Question on radiation from an emitter at range.

Post by **NecronLord** » 2005-03-09 03:35pm

How does one calculate the energy intensity given an omnidirectional source (such as a nuclear weapon) at a given range. I suspect it's a variant of X/4πR² where X is the energy released, and R the range in meters, but for some reason I can't seem to figure out how one gets a value in joules per m² from that.

Lord Revan · Post by **Lord Revan** » 2005-03-09 03:48pm

well if range=R then the area is 4*(3,14...)*R^2=A if total ener released is X
then intensity energy is I=X/4π(R^2). resault is in Jm^-2 as R is in meters, X is joules and 4 and π are with out an unit frankly I don't understand how somebody could get anything other from it.

Post by **Darth Wong** » 2005-03-09 04:00pm

It depends on the mode of energy dispersal. If it's straight radiation in vacuum, it's simple inverse-square law based on the surface area of the sphere of radius R. In atmosphere, it's much more complicated.

Lord Revan · Post by **Lord Revan** » 2005-03-09 04:12pm

Darth Wong wrote:It depends on the mode of energy dispersal. If it's straight radiation in vacuum, it's simple inverse-square law based on the surface area of the sphere of radius R. In atmosphere, it's much more complicated.

doesnät the basic formula still remain the, you add the extra term needed for atmospheric interaction?

Post by **NecronLord** » 2005-03-09 04:14pm

So a device of 5.0208e18J has an energy density of around 399.5 Gj/m² in a vacuum at one kilometer?

I have no idea how I got the... other value now... Could have sworn I was doing the same thing...

Post by **Darth Wong** » 2005-03-09 04:16pm

Lord Revan wrote:
Darth Wong wrote:It depends on the mode of energy dispersal. If it's straight radiation in vacuum, it's simple inverse-square law based on the surface area of the sphere of radius R. In atmosphere, it's much more complicated.
doesnät the basic formula still remain the, you add the extra term needed for atmospheric interaction?

No. Atmospheres tend to absorb energy volumetrically, hence not following the inverse-square law.

Lord Revan · Post by **Lord Revan** » 2005-03-09 04:18pm

NecronLord wrote:So a device of 5.0208e18J has an energy density of around 399.5 Gj/m² in a vacuum at one kilometer?

I have no idea how I got the... other value now... Could have sworn I was doing the same thing...

Well assuming you the correct formula that quite right.

Lord Revan · Post by **Lord Revan** » 2005-03-09 04:18pm

Darth Wong wrote:No. Atmospheres tend to absorb energy volumetrically, hence not following the inverse-square law.

thanks