Looks like a great cell, possibly the highest capacity 26650 cell you’ve tested. It looks like it’s the 5500mAh version from PLB:
http://www.powerlongbattery.com/battery-cells/
This manufacturer also most likely makes the cell in the Liitokala 26650.
So it was not a protection circuit that blew, correct? Do you think it was something like a current interrupt device (CID) tripping, from getting too hot and internal pressure increasing?
Correct
Yes, but the cell do not look like it vented, the top do still look pristine (usual there will be some traces when the cell vent).
Nice review! Looks like a best performer, all around. For 26650's, the iJoy 26650 4200 might have to take 2nd place now, accept if running at 30A.
I'd like to buy a couple. Just emailed Shockli - can't find a retailer.
What happens after the fuse is blown? Is the battery usable?
No, it is dead. It is not a protection circuit or PTC that can recover.
Hi HKJ, the coolest tester in the world 
Do you know where that fuse is located?
Is it possible to the throw it away and use the cell “unfused”?
It is part of the safety inside the can, i.e. you cannot get to it without destroying the cell.
The fuse is there to prevent the cell from doing something nasty (like explode).
Wondering if they can send you replacements - love to see the internal resistance result. I'd suspect it's below 0.4 from your discharge curves.
HKJ
That’s what I mean - pry it open, short the fuse and use it again. But not at 30 amps
I doubt you can put it together again if you pry it open. The cell may also take damage from the air.
Oh, it’s somewhere inside the chemistry. Now I get you. Thanks.
Excellent review. Like the fuse for safety (of which the cut off point is way higher than general flashlight use). Just a shame that the cell is so wide, almost a 27650.
Comparing the discharge curves to the liitokala 26650, the IR looks just a bit higher for this shockli cell.
It is a fat cell, like the liitokala cell. Makes sense if they are from the same manufacturer.
Average internal resistance looks slightly below 25mΩ, just a gnat's ass higher than that of a LiitoKala 26650-50A. Definitively an improvement in energy density, with an actual power delivery rating. Verrry good! 
About the fuse, a bit sensitive I'd say. Temperature monitoring definitively recommended/advisable. 
Cheers ^:)
Originally posted on Mon, 04/24/2017 - 01:33; IR estimation fixed, content added and make upped.
Maybe I don't understand it. In the comparator looking at 3A to 10A, the LK looks worse in 98% of the discharge curve.
http://lygte-info.dk/review/batteries2012/Common26650comparator.php
The Shockli has a higher V for just about the entire discharge. I thought that was better?
The shockli would perform better over time because of its higher capacity, not because of a lower IR. At a given resting voltage, like fully charged at 4.2V for example, the cell with the lower IR will provide more current in a direct drive situation, but a cell with a lower capacity will drop its resting voltage quicker over time.
It seems like a compromise between IR and capacity oftentimes. The VTC5A would give more amps at start than this shockli, but obviously the capacity is in a different ballpark and would not be as suitable in a high power light.
Here's the 5A comparison below. So accept for the first 0.25 discharge, the Shockli performs better - the blue line is clearly well above the red line. I though it should produce higher amps in a DD/FET mod, accept for the first 250 mAh. Am I reading this right? Does that first small advantage for the LK mean it will have a lower internal resistance?
I'm confused...
The internal resistance is measure at 15A (if the cell can deliver that), but higher voltage at the desired current is always a better metric to look at.
OK, thanks! Didn't realize that. Still at 15A, the initial drop is steeper than the LK, but maintains a higher voltage after that first 250 mAh.
If we are to substract the 20A from the 1A curves, we attain a dV/Ah one which reveals the average internal cell voltage drop for that 19A rate. This reveals close effective average internal resistance. Both first and last parts of the curve shouldn't be used, as there usually steep slope shifts happen because of dramatic changes in the cell's available energy and internal resistance.
This is as far as I understand for now.
Cheers ^:)