Calculating Bird Flock RCS for Comparable Detection to a Fighter Jet

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Homework Help Overview

The discussion revolves around calculating the radar cross-section (RCS) of a flock of birds in comparison to a fighter jet, specifically focusing on how far the birds would need to be located to achieve similar detectability. The subject area includes concepts from radar technology and geometry.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between RCS and distance using ratios and spherical area calculations. Some question whether a fourth root equation for radar range is applicable, while others consider the percentage of RCS relative to the surface area of a sphere.

Discussion Status

The discussion includes various mathematical approaches and interpretations of the problem. Some participants provide calculations and insights, while others express understanding or seek clarification. There is no explicit consensus on a single method or solution.

Contextual Notes

Participants are working within the constraints of radar detection principles and the specific RCS values provided. There is an emphasis on the assumptions regarding spherical geometry and the nature of RCS in relation to distance.

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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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