RM2 Reomizer Building questions

[bittermelon]

<snip>

Oh! Dont forget quality batteries and charger. Hit up rtd vapor. Good prices and fast shipping. Know your battery limits and STAY SAFE.
and we love questions, so dont be afraid to ask

Ok! I'm still reading and rereading this thread and others so that I understand what to do & how it works when I attempt making my first coil. But I've always found math challenging (that's putting it lightly) and I'm kinda phobic about it (sub-par schooling in jr & sr high school ruined me ). So I'm struggling to understand and comprehend how the ohms law stuff works. What is the battery limit for the AW 14500 600mAh, 3.7v and what's the resistance range of coils safe with it??? (i read the battery info post by Baditude but the 14500 is not listed. Can anyone here chime in? I asked BT directly also.)

Did I get this right: the diameter of the finished coil + wire size gauge + number of wraps determines its resistance ohms??

 

[CMD-Ky]

Don't miss Super_X_Drifter's video in the stickies on the RM2.

 

[bittermelon]

Don't miss Super_X_Drifter's video in the stickies on the RM2.

Yes, thanks!! Watched it twice already.

 

[Papa_Lazarou]

Ok! I'm still reading and rereading this thread and others so that I understand what to do & how it works when I attempt making my first coil. But I've always found math challenging (that's putting it lightly) and I'm kinda phobic about it (sub-par schooling in jr & sr high school ruined me ). So I'm struggling to understand and comprehend how the ohms law stuff works. What is the battery limit for the AW 14500 600mAh, 3.7v and what's the resistance range of coils safe with it??? (i read the battery info post by Baditude but the 14500 is not listed. Can anyone here chime in? I asked BT directly also.)

Did I get this right: the diameter of the finished coil + wire size gauge + number of wraps determines its resistance ohms??

Okay, I'm hoping this will help you some....

Ohm's Law says:

Votage = Current x Resistance, shortened to V = I x R

This can be restated as:

I = V / R

In other words, the current utilized equals the battery voltage divided by the coil resistance.

The batteries we use are 4.2 volts at full charge, so if you had a 1.0 ohm coil, the current = 4.2 / 1.0 or 4.2 amps (the unit of current). If you had a .5ohm coil, the current would be 4.2 / 0.5 or 8.4 amps.

Basically, just take 4.2 and divide it by the resistance of the coil you're building to find out the amps you'll be utilizing.

Now, the battery you have has a maximum amp rating (the most you should be drawing) of 4 amps. This means the lowest coil resistance you could use is about 1.05 ohms. 4 amps = 4.2V / 1.05 ohms.

Now, you want to factor in a safety buffer of, say, 20 - 50%, so really you should consider about 1.35 ohms as your lower limit.

Okay, so you want to make coils with 1.35 ohms resistance or higher. How do you know what to make in order to get that?

Well, it's about the length of the wire used and it's resistance per foot of wire used, and it's hard to do that math in your head. There are, however, some good tools to calculate this for you. Here's one...

Coil wrapping | Steam Engine | free vaping calculators

You just use the gadgets in the tool to input your wire type, target resistance, and coil diameter - it will tell you how many wraps of that wire you need. There are more bells and whistles in the tool for more advanced coil building, but keep it simple to start with.

For example, if you used 28 gauge wire and a 1/16" coil diameter (very common in the rebuilding scene) and we wanted our 1.35ohm coil, the tool tells me that you need 10 wraps.

Hoping like heck that some of this helps you. If not.....



tl;dr - Your battery is good to 4 amps, you should build coils >1.35 ohms, 10 wraps of 28g kanthal around a 1/16" mandrel will get you there.

 

[ancient puffer]

And just to clarify (just my opinion, really), yes, if you're vaping 1.8 ohms now, then that's what I'd shoot for to *start* with. That will give you an idea of the difference between the disposables you're used to, and a rebuildable. Then you can adjust up or down (resistance) for warmer or cooler vape....if you want to play around.

I started out using 1.8 ohm atomizers and cartos on my first REO, and built 1.8 ohm coils on my first rebuildable (RM2). I tried higher and lower resistance, but always came back to 1.8. But eventually I succumbed to the siren song of dual coils, so now I'm happy running an Odin at .6 ohms. Once you're comfortable with rebuilding, it's just a matter of when or if you ever feel the need to experiment. Truthfully, rebuilding is so easy once you get your feet wet, most of us wind up experimenting a LOT at first until we find our "sweet spot".

 

[super_X_drifter]

Here's a vid to explain ohms law so even I can understand stand it if you can tolerate me for that long

http://youtu.be/QQbWVHSiltQ

 

[bittermelon]

Okay, I'm hoping this will help you some....

Ohm's Law says:

Votage = Current x Resistance, shortened to V = I x R

This can be restated as:

I = V / R

In other words, the current utilized equals the battery voltage divided by the coil resistance.

The batteries we use are 4.2 volts at full charge, so if you had a 1.0 ohm coil, the current = 4.2 / 1.0 or 4.2 amps (the unit of current). If you had a .5ohm coil, the current would be 4.2 / 0.5 or 8.4 amps.

Basically, just take 4.2 and divide it by the resistance of the coil you're building to find out the amps you'll be utilizing.

Now, the battery you have has a maximum amp rating (the most you should be drawing) of 4 amps. This means the lowest coil resistance you could use is about 1.05 ohms. 4 amps = 4.2V / 1.05 ohms.

Now, you want to factor in a safety buffer of, say, 20 - 50%, so really you should consider about 1.35 ohms as your lower limit.

Okay, so you want to make coils with 1.35 ohms resistance or higher. How do you know what to make in order to get that?

Well, it's about the length of the wire used and it's resistance per foot of wire used, and it's hard to do that math in your head. There are, however, some good tools to calculate this for you. Here's one...

Coil wrapping | Steam Engine | free vaping calculators

You just use the gadgets in the tool to input your wire type, target resistance, and coil diameter - it will tell you how many wraps of that wire you need. There are more bells and whistles in the tool for more advanced coil building, but keep it simple to start with.

For example, if you used 28 gauge wire and a 1/16" coil diameter (very common in the rebuilding scene) and we wanted our 1.35ohm coil, the tool tells me that you need 10 wraps.

Hoping like heck that some of this helps you. If not.....



tl;dr - Your battery is good to 4 amps, you should build coils >1.35 ohms, 10 wraps of 28g kanthal around a 1/16" mandrel will get you there.

Hey Papa (or anyone),

If my battery (aw 14500) is good @ 4.2 volts full charge, then why does the label say "3.7v"??? Confused still.

 

[Papa_Lazarou]

Hey Papa (or anyone),

If my battery (aw 14500) is good @ 4.2 volts full charge, then why does the label say "3.7v"??? Confused still.

3.7V is the average discharge voltage during use from fully charged to fully depleted. This is how bats are typically rated - they start off higher and deplete lower than this rating.

Your bats will start at 4.2V and then decrease as you vape. Most folks swap them out at about 3.8V. Never let them get too low (say, below 3.4 or so as a guide, just to be safe).

 

[Rickajho]

Hey Papa (or anyone),

If my battery (aw 14500) is good @ 4.2 volts full charge, then why does the label say "3.7v"??? Confused still.

4.2 volts is the charge termination voltage for most Li-On chemistry batteries. 3.7 volts is considered their nominal working voltage.

Didn't anyone point you at this thread yet? ~> http://www.e-cigarette-forum.com/forum/reos-mods/413000-rm2-rba-2-0-reomizer-2-rebuild-tutorial.html

It sounds like you may be over thinking this whole thing a bit. Keep in mind that once you have a coil on it that's working right and gives you what you want you may not have to replace that coil for months - or until you are really bored and just wanna do it. You will need to clean it and dry burn on occasion, but actually removing it, coiling and replacing it - probably not so much. The wick will have to come and go often depending on which wick material you settle on. If you decide you like silica even you don't have to change that - you dry burn it in place to clean it. But cotton goes mooshy and gets... not so good after a couple of days and has to be replaced. Still, removing and replacing cotton is pretty easy.

All I'm saying is coil building and coil replacement probably won't happen as often as you are thinking right now.

P.S. The new AW 650 mAh 14500's are good for up to 9 amps continuous discharge! Woo! Gonna get me some of those. You can find some places closing out the old 600 mAh version at bargain prices.

 

[bittermelon]

<snip>
Didn't anyone point you at this thread yet? ~> http://www.e-cigarette-forum.com/forum/reos-mods/413000-rm2-rba-2-0-reomizer-2-rebuild-tutorial.html

Actually, yes I am familiar with that thread--bookmarked. Watched it twice already and that's certainly not my last.

It sounds like you may be over thinking this whole thing a bit.
All I'm saying is coil building and coil replacement probably won't happen as often as you are thinking right now.

.

Yeah, I am overthinking it somewhat--it's my tendency to OCD on anything that really interests me at the time. It's just that the battery safety stuff freaks me out some--ya know, the graphic maimed vaper's photos on Baditude's blog. 8-o Not to mention my eighth-grade self going, "oh YUCK, it's MATH (my worst subject re: ohms law)." IF it weren't for the fact that I prefer high VG and like the taste of dripping, not to mention admiring the simplicity & durability of the REOs, I would have jumped on the vvMod/iStick/iPV/Mv3 etc. w/Tank-du-jour train instead of moving to Reoville in order to not mess with that stuff.

 

[bittermelon]

3.7V is the average discharge voltage during use from fully charged to fully depleted. This is how bats are typically rated - they start off higher and deplete lower than this rating.

Your bats will start at 4.2V and then decrease as you vape. Most folks swap them out at about 3.8V. Never let them get too low (say, below 3.4 or so as a guide, just to be safe).

Papa,

by "swapping the battery out" you mean for a freshly charged batt, right? I tested the one currently in my REOm and its @ 4.02. How long do these type batteries usually last (when will I have to replace them with new)?

VAPE MAIL!!! I just got my pack of Labo Cotton puffs, so now I have a lifetime supply of wicking. Decided to go with cotton for the taste and because it seems safer than silica to me. Now all I need is my Kanthal wire!! *stalking the postal carrier & mailbox*

A big Thank You and Hug to all who've chimed in with advice. It's starting to make more sense to me!!

 

[Sloth Tonight]

Papa,

by "swapping the battery out" you mean for a freshly charged batt, right? I tested the one currently in my REOm and its @ 4.02. How long do these type batteries usually last (when will I have to replace them with new)?

By swapping out, yes, he just means for a freshly charged batt. Most of us swap a battery by or before 3.7v (letting it drain past 3.7 can shorten the life cycle of the battery). Some of us swap sooner, because we notice the vape performance drops off. I pull mine around 3.8 or 3.9, whenever I feel like it.

A well taken care of battery will last >500 charge cycles supposedly

 

[Papa_Lazarou]

sloth's got you covered

 

[bittermelon]

Okay, I'm hoping this will help you some....

Ohm's Law says:

Votage = Current x Resistance, shortened to V = I x R

This can be restated as:

I = V / R

In other words, the current utilized equals the battery voltage divided by the coil resistance.

The batteries we use are 4.2 volts at full charge, so if you had a 1.0 ohm coil, the current = 4.2 / 1.0 or 4.2 amps (the unit of current). If you had a .5ohm coil, the current would be 4.2 / 0.5 or 8.4 amps.

Basically, just take 4.2 and divide it by the resistance of the coil you're building to find out the amps you'll be utilizing.

Now, the battery you have has a maximum amp rating (the most you should be drawing) of 4 amps. This means the lowest coil resistance you could use is about 1.05 ohms. 4 amps = 4.2V / 1.05 ohms.

Now, you want to factor in a safety buffer of, say, 20 - 50%, so really you should consider about 1.35 ohms as your lower limit.

So, plugging in my preferred coil resistance of 1.8o I get:

I = V / R

4.2/1.8=2.33 amp. Now I read this to mean that I'm well UNDER the amp limit for the battery (4 amp), therefore this coil would be safe to use. And anything over 4 amps is dangerous. Good??

does I=amp=current (do they all mean the same thing)?? I apologize for these stupid questions, but vaping has opened up a whole new world to me, making me become a bit familiar with concepts that really intimidated me, and therefore habitually avoided. Thanks again muchly, everyone!!

 

[Papa_Lazarou]

So, plugging in my preferred coil resistance of 1.8o I get:

I = V / R

4.2/1.8=2.33 amp. Now I read this to mean that I'm well UNDER the amp limit for the battery (4 amp), therefore this coil would be safe to use. And anything over 4 amps is dangerous. Good??

does I=amp=current (do they all mean the same thing)?? I apologize for these stupid questions, but vaping has opened up a whole new world to me, making me become a bit familiar with concepts that really intimidated me, and therefore habitually avoided. Thanks again muchly, everyone!!

A 1.8ohm coil would be fine with your 4amp rated battery (yup, it's gonna draw 2.33amps, so you've got a good safety margin). Yes, you do not want to build a coil that will draw more amps than your battery is rated for, plus you should account for a 20-50% safety margin, so nothing more than, say, 3 amps (just to use round, safe numbers).

In the Ohm's Law formula, "I" is used to represent current, which is measured in amps (or milliamps, sometimes, which is 1/1000th of an amp). So...

amps = the unit name of current
I = the symbol used to represent current in Ohm's Law, so
I = Current (measured in amps)

The same is true for the other parts of the Law, they're just less confusing...

V = Voltage (measured in volts)
R = Resistance (measured in ohms)

 

[Sloth Tonight]

Think of a water pipe with a valve-like shut off thing in it. The water running through the pipe is your current. The current is the water running through the pipe. The resistance is that valve thingy. Depending on how closed or shut that valve is, the current will change, right? If the valve is mostly shut (ie, high resistance) the current is low. If the valve is mostly open (low resistance) the current is strong, water comes gushing through.

Not sure how to throw voltage into this analogy, though I'm sure it could be done by someone smarter than me

 

[penguiness]

Think of a water pipe with a valve-like shut off thing in it. The water running through the pipe is your current. The current is the water running through the pipe. The resistance is that valve thingy. Depending on how closed or shut that valve is, the current will change, right? If the valve is mostly shut (ie, high resistance) the current is low. If the valve is mostly open (low resistance) the current is strong, water comes gushing through.

Not sure how to throw voltage into this analogy, though I'm sure it could be done by someone smarter than me

Voltage -- How large is the pipe, so how much water can flow through without bursting it?

 

[coaldriller]

Sent from my iPhone using Tapatalk

 

[bittermelon]

 

[slappy3139]

In the water pipe analogy, voltage would be the pressure(or force) pushing the water through the pipe, such as a hydrostatic column or pump.

ETA: The size of the pipe would be more a function of resistance to the flow of water, ie. bigger pipe more flow, smaller pipe less flow.