired to work out the math in my head right now. But I can tell you that I have two Kayfuns. One has a 1.8 ohm coil in it and the other has a 1.1. One is a 28 gauge coil and the other 26. I find that both are perfect at 14.5 watts and that both tend to start tasting burned over 15 watts. that amperage is different, but it tastes the same to me at the same wattage. I'll do the math tomorrow.
No need, you already did. You said both of them taste fine at 14.5 watts and taste burnt over 15 watts. True enough, perhaps, but you didn't change wattage. You are holding both of them at 14.5 watts. The 1.1 Ohm coil is using 3.63A at 3.99V; while the 1.8 Ohm coil is drawing 2.8A at 5.1V.
 

suprtrkr

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Sure, which would increase wattage. Voltage increase would increase increase current, so both variables in the W=V*I equation would increase, increasing wattage.
Lol, we're arguing in circles, my friend. If wattage increases, so does the power. It can't remain constant; wattage *is* power.
 

GoBlue88

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Lol, we're arguing in circles, my friend. If wattage increases, so does the power. It can't remain constant; wattage *is* power.

My response was to whether the poster should think of voltage or wattage as power. My point was that wattage is power, and for the purposes of a vaper using a constant resistance coil, increasing voltage will increase power (wattage).
 
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suprtrkr

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My response was to whether the poster should think of voltage or wattage as power. My point was that wattage is power, and for the purposes of a vaper using a constant resistance coil, increasing voltage will increase power (wattage).
Now that I agree with. That's how regulated mods work. You set the wattage requirement, it measures atty resistance and provides whatever voltage may be required to produce the set wattage, letting amps fall where they may (within the limits of the board.)
 

roxynoodle

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ETA: current does come into play on the limitations of the mod. My Sigelei 100W can't reach 100W with any resistance I throw on it. There are both current and voltage limitations. So one needs to look at their mods specs and do the math to determine what resistances can be used to reach maximum power, if that matters to you. I don't vape at 100W so I don't care much. I might check when I'm bored :D
 
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