I don’t see that the blast radius would be linear as that maps suggests.
For one thing, the strength of the pulse would be inversely proportional to the square of the distance from the earth, just like any electromagnetic force.
The other thing is, the earth is not flat. The curvature of the earth would make both the east and west coasts even further away from the center than the map suggests and the pulse would be further weakened by then.
For one thing, the strength of the pulse would be inversely proportional to the square of the distance from the earth, just like any electromagnetic force.
The other thing is, the earth is not flat. The curvature of the earth would make both the east and west coasts even further away from the center than the map suggests and the pulse would be further weakened by then.
Sounds like there is no need for concern then.
EMP blast radius is line of sight. So areas behind mountains might be more protected than those which are not.
The map in question takes into account the curvature of the earth.
Iran currently has missiles with a range of 2500km able to carry a 1 ton payload. Those certainly won’t reach the high orbital altitude but are a threat.
Currently reports have them orbiting a 65kg payload into orbit.
“I don’t see that the blast radius would be linear as that maps suggests.”
It would. it’s based on Line-of-Sight - not a “blast” per se, more of a zap.... Near the center, per unclassified models, it’s 50,000 V/m, near the edge, it’s on the order of 20,000 V/m.
“For one thing, the strength of the pulse would be inversely proportional to the square of the distance from the earth, just like any electromagnetic force”
This would be true if it were a point source - but the energy comes from a very large area in the atmosphere above the earth, so it’s “near-field” and inverse square does not apply.
“The curvature of the earth would make both the east and west coasts even further away from the center than the map suggests and the pulse would be further weakened by then.”
The map does account for that. The decrease is more linear, rather than inverse square because it is a highly distributed source.
Good comments, though.