Tuesday, October 30, 2007

Movie plot weapon

You heard it here first kids,

High powered microwave weapon ... FROM SPACE.

Use it to start fires on the ground of a country that has been giving us too much lip.

Let's see here.

a match is what, about 1BTU?

via google, to metric is:
1055.05585 Joules

lets check that
1 gram of cellulose releases 1.5cal/mg in burning, since flame spreads,
6279 J/g would be more than enough to ignite 1 gram of wood.
Once a small patch is ignited, it should spread.

on the low end of density of something like oak (0.6g/cc) that means 3767.4 J/cc.
spreading that out to a 1mm thick slice, thats what, 10cm^2 area that needs the 3767 J.

lets double it to 7534.8 J required due to losses from the delivery of the energy over time.

A LEO satellite orbits the earth every 90 min.
Lets say it is visible, um, 1/9th of the time. 10 min. (the ISS is visible here ~5 min)
so lets give you 1 minute of time to deliver the required energy.

7534.8/60 = 125.58 watts
for things of this nature. but, the energy doesn't have to "return" so the R_r^2 squared term is removed. further simplifying, lets just say that P_r = P_t*o*a
0 = (5% absorbed*)
a = power in/power delivered = 0.1

so P_r = P_t * 0.005

sweet. P_t needs to be ~ 25116 watts

thats 25.116 kw. The Boeing 702 satellite bus can provide 25kw.

* from "An experiment with microwave radiation to determine absorption
coefficients and defects of wood - Physics Education,Number 6,Volume 28"
It is really low, it seems... if i was serious i would look to see what frequency wood absorbs the most. scotts says that it CAN happen...

I'm going to claim that it is possible to start a fire from space with a satellite using electromagnetic radiation.

Who's with me?


Anonymous said...

Kerr is not with you.

The R^4 term that you dropped should instead become R^2 when we're only interested in power received by the scatterer.

Gauss's law tells us that radiation arriving on a scatterer with radar cross section o is equal to the following:

P_s = P_t o / (4pi R^2).

In the case of a monostatic radar, the transmitter and receiver at colocated, radiation reflected by the scatterer may be modeled as simply another transmitter, so power at the receiving station is:

P_r = P_s A / (4pi R^2).

This P_r is roughly the radar equation (assuming transmitting antenna gain of 0 dB). P_s is the power received by the wood target we're trying to ignite.

Returning to your scenario, at an altitude of 200 nautical miles (370.4 km), absorbed power will be many orders of magnitude less than what you've computed.

This complicates matters as during the course of a single minute, the range from transmitter to the square meter of wood or so that you are targeting changes dramatically and nonlinearly. Even if you computed power with an average range, the result would have significant error.

Anonymous said...

It would be cool if blogger supported inline \TeX and would automatically typeset it and produce internally managed images of the markup the way Wikipedia does.

Anonymous said...

..since I had a few minutes, I decided to code what I believe to be a useful tool in solving this problem.

It computes the energy delivered to some target on the surface of the earth from a passing satellite in circular orbit. Assumes lossless medium, spherical earth, and the target is in the illuminator's far field.

If my calculations are correct, a 25kW weapon with 30 dB gain placed in a 200 nm circular orbit would deliver 13 microJoules to 1 square meter of the Earth's surface in 1 minute.

Hope this helps. It was an interesting problem.

sstc said...


Using your code, with a 200km orbit instead of 200nm,
I get 0.002236 J ... which still sucks. It only gets worse with 10cm^3.

Do you think that we can't get more gain that 30dB?

I'm just thinking that less than 10% of the transmitted power hitting a target from space seems a bit low.

Radar cross section is measuring reflection, not absorption.

Doesn't that help us out?

I think the much bigger issue is keeping a satellite pointed at an object that is 10 cm^2 for that long and from that altitude.

Anonymous said...

What I called 'radar cross section' should be the effective cross sectional area of the target making P(t) the power incident to it. This is an upper bound on power delivered. With an absorption coefficient strictly less than 1, we cannot obtain even this. (also, heat is dissipated over one minute, etc..)

My MATLAB code may have had an error when you downloaded it. I forgot to multiply by 60 since time units were in minutes, not seconds, and the integral should have been:

E = int(60*P(t), dt, t_0, t_1)
with t in minutes

I fixed it at around 8:40am ish so check to verify the multiplier in the expression for E.

The gain I chose was an arbitrary conservative selection, and we could almost certainly find a phased array that could do better. How much better depends on the system, but I'd be surprised to find a beamwidth so narrow that we'd achieve four to six orders of magnitude higher gain.

10% striking a target doesn't sound so unrealistic if the target is hundreds of kilometers away and the size of a matchbook. At a range of 200 km, that's 0.001/(200e3)^2 = 2.5e-14 steradians.

So, we need a fleet of weapons in orbits such that they may illuminate the same target at the same time and be as near as possible. More power.