Understanding rim testing impact energy and resulting speed

What am I doing wrong? Road.cc explains Specialized new test standard where the highest impact energy is 70 joules (Sagan's safety standards: Roval revives tubeless for Rapide II and Alpinist II | road.cc). I wanted to understand how fast that is with 190lb (86.2kg) rider + bike weight and come up with 2.85 mph (4.59 kph) which doesn’t sound right to me. I used v = sqrt((KE * 2)/m). Not believing my own answer I found this calculator which gave me the same result: Kinetic Energy Calculator

The same road.cc article captions the impact video as 180lbs, 20mph, 40joules. But plugging those values into the KE calculator results in 3263 joules.

70 J refers to the drop test where the full KE is transferred to wheel - eg. 7 kg dropped from 1 m, 3.5 kg from 2 m, etc. In the roll over test, 180 lb rider @ 20 mph, they estimate 29 J transferred to wheel - rider continues afterwards with 29 J less KE. Really, a lot of this will depend on tire widths, pressures, shape of edge, etc. You can do much much fudging. Also not a great test to start. What would be much better if they, for example, repeatedly dropped until failure, over a range of drop energy, with/without tire, etc. IDK - CF entirely not worth the bother for me where I can avoid it - brittle fracture not a good fail mode. Forged metal so much better - eg. F1 rims still like that.

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I mean that number you calculated (if correct, idk I didn’t check it) would be like if you took the tires off the wheels and then rode them straight into a wall to a full stop. So the tires would dissipate a lot of that energy before the impact reached the rim and you also would most likely not be coming to a complete stop (you probably have a bigger problem than a cracked rim if you do).

So that 70J is the max amount they estimate the rim will see when hitting an object (like a curb).

Ok, Thanks. That makes sense: either energy transferred from bike or 7kg drop.