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#31 (permalink)      1/4/2017 1:34:20 AM US Central   quote/reply + tips
JPBel
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You don't need to work with bazillions amps to understand where the "recovery" you are talking about come from.
No, it is not a recovery at all.
And the "ionic recovery" i wrote about is something completly different.
You will have more luck searching for "recovery effect"

The IR of a battery in open circuit is not significantly high enough compair to your voltmeter internal resistance to affect the measurement in a significant manner.
There is a up to 9 orders of magnitudes difference between the two.

If you apply 3V accross two resistors of same value, you will measure 1.5V across each one of them and 3V accross both.
It does'nt matter if they are both 1Ω or both 10KΩ, you will measure 1.5V across each one of them and 3V accross both.
This will remain true untill the power source begin to limit the current below what the two resistances do given the voltage across them.

A resistance is not a voltage dropper, it is a current limiter.
If you keep viewing them as voltage droppers, you will never be able to understand what you are measuring.

The voltage you measure accross the load is the power(watts) the load is dissipating, not the power flowing through the whole circuit.
The remaining volts you can't measure are dissipated elsewhere in the circuit, most likely in the power source IR.
With dc to dc buck, buck/boost, converters, those volts are simply not provide to the output, and are not waste in the battery neither.

If you don't measure a voltage accross two points of a 0000ga, does it mean no power is flowing through it ?

I might have an incorrect terminology with "Chemical resistance".
But no matter how you explain it, it remain a resistance at providing the charges.
A more correct word is probably "reluctance", a ESR(Equivalent Series Resistance).

Edited on 1/4/2017 at 2:00 AM. Reason:
#32 (permalink)      1/4/2017 2:36:20 AM US Central   quote/reply + tips
JPBel
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I do all my calcs for my mech with nominal 3.7V, because under high loads, that is what the battery actually supplies, at the batteries, and at the atomizer, and why it doesn't say 4.2V on the side of a battery.


3.7V is the nominal voltage.
The average voltage needed to calculate the battery's stored energy(Watt-hour) when full, given the battery's charges capacity.

3.7V is only a mid voltage when used for vaping on mech.
The voltage "drop" you will measure change with the load resistance value, the battery's IR, and the battery actual voltage.
The voltage accross the load will never be 3.7V, except in one case : the load resistance value, the battery's IR, and the battery actual voltage are the right values to give 3.7V as a result.

#33 (permalink)      1/4/2017 6:14:06 PM US Central   quote/reply + tips
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wilderbeast
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https://www.e-cigarette-forum.com/forum/attachments/image-png.615475/

This is similar to the test I told you to do with the spoons, and this is only 10A. I may adjust slightly for the battery type and amps I am aiming for, but if this isn't a basic enough example of why 3.7V is a good place to start calculating real watts vs ideal like the grade 7 science class explanations you are giving that are either strawmen or just random and unrelated, I fear your head may be trapped in your rectum for life.
#34 (permalink)      1/4/2017 6:26:27 PM US Central   quote/reply + tips
Hawki
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This works well from .5 to .7
Fish are dum dum dum. They chase anything that glitters(beginners)
#35 (permalink)      1/4/2017 6:27:38 PM US Central   quote/reply + tips
Hawki
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Maybe .4 to .8
Fish are dum dum dum. They chase anything that glitters(beginners)
#36 (permalink)      1/4/2017 6:28:19 PM US Central   quote/reply + tips
Hawki
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Never over .9
Fish are dum dum dum. They chase anything that glitters(beginners)
#37 (permalink)      1/4/2017 6:29:32 PM US Central   quote/reply + tips
Hawki
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Unless it's real thin wire maybe .99
Fish are dum dum dum. They chase anything that glitters(beginners)
#38 (permalink)      1/4/2017 7:10:36 PM US Central   quote/reply + tips
JPBel
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wilderbeast wrote:

https://www.e-cigarette-forum.com/forum/attachments/image-png.615475/

This is similar to the test I told you to do with the spoons, and this is only 10A. I may adjust slightly for the battery type and amps I am aiming for, but if this isn't a basic enough example of why 3.7V is a good place to start calculating real watts vs ideal like the grade 7 science class explanations you are giving that are either strawmen or just random and unrelated, I fear your head may be trapped in your rectum for life.


I don't fear.
I know your's is stuck in BS.

You wrote in a previous post that i don't calculate watts properly, and wrote it again.
It has'nt occured to you that you are the one not doing it right ?

Stop your mumbo jumbo rambling on how stupid i am and show me mathematical proofs.

I don't want to see discharge curve, i know what they are.
They proves only one thing, that such thing happen, not why, nor can they be used to get precise values for all batteries.
And screw the "it is because of the battery chemistry magic" kind of answers.

I want you to explain to me why the voltage drop lower as the current increase.
With mathematical proofs that can be traced back to self-evident.
Yes, it is possible to do it using math, i have tried to do it for you.
But you can't be arsed to look into it.


If you can't do that, STFU and GTFO.


Edited on 1/4/2017 at 7:24 PM. Reason:
#39 (permalink)      1/5/2017 1:20:41 AM US Central   quote/reply + tips
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wilderbeast
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Look into it? You mean like you have done with my arguments that are going straight over your head? Doing your simple math would be an admission that it is correct, not a question of can or can't. Now how much calculus have you done? You don't even need to be at engineering levels of understanding to get this. Here is a math problem for you, how do simple math equations graph out? Answer: linearly. But the graphs for output are curved, therefore we can conclude the simple equation cannot accurately predict real world results. A lot of engineering is done by reading off graphs and tables for input vs. output, and those are produced by the manufacturers own bench tests, and they produce the most accurate results when the components get put together (if precision is required beyond maximums for safety). Examples, electric pump output vs pressure head, or flow rate through a valve with pwm signal vs pressure. I have provided reasons why the curves exist as well and your argument is they are bs or magic... so due to your ignorance there is nothing left to do but throw tomatoes. Boooo!
#40 (permalink)      1/5/2017 1:51:18 PM US Central   quote/reply + tips
JPBel
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1- 3.7V is not the voltage a battery output under a load.
It is a voltage you can actually get at a specific load resistance value tho.

2- I am not saying that using 3.7V do not kinda do the job.
What i am saying is that it stop "kinda doing the job" as the current increase.

3- If you are alergic to maths, i will do them for you.

Ideal scenario :

Battery's voltage: 4.2V
Battery's IR : 10mΩ
Coils resistance : 100mΩ
Total circuit resistance : 110mΩ
Current flowing through : ~38.2A
Total power : ~160.4W

The 100mΩ resistance will dissipate : ~145.8W
The 10mΩ resistance : ~14.6W

------------------------------------------------------------

Closer to reality :

Battery's voltage: 4.2V
Battery's IR : 100mΩ
Coils resistance : 100mΩ
Total circuit resistance : 200mΩ
Current flowing through : 21A
Total power : 88.2W

The 100mΩ resistance will dissipate : 44.1W
The voltage you will measure accross it will be : 2.1V

The 100mΩ battery's IR : 44.1W
And if we were able to measure the voltage across the IR, it would measure 2.1V.

-----------------------------------------------------

Same math with 0.5Ω coil :

Battery's voltage: 4.2V
Battery's IR : 50 mΩ
Coils resistance : 500mΩ
Total circuit resistance : 550mΩ
Current flowing through : ~7.64A
Total power : ~32.07W

The 500mΩ resistance will dissipate : ~29.2W
And the voltage you will measure across it : ~3.82V

The 50mΩ IR : ~2.92W



Can you see how series resistances are affecting your volts measurements now ?
Can you see where your graphs came from ?
#41 (permalink)      1/5/2017 2:28:33 PM US Central   quote/reply + tips
JPBel
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You can even calculate your battery's true IR fairly accurately by measuring the current(with the coil as a load), the coil's resistance, and the battery's voltage under a very light load(10KΩ or more).

Edited on 1/5/2017 at 2:30 PM. Reason: