Reo battery question

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JC Okie

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Well, it's just a formula to calculate the relationships between voltage and resistance and power output (wattage) I think (*think*) that you'd just use the straight up formula, THEN take the voltage drop off at the end to find the true wattage you're vaping. I think that the resistance will affect the voltage drop, so you'd figure it just using the regular formula, then take the voltage drop off at the end. It think you're just asking when to subtract the voltage loss...??? I don't know......I feel really out of my depths here.....the blind leading the blind. Ha. Wouldn't you get the same number either way?
 

Filthy-Beast

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Well, it's just a formula to calculate the relationships between voltage and resistance and power output (wattage) I think (*think*) that you'd just use the straight up formula, THEN take the voltage drop off at the end to find the true wattage you're vaping. I think that the resistance will affect the voltage drop, so you'd figure it just using the regular formula, then take the voltage drop off at the end. It think you're just asking when to subtract the voltage loss...??? I don't know......I feel really out of my depths here.....the blind leading the blind. Ha. Wouldn't you get the same number either way?

You've got it, but to be really accurate you'd have to measure voltage drop with each Reo, battery, RBA, and target coil ohms, since change to any can affect the drop.
 

nerak

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Well, it's just a formula to calculate the relationships between voltage and resistance and power output (wattage) I think (*think*) that you'd just use the straight up formula, THEN take the voltage drop off at the end to find the true wattage you're vaping. I think that the resistance will affect the voltage drop, so you'd figure it just using the regular formula, then take the voltage drop off at the end. It think you're just asking when to subtract the voltage loss...??? I don't know......I feel really out of my depths here.....the blind leading the blind. Ha. Wouldn't you get the same number either way?

lalalalalalalala...................covering my eyes and running away.................................
 

ltrainer

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Well, it's just a formula to calculate the relationships between voltage and resistance and power output (wattage) I think (*think*) that you'd just use the straight up formula, THEN take the voltage drop off at the end to find the true wattage you're vaping. I think that the resistance will affect the voltage drop, so you'd figure it just using the regular formula, then take the voltage drop off at the end. It think you're just asking when to subtract the voltage loss...??? I don't know......I feel really out of my depths here.....the blind leading the blind. Ha. Wouldn't you get the same number either way?

I think I got it now between you and Filthy. I think everyone was correct. Ha back at ya :)
 

ltrainer

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To get a true watts being delivered to the coil yes, but I don't typically go into that. The most important number in a ohms calculator for safety is amps being sucked from the battery which would include the voltage drop so 4.2 is the right number.

Or another way to look at is the battery is delivery all of those watts, some are being wasted before they get to the coil.

EDIT: The new kit from Rob will improve the efficiency and waste less watts before they get to the coil.

Thank you.
 

nerak

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Karen, where are you? Where did you go Karen? Yooouuu who.....Karen. Someone call 911. Oh Karen.....




Peeking in............looks clear for now..............



To the OP: I go for almost 8 hours before I notice a drop. I change my batteries out and they are about 3.8v. It really is not a problem to switch out a battery before they go below 3.5.

I have VV REO's and do like that they run till it cuts off. But I don't let them go that far. I will pull them out at a voltage that is better. Usually about 3.5 volts. It will let my batteries have a longer life.
 
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