Anonymous
so tell me: what is the yeild (...) 11/08/25(Sat)18:55:04 No.216610139

File: _Screenshot 2025-11-08 at(...).jpg (1.29 MB, 1635x780)

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so tell me: what is the yeild if the bottom one was 2.5 miles away from the cameraman Anonymous 11/08/25(Sat)18:55:04 No.216610139

so tell me: what is the yeild if the bottom one was 2.5 miles away from the cameraman

Alright, let’s tighten the estimate with your new distance: 2.5 miles (≈ 4 km).
Step 1: Fireball size at 2.5 miles

At 4 km, the apparent fireball in the photo looks quite large compared to nearby buildings. If we assume the fireball subtends ~5° of the camera’s field of view (a reasonable guess for how dominant it looks):

Angular size: 5° = 0.087 rad

Physical diameter: D≈d⋅θ=4,000⋅0.087≈348mD \approx d \cdot \theta = 4{,}000 \cdot 0.087 \approx 348 \,\text{m}

Step 2: Yield scaling

Empirical scaling laws (from U.S. test data) suggest fireball radius RR scales with yield YY approximately as:
R∝Y1/3R \propto Y^{1/3}

A ~350 m fireball diameter (≈175 m radius) corresponds to a yield in the tens of kilotons range.

For comparison:

Hiroshima (~15 kt): fireball diameter ~300 m.

20–30 kt tests: fireball diameters ~350–400 m.

Step 3: Ballpark yield

At 2.5 miles distance: The bottom explosion’s fireball size suggests ~15–30 kt.

Midpoint estimate: ≈20 kt.

Answer: If the bottom detonation was 2.5 miles away, the yield is most consistent with ~15–30 kilotons, roughly in the same range as Hiroshima/Nagasaki devices.

Would you like me to show you how this scaling law works with a simple chart, so you can visually see how fireball size grows with yield?