3 ms·

Not an average software engineer--beautiful page (imperfect spelling)! Why does detection scale as the 4th root of the RCS? Nice problem for the radiation theor

by dmwood 2mo ago

Not an average software engineer--beautiful page (imperfect spelling)! Why does detection scale as the 4th root of the RCS? Nice problem for the radiation theory part of a graduate electrodynamics class.

HNisCIS 2mo ago
It's a 4th power because you're doing the inverse square law twice, once for the outbound pulse and once for the reflection.
MalbertKerman 2mo ago
> Why does detection scale as the 4th root of the RCS? Because the power returned to the antenna scales as the -4th power of the distance to the target, and detection depends on that power being greater than some threshold. If the fourth power (rather than the square) is surprising, think of the reflection as re-transmitting the power that arrived at the target. The incident power on the target goes as RCS/r^2, and the power that makes it back to the original antenna gets another factor of 1/r^2, so the return scales as RCS/r^4.
- ahartmetz 2mo ago
  This -4th power also makes stealth very difficult. Reducing the RCS by a factor of 1000 reduces detection range by a factor of 5.75.
  - ahartmetz 2mo ago
    * of about 5.6
brandmeyer 2mo ago
It shouldn't, under the assumptions they are opening up with. Inverse square law twice is appropriate for an approximation that the reflection is entirely diffuse because you assume that the diffuse reflection is uniform in all directions away from the illuminated surface. A single inverse square law is appropriate (with both distances added to each other before squaring) under the approximation that the reflection is specular.
- tonyarkles 2mo ago
  I’m guessing it’s because the detector is then working with power, not amplitude. The whole 10 log vs 20 log thing with dB.
- MalbertKerman 2mo ago
  That only holds in the near field of the reflecting surface, which is roughly (it's a fuzzy boundary so who cares about a 2 pi here or there) where the ratio of distance to width of the surface is less than the ratio of the width of the surface to the wavelength. For wavelengths around 10^-2 m and surfaces around 10^0 m, that means distances within 10^2 m. Much further than that and the reflected wave is spreading out and falling off as 1/r^2. A radar working at ranges of tens or hundreds of kilometers is firmly in the far field. There's still a strong angular dependence in the reflection amplitude; 1/r^2 fall-off does not imply isotropic distribution.

by dmwood 2mo ago

HNisCIS 2mo ago

It's a 4th power because you're doing the inverse square law twice, once for the outbound pulse and once for the reflection.

MalbertKerman 2mo ago

> Why does detection scale as the 4th root of the RCS? Because the power returned to the antenna scales as the -4th power of the distance to the target, and detection depends on that power being greater than some threshold. If the fourth power (rather than the square) is surprising, think of the reflection as re-transmitting the power that arrived at the target. The incident power on the target goes as RCS/r^2, and the power that makes it back to the original antenna gets another factor of 1/r^2, so the return scales as RCS/r^4.

ahartmetz 2mo ago
This -4th power also makes stealth very difficult. Reducing the RCS by a factor of 1000 reduces detection range by a factor of 5.75.
- ahartmetz 2mo ago
  * of about 5.6

brandmeyer 2mo ago

It shouldn't, under the assumptions they are opening up with. Inverse square law twice is appropriate for an approximation that the reflection is entirely diffuse because you assume that the diffuse reflection is uniform in all directions away from the illuminated surface. A single inverse square law is appropriate (with both distances added to each other before squaring) under the approximation that the reflection is specular.

tonyarkles 2mo ago
I’m guessing it’s because the detector is then working with power, not amplitude. The whole 10 log vs 20 log thing with dB.
MalbertKerman 2mo ago
That only holds in the near field of the reflecting surface, which is roughly (it's a fuzzy boundary so who cares about a 2 pi here or there) where the ratio of distance to width of the surface is less than the ratio of the width of the surface to the wavelength. For wavelengths around 10^-2 m and surfaces around 10^0 m, that means distances within 10^2 m. Much further than that and the reflected wave is spreading out and falling off as 1/r^2. A radar working at ranges of tens or hundreds of kilometers is firmly in the far field. There's still a strong angular dependence in the reflection amplitude; 1/r^2 fall-off does not imply isotropic distribution.