A quick Will Mcdaniel CRT

[Oiguana2701]

I recently got this calculation accepted, getting Class M results. This was a pretty casual feat by Will, and thus, I believe the new LS rating should be ''At least Class M''. This will affect Will Mcdaniel himself, as well as The Ringleader and the Lock Down Buddy, both of which scale directly to Will.

That's it. This is pretty straightforward.

 

[IDK3465]

Can’t you also get AP from this by multiplying the force in newtons by the displacement in meters?

 

[Oiguana2701]

Can’t you also get AP from this by multiplying the force in newtons by the displacement in meters?

Yeah, but also, finding a value for the displacement when the computer wasn't ''evenly'' crumpled up feels weird.

 

[IDK3465]

Yeah, but also, finding a value for the displacement when the computer wasn't ''evenly'' crumpled up feels weird.

It seemed to mainly be from the corner of the screen to the center of the hinge, though only half of the work was applied there since the MacBook was still open, so I guess something like:
sqrt(21.24^2 + (30.41/2)^2) = 26.1214399488 cm
(62637312.52800001/2) * 0.261214399488 = 8180883.9888 joules (almost 9-B+)

 

[DarkDragonMedeus]

Looks simple enough

 

[Oiguana2701]

Bump

 

[Mr. Bambu]

I don't know. The math works but this seems basically impossible to be accurate, no? It seems like something literally must be slipping past us. We're getting Class M for crushing 1.4 kg of aluminum and plastic where the chief deciding factor of that is that it has high surface area. This seems like like we must be egregiously and heinously misapplying the tools at our disposal here.

 

[IDK3465]

I don't know. The math works but this seems basically impossible to be accurate, no? It seems like something literally must be slipping past us. We're getting Class M for crushing 1.4 kg of aluminum and plastic where the chief deciding factor of that is that it has high surface area. This seems like like we must be egregiously and heinously misapplying the tools at our disposal here.

I think the issue is that we're treating this like he's applying pressure to the entire surface area when really it's just the area under his hands that's getting crushed, though I still kinda feel weird about the feat

 

[Oiguana2701]

I don't know. The math works but this seems basically impossible to be accurate, no? It seems like something literally must be slipping past us. We're getting Class M for crushing 1.4 kg of aluminum and plastic where the chief deciding factor of that is that it has high surface area. This seems like like we must be egregiously and heinously misapplying the tools at our disposal here.

I cannot speak too much about the subject as I am not too experienced with crushing calculations, but aluminum still has a high compressive strength, which, combined with its high surface area, would naturally elevate the result. I do agree with it being a strange result, but unless there's another formula or something similar, I'm not sure what to do about this.

 

[Mr. Bambu]

To be quite honest, I'm also unfamiliar. A lot of calc methods sort of creep in under the doorway, y'know-- a user presents it, and a CGM accepts it ("I don't know", they'll say, "the math was right, at least") and it becomes common practice. The result here is so obviously silly that I cannot help but feel this is probably a case of that, and the philosophy behind it is also nonsensical. If I shape a 2 kg mass of steel into a cube, it's going to yield exponentially less than the same mass formed into a heatsink, because a heatsink is deliberately designed to have a really high surface area. But in practice, we know it is easier to crush these thin planes of material. So it just doesn't hold up to the base principals that I do understand.

 

[Oiguana2701]

To be quite honest, I'm also unfamiliar. A lot of calc methods sort of creep in under the doorway, y'know-- a user presents it, and a CGM accepts it ("I don't know", they'll say, "the math was right, at least") and it becomes common practice. The result here is so obviously silly that I cannot help but feel this is probably a case of that, and the philosophy behind it is also nonsensical. If I shape a 2 kg mass of steel into a cube, it's going to yield exponentially less than the same mass formed into a heatsink, because a heatsink is deliberately designed to have a really high surface area. But in practice, we know it is easier to crush these thin planes of material. So it just doesn't hold up to the base principals that I do understand.

I'd say this specific calculation type deserves a deeper discussion, since this isn't the first time these calcs bring strange results. For example, this FNaF calculation got 16510000 Newtons of force for crushing a very small amount of mild steel, when drum crushers, which are able to crush entire barrels made out of steel, only use around 164500 Newtons.

 

[Mr. Bambu]

I'd say this specific calculation type deserves a deeper discussion, since this isn't the first time these calcs bring strange results. For example, this FNaF calculation got 16510000 Newtons of force for crushing a very small amount of mild steel, when drum crushers, which are able to crush entire barrels made out of steel, only use around 164500 Newtons.

Agreed, whole thing seems ******.