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Subject: Ways to defeat an F22?
JTR~~    12/10/2010 1:36:21 PM
Now i was thinking about this a few days ago and it struck me. Now as far as i can remember i think someone compared the radar signature of the F22 to that of a small bird, as the plane is not invisible to radar but merely VLO. Now (someone please correct me if i am wrong here, if so this comment is void and we can all sleep happily again) if i am correct radar can detect the speed, altitude and distance of enemy aircraft right? if this is the case surely wouldn’t it be sane to believe that if the F22 has the RCS of a small bird then surely if you were to see something on your own radar travelling at MACH 2+ at 65000f t you would quickly come to the decision that it in fact was not a small bird but rather a very stealthy fighter jet (as far as i know there are not any birds that fly that high and fast, ha). Now if I am wrong or have missed out some blatantly obvious fact then I’m going to look very stupid, but if i am correct i might have something here. So that is what i wish to find out hence my question. also surely if the plane is operating without the support of an AWACS it would have to utilise its own radar, and while efforts have been made to reduce its emissions thereby not betraying its position surely looking for its radar signature would make the plane easier to target and defeat than trying to track the thing itself?? any insight or info would be most helpful, regards JTR~~
 
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Mike From Brielle    I believe the ....   12/20/2010 7:30:05 PM
Doppler equation is the difference between the high and the low frequency devided by a third parameter.  But you have to get the two original frequencies first.
 
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Mike From Brielle    I believe the ....   12/20/2010 7:34:28 PM
Doppler equation is the difference between the high and the low frequency devided by a third parameter.  But you have to get the two original frequencies first.
 
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warpig       12/20/2010 9:02:39 PM
I suggest that if anyone wishes to post in this topic, that they go to the Fighter Board and post in that thread by the same name that exists over there as well, and already has several responses in it.
 
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