Understanding Radar Signature Data (Full Version)

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 obrien979 -> Understanding Radar Signature Data (9/24/2019 10:33:57 AM) I am struggling to understand the Radar Signature information for Aircraft. I am probably overthinking this and have missed the boat somewhere. Based on the photo attached, it appears that the F-117A has a lower radar picture then the newet F-22 Raptor. Meaning that it is more stealthier when up against A-D Band Radars and more likely to be seen using E-M Band Radars. Am I correct on this? [image]local://upfiles/59654/8AC806D8535B419092A6FC5A65BE6A12.jpg[/image]

 Dragon029 -> RE: Understanding Radar Signature Data (9/24/2019 11:00:44 AM) Radar signature is measured in dBsm, which is a decibel / logarithmic scale. To convert from decibel square metres to normal square metres, you use the formula: area in square metre = 10 ^ (dBsm / 10) Then to convert an area in square metres back into dBsm, you use the formula (where the log is using base 10): area in dBsm = 10 x log(area in square metres) Due to the nature of logarithms and powers, the more negative a value is in dBsm, the smaller an area it represents. So according to that chart (which is going to be speculation or from an incomplete simulation keep in mind), the F-117A has an average radar signature of -14dBsm in radar bands A-D, which roughly equals 0.04m^2, and a radar signature of -24dBsm in radar bands E-M, which roughly equals 0.004m^2. The F-22 meanwhile apparently has an average radar signature in the A-D bands of -20dBsm or 0.01m^2 (4x smaller than the F-117), and an average radar signature in the E-M bands of -30dBsm or 0.001m^2 (again 4x smaller than the F-117).

 obrien979 -> RE: Understanding Radar Signature Data (9/24/2019 11:29:36 AM) quote:ORIGINAL: Dragon029 Radar signature is measured in dBsm, which is a decibel / logarithmic scale. To convert from decibel square metres to normal square metres, you use the formula: area in square metre = 10 ^ (dBsm / 10) Then to convert an area in square metres back into dBsm, you use the formula (where the log is using base 10): area in dBsm = 10 x log(area in square metres) Due to the nature of logarithms and powers, the more negative a value is in dBsm, the smaller an area it represents. So according to that chart (which is going to be speculation or from an incomplete simulation keep in mind), the F-117A has an average radar signature of -14dBsm in radar bands A-D, which roughly equals 0.04m^2, and a radar signature of -24dBsm in radar bands E-M, which roughly equals 0.004m^2. The F-22 meanwhile apparently has an average radar signature in the A-D bands of -20dBsm or 0.01m^2 (4x smaller than the F-117), and an average radar signature in the E-M bands of -30dBsm or 0.001m^2 (again 4x smaller than the F-117). Thank you for this outstanding explanation in Layman's terms. I think I have an understanding of this now. I suspected that the lower the negative value, the better its stealthiness. Time to put this new knowledge to test on some missions.

 c3k -> RE: Understanding Radar Signature Data (9/24/2019 6:21:09 PM) Or, in simpler terms, the more negative the value, the LESS energy gets reflected BACK to the radar antenna. You want a BIG REDUCTION in energy being reflected back to maintain stealth.

 SeaQueen -> RE: Understanding Radar Signature Data (9/25/2019 10:33:29 PM) Electrical engineers like to work in decibels (dB) rather than simpler to understand linearly scaled units. The reason is that when you measure things on a logarithmic scale (i.e. dB) small changes in the underlying quantity that don't matter much get washed out. In that sense, dBsm are actually a much more convenient way of measuring rcs. ```rcs (m^2) | rcs (dBsm) ------------------------ 1000 | 30 100 | 20 10 | 10 1 | 0 0.1 | -10 0.01 | -20 0.001 | -30 ```

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