Help with a small calc.

[vivftp]

Howdy there, I'm just trying to work out a bit of math unfamiliar to me, and not too sure if I'm on the right track. I was hoping you guys might be able to help me sort it out.

Essentially I'm trying to work out what portion of an explosion an object a certain distance away will be hit with. ie. if a bomb of a certain yield explodes 2km away from a ship in an omni-directional explosion, how much energy will hit the ship 2km away.

I know it has to do with the inverse square law, but googling it hasn't helped me much since it's always referring to comparing 2 points at a distance from the source, rather than comparing the source to 1 point.

One approach I was taking was getting the surface area of a sphere where the distance from the bomb to the ship would be the radius (4km diameter), then getting the area of the ship (in this case an ellipsoid due to shields), dividing that into the surface area of the larger sphere and then multiplying by the total energy of the bomb.

I think it's flawed, but I need a more informed opinion.

Thanks!

[Kuroneko]

If your distance is very large compared to the size of the bomb (which 2km should be), just treat the blast as point-like. If you're dealing with a uniform omnidirectional blast in space (so you have uniform spread of energy), it means the normal intensity of the blast at distance r is the yield divided by the surface area of the sphere of that radius, assuming no blockage occurs before the blast reaches that distance. In other words, flux is [E/(4πr²)]. If the source is far enough away, you can calculate how much your ship receives simply by the cross-section in the plane perpendicular to the direction of the blast (if it's closer, it will no longer be approximately planar), while the true intensity will be proportional to the cosine of the angle of incidence.

[vivftp]

Thank you for your reply.

So just to make clear, say yield is 1 megaton, so 4184000000000000 joules. The surface area of a sphere with a 2,000 meter radius is 50240000 meters square.

That means the energy per square meter at that distance is 83280254.777 joules?

I admit, the last part of your statement went zooming above my head...

*kicks self for not being more interested in math as a young lad*

[Kuroneko]

Thank you for your reply. So just to make clear, say yield is 1 megaton, so 4184000000000000 joules. The surface area of a sphere with a 2,000 meter radius is 50240000 meters square. That means the energy per square meter at that distance is 83280254.777 joules?

Well, close enough, anyway, and neglecting the number of significant digits you have--83 million joules per square meter.

I admit, the last part of your statement went zooming above my head...

It just means that the total energy your ship receives depends on the cross-sectional area, not on the surface area--the actual inensity on the ship surface will be less unless unless you ship has a flat face completely normal (perpendicular) to the blast.

[vivftp]

Well, close enough, anyway, and neglecting the number of significant digits you have--83 million joules per square meter.

Ahh, perfect, thank you.

It just means that the total energy your ship receives depends on the cross-sectional area, not on the surface area--the actual inensity on the ship surface will be less unless unless you ship has a flat face completely normal (perpendicular) to the blast.

Hmm, I see. So if this were say, an egg shaped shield bubble, the intensity difference would come from the fact that the blast wave has to travel outwards to hit further up the shield, and thusly as it move out, its intensity drops off even further.

Heh, a quick google on that showed it's enough math to make my head hurt, but it's ok, the energy per square meter is more than enough for what I wanted to find out


Thanks again!

[Kuroneko]

Hmm, I see. So if this were say, an egg shaped shield bubble, the intensity difference would come from the fact that the blast wave has to travel outwards to hit further up the shield, and thusly as it move out, its intensity drops off even further.

Not quite. If the distance is small relative to the ship size, that won't matter. Consider a flat horizontal rectangle with a distant light source (say, Sun) shining directly down uniformly on it and the surrounding area. If you rotate the rectangle so one edge of it is lower than another, there will be less light shining down on it, since you've just decreased the rectangle's profile--the rest will simply miss it(see diagram). Since the area of the rectangle is unchanged, the intensity of the incident light on it will be lower. One can show that it is proportional to the cosine of angle it makes with the incident light rays. That is why the area of the ship is not as important as the cross-section in the plane perpendicular to the incident radiation/blast/energy.

Code: Select all

| | | light  | |
| | |        | v 
v v v        | /
----- vs.    v/  2 units
3 units      /

[vivftp]

Not quite. If the distance is small relative to the ship size, that won't matter. Consider a flat horizontal rectangle with a distant light source (say, Sun) shining directly down uniformly on it and the surrounding area. If you rotate the rectangle so one edge of it is lower than another, there will be less light shining down on it, since you've just decreased the rectangle's profile--the rest will simply miss it(see diagram). Since the area of the rectangle is unchanged, the intensity of the incident light on it will be lower. One can show that it is proportional to the cosine of angle it makes with the incident light rays. That is why the area of the ship is not as important as the cross-section in the plane perpendicular to the incident radiation/blast/energy.

Code: Select all

| | | light  | |
| | |        | v 
v v v        | /
----- vs.    v/  2 units
3 units      /

Ahh ok, I get it now. Thanks for the clarification