huge heat difference? just me?

[Ryan Kelly]

so is it just me or is there a huge difference in the heat that a .2 coil produces verses a .3 coil? i don't understand how .1 could make such a difference. i usually build coils around .2 ohms and they are hot. really hot. i have my TOBH atty drilled out to keep the vapor cool enough to inhale and i love the heat. but today i built a .3 ohm coil to try and increase my battery life a little bit without sacrificing a lot of vapor and heat but it is literally cold...like i don't feel any heat at all. i even closed my airflow a little bit and there is still no heat. i would imagine that subohming so low would be a very hot vape no matter what but the vapor from this setup is literally cold. am i doing something wrong or am i just weird?

 

[Tangaroav]

Let me start by saying that I do not sub ohm, (yet ).

The way I understand, from .2 to .3 you have about 33 % less wattage to the coil. Still as you say at .3 ohm with restricted air flow. I would expect the vapor to be hot.

I can't wait for more experienced sub OHMers to comment.

 

[VapingTurtle]

Are you using a protected Trustfire battery?

 

[k702]

as you go down in resistance the amps being put out will increase more and more.

for instance.

a 2 ohm coil on a fully charged battery will push 2.1 amps

a 1.5 ohm coil on a fully charged battery will push 2.8 amps

only a .7 difference in amps over a .5 difference in resistance...


But you compare

a .3 ohm coil on a fully charged battery pushing about 14 amps

a .2 ohm coil on a fully charged battery pushing around 21 amps.

a 7 amp difference over a .1 difference in resistance.



Taking that into account it's pretty easy to understand why it would be so much hotter with only a .1 difference. Add in the fact that most of the ohmmeters we use aren't 100% accurate (being +/- .1 or.2 is pretty common) meaning you could actually be dangerously low for even a sony vtc5 and it's no wonder at all.



If you have to ask this question you probably shouldn't be running a .2 ohm coil. Just saying..

 

[Ryan Kelly]

as you go down in resistance the amps being put out will increase more and more.

for instance.

a 2 ohm coil on a fully charged battery will push 2.1 amps

a 1.5 ohm coil on a fully charged battery will push 2.8 amps

only a .7 difference in amps over a .5 difference in resistance...


But you compare

a .3 ohm coil on a fully charged battery pushing about 14 amps

a .2 ohm coil on a fully charged battery pushing around 21 amps.

a 7 amp difference over a .1 difference in resistance.



Taking that into account it's pretty easy to understand why it would be so much hotter with only a .1 difference.



If you have to ask this question you probably shouldn't be running a .2 ohm coil. Just saying..

I'm perfectly capable of running a .2 ohm coil thanks. i understand ohms law and i know the difference in amp and wattage outputs because i do the math before actually firing any coils that i build i just didn't think there would be THIS big of a difference. I'm not like confused about why its a cooler vape i just didn't think it would go from scorching hot to cool. i had an observation and posted about it. has nothing to do with my knowledge or capability of vaping sub ohm builds

 

[Tangaroav]

Hopefully, we may get factual comments to your observation ... I, for one am still curious.

I would have though that + 50 watt, of power to the .3 ohm coil would produce a scorching vapor if the air flow is restricted. As I mentionned I have never sub ohm, and I am curious and interested for an explanation.

I am sure you will also get a bunch of warnings along with the answers. I guess they are often addressed more to the less initiated to sub Ohm practice, (like me), than to you directly.

 

[Richard75]

This easy, and is pure mathematics:

P = V^2/R, where P is watts, V is voltage, and R is resistance.

This is an exponential formula, and here's what happens when it's graphed in Excel (with V set at 3.7):



As you can see, as R gets closer and closer to zero, P spikes up at an ever increasing rate.

Thus, it stands to reason that you'd notice a larger difference in heat (watts) between 0.3 and 0.2 (~22 watt difference) compared to 0.4 and 0.3 (~11 watt difference). Imagine if you went to 0.1, that an ~68 watt difference!

Hope this helps!

PS: apologies for the picture, didn't think it would turn our that small.

PPS: I realize I didn't exactly answer your question. My guess is that you got used to the 68 watts, and 22 watts lower is a pretty decent spike.

 

[Tangaroav]

Very telling graph, ( frightening too), thank you.

The op indicates she/he understands ohm's law and measures the resistance of coils before usage. she/he is surprised after having experienced the following:

The vapor is ''really'' hot with a fresh battery with a .2 ohm coil, the vapor is '' literally'' cold with fresh battery with a .3 ohm coil and with air flow cut down.

To rephrase the question for MY understanding: Should the vapor produced at .3 ohm with less air flow be cold ? Is it only a matter of too much air ? I wish active sub ohmers, (to that level), can tell us what they experienced.

 

[Dampmaskin]

so is it just me or is there a huge difference in the heat that a .2 coil produces verses a .3 coil? i don't understand how .1 could make such a difference.

.1 may not sound like much, but the reduction in resistance is 1/3, in other words 33%.

So a .2 ohm coil will give you 150% the current and 150% the power of a .3 ohm coil (disregarding the internal resistance of the battery).

 

[Tangaroav]

.1 may not sound like much, but the reduction in resistance is 1/3, in other words 33%.

So a .2 ohm coil will give you 150% the current and 150% the power of a .3 ohm coil (disregarding the internal resistance of the battery).

So, does cutting the air flow 33% get you similar temperatures, if not volume, to a .2 ohm coil , ( all other conditions remaining the same) ?

 

[Dampmaskin]

So, does cutting the air flow 33% get you similar temperatures, if not volume, to a .2 ohm coil , ( all other conditions remaining the same) ?

I have no idea, turbulent fluid dynamics is much more complicated than DC electronics, and way over my head.

 

[zoiDman]

This easy, and is pure mathematics:

P = V^2/R, where P is watts, V is voltage, and R is resistance.

This is an exponential formula, and here's what happens when it's graphed in Excel (with V set at 3.7):

View attachment 359791

As you can see, as R gets closer and closer to zero, P spikes up at an ever increasing rate.

Thus, it stands to reason that you'd notice a larger difference in heat (watts) between 0.3 and 0.2 (~22 watt difference) compared to 0.4 and 0.3 (~11 watt difference). Imagine if you went to 0.1, that an ~68 watt difference!

Hope this helps!

PS: apologies for the picture, didn't think it would turn our that small.

PPS: I realize I didn't exactly answer your question. My guess is that you got used to the 68 watts, and 22 watts lower is a pretty decent spike.

Actually, P = V^2/R is Not an Expediential Function. Because the Independent Variable is Not the Exponent of a Term.

P = V^2/R is a Garden Variety 2nd Degree Function.

With the R in the Denominator of a Quotient, and a Fixed Value in the Numerator, math people sometimes refer to these as "Landmines" when R approaches 0. Because they Blow Up.

Also Notice since there is a 2 in the Exponent of the Voltage, that if R is Fixed and V becomes the Independent Variable, that the Rate of Change of P is Not plodding along in a Straight Line, but Changes (Increasing at a Increasing Rate) with the Square of V.

 

[Tangaroav]

Actually, P = V^2/R is Not an Expediential Function. Because the Independent Variable is Not the Exponent of a Term.

P = V^2/R is a Garden Variety 2nd Degree Function.

With the R in the Denominator of a Quotient, and a Fixed Value in the Numerator, math people sometimes refer to these as "Landmines" when R approaches 0. Because they Blow Up.

Also Notice since there is a 2 in the Exponent of the Voltage, that if R is Fixed and V becomes the Independent Variable, that the Rate of Change of P is Not plodding along in a Straight Line, but Changes (Increasing at a Increasing Rate) with the Square of V.

Although I had to google a few terms, the graph and this info not only increased my vocabulary but also my comprehension of the risks associated with really low sub ohming.

...but I still would like to have a hands on answer concerning the temperature of the vapor. Although I will in all likelyhood never vape to less than 1.0 ohm.

This is an interesting discussion.

 

[anumber1]

10 Characters

 

[zoiDman]

Although I had to google a few terms, it not only increased my vocabulary but also my comprehension of the risks associated with really low sub ohming.

Math, on Most Levels, is a Very Pure Discipline.

There are No "I" before "E" except after "C". Or Silly Capitalization Rules.

But I think what is Lost to Many People is Math, on Most Levels, is not just a Bunch of Numbers. It is the Way of Quantifying "Real" events.

When I look at P = V^2/R and someone tells me that R is going to Move Towards Zero (Doesn't Get to Zero, just Approaches it in Every Closer and Closer Increments), then I know that P is going to Get Large. As Long as V^2 is > R

Don't care about Numbers. Just Thinking about What Happens to P.

 

[rolygate]

Also as coil resistance falls and more power is drawn, it is likely that resistance through the APV will become more of a factor in heat generation. In addition there may be some inductive load factor that is not calculated for, the circuit may not be a purely resistive load.

 

[zoiDman]

BTW - What Happens to P in P = V^2/R when R moves the Other Way? When R > V^2 and R continues to become Ever Larger?

P becomes Small, Right? Does that Make Sense?

As Resistance become Larger and Larger, that the Power become Smaller and Smaller. And that when Resistance is Large Without Bound (Infinite) that Power should Zero.

 

[zoiDman]

Also as coil resistance falls and more power is drawn, it is likely that resistance through the APV will become more of a factor in heat generation. In addition there may be some inductive load factor that is not calculated for, the circuit may not be a purely resistive load.

Ohms Law describes things on a Fundamental Level. But does not take into Count Many Factors that occur in the "Real" World.

A Good Example is when One Mashes Down the Power Button on their APV and does a Dry Burn, the Coil will turn an Angry Red. And you look down thru the Smoke and see Watts, so to Speak.

But the Red Color you see is the Results of Coil Getting Hot. Doesn't Resistance Change in a Wire when the Wire Gets Hot?

 

[Tangaroav]

'' But I think what is Lost to Many People is Math, on Most Levels, is not just a Bunch of Numbers. It is the Way of Quantifying "Real" events. ''

I like that sentence.

That one too

''With the R in the Denominator of a Quotient, and a Fixed Value in the Numerator, math people sometimes refer to these as "Landmines" when R approaches 0. Because they Blow Up.''

and that one frightens me. ( how does it change ? up or down ?

''But the Red Color you see is the Results of Coil Getting Hot. Doesn't Resistance Change in a Wire when the Wire Gets Hot? ''

 

[Richard75]

Actually, P = V^2/R is Not an Expediential Function. Because the Independent Variable is Not the Exponent of a Term.

P = V^2/R is a Garden Variety 2nd Degree Function.

With the R in the Denominator of a Quotient, and a Fixed Value in the Numerator, math people sometimes refer to these as "Landmines" when R approaches 0. Because they Blow Up.

Also Notice since there is a 2 in the Exponent of the Voltage, that if R is Fixed and V becomes the Independent Variable, that the Rate of Change of P is Not plodding along in a Straight Line, but Changes (Increasing at a Increasing Rate) with the Square of V.

You're right, that was my mistake. It's been a couple years since I took some calculus classes.