J-20... The New Generation Fighter II

Re: Final estimate for J-20 canards' radar return energy is 3.276x10^-19

Martian said:

Since you're a new member and appear totally clueless, I would appreciate it if you clicked on the link that I provided before giving me an offhand and flippant reply. The Answers.com link is to an article by McGraw-Hill encyclopedia. McGraw-Hill is one of the largest publishers in the United States and they were the owners of Businessweek, until its recent sale to Bloomberg. My citation on RAM effectiveness is McGraw-Hill Science and Technology Encyclopedia, not Answers.com.

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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...

To produce such absorbers, it is necessary in practice to taper the material over distances which are large compared with the wavelength of the frequencies to be absorbed. Therefore, practical absorbers of this type giving greater than 20dB absorption vary in thickness from about 0.8 in. (2 cm) at 10 GHz and above to 6 ft. (2 m) at 100 MHz and above. The absorber performance improves with increasing thickness until the point is reached where all of the energy that enters the material is absorbed and only the front-face reflection is left. While this type of absorber is capable of producing a high degree of absorption over a broad bandwidth, it is at the same time a relatively thick material.

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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...

Silicone and polyurethane sheets are a thin flexible resonant absorber available for frequencies within 1.5 to 18 GHz. The typical reflectivity is – 25 dB at the centre frequency and the band width is 10%.

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The only type of absorber that can affect up to five-9s of the impinging signal would be the pyramidal type...

The performances of pyramidal absorbers are better than flat absorbers particularly at higher frequencies where any reflections from the surface of the pyramids tend to be channelled down into the absorber. When the height of the pyramids is greater than 10 wave lengths the reflectivity may reach –52 dB.

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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?

Re: Final estimate for J-20 canards' radar return energy is 3.276x10^-19

challenge said:

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assessment by the Russian aviation writer,good read.

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I have brought up the center of gravity before. However, this article lost its credibility when it claims F-35 is optimized for supercruise.

And the flight photos (the clearest one) show the canards are not tilted.

There is a paper on the net, written by Chinese researchers, indicating that the canards/lex/lifting body/delta combination generates up to 80% more lift than regular delta. For comparison, canards/delta like J-10 give about 20% more lift. The center-of-gravity shows that not only the paper's analysis is correct, but also the additional lift is mainly on the front of the body.

So, imagine a fighter, with 80% bigger wing area as J-20 (without the drag). This is an extremely agile fighter, with very good supercruise performance. The sacrifice is the weapon loading capacity. I am 80% sure it can only carry 6 missiles.

Re: Final estimate for J-20 canards' radar return energy is 3.276x10^-19

delft said:

Your reference to the McGraw-Hill Encyclopedia gives 50 dB for the pyramid field type of absorber, more than 25 dB for a ram coating. I'm no expert in this field, but I would then expect 99.9 % absorption rather than 99.999 %.

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50 dB would give 99.999% absorption. 25 dB would give 99.684%. The difference in reflected radar energy is actually more than 300 times between the two cases.

Hi does anyone understand this:
slideshare.net/AmicusCuriae/fifth-generation-jet-fighter-comparison
i failed to get the point of this presentation....

Seems like they're just trying to say that the J-20 has the potential to tussle with the best of the 5th gens, but the proper powerplant is critical. Otherwise it'll only be an F-35 killer, and not much else.

Re: Final estimate for J-20 canards' radar return energy is 3.276x10^-19

johnqh said:

I have brought up the center of gravity before. However, this article lost its credibility when it claims F-35 is optimized for supercruise.

And the flight photos (the clearest one) show the canards are not tilted.

There is a paper on the net, written by Chinese researchers, indicating that the canards/lex/lifting body/delta combination generates up to 80% more lift than regular delta. For comparison, canards/delta like J-10 give about 20% more lift. The center-of-gravity shows that not only the paper's analysis is correct, but also the additional lift is mainly on the front of the body.

So, imagine a fighter, with 80% bigger wing area as J-20 (without the drag). This is an extremely agile fighter, with very good supercruise performance. The sacrifice is the weapon loading capacity. I am 80% sure it can only carry 6 missiles.

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I think the J-20 is kinda like a case of "reverse MIG-25". The capabilities of the former was greatly exaggerated while the capabilities of the latter are severely underestimated.

Re: Final estimate for J-20 canards' radar return energy is 3.276x10^-19

siegecrossbow said:

I think the J-20 is kinda like a case of "reverse MIG-25". The capabilities of the former was greatly exaggerated while the capabilities of the latter are severely underestimated.

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the former Soviet Union for Mig-25 a high degree of confidentiality, the Western only found that the Mig-25 super fast speed, they were shocked.
J-20 only a prototype, just had it's maiden flight, the different kind of " experts" just to say what they like to want,they can ignore the science.

I'm wondering how the J-20 would perform if the attached tailfins were removed. I'm assuming they can't be removed right now due to a lack of TVC, but would it be viable in the future to run a tailfin-less J-20 with rudder controls substituted for by TVC?

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.

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To my eyes, the only obvious weakness in China's J-20 front-aspect stealth is the canard. There is a total of three radar-signature contributors from the canards.

1. By far, the largest radar-reflecting source is the initial scatter from the canard. I will calculate this today.

2. There is a much smaller secondary scatter (which I'll calculate later to prove that it's inconsequential) as the radar waves bounce off the canard, bounce off the fuselage, and return to the enemy aircraft's receiver.

3. There is also another secondary scatter, which should equal the secondary scatter from the canard, as radar waves bounce off the fuselage, bounce off the canard, and return to the enemy aircraft's receiver.

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I have revised upward the size of the J-20's canards in the following calculations.

The J-20 has two canards. I'll estimate that each canard is 2m (or 6 feet) long. I'll further estimate that each canard is 0.75m (or 2.25 feet) high. The total canard area facing an enemy radar is 2m x 0.75m x 2 [canards] = 3 m^2.

I don't have the faintest idea of the illumination cone for a directed military radar. However, to be of any use, I'll estimate that the illumination cone is 1km in radius. If the illumination cone is significantly smaller than 1km, I don't see how you can find an enemy fighter within a reasonable amount of time.

Area of a circle = pi * r^2 = 3.14 * (1km*1km) = 3.14 * (1,000m*1,000m) = 3.14 x 10^6 m^2

Initial scatter ratio = (3 m^2) / (3.14 x 10^6 m^2) = 9.554 x 10^-7 = 0.0000009554. This is the ratio of the emitting radar energy that hits the J-20's canards.

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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.

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However, only a tiny fraction of the radar energy that is scattered by the RAM-covered canards will make it back to the enemy aircraft's transmitter/receiver radar.

The total energy that is scattered from the canards' surface will radiate into a spherical shell that is determined by the following formula:

Surface area of a sphere = 4 * pi * r^2

For the radius, we will use 13.5 km from the detection range of an EADS DR 174 ground-based mobile 3D radar.

Total surface area of spherical shell = 4 * 3.14 * (13.5km x 13.5km) = 2289 km^2 = 2289 (1,000m x 1,000m) = 2289 x 10^6 m2 = 2.289 x 10^9 m2

Based on the picture shown at the bottom of the post, I will estimate the radius of the receiving radar to be 0.5m.

Total surface area of receiving radar = pi * r^2 = 3.14 * (0.5m x 0.5m) = 0.785 m^2

The fraction of the reflected radar energy from the RAM-coated canards that is seen by the receiver is:

Reflected radar energy from canards seen by receiver = 0.785 m^2/2.289 x 10^9 m2 = 3.429 x 10^-10

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Putting it all together, the amount of the emitting radar energy that returns to the receiver from two RAM-coated canards located 13.5 km away is:

Fraction of emitting radar energy returning to receiver = 9.554 x 10^-7 x 0.00316 x 3.429 x 10^-10 = 1.035 x 10^-17 or 0.00000000000000001035.

QED: The J-20's canards have an insignificant impact on its stealthiness.

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Figure 5 shows the phased array radar used on the F/A 22 Raptor. This radar employs approximately 2000 microwave transmitter/receiver pairs, each the size of a pack of chewing gum. (courtesy USAF)

Inst said:

I'm wondering how the J-20 would perform if the attached tailfins were removed. I'm assuming they can't be removed right now due to a lack of TVC, but would it be viable in the future to run a tailfin-less J-20 with rudder controls substituted for by TVC?

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I think TVC is only turned on for high maneuvering demands. Turning them on all the time will wear parts out much faster. A fighter plane should be able to fly normally with TVC switched off, also it will affect with giving the plane a sustained, stable thrust. Rudders changes the pointing orientation of the plane more, TVC changes the orientation of the velocity vector more. They complement each other.

Tail-less designs I think right now is still more suitable for platforms not demanding in maneuverability, like the B-2, UAV strike aircraft and the F-22 fighter-bomber version.