What's new

Top 10 future weapons of CHINA

That is scary.

Not as scary as this.

14lug7s.jpg
 
In this video, you can see a J-10B Vigorous Dragon with short flexible petals (which indicates an indigenous WS-10 engine). The video for the J-10B with WS-10 engine starts at 4:33 and the takeoff occurs at 10:00.

Starting at 12:24, it is indisputable that the J-10B is equipped with a WS-10 engine. The flexible petals are clearly short.

 
Last edited by a moderator:
First...Composite materials does not guarantee absorbance. So the composite materials can be tossed.
True. Plywood is a composite. Concrete is a composite.

Second...Absorbers are not %100 effective. There are always trace EM reflections from the surface. And it is their behaviors that are unpredictable.
True. Passive absorbers are generally restricted to certain bandwidths. The wider the bandwidth, the thicker the material, and the more costly the weight. In RCS control, the first law is to be mindful of the threat frequency. Meters length freqs are long range search and usually ground based so they are not considered 'threat' frequencies. It is the centimetric and some milimetric systems with high pulse repetition freq (PRF) characteristics like the missile guidance/homing that are considered 'threat' frequencies.
 
All moving control surfaces increase RCS, not just canards.

And the only real difference between canards and stabilators (all moving tailplanes) is the location on the aircraft.

Take a look at the F-22.

2qtlc41.jpg
Yes...But the issue is the reduction of scattered signals interactions with each other. Canards produces those interactions while the conventional and supposedly 'inferior' horizontal rear stabs do not.
 
In your picture, you claim the "tail fins and booms" are an inferior design. I disagree. The tiny ventral fins follow planform alignment with the large vertical stabilizers and add virtually nothing to RCS. To the contrary, the "tail fins and booms" dramatically increase the J-20 Mighty Dragon's side infrared-stealth.
Excellent example of 'Chinese' physics. Selective RCS contributorship based upon mere 'I say so'. Guess Chinese engineers must have missed that section from Skolnik's and Knott's books. :lol:

By the way, I have not forgotten about those two posts that will show everyone what a fool you are about this issue.
 
Not as scary as this.
Disintegrating engine blades or oil leaks are worse. If we are to take the Russians at their words about what happened in that demo, then that particular engine will be fine. What happened was a case of erroneous air data inputs to the engine control, not of engine manufacturing.
 
attachment.php


As far as the canards on the J-20 is concerned, any canard would generally require a junction between the wing and the fuselage. This junction is hard to hide on radar.

I believe the Americans with their F-22 preferred to hide this behind main wing on an elevator. Ever wonder why the Americans never went down the path of delta/canard configuration?

Canards also tend to have large, angular surfaces that reflect radar signals. To offset this problem, the Eurofighter Typhoon for example (4th generation) uses software to control its canards. This similarly applies to the Rafale and Gripen. Question is: Is this applicable to a truly stealth aircraft?

Are the designers of the J-20 choosing to compromise some stealth, or am I missing something?

There are however aerodynamic advantages of the delta/canard configuration.
-It allows more lift than the traditional arrangement whenever considering total lift.

-During maneuvers, the forces on the canards mirror towards the main wing. Thus, adding more lift when climbing and decreasing the lift to descend. Such characteristics allow the bird to fly faster and tighter compared to the traditional set up.

-Since the canard generates and upward lift, a traditional tail plane produces a downward lift. Hence, we can say that there is a reduction in lift required from the main wings. The overall drag on the aircraft are reduced.

-At times, the canard is designed to stall prior to the main wing. Once the canard stalls, the nose pitches down. And thus, lowering the angle of attack of the main wings.

Disadvantages:
-The wing root would show that it experiences a down wash from the surface of the canards. This reduces the overall efficiency.The traditional tail-plane set-up doesn't suffer from this.

-The wing tips on the other hand can experience an up wash from the surface of the canards. This increases the angle of attack on the tips. This enables premature separation of the air flow over the wing tip. What does this mean? It means that this premature separation of the air flow on either of the wing tips can enable a wing drop (wing drop: Definition from Answers.com) at the approach of stall. This can potentially lead to a spin. To avoid this, the wing must be carefully designed and may require more weight on the wing structure to support this, coupled with a wing root able to support the structure. That means, the solution is expensive. Some things are better spent on instead.

-Since the canards must be designed to stall before the main wings, the main wing never stalls and hence never achieves the maximum lift coefficient. To overcome this, it requires larger wings to provide extra wing area such that the aircraft can achieve the optimum take-off and landing distance performance.

-It is difficult to use flaps on a canard/delta configuration. Deploying flaps causes a nose down pitching moment. In the conventional design, this effect is reduced since the down wash on the tail planes increase. This results in the aircraft to result in a nose-up pitching moment. The canard/delta design cannot alleviate this since there are no tail planes.

-For the aircraft to achieve longitudinal static stability, the small surface of the canards produce a relatively high lift-coefficient. Whereas the main wings - despite having a much larger surface, operates at a smaller lift coefficient and therefore hardly reaches the full potential. A waste of resources and adding up more redundancy.

-Since maximum lift potential of the main wings are typically unavailable, and flaps are absent (if they are present, they'd be hard to use), typically - take off and landing distances, coupled with speeds are often higher than the conventional design.

Some of the advantages and disadvantages can apply to all situations. As far disadvantages are concerned, they have been taken into account when designing high performance aircraft such as the Rafale and the Typhoon. It could be where potential aerodynamic stability in the designs allows more opportunities to improve the overall maneuverability of the aircraft. Maybe that's why they're so expensive.

However, I personally think that maneuverability is not the most important thing in today's standards.

Bearing in mind that the J-20 has TVC, are canards necessary? It is a redundant feature if you ask me. Albeit flawed. Both the Raptor and T-50 has TVC. None of them have canards.
 
I proved him wrong in his assertion that a ballistic missile warhead only falls vertically onto its target. He mumbled something about "vertical" not being "vertical" and that I misunderstood him. I've given up hope that he will muster the courage to ever admit a mistake. I've made my share of mistakes, but I admit to them.

Anyway, Gambit is Gambit. <shrug>
You will be seriously spanked about that. And I will use a Popular Science article from forty years ago and the Space Shuttle of today to do it. :lol:
And here we go...

In a March 1958 Popular Science issue, Frank Harvey explained the basics of an ICBM trajectory, from launch to descent.

popsci_1958_icbm_trajec_f-harvey.jpg


On the left of the illustration, we have the launch point, but for our purpose, we will focus only from apogee (center) to impact (right).

Long range radars designed to look for ICBM warheads do not look 'up', as in vertically straight up from ground perspective. Rather, they look only mildly angular towards the horizon. Like this...

http://upload.wikimedia.org/wikipedia/commons/9/9b/PAVE_PAWS_Radar_Clear_AFS_Alaska.jpg

The reason the radar (TARGET) look slightly over the horizon is because of the incoming warhead's orbital altitude, which will appear to be just right over the horizon. A rising or setting moon do not force you to look 'up', does it? No, a rising or setting moon will have you look quite straight ahead AT THE HORIZON. It is a matter of perspective. The moon is still about 1/4 million miles away in space. No different than for the warhead when it is just slightly above the atmosphere and just broke horizon and came into view. By the time the warhead enter the 'RE-ENTRY' phase in the Harvey's illustration, its descent mode will be quite vertical.

Now we come to the difficult part, at least for you anyway. To make it easier for you, we will assume the Earth's rotation is right-left. That would make the warhead and the target approaches each other. Try -- TRY -- to envision the Earth in that illustration as rotating. Then it will be obvious as to why at the target, from the target's ground perspective, the warhead will descend vertically. Not only is it perception, but it is quite true that the descent angle will be quite steep, as in approaches perpendicular.

If we assume the Earth's rotation as left-right in that same illustration, what will happen is that the warhead travels 'ahead' of the target -- SPATIALLY SPEAKING -- then proceed to descend at a calculated point in space. The Earth's rotation will move the target and eventually the two paths will intersect. The same result, the warhead will descend at a quite vertical angle.

Now we come to the 'evidence' that you supposedly used to 'proved' me wrong...

http://upload.wikimedia.org/wikipedia/commons/5/5f/Peacekeeper-missile-testing.jpg

Are you that lacking in critical thinking skills?

These are nuclear warheads. Targets that deserve nuclear warheads are not so near each other.

What are the altitudes of those lines?

What is the visual horizon distance at ground level, which is this photograph? Keyword search for you 'ground visual horizon distance'.

Do you really believe that the actual ground impact points are at the very ends of those straight lines?

What if those actual ground impact points are a few hundred miles further from the horizon?

Let us take the bright line furthest right.

Do you really believe that is the true descent angle?

What if YOU -- the observer -- shift your position 90 deg?

Is it possible that you will see a different descent angle than the one you see in this image?

Is it possible that the final targets are actually several hundreds miles apart and therefore in order to effectively capture as many of the demonstration warheads as possible in one image, we have to resort to a wide angle perspective and that resulted in these defined angles?

More keyword search for you 'atlantis iss deorbit' which would give you this NASA source...

Photo Index 24

But to save everyone the hassle...

atlantis_iss_deorbit.jpg


We can see the 'forward' motion of the views by looking at the Earth's horizon and check the stars' movements. From them we can see Atlantis is deorbiting AGAINST the Earth's rotation. The deorbiting trajectory looks quite steep, no? Just like the Space Shuttle, the ICBM warhead will bounce off the atmosphere if the deorbit angle is sufficiently shallow. Unlike the Space Shuttle, the warhead will not exploit aerodynamic forces to maneuver and enter into glide mode when the atmosphere is sufficiently dense.

You are wrong on so many levels about this subject. I tried to give you the benefits of reexamining your own arguments to save you the embarrassment, especially about the MIRV photograph and viewer's perspective but your ego got in the way of your sensibility. The problem now for you is how to reconcile your own flawed interpretation of that MIRV photo versus the Shuttle's deorbiting photo. For the Shuttle, the viewer's perspective is that of the "God's Eye" position where we can see its true deorbiting mode -- very close to vertical so that the craft would not bounce off the atmosphere. If you insist that the Shuttle's deorbiting photo is open to interpretation, then so is the MIRV photo that you used to supposedly 'proved' me wrong.

We now have a time span of sixty years, from Popular Science in 1958 to the Space Shuttle of 2011 that says YOU are wrong.
 
And here we go...

In a March 1958 Popular Science issue, Frank Harvey explained the basics of an ICBM trajectory, from launch to descent.

On the left of the illustration, we have the launch point, but for our purpose, we will focus only from apogee (center) to impact (right).

Long range radars designed to look for ICBM warheads do not look 'up', as in vertically straight up from ground perspective. Rather, they look only mildly angular towards the horizon. Like this...

The reason the radar (TARGET) look slightly over the horizon is because of the incoming warhead's orbital altitude, which will appear to be just right over the horizon. A rising or setting moon do not force you to look 'up', does it? No, a rising or setting moon will have you look quite straight ahead AT THE HORIZON. It is a matter of perspective. The moon is still about 1/4 million miles away in space. No different than for the warhead when it is just slightly above the atmosphere and just broke horizon and came into view. By the time the warhead enter the 'RE-ENTRY' phase in the Harvey's illustration, its descent mode will be quite vertical.

Now we come to the difficult part, at least for you anyway. To make it easier for you, we will assume the Earth's rotation is right-left. That would make the warhead and the target approaches each other. Try -- TRY -- to envision the Earth in that illustration as rotating. Then it will be obvious as to why at the target, from the target's ground perspective, the warhead will descend vertically. Not only is it perception, but it is quite true that the descent angle will be quite steep, as in approaches perpendicular.

If we assume the Earth's rotation as left-right in that same illustration, what will happen is that the warhead travels 'ahead' of the target -- SPATIALLY SPEAKING -- then proceed to descend at a calculated point in space. The Earth's rotation will move the target and eventually the two paths will intersect. The same result, the warhead will descend at a quite vertical angle.

Now we come to the 'evidence' that you supposedly used to 'proved' me wrong...

Are you that lacking in critical thinking skills?

These are nuclear warheads. Targets that deserve nuclear warheads are not so near each other.

What are the altitudes of those lines?

What is the visual horizon distance at ground level, which is this photograph? Keyword search for you 'ground visual horizon distance'.

Do you really believe that the actual ground impact points are at the very ends of those straight lines?

What if those actual ground impact points are a few hundred miles further from the horizon?

Let us take the bright line furthest right.

Do you really believe that is the true descent angle?

What if YOU -- the observer -- shift your position 90 deg?

Is it possible that you will see a different descent angle than the one you see in this image?

Is it possible that the final targets are actually several hundreds miles apart and therefore in order to effectively capture as many of the demonstration warheads as possible in one image, we have to resort to a wide angle perspective and that resulted in these defined angles?

More keyword search for you 'atlantis iss deorbit' which would give you this NASA source...

Photo Index 24

But to save everyone the hassle...

We can see the 'forward' motion of the views by looking at the Earth's horizon and check the stars' movements. From them we can see Atlantis is deorbiting AGAINST the Earth's rotation. The deorbiting trajectory looks quite steep, no? Just like the Space Shuttle, the ICBM warhead will bounce off the atmosphere if the deorbit angle is sufficiently shallow. Unlike the Space Shuttle, the warhead will not exploit aerodynamic forces to maneuver and enter into glide mode when the atmosphere is sufficiently dense.

You are wrong on so many levels about this subject. I tried to give you the benefits of reexamining your own arguments to save you the embarrassment, especially about the MIRV photograph and viewer's perspective but your ego got in the way of your sensibility. The problem now for you is how to reconcile your own flawed interpretation of that MIRV photo versus the Shuttle's deorbiting photo. For the Shuttle, the viewer's perspective is that of the "God's Eye" position where we can see its true deorbiting mode -- very close to vertical so that the craft would not bounce off the atmosphere. If you insist that the Shuttle's deorbiting photo is open to interpretation, then so is the MIRV photo that you used to supposedly 'proved' me wrong.

We now have a time span of sixty years, from Popular Science in 1958 to the Space Shuttle of 2011 that says YOU are wrong.

It's unbelievable how stupid you can be. Who cares what Frank Harvey what's-his-name wrote in Popular Science. Are you blind? Look at the following photographic and video evidence. Do the warheads look like they're falling vertically? Idiot.

ebHD0.jpg

Splash image showing the Peacekeeper MIRV

 
Last edited by a moderator:
It's unbelievable how stupid you can be. Who cares what Frank Harvey what's-his-name wrote in Popular Science. Are you blind? Look at the following photographic and video evidence. Do the warheads look like they're falling vertically? Idiot.
And it is unbelievable how obtuse YOU are. The questions I posed about that wide angle MIRV photo are legitimate. Do you really believe that those points on the horizon are exactly where you perceive the warheads will land? No one sane will say 'Yes' because of the lack of altitude information about those lines.
 
It's unbelievable how stupid you can be. Who cares what Frank Harvey what's-his-name wrote in Popular Science. Are you blind? Look at the following photographic and video evidence. Do the warheads look like they're falling vertically? Idiot.

ebHD0.jpg

Splash image showing the Peacekeeper MIRV

(insert video)

Yeah, who cares about frank...? When we have the awesome photographic evidence from martian2...a self proclaimed internet douche bag.

You know, i used to think that you were capable of some form of intelligent thought and possessed a fairly good scientific background. Reading your recent posts, i dont think you're anything more than a fanboy with too much time on his hands.

I notice something peculiar about your posts...you screw up on an epic scale when attempting to explain things. However, the copy/paste information database is alright.
 
And it is unbelievable how obtuse YOU are. The questions I posed about that wide angle MIRV photo are legitimate. Do you really believe that those points on the horizon are exactly where you perceive the warheads will land? No one sane will say 'Yes' because of the lack of altitude information about those lines.

I give up. You're a moron. Oh look, some no-name person wrote an article somewhere. This must be the truth! My eyes deceive me! My senses deceive me! Look Martin, I found a book where someone said the sky is green!!! You are wrong!

We had this discussion before. I am not going to repost all of the ballistic charts from RAND and other mainstream publications again. Well, maybe just one. Link is here if you want to read all of the old arguments again: http://www.defence.pk/forums/china-defence/59274-chinas-missile-defense-system-3.html

----------

My post from May 29, 2010:

What do the experts say? The experts are in agreement that a ballistic missile and warhead come in at an angle. Look at the dotted yellow, pink, blue, and purple parabolas.

You can find the same altitude vs. distance graph in Figure 3.7 (that shows a range of 1,200 km) on page 28 in the RAND report on "Estimation and Prediction of Ballistic Missile Trajectories" (see http://www.rand.org/pubs/monograph_reports/2006/MR737.pdf).

http://hadmernok.hu/archivum/2007/2/2007_2_balajti.html

2007_2_balajti_clip_image038.gif

Figure 19. Ballistic missile trajectories and its detection requirements
 
I give up. You're a moron. Oh look, some no-name person wrote an article somewhere. This must be the truth! My eyes deceive me! My senses deceive me! Look Martin, I found a book where someone said the sky is green!!! You are wrong!
And who are YOU? Someone who needs to take a basic photography class.

Here is another one...

peacekeeper_usaka.jpg


Look at the single vertical line and try to imagine a different viewer perspective. How do you know that a shift in viewer's angle will not produce a shift in supposedly descent angle?
 
And who are YOU? Someone who needs to take a basic photography class.

Here is another one...

Look at the single vertical line and try to imagine a different viewer perspective. How do you know that a shift in viewer's angle will not produce a shift in supposedly descent angle?

Read the old thread. I've already explained to you that it's not an optical illusion. You cannot reproduce the extreme angles in the Peacekeeper photograph. Also, if you watch the MIRV video, the unarmed warheads create a debris pattern in a particular direction. The non-circular debris pattern is consistent with an incoming MIRV that struck the island at an angle. Simple physics.
 
Read the old thread. I've already explained to you that it's not an optical illusion. You cannot reproduce the extreme angles in the Peacekeeper photograph. Also, if you watch the MIRV video, the unarmed warheads create a debris pattern in a particular direction. The non-circular debris pattern is consistent with an incoming MIRV that struck the island at an angle. Simple physics.
And you are confused between a different viewing perspective and an illusion. Do you even know the definition of an 'illusion'? As for that Figure 3.7 in that RAND source. The laugh is still upon YOU for failing to consider the graduations of the vertical graph, which is altitude. The graduations are very coarse so of course the entire trajectory graph will have the typcial parabolic path. But at the target's position, if we refine the altitude graph, the descent trajectory will be close to vertical.

And the 'round nose' post to prove how wrong you are is coming.
 

Country Latest Posts

Back
Top Bottom