In the first Samurai Shodown, the final boss Amakusa is shown rending a tenshu apart.
We don't know what castle this is, and its structure is even rather unconventional, as most tenshu have 5 stories, while this has 6. Also, tenshu are comprised of a stone foundation and then the various floors, but the former and part of the first floor are covered by the boundary wall. The only castle for which I could find the height of both the base and the tower (instead of the whole thing) is the Himeji castle (15m for the stone base and 31.5m for the tower).
Even though it's pretty tall when compared to other tenshu, I think it kinda checks out with this one being abnormally tall, due to aforementioned extra floor.
The formula for cutting is 0.6 * Tensile Strength * Area * width of the blade.
Tensile strength of wood appears to be 70 to 140 MPa (source here), I'll go with 70 for safety.
As we don't know the exact width of whatever performed the cut, I'm going with the average width/thickness of a katana (0.7 cm, source: Google) because this is Samurai Shodown.
- Tenshu: 31.5 m = 290 px = 0.11 m/px
- Height: 58 px = 6.38 m
- Width: 213.68 px = 23.50 m
- Area: 149.93 m2 = 1499300 cm2
0.6 * 70 MPa * 1499300 cm2 * 0.7 cm = 44079420 J or 0.010 tons (Small Building level)
I guess we can also use this feat for speed. The three white flashes that mark the cut occur in 15 frames at 60 fps. 23.50/(15/60) = 94 ms/s (Subsonic)