hey this is not too believable….
has anyone tested these ??? thank you blf
No. They are not. Probably more like 1200-1600mAh.
Edit: There is a reference to GTL in the eBay description. Probably more like 800-1000mAh.
No, they are actually more than 5000 mAh. And, perhaps, would like to buy a nice suspension bridge? I have a rather nice one for sale in San Francisco. 
There’s and even nicer one in Michigan, but it’s not for sale.
I can’t sell you London Bridge, that was sold to a US tourist in 1970’s, though I do know Brooklyn bridge was sold more than once, as well as Eifel tower, and the Great Clock Of Westminster, aka BIG BEN (big ben is the name of the largest bells inside, NOT the tower).
tabs
is this a joke?
of course they are
and those red/blue 3000/2400 are all real of course too
soon 6000 will be released, I believe, effectively doubling brand Panasonics
Wiki says the volumetric energy density is 250 to 730 W·h/liter for this type battery.
The battery volume is 0.0165 liter and the battery energy is 18.5 W-h, which comes to 1120 W-h/liter.
No, it is not very credible.
5,000 mah 18650’s? Yeah!… And I’m Elvis! (Not)
The 3400mah 18650 Panasonics is the highest that have been released in loose cells that I’am aware of.
Sanyo 14500 has around 800mah.
By the same standard and based on size and technology, between 0.6 and 1.8 W-h of energy.
Ahhh. Chinese advertising and using Japanese cells. Ah. I see said the blind man! 
But only 1000mAh’s. WOW!
Actually Sanyo UR14500P has 3.1 Wh of energy 
Wh = Ah * Average(Voltage over discharge cycle)
mWh = mAh * Average(Voltage over discharge cycle)
So, lets say a typical 18650 will maintain an AVERAGE of 3.5V (Range From 4.0 to 3.0 ) over it’s discharge cycle. If it’s a 2900mAh Cell.
2900mAh * 3.5V = 10,150 mWh
or
2.9Ah * 3.5V = 10.15Wh
Guys, you actually CAN get 5000mAh from these cells.
Just need to stick ~6 diodes in forward-bias before your load.
Thus you get 5000mAh...for ~1V .
Still only get <5Wh.
EDIT: Whoops, horrible math. 0.7V x 6diodes =/= ~3V. Sorry, I guess I meant 4/5.
That’s a joke, right?
Maybe, maybe not. ;)
I’ll just assume it’s a joke since diodes will not change the mAh in the cell…
PPtk
lol* not even half that. I wish Ebay had better control and monitoring of these fake and over-claiming sellers selling somtimes unsafe crap. 
5000mAh?. wow. goodbye AW.
:-p
I'm not talking about the mAh in the cell. I'm talking about the mAh going across the load.
What I'm basically saying (and I'm also assuming my diodes are ideal) is that I'm basically transforming the voltage from the battery to a voltage a fraction of that across the load. Since energy, and therefore power, must be conserved, the current is multiplied by the inverse of that fraction.
So if the voltage across my load is a fifth of the battery's then current supplied to the load will be five times actually coming out of the batt. No physics defying magic here.