That slow-mo is rather impressive to watch!

pooter
i see you are sadist
but still, that gives me a shiver everytime i watch it.
ugggghhhhhh~~ shiver!!!
oh my god.

The guy in red was talking or something, all his fault!

Vote slay him!

[Spartacus]

someone did the maths, it was impossible for spotters to catch the bar.

Even Mark Rippletoe agreed in the forums that the spotters had no chance. The guy lifting had to have a firmer grip on the bar.

[SIZE=13px]For a while, I thought "Well, yeah, it's possible to at least get hands on the bar". But we're talking about a bar that is travelling at 8ft/sec and accelerating at 30+ft/sec^2 before we can lay hands on it. In a deadlift, the acceleration most people can get on 400lb + deadlift is small - I'd put it at maybe 1ft/sec^2, and that's in a position of maximal efficiency.[/SIZE]

[SIZE=13px]Even if the spotters were near superhuman, once the bar is moving at that speed, it going to be tough to stop it before contact with the chest. How hard? Charitably, I'm going to give 8inches of "catching room" (20cm) for a 180kg deadlift.[/SIZE]

[SIZE=13px]How long does the bar take to fall 30cm?[/SIZE]
[SIZE=13px]s = 0.3m[/SIZE]
[SIZE=13px]a = 9.8m/s^2[/SIZE]
[SIZE=13px]s = 1/2 a t^2[/SIZE]
[SIZE=13px]0.3 = 4.9 t^2[/SIZE]
[SIZE=13px]t = 0.25s[/SIZE]

[SIZE=13px]How fast is the bar going after 30cm (= 1 foot)?[/SIZE]
[SIZE=13px]v = at [/SIZE]
[SIZE=13px]v = 0.25 x 9.8 = 2.45m/s = 8ft 4 inches / sec[/SIZE]

[SIZE=13px]Now, what force is required to stop it hitting the lifter?[/SIZE]
[SIZE=13px]First we calculate the time in which the bar must stop[/SIZE]
[SIZE=13px]t = s / average velocity = s / (u+v)/2 = 2s/u[/SIZE]
[SIZE=13px]s = 0.2[/SIZE]
[SIZE=13px]u = 2.45m/s[/SIZE]
[SIZE=13px]t = 0.16s[/SIZE]

[SIZE=13px]Now we can calculate the acceleration on the bar required by the spotting team.[/SIZE]

[SIZE=13px]Let x = acceleration generated by the spotting team.[/SIZE]
[SIZE=13px]v = u + at[/SIZE]
[SIZE=13px]v = 0[/SIZE]
[SIZE=13px]u = -2.45m/s[/SIZE]
[SIZE=13px]a = -9.8m/s^2 + x m/s^2[/SIZE]
[SIZE=13px]t = 0.16s[/SIZE]
[SIZE=13px]0 = - 2.45 + (-9.8 + x ) 0.16[/SIZE]
[SIZE=13px]2.45 /0.16 = x - 9.8 [/SIZE]
[SIZE=13px]x = 25.11m/s^2[/SIZE]
[SIZE=13px]That's a net upward acceleration of 25 - 9.8 = 15.2m/s^2[/SIZE]

[SIZE=13px]Now, let's convert to the units of force, Newtons:[/SIZE]
[SIZE=13px]N = kg m/s^2[/SIZE]
[SIZE=13px]N = 180 x 15.2 = 2736 Newtons[/SIZE]
[SIZE=13px]What mass implies gravity has the same force?[/SIZE]
[SIZE=13px]N = x . 9.8[/SIZE]
[SIZE=13px]2736/9.8 = 279kg = 615lbs[/SIZE]

[SIZE=13px]In other words, to catch the bar, that pair of guys would need to be exert enough force to lift 615lb between them in that mechanically compromised position right from the instant they touch the bar. You could have Kirk Karwoski and Ed Coan in that spot, and that bar is still hitting the kid. And we are expecting the super-human feat of strength at a random time half-way through a long day. They have to be prepared for this for maybe 105 attempts through the meet.[/SIZE]

http://startingstrength.com/resources/fo...91&page=19