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: