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I don't want to crash the party here in regards to the AP (rest is fine enough, at a glance), but I have a few questions about the method (it's not about the steel bars not being there when the feats were performed or anything, it's a calc concern).

Specifically, I'm not sure I understand how the volume of the steel beams was calculated. If I understand correctly from this blog, were they calculated as solid rectangular prisms? Because I think they'd be hollow, and quite thin at that. If I'm wrong I apologize but generally to my understanding that's what steel beams are like. There's also the assumption that the entire beams are being shattered, and not only the section where the crater happens, which itself would lower the result quite a bit.

I also have secondary concerns. It's a bit weird to say that (using the blog I quoted as an example) 2.6 beams were destroyed instead of just saying it was 2 or 3, realistically either you hit the whole thing from the side, or miss it completely. There's also the 101.6 cm pixel scaling which I think may be wrong since I don't see the floor in the scan used, but that may just be me not knowing the video the scan is from.
 
Hey, something I can finally respond to!
Specifically, I'm not sure I understand how the volume of the steel beams was calculated. If I understand correctly from this blog, were they calculated as solid rectangular prisms?
Were they? That's something you should ask @Kachon123 about, as I'm not too sure. Do you want an image of what the steel beams look like?

Because I think they'd be hollow, and quite thin at that. If I'm wrong I apologize but generally to my understanding that's what steel beams are like.
No problems. I recommend you ask Kachon about it first.

There's also the assumption that the entire beams are being shattered, and not only the section where the crater happens, which itself would lower the result quite a bit.
I mean, didn't Kachon calculate the amount of steel beams that were affected? I'm confused.

I also have secondary concerns. It's a bit weird to say that (using the blog I quoted as an example) 2.6 beams were destroyed instead of just saying it was 2 or 3, realistically either you hit the whole thing from the side, or miss it completely.
So what would we do here? Round it to the nearest tenth?

There's also the 101.6 cm pixel scaling which I think may be wrong since I don't see the floor in the scan used, but that may just be me not knowing the video the scan is from.
I'm pretty sure the floor scaling is fine. I think we had a talk about this with a different calc member iirc.
 
Were they? That's something you should ask @Kachon123 about, as I'm not too sure. Do you want an image of what the steel beams look like?

No problems. I recommend you ask Kachon about it first.
@Kachon123
I mean, didn't Kachon calculate the amount of steel beams that were affected? I'm confused.
He did, but he's assuming the entire things were destroyed even though you'd only need to smash them for the hole's length. Say if a hole is 1 meter wide, you'd only need to destroy one meter of beam
So what would we do here? Round it to the nearest tenth?
The lower tenth if you wanna be safe but yes, basically.
I'm pretty sure the floor scaling is fine. I think we had a talk about this with a different calc member iirc.
If it's fine it's fine, it's not my primary concern.
 
He did, but he's assuming the entire things were destroyed even though you'd only need to smash them for the hole's length.
Hmm, so what would we do here then if that's the case? This better not significantly lower the results.

The lower if you wanna be safe but yes.
I mean, if it's 2.6 like you said, rounding to the nearest tenth would make it 3. We can use it as a highball.
 
Bruh I'm just tryna rest man. I got in an accident a week ago and I want to ease my way back into stuff like this. Just give me a while.
Damn.

My bad bro. Rest easy.

Your lost will not be in vain my fellow comrade.

In all seriousness, take a good break. Get better.
 
Okay. Let's try not to sour the mood and let's continue.
He did, but he's assuming the entire things were destroyed even though you'd only need to smash them for the hole's length. Say if a hole is 1 meter wide, you'd only need to destroy one meter of beam
Hmm, so what would we do here then if that's the case? This better not significantly lower the results.
@Armorchompy
 
Well if you want to know how much it'd lower the results, it's effectively (this is a bit of a highball but i don't wanna try and bust out harder math) result divided by beam length multiplied by hole radius, so it's gonna drop it by like, 2/3x i assume. The beams being potentially hollow may lower it by much more than that.
 
Well if you want to know how much it'd lower the results, it's effectively (this is a bit of a highball but i don't wanna try and bust out harder math) result divided by beam length multiplied by hole radius, so it's gonna drop it by like, 2/3x i assume. The beams being potentially hollow may lower it by much more than that.
And what calc are you referring to? Or is it all of them?
 
I may be wrong on this, we'd also have to see if the beams are hollow which may end up being more important than this.
 
Hmm, I'm not an expert but I actually think those welding marks do mean the beams may be hollow. When I say I'm not an expert I mean I only have the faintest clue here so if anyone can prove or even just confidently tell me I'm wrong I'll concede, but I do think they'd be hollow, this type of beam.

This confirms the pixel scaling is good, though, that is indeed the floor.
 
I don't want to twist the knife, but it'd decrease by a lot.
 
Armor we can talk about this. We don't have to downgrade the verse significantly once again...
 
Armor we can talk about this. We don't have to downgrade the verse significantly once again...
Not really a downgrade less prevention of an upgrade

But yeah hollowness usually decreases the volume of beams by 80% iirc, that kills the calc so hard it’s not even funny

I’ll look into the properties of the steel being used here
 
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