Quick question about twisted kanthal/dual coils

[Dissonance]

I know with dual coils, the ohms reading is 1/2 the resistance of one coil... I'm assuming that means you need 4x the regular voltage to get the same watt vape? (Ex: my IGO-W is reading 1.1 ohms; would giving it 4.4 volts equal out to each coil operating at ~ 4.4 watts?)

Same question with twisted kanthal (in a single coil setup)... Is there any 'multiplier' that comes into play?

 

[Jazzu]

Volts to ohms never yields the same amount of watts as the volts you fire. There are charts out there that you can find that will tell you how many watts would be pumping through the tank dependent upon the ohms and volts you are firing at. Ex: My .6 ohm coil at 3.7v would produce around 25 watts. Now I didn't use math in that, it's just an estimate on how many watts would be produced when firing it.

 

[Dissonance]

Uhh... Unless you're talking about a chart for dual coils I use a voltage chart... But since the power is split 2 ways, and the resistance shown (1.1 ohm in my case) is 1/2 the resistance of each coil, do I need to supply 4x the power (Or use the chart as if I have a 4.4 ohm atty) to get the same wattage results as in the chart?

 

[SoaFung]

Uhh, electricity isn't that complicated. Doesn't matter how a certain resistance is achieved (dual twisted figure 8 herp derp coils). To the electrons involved, 1.1 ohms is 1.1 ohms.


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[BobC]

Uhh, electricity isn't that complicated. Doesn't matter how a certain resistance is achieved (dual twisted figure 8 herp derp coils). To the electrons involved, 1.1 ohms is 1.1 ohms.


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Woah buddy, not a very nice response, as much an expert you may be, some folks may not be up to your speed, that's why they come here and ask questions, they don't come to be berated.
If you're here to act like that, maybe you should take it to another 'more sophisticated' forum
Thanks

 

[dripdaze]

I know with dual coils, the ohms reading is 1/2 the resistance of one coil... I'm assuming that means you need 4x the regular voltage to get the same watt vape? (Ex: my IGO-W is reading 1.1 ohms; would giving it 4.4 volts equal out to each coil operating at ~ 4.4 watts?)

Same question with twisted kanthal (in a single coil setup)... Is there any 'multiplier' that comes into play?

No multiplier needed. watts=volts x volts/resistance 4.4x4.4/1.1=17.6 watts split across 2 coils = 8.8 watts per coil
Or amps= volts/resistance 4.4/1.1=4.4 amps and watts=volts x amps 4.4 x4.4=17.6 watts = 8.8 watts per coil

If the 2 coils = 1.1 ohms each coil =2.2 ohms and you can repeat the comps for 1 coil ?????

 

[SoaFung]

Woah buddy, not a very nice response, as much an expert you may be, some folks may not be up to your speed, that's why they come here and ask questions, they don't come to be berated.
If you're here to act like that, maybe you should take it to another 'more sophisticated' forum
Thanks

it was not my intention to come off belittling. Just trying to point out that electrons have no concern to the path they take. All these crazy coil configurations make no difference to the equations used to calculate watts or amps. A 1.1 ohm load is just that.





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[KenD]

The whole wattage-split-between-two-coils makes it sound like a 1.5 dual coil will give you half the vape of a 1.5 single coil, though. It's not quite that simple. The power used may be split to two coils, but you're compensated by get heat from two coils. Yes, the power's split between two coils but they also add up. You don't really need to double the power. My dual coil Aqua at 0.9 ohms (19.6w at 4.2v, mathematically less than 10w per coil) gives more vapour than my single coil Kayfun at 1 ohm (17.64w at 4.2v). Yes, they are different RBAs and that needs to be taken into account, but still...

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[suspectK]

The question wasn't directed if a different wrapping technique plays a role in the power to the coil.

You do get 4x's the surface area, but I only really consider that with actual dual coils..twisted, you're really just increasing the surface area, with some texture added.. With a dual coil, you'll have two wicks in that extra surface area. You do have 4x's the surface area, but look at it like this..

With a dual coil you have two coils of the same resistance, which equals half of one coils resistance...so two 2ohm coil's equivalent resistance/total resistance is 1ohm. You have 4 times that surface area, but instead of hitting it with the same amount of power, you're applying twice the amount of power if it was just a single 2ohm coil. So you're only really having to figure out what you can do with twice the surface area. I wrap singles in the 0.6-1ohm range, and I keep dual coils around 0.5ohm...I sometimes go down to 0.3ohm, but meh..too hot for my batteries and me..

 

[SoaFung]

The question is about voltage needed for dual coil or single twisted wire coil. Without more clarification from the OP, we are all grasping at straws here. I am simply stating the coil confirmation makes no difference to the use of ohms law.





My reading comprehension might be sub par.



Ultimately he/she needs to find the volts/watts he/she likes.



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[BobC]

The question wasn't directed if a different wrapping technique plays a role in the power to the coil.

You do get 4x's the surface area, but I only really consider that with actual dual coils..twisted, you're really just increasing the surface area, with some texture added.. With a dual coil, you'll have two wicks in that extra surface area. You do have 4x's the surface area, but look at it like this..

With a dual coil you have two coils of the same resistance, which equals half of one coils resistance...so two 2ohm coil's equivalent resistance/total resistance is 1ohm. You have 4 times that surface area, but instead of hitting it with the same amount of power, you're applying twice the amount of power if it was just a single 2ohm coil. So you're only really having to figure out what you can do with twice the surface area. I wrap singles in the 0.6-1ohm range, and I keep dual coils around 0.5ohm...I sometimes go down to 0.3ohm, but meh..too hot for my batteries and me..

I think you're missing something here re surface area, there are 3 components to resistance, conductivity, length and width, which is surface area, so whether it's twisted, dual coil etc. the wider the path, the more electrons that flow, the lower the resistance

 

[BobC]

it was not my intention to come off belittling. Just trying to point out that electrons have no concern to the path they take. All these crazy coil configurations make no difference to the equations used to calculate watts or amps. A 1.1 ohm load is just that.

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maybe you should reread your response then, starting with "uhh," is derogatory

 

[CloudZ]

maybe you should reread your response then, starting with "uhh," is derogatory

The OP's reply before this post also started with "uhhh...".

I know this is probably one of the most polite forums around, but it is growing rapidly and it is still the internet. Brace yourself, as its only going to get more hostile around here. That's just the way it is. The moderators won't be able to keep up. I actually hope I am wrong on this, but I doubt it.

 

[SoaFung]

maybe you should reread your response then, starting with "uhh," is derogatory

I have. Unfortunately the internet does not reflect tone of voice well.





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[jersey_emt]

I think you're missing something here re surface area, there are 3 components to resistance, conductivity, length and width, which is surface area, so whether it's twisted, dual coil etc. the wider the path, the more electrons that flow, the lower the resistance

Actually it's not surface area that matters, it is sectional density.

Let's compare a solid round wire, and a piece of wire that is the same diameter, but has grooves running the entire length. An extruded star so to speak.

Like this but stretched out:

This would have much more surface area than the equal diameter round wire, but the round wire would have lower resistance, since there is much more mass of wire for the electricity to flow through.

Now, if we're talking about high-voltage AC, things get different because of the skin effect. When I visited the Hoover Dam, in the gift shop they had sections of the original transmission lines from the hydroelectric power plant. These were not solid copper cables, but thin hollow tubes.

With high-voltage AC, the current tends to be concentrated near the surface of the conductor, so a hollow cable is much more efficient because of its higher surface area (both the outside and inside surfaces).

But with low-voltage DC like we use when vaping, skin effect doesn't come into play. So surface area isn't a factor with resistance.