Chengdu J-20 5th Generation Aircraft News & Discussions

[amalakas]

But do not forget, his video has over 80k views while we who tried to inject a little bit is scientific integrity into the discussion have no videos therefore our challenges to his nonsense are based upon racism and jealousy.

i am jealous indeed.... LM engineers are also thankful because they finally acknowledged all the things they couldn't figure out all this time with the F-35 if only he had gotten his hands on these issues earlier the USAF would now have finished with this monumental project.

 

[Martian2]

oh boy u are trully a fanbouy ,no it cant as F22 wont turn it's radar as it would depend upon passive detection for detection of
J20 so how can j20 aesa radr detect f22 if it doesnt turn on it's radar ,

Use KJ-2000 AWACS AESA radar to search for F-22s.

KJ-2000 AWACS (i.e. Airborne Warning and Control System)

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Calculating the subtended surface of 3 inches by 1 foot from 50 miles away.

The surface of a sphere is 4*pi*R^2.

The surface of a sphere with a fifty mile radius is 4*3.14*(50 miles * 5,280 feet per mile)^2

The surface is therefore 875,381,760,000 square feet.

The radar covers a semi-circle. Therefore the coverage area is 437,690,880,000 square feet.

The assumed little canard gap is 3 inches by say 1 foot. The square footage is 0.25 foot x 1 foot = 0.25 square foot.

The radar will be looking at an additional reflection (after RAM absorption) of 0.25 square foot in a sphere with surface area of 437,690,880,000 square feet.

The additional ratio (after discounting 99.99% or so RAM absorption) is 0.25 square foot / 437,690,880,000 square feet = 0.0000000000005712 return with respect to the power of the emitted radar upon impact

However, the reflected radar signal must now head back to the radar receiver. That's another 50 miles journey back.

Hence, the surface of a half sphere is again 437,690,880,000 square feet.

The reflected signal is 0.0000000000005712.

Let's say the radar receiver has a 3 feet by 3 feet area. This means the radar receiver is 9 square feet in size.

The formula for reception is size of reflected signal * proportion of returned energy.

The signal received by the radar receiver is 0.0000000000005712 * (9 square feet / 437,690,880,000 square feet) = 0.0000000000005712 * 0.00000000002056 = 0.000000000000000000000011745 (* original radar emitting power)

I don't think a reflected signal off a canard gap (after 99.99% RAM absorption, which I did not subtract from the calculated figure because I want to illustrate the effect of distance in mitigating the signal) with a ratio of 0.000000000000000000000011745 to the original power emission is detectable.

However, you guys can feel free to claim otherwise. As far as I can tell, I'm the only person doing the calculations. You guys just keep harping there is a little canard gap. So what? It's trivial.

 

[Peaceful Civilian]

Man here we go again.

Martian2 is slipping more and more into madness. First came the photo of the planform alignment of the J-20 which is I suspect much worse than that of the T-50, then came the arguments about the canards, then instead of silencing his mouth after
been exposed for double standards on the T-50 IRST sensor , he carried on.

The final nail on the coffin is his ability to guest-imate percentages, 2% this , 10% that... man, the collective pride and dignity of all my respected lecturers and professors is going down the drain by me reading what this guy writes.

He gets pleasure from writing "mighty dragon" over and over again, ... I suggest you should write fighting falcon and super hornet and raptor every time you write.
also fulcrum and flanker B etc etc etc ... man ....

i can say one thing for sure, science has taken quite a hit on this thread despite some brave efforts by some to maintain some reasonable scientific level ...

where does one begin with this one?

Martian have you ever been involved with the tech you so easily talk about ?

I will give you an example, a 50 year old design, the venerable MiG-21 (fishbed i think) since you like the nicknames so much, is almost impossible to detect head on from most modern day fighters, even F-16s blk 52+ cannot detect it head on at about 100km or approx 60miles. Do you know what kind of a pain in the butt that is ?

I dare say you have never looked down a radar screen in a fighter plane trying to test the radar for your country's particular terrain and see how it behaves in various modes and what kind of performance it provides in different scenarios.

to you a radar sees all and stealth absorbs all !!!

I got news buddy ... it ain't so !

There is a hint to break F22 stealthy features. When F22 penetrate in your Country, Hack the F22 AN/APG-77 signals from ground. Same signals and deplete its instructions from CIP(Common Integrated Processor) of F22.. Now virus has penetrated and CIP instructions are alternated. Now we have control over F22 than the US airbase. Now, These alternated instructions will almost Malfuntion the F22.
Same virus that iran has used to control the drone.
Tomorrow is World of Hackers

 

[amalakas]

Use KJ-2000 AWACS AESA radar to search for F-22s.

KJ-2000 AWACS (i.e. Airborne Warning and Control System)

----------

Calculating the subtended surface of 3 inches by 1 foot from 50 miles away.

The surface of a sphere is 4*pi*R^2.

The surface of a sphere with a fifty mile radius is 4*3.14*(50 miles * 5,280 feet per mile)^2

The surface is therefore 875,381,760,000 square feet.

The radar covers a semi-circle. Therefore the coverage area is 437,690,880,000 square feet.

The assumed little canard gap is 3 inches by say 1 foot. The square footage is 0.25 foot x 1 foot = 0.25 square foot.

The radar will be looking at an additional reflection (after RAM absorption) of 0.25 square foot in a sphere with surface area of 175,076,352 square feet.

The additional ratio (after discounting 99.99% or so RAM absorption) is 0.25 square foot / 175,076,352 square feet = 0.000000001428 return with respect to the power of the emitted radar upon impact

However, the reflected radar signal must now head back to the radar receiver. That's another 50 miles journey back.

Hence, the surface of a half sphere is again 175,076,352 square feet.

The reflected signal is 0.000000001428.

Let's say the radar receiver has a 3 feet by 3 feet area. This means the radar receiver is 9 square feet in size.

The formula for reception is emitted signal * size of reflected signal.

The signal received by the radar receiver is 0.000000001428 * (9 square feet / 175,076,352) = 0.000000001428 * 0.0000000514 = 0.0000000000000000734 (* original radar emitting power)

I don't think a reflected signal off a canard gap (after 99.99% RAM absorption, which I did not subtract from the calculated figure because I want to illustrate the effect of distance in mitigating the signal) with a ratio of 0.0000000000000000734 to the original power emission is detectable.

However, you guys can feel free to claim otherwise. As far a I can tell, I'm the only person doing the calculations. You guys just keep harping there is a little canard gap. So what? It's trivial.

So the chinese engineers used a super computer to calculate the DSI et al but you just pulled out a pocket calculator to enlighten us all?

wow !

 

[gambit]

So the chinese engineers used a super computer to calculate the DSI et al but you just pulled out a pocket calculator to enlighten us all?

wow !

More like an abacus. That is how amazing he is.

 

[Martian2]

I have explained that the reflected signal from the little canard gap is trivial.

The exact return from the little canard gap is 0.000000000000000000000011745 * 0.00001 (approximate reduction from RAM coating) = 0.000000000000000000000000000011745 of original radar signal.

Now that I have completed the calculations, it should be obvious to everyone that the additional radar reflection is far less than 2%. It's negligible at 50 miles away.

 

[gambit]

I have explained that the reflected signal from the little canard gap is trivial.

The exact return from the little canard gap is 0.000000000000000000000011745 * 0.00001 (approximate reduction from RAM coating) = 0.000000000000000000000000000011745 of original radar signal.

Now that I have completed the calculations, it should be obvious to everyone that the additional radar reflection is far less than 2%. It's negligible at 50 miles away.

What you have done is nothing more than basic radar range equation which I used to do in my head when I was active duty over 20 yrs ago. The point here is what I pointed out in post 1247. For the J-20, the angles and gaps from the canards, the inlets, the wing leading edges, and everything else contributed to the overall RCS and this is something not even APA can do on their consumer grade PERSONAL COMPUTERS. Complex bodies can be broken down into individual components and run through these algorithms but not when these components are assembled into the original complex body. Why the hell do you think the designers uses supercomputers in the first place? Because the interactions between the diffracted signals from these discrete structures are extremely difficult to predict and model. But here you are using the consumer grade equivalence of abacus to tell us how 'stealthy' is the J-20. No wonder only the Chinese boys take you seriously.

 

[amalakas]

I have explained that the reflected signal from the little canard gap is trivial.

The exact return from the little canard gap is 0.000000000000000000000011745 * 0.00001 (approximate reduction from RAM coating) = 0.000000000000000000000000000011745 of original radar signal.

Now that I have completed the calculations, it should be obvious to everyone that the additional radar reflection is far less than 2%. It's negligible at 50 miles away.

you don't even know the dimensions of the canard gap!

 

[FairAndUnbiased]

Then it is equally insane to make baseless assumptions as to the J-20's low radar observability if it is comparable to the F-22 within any statistical significance. In that case, saying the J-20's RCS is less than 5% to the F-22 is pretty insane.

I am making no judgments here, just saying that the J-20's RCS and the F-22's should be comparable, due to similar wing shaping and many of the same design features. Perhaps the J-20 has a larger RCS. Perhaps not.

I'm just going to throw out a wild guess at an upper bound. The J-20's RCS is smaller than a clean EF-2000. Both planes have canards and 2 engines, but the J-20 at the very least has radar absorbing paint. Since the J-20 has internal weapons bays, putting weapons in it won't make its RCS bigger than the EF-2000. Do you dispute this?

 

[amalakas]

I am making no judgments here, just saying that the J-20's RCS and the F-22's should be comparable, due to similar wing shaping and many of the same design features. Perhaps the J-20 has a larger RCS. Perhaps not.

I'm just going to throw out a wild guess at an upper bound. The J-20's RCS is smaller than a clean EF-2000. Both planes have canards and 2 engines, but the J-20 at the very least has radar absorbing paint. Since the J-20 has internal weapons bays, putting weapons in it won't make its RCS bigger than the EF-2000. Do you dispute this?

it would be funny if it turns out to be just black paint

 

[yyetttt]

^ You'd have to be stupid. RAM doesn't shine.

 

[houshanghai]

New Angle of J20

a good pic display J20's weapon-bay

thx to Henri K

 

[amalakas]

^ You'd have to be stupid. RAM doesn't shine.

you have heard of matte paints right?

 

[marshall]

First.. The thread is on the J-20.. so ANYONE can discuss the J-20.
Second.. Pak-Fa CAN be compared to the J-20 as a test aircraft ..
Third.. That is a very lame attempt at being cynical..

Couldn't agree more with your Third observation. How many times have we heard that RCS cannot at the least be roughly estimated from observations to shut down a losing argument? LOL That's the whole point of these speculations and debates, not to troll and pursue agendas.

 

[rcrmj]

it would be funny if it turns out to be just black paint

the CAC guy says it is radar absorbing paint, do you have any problem with that?