Calculating Bird Flock RCS for Comparable Detection to a Fighter Jet

AI Thread Summary
To achieve comparable detectability to a fighter jet with a radar cross-section (RCS) of 1 m² at 200 nautical miles (nmi), a flock of birds with an RCS of 0.0015 m² would need to be located at approximately 24.32 nmi. The calculation involves using the ratio of their RCS values and the spherical area formula. The discussion confirms that the radar range can be derived from the fourth root equation. The RCS of the jet is analyzed as a percentage of the total surface area of the sphere at the given distance. The final conclusion is that the birds must be significantly closer to be detected similarly.
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A fighter jet has 1 m^2 rcs and is located at 200nmi. Where would a flock of birds with 0.0015 m^2 rcs should be located in order to have the same detectability?
 
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isn't this a simple ratio with spherical areas?

1 / ( 4 pi (200)^2 ) = 0.0015 / ( 4 pi x^2 )
 
jedishrfu said:
isn't this a simple ratio with spherical areas?

1 / ( 4 pi (200)^2 ) = 0.0015 / ( 4 pi x^2 )

Are you referring to the 4th root equation for radar range ?

((0.0015/1)^(0,25))*370,4=72,89 km or 32 NMI
 
I was thinking that the RCS is a % of a larger spherical surface. So the jet has an RCS 1sq meter at 200 nmi. Figure what % is 1 sq meter to the total surface area of the sphere of radius 200nmi (converted to 370636.8 meters).

1 / ( 4 * pi * ( 370636.8 )^2 ) = 5.80e-13 percent

0.0015 / (5.80e-13 % ) =4 * pi * r^2

and r = 45068.86 meters = 24.32 nmi
 
Ok , got it . Thank You.
 
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