DC
Banned
- Joined
- Mar 20, 2008
- Messages
- 23,064
Would 1908 kg of CHARGE CUTTING LINEAR be enough to bring down a WTC Tower?
Assumptions:
(Simplified and based on this info)
Core Columns
52" by 22" (ca. 1.32m x 0.31m)
4" thick (ca. 100mm)
3.7m long
48 of them
CHARGE CUTTING LINEAR - CCL
2.7 kg/m
Circumference with the 52" sides in a 35° angle = 3.85m
48 * 2.7 * 3.85 = 498.96 kg
we need 2 cuts (top of column and bottom)
498.96 * 2 = 997.92 kg
Perimeter Columns
14" by 14" (ca. 0.36m x 0.36m)
2.3" thick (ca. 60mm)
3.7m long
240 of them
CHARGE CUTTING LINEAR - CCL
1.2 kg/m
Circumference with 2 sides in a 35° angle = 1.58m
240 * 1.2 * 1.58 = 455.04 kg
we need 2 cuts (top of column and bottom)
455.04 * 2 = 910.08 kg
997.92 + 910.08 = 1908 kg (ca.4206.4 lb)
would that be enough?
Edit:
no airplane damage. just a hypothetical assumption.
with Airplane damage and Fires, it is claimed to not need any explosives.
but when talking about CT's i often hear it would need a huge amount of explosives. so i wondered how much would that be?
Assumptions:
(Simplified and based on this info)
Core Columns
52" by 22" (ca. 1.32m x 0.31m)
4" thick (ca. 100mm)
3.7m long
48 of them
CHARGE CUTTING LINEAR - CCL
2.7 kg/m
Circumference with the 52" sides in a 35° angle = 3.85m
48 * 2.7 * 3.85 = 498.96 kg
we need 2 cuts (top of column and bottom)
498.96 * 2 = 997.92 kg
Perimeter Columns
14" by 14" (ca. 0.36m x 0.36m)
2.3" thick (ca. 60mm)
3.7m long
240 of them
CHARGE CUTTING LINEAR - CCL
1.2 kg/m
Circumference with 2 sides in a 35° angle = 1.58m
240 * 1.2 * 1.58 = 455.04 kg
we need 2 cuts (top of column and bottom)
455.04 * 2 = 910.08 kg
997.92 + 910.08 = 1908 kg (ca.4206.4 lb)
would that be enough?
Edit:
no airplane damage. just a hypothetical assumption.
with Airplane damage and Fires, it is claimed to not need any explosives.
but when talking about CT's i often hear it would need a huge amount of explosives. so i wondered how much would that be?
Last edited: