Martian
Senior Member
J-20 Mighty Dragon DSI and S-duct design is superior to F-22 S-duct-only design
The J-20 Mighty Dragon is a truly modern design. The old F-22 Raptor really shows its age when we compare the S-ducts for the world's two premier stealth fighters.
J-20 Mighty Dragon incorporates advanced DSI to dramatically increase the stealthiness of its S-duct design.
J-20's combined DSI and S-duct design forces radar waves to travel a circuitous path of five reflections before they can strike the engine fan blades.
Lacking an advanced DSI, the F-22 is helpless from preventing radar waves in traversing the shortest path to reach the F-22 engine fan blades in only two reflections.
The following calculations will show the massive advantage in the stealthiness of the J-20 S-ducts, when compared to the F-22.
The effectiveness of RAM coating is a 99.684% reduction (or 0.00316 left) in radar energy (see citation below). For radar energy entering the J-20 S-duct, it must undergo a minimum of five reflections before it strikes the engine fan blades. The shortest path to exit the J-20 S-duct is to reverse course and escape in another five reflections. The total minimum number of reflections is ten.
To calculate the amount of the original radar energy that entered the J-20 S-duct and was able to egress/reflect out of the S-duct after striking the engine fan blades, we have to reduce the original radar energy by 10 reflections from the RAM coated S-air duct walls.
Amount of original radar energy that can escape J-20 S-duct = (0.00316) ^ 10 = 9.93 x 10E-26
Since incoming radar waves can escape the F-22 S-duct in only four reflections, the attenuation of the enemy radar waves is inferior to the J-20 by many magnitudes.
Amount of original radar energy that can escape F-22 S-duct = (0.00316) ^ 4 = 9.97 x 10E-11
The effectiveness of the J-20 DSI and S-duct design is approximately 1 x 10E15 times better than the F-22 S-duct-only design or 1,000,000,000,000,000 times stealthier.
Admittedly, the amount of radar energy escaping from the S-ducts of the J-20 and F-22 are both extremely low and virtually undetectable. Nevertheless, from a stealth design standpoint, the J-20 is clearly far superior.
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Citation for RAM coating reduction of 99.684% reduction in radar energy.
From my February 12, 2011 post:
Revised final estimate for J-20 canards' radar return energy is 1.035 x 10^-17
I find Gambit's arguments for a -25 dB reduction, instead of -50 dB, from RAM coating to be convincing. I have revised my calculations for the effect from China J-20's canards. Quickie is correct that -25 dB is equivalent to 99.684% reduction (e.g. 10^2.5; take inverse; and convert to percentage). Thank you to Delft for highlighting the issue.
...
After hitting the canards, we know that 99.684% of the reflected energy is reduced by the military-grade RAM. (See ) This means that only 0.00316 (e.g. 1 - 0.99684 = 0.00316) of the impacting radar energy survives contact with the canard's RAM surface.
The J-20 Mighty Dragon is a truly modern design. The old F-22 Raptor really shows its age when we compare the S-ducts for the world's two premier stealth fighters.
J-20 Mighty Dragon incorporates advanced DSI to dramatically increase the stealthiness of its S-duct design.
J-20's combined DSI and S-duct design forces radar waves to travel a circuitous path of five reflections before they can strike the engine fan blades.
Lacking an advanced DSI, the F-22 is helpless from preventing radar waves in traversing the shortest path to reach the F-22 engine fan blades in only two reflections.
The following calculations will show the massive advantage in the stealthiness of the J-20 S-ducts, when compared to the F-22.
The effectiveness of RAM coating is a 99.684% reduction (or 0.00316 left) in radar energy (see citation below). For radar energy entering the J-20 S-duct, it must undergo a minimum of five reflections before it strikes the engine fan blades. The shortest path to exit the J-20 S-duct is to reverse course and escape in another five reflections. The total minimum number of reflections is ten.
To calculate the amount of the original radar energy that entered the J-20 S-duct and was able to egress/reflect out of the S-duct after striking the engine fan blades, we have to reduce the original radar energy by 10 reflections from the RAM coated S-air duct walls.
Amount of original radar energy that can escape J-20 S-duct = (0.00316) ^ 10 = 9.93 x 10E-26
Since incoming radar waves can escape the F-22 S-duct in only four reflections, the attenuation of the enemy radar waves is inferior to the J-20 by many magnitudes.
Amount of original radar energy that can escape F-22 S-duct = (0.00316) ^ 4 = 9.97 x 10E-11
The effectiveness of the J-20 DSI and S-duct design is approximately 1 x 10E15 times better than the F-22 S-duct-only design or 1,000,000,000,000,000 times stealthier.
Admittedly, the amount of radar energy escaping from the S-ducts of the J-20 and F-22 are both extremely low and virtually undetectable. Nevertheless, from a stealth design standpoint, the J-20 is clearly far superior.
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Citation for RAM coating reduction of 99.684% reduction in radar energy.
From my February 12, 2011 post:
Revised final estimate for J-20 canards' radar return energy is 1.035 x 10^-17
I find Gambit's arguments for a -25 dB reduction, instead of -50 dB, from RAM coating to be convincing. I have revised my calculations for the effect from China J-20's canards. Quickie is correct that -25 dB is equivalent to 99.684% reduction (e.g. 10^2.5; take inverse; and convert to percentage). Thank you to Delft for highlighting the issue.
...
After hitting the canards, we know that 99.684% of the reflected energy is reduced by the military-grade RAM. (See ) This means that only 0.00316 (e.g. 1 - 0.99684 = 0.00316) of the impacting radar energy survives contact with the canard's RAM surface.
gambit said:Even if you do not have relevant experience in the field, IF you actually read your source carefully, scant as it is, you would not have made the ridiculous claim that an airborne absorber would affect up to five-9s of the impinging signal.
Your quite general source reads...
The quarter wavelength rule is quite applicable to airborne absorber. As material DECREASING thickness approaches quarter wavelength of the targeted freq, absorber performances decreases. In most cases, the targeted freq is the X-band, which is the centimetric (cm) band. We found out a long time ago that increasing thickness to greater than quarter wavelength would incur an unacceptable weight penalty, especially if the absorber is of the magnetic type, which are ferrite particles in a dielectric containment, aka sheet or liquid applique.
Here is a source to prove that...
The only type of absorber that can affect up to five-9s of the impinging signal would be the pyramidal type...
Absorber performance is highly dependent upon the targeted freqs, even if it is 'wideband'.
Here is an F-22 in an EM anechoic chamber...
All those pyramidal absorbers would give us the most accurate RCS measurement of any object since they will absorb any chamber walls reflections that could constructively interfere with the reflections off the aircraft.
If absorber in general would affect five-9s of the impinging radar signal as you (falsely) claimed, there would be no need for shaping at all since whatever left of the signal -- the echo -- would lose even more energy on the way back to the seeking radar. What is that about energy loss to the square of the distance rule? Why not coat the whole aircraft with the stuff instead of just the canards? If this is true, we would have never built the F-117 in the first place looking funky as it is?
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