How do you know that if you don't know the weight of steel and concrete on each level. I am admitting that I do not know where the center of gravity was and that the top of the tilted portion moved 121 ft horizontally relative to the bottom to the tilted portion and that the NIST admits that the core supported 53% of the weight.
The core is smack bang through the centre. The truss sits on top of it, overall if you spread that weight out, it is less than one floor and has none of the concrete reinforcement, office equipment, furniture, fire cladding and all sorts of other weight extras.
Since the core was rectangular and narrower than the width of the building the CoG would not have to move very far to be beyond the core of the lower stationary portion.
But that is not even accurate speculation. You suggest that the support structure largely made of steel, represents the entire weight of each relative floor. You do realise it makes up what looks like 30% of the overall floor? The exterior walls alone amounted to 30% of the downward force of the building.
I do not care about the drop distance. Greening's formula is the same as mine he is just computing a distance perpendicular to the one I am talking about.
You aren't making any sense. You used the wrong formula for the redistribution of COG. The equation you made shows the drop, not where the new COG is. That cannot be computed from a single formula - it needs a purpose built one that takes in the shape, weight, distribution and sizes.
All you can do is accuse me of being stupid and therefore accurate data is irrelevant because YOU say so.
Well no. I don't accuse you of anything so terse. You are just mistaken, your equation simply shows the drop.
I am thoroughly impressed!
Stop waving your arms about. You aren't the font of all knowledge. I learn from other people, things I don't know and show other people things I do know.
The truss is not a significant weight, shape or size to alter the COG.