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Firehed 525 days ago | link | parent

The intensity of that heat is going to drop off rapidly with distance. I'm no physicist, but I'd bet it follows the Inverse-Square Law (https://en.wikipedia.org/wiki/Inverse-square_law) just as light does.

Meaning that at 20m from the center of the explosion (the fireball was 20m diameter, or 10m radius) it should already be at 1/4 the intensity, already less than the surface of the sun. At 7 miles or 11200m, you're talking 1/(1120^2) or 7.97e-7 of the original intensity - effectively zero - if I've done that math right.



sliverstorm 525 days ago | link

Let's not forget, apart from distance, the fireball is expanding in all directions. At 1" from the center, you're exposed to half the fireball. At 1 mile from the center, you're exposed to a tiny spherical arc of the fireball.

The energy is going in every direction, not just at you. With distance, you will receive a smaller portion of whatever energy made it 7 miles out.

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gnaritas 525 days ago | link

That's what he just said, the inverse square law. You're describing the same phenomenon.

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sliverstorm 525 days ago | link

Oh, I thought he was talking about energy dropping off with distance the way a bullet loses speed with distance.

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