tag:blogger.com,1999:blog-29731691.post2807308671835814880..comments2023-03-26T19:40:27.134-05:00Comments on Uniquely Joe: Movie plot weaponsstchttp://www.blogger.com/profile/00821202515283762603noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-29731691.post-37194723865865324072007-11-02T09:11:00.000-05:002007-11-02T09:11:00.000-05:00What I called 'radar cross section' should be the ...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..)<BR/><BR/>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:<BR/><BR/>E = int(60*P(t), dt, t_0, t_1) <BR/> with t in minutes<BR/><BR/>I fixed it at around 8:40am ish so check to verify the multiplier in the expression for E.<BR/><BR/>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.<BR/><BR/>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.<BR/><BR/>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.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-29731691.post-77266614846701941752007-11-02T08:22:00.000-05:002007-11-02T08:22:00.000-05:00Andykerr:Using your code, with a 200km orbit inste...Andykerr:<BR/><BR/>Using your code, with a 200km orbit instead of 200nm, <BR/>I get 0.002236 J ... which still sucks. It only gets worse with 10cm^3.<BR/><BR/>Do you think that we can't get more gain that 30dB? <BR/><BR/>I'm just thinking that less than 10% of the transmitted power hitting a target from space seems a bit low.<BR/><BR/>Radar cross section is measuring reflection, not absorption. <BR/><BR/>Doesn't that help us out? <BR/><BR/>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.sstchttps://www.blogger.com/profile/00821202515283762603noreply@blogger.comtag:blogger.com,1999:blog-29731691.post-85634852118301608492007-11-01T23:27:00.000-05:002007-11-01T23:27:00.000-05:00..since I had a few minutes, I decided to code wha.....since I had a few minutes, I decided to code what I believe to be a useful tool in solving this problem.<BR/><BR/>http://akerr.net/programming/space.m<BR/><BR/>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.<BR/><BR/>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.<BR/><BR/>Hope this helps. It was an interesting problem.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-29731691.post-90958192259841761642007-11-01T12:23:00.000-05:002007-11-01T12:23:00.000-05:00It would be cool if blogger supported inline \TeX ...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.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-29731691.post-27419132032994647732007-11-01T12:01:00.000-05:002007-11-01T12:01:00.000-05:00Kerr is not with you.The R^4 term that you dropped...Kerr is not with you.<BR/><BR/>The R^4 term that you dropped should instead become R^2 when we're only interested in power received by the scatterer.<BR/><BR/>Gauss's law tells us that radiation arriving on a scatterer with radar cross section o is equal to the following:<BR/><BR/>P_s = P_t o / (4pi R^2).<BR/><BR/>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:<BR/><BR/>P_r = P_s A / (4pi R^2).<BR/><BR/>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.<BR/><BR/>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.<BR/><BR/>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.Anonymousnoreply@blogger.com