F-22 and F-35 RCS revealed by USAF

[mityan]

Closer manufacturing tolerances will result in lower overall RCS since it reduces the effects of surface discontinuity.

Closer manufacturing tolerances give a practical results closer to a simulated ones. But what about metallized canopy? In that case the tolerance, which is negligible to wavelength, gives a negligible advantage. Do you think the application case of the formula isn't quite correct?

[hornetfinn]

Why flat? We just cut a plane from a frontal part of canopy - 1 m long and 0.1 m wide - such a strip I think can be well approximated by flat plane.
Then we may remove all the rest of the canopy and all the rest of the airframe (though they also have a non-zero reflections)
and put the values in the formula (it seems the incidence angle is near 60 deg. to normal, A = the square of the strip)
So, we get the value of 0.00058 which is 5.8 times greater than reported overall RCS.

The result is heavily influenced by wavelength used in calculations and also on exact incident angle. Of course the canopy is also quite a bit more complicated object than your approximation and that can result in quite significant differences. Also it might be that the canopy does not have perfect reflectivity but rather attenuates the radar waves with some mechanism.

[wrightwing]

Detected at 28km, tracked at an even shorter range. Meanwhile, the F-22 detected the Flanker at 200+km, and T50 at 100+km, well before their optical systems would be of use.

And that is all? Is it so easy in your way.

Which is where having significant RCS advantages, also provide enormous tactical advantages.

[wrightwing]

Closer manufacturing tolerances will result in lower overall RCS since it reduces the effects of surface discontinuity.

Closer manufacturing tolerances give a practical results closer to a simulated ones. But what about metallized canopy? In that case the tolerance, which is negligible to wavelength, gives a negligible advantage. Do you think the application case of the formula isn't quite correct?

They didn't spend all the time, effort, and money on RCS reduction of the airframe, to haphazardly throw a canopy that negates these effects. The design is intentional, to prevent RF radiation from reflecting off the non-stealthy contents of the cockpit. Stealth isn't just about eliminating reflectivity, but controlling it, and using it in your favor.

[mityan]

The result is heavily influenced by wavelength used in calculations and also on exact incident angle. Of course the canopy is also quite a bit more complicated object than your approximation and that can result in quite significant differences. Also it might be that the canopy does not have perfect reflectivity but rather attenuates the radar waves with some mechanism.

1. Just vary wavelength in fighter radar band (X), vary angles and see "the heaviness" of this influence.
2. I took just a little cut from the canopy assuming all other reflection are zero. So the result figure is less than real. The formula in the book represents a kind of continuous function
https://en.wikipedia.org/wiki/Continuous_function
so a little variation cannot result in "quite significant" difference.
3. "it might be" together with "most likely" are your favourite sentences in any analysis. What mechanism?

[mityan]

Which is where having significant RCS advantages, also provide enormous tactical advantages.

Raptor average frontal RCS is close to 0.01 sq.m.
It can be tracked far enough by means of aerial or ground radar.
Why?
The answer is in the first post in this thread:

Nov 2005: The U.S. Air Force, in it’s effort to get money to build more F-22s, has revealed just how “stealthy” the F-22 is. It’s RCS (Radar Cross Section) is the equivalent, for a radar, to a metal marble.

TO GET MORE MONEY.

Very, very nice try! What a wonderful fairytales can we invent in order to get more money!
They revealed the local narrow angle minimum RCS and made everyone to believe it's actual value.
(Special thanks to Fulghum from AW journal - he wrote that -40 dBsm is ALL-ASPECT RCS).

[mityan]

They didn't spend all the time, effort, and money on RCS reduction of the airframe, to haphazardly throw a canopy that negates these effects. The design is intentional, to prevent RF radiation from reflecting off the non-stealthy contents of the cockpit.

Metallized canopy reduces RCS, because it has smoothed surface and it covers the cockpit interior, the elements of which might have significantfy higher RCS.

Stealth isn't just about eliminating reflectivity, but controlling it, and using it in your favor.

How the glass or metal film controles reflectivity? You are free to hope so.

[wrightwing]

They didn't spend all the time, effort, and money on RCS reduction of the airframe, to haphazardly throw a canopy that negates these effects. The design is intentional, to prevent RF radiation from reflecting off the non-stealthy contents of the cockpit.

Metallized canopy reduces RCS, because it has smoothed surface and it covers the cockpit interior, the elements of which might have significantfy higher RCS.

Stealth isn't just about eliminating reflectivity, but controlling it, and using it in your favor.

How the glass or metal film controles reflectivity? You are free to hope so.

You might want to do some more reading on the subject, before you embarrass yourself any more. The metal oxide finish on the canopy glass not only prevents RF energy from reflecting off the surfaces inside the cockpit (which product large returns), but attenuates the RF energy, and ensures that it isn't reflected back to the source. This is the principle of the entire airframe- to direct reflections away from source emitters, and in controlled shapes. There's no hoping involved. Merely physics.

Your RCS rebuttal is equally laughable. You're attempting to draw empirical conclusions, based upon an analogy. Stick to youtube comments. That's a better forum for this level of discourse.

[mityan]

This is the principle of the entire airframe- to direct reflections away from source emitters, and in controlled shapes. There's no hoping involved. Merely physics.

Yes. It is physics. Some energy (a very small amount) is reflected back according to formulas. And this gives the result that is greater than you want it to be.
You dont know physics, you dont know the principles - how metal plate or glass can absorb wave, but to insist on this - is your duty.
Not hope, but faith.
I've posted the formula, the link to a good american book on RCS.
I've posted US patents on RAM - with the absorbtion figures.
And what is your ground? What are you based on, what should I learn to know your point?

[archangel117]

It's because the cockpit glass is such a shape that it redirects the reflections away from the radar.
... The reflecting area of the cockpit glass is extremely small compared to the geometrical area as only very small amount of the radar energy hitting it will go back towards the radar.

I think it's not quite true because we are not in the region where the laws of optic propagation act.

In general, the the waves of any length (not only optic band) undergo the Huygens–Fresnel principle
https://en.wikipedia.org/wiki/Huygens%E2%80%93Fresnel_principle

According to it, every point which a luminous disturbance reaches becomes a source of a spherical wave; the sum of these secondary waves determines the form of the wave at any subsequent time.

This means that we may consider the canopy metal layer as a great antenna, which reradiates incident power.
It has its main lobe and side lobes just like any anttenna.

The mainlobe width is dependent on relatioship between antenna size (aperture) and wavelength.
The mirror in optic band, for example 1 meter long, will give the ratio near 2000000 times (optic waves are about 0.5 micrometer), and the beam is extremely narrow. And the sidelobes far from the main are extremely negligible (due to sin(x)/x function)

But what about X-band? For 1 m aperture and 0.03 m wave the ratio is only 33.
Then we get a picture like this:

angles.png

And what is the exact value of reflection? Look at this picture.

canopy.png

For tilted flat plane here is the RCS formula:

formula.png

It is taken from here:
http://books.google.by/books?id=XDs04HdQ4-gC&pg=PA183&lpg=PA183&dq=tilted+flat+plate+RCS&source=bl&ots=dol7wl54FC&sig=27W3hnnBtGpPkvwbuF5zzoA2oek&hl=ru&sa=X&ei=obZhVKDtG4HkOKL0geAC&ved=0CCcQ6AEwAg#v=onepage&q=tilted%20flat%20plate%20RCS&f=false
Why flat? We just cut a plane from a frontal part of canopy - 1 m long and 0.1 m wide - such a strip I think can be well approximated by flat plane.
Then we may remove all the rest of the canopy and all the rest of the airframe (though they also have a non-zero reflections)
and put the values in the formula (it seems the incidence angle is near 60 deg. to normal, A = the square of the strip)
So, we get the value of 0.00058 which is 5.8 times greater than reported overall RCS.
I want to remind that we nullified all other reflections from the airframe (!) and took just a narrow strip from canopy.

Everybody can calculate the square of circle, and that was done in 2005.
But also everyone can put the values in formula given above.
What you think guys?

My guess is that we've been reported a true minimal RCS value

value.png

which is fair for very narrow angle range. And a true average value is a definitely classified information and it should be much greater evidently.

Let me get this straight, you are trying to cut a flat plane out of an object designed with continuous curvature? Seems to me it would be obvious that you would get a higher RCS from a flat plate than a continuous curve designed to "obsorb", attenuate and "reflect" RF energy away from the receiver. This isn't an F-117 it has very few flat surfaces, even it's saw-tooth edges are not completely flat. So I suggest you figure out how to fit the curvature of the canopy into the equation.

[wrightwing]

This is the principle of the entire airframe- to direct reflections away from source emitters, and in controlled shapes. There's no hoping involved. Merely physics.

Yes. It is physics. Some energy (a very small amount) is reflected back according to formulas. And this gives the result that is greater than you want it to be.
You dont know physics, you dont know the principles - how metal plate or glass can absorb wave, but to insist on this - is your duty.
Not hope, but faith.
I've posted the formula, the link to a good american book on RCS.
I've posted US patents on RAM - with the absorbtion figures.
And what is your ground? What are you based on, what should I learn to know your point?

You should learn that shaping is the most important aspect of stealth, as it minimizes the amount of energy that is reflected back to the source. RAM works in conjunction with shaping, to further reduce reflectivity. The interior of a cockpit has a lot of highly reflective sources, so.....the goal of signature reduction, is to prevent RF energy from entering/reflecting off of these surfaces. The SHAPE and composition of the canopy minimizes any returns, to the source. You can't gain any useful information, regarding signature reduction, if you're trying to base it off of a chunk of material. For your sake, and for the rest of us, put your google skills to use, and read about signature reduction measures incorporated in canopies.

[arl8733]

Anybody here know how long the coatings on the transparency are lasting these days? Knew it was a problem in the early days.

[mityan]

Let me get this straight, you are trying to cut a flat plane out of an object designed with continuous curvature? Seems to me it would be obvious that you would get ...

Sure!
And I do it with success, like all mathematicians in the world.

surface.png (21.16 KiB) Viewed 114705 times

You must be surprized that delta Si at the picture is a flat square.
I just cut this:

canopy_cut.png (47.99 KiB) Viewed 114705 times

This is flat enough for 3-cm wave dont you think?
Or do you think 1-mm inaccuracy (difference between flat and curve) may result in 10 or even 100 times in reflection?

Let me get this straight too. You are trying to apply optical propagation laws to X-band radiowave in spite of incomparable wavelength/object size ratio? Seems to me it would be obvious that you would get a much lower RCS just like radar signal is a laser beam and canopy is usual mirror.

The radiated energy, in fact, takes a pattern like a typical reflected wave structure. The width of the main forward scattered spike is proportional to the ratio of the wavelength to the dimension of the reradiating surface, as are the magnitudes of the secondary and tertiary spikes. The classical optical approximation applies when this ratio approaches zero. Thus, the backscatter - the energy radiated directly back to the transmitter increases as the wavelength goes up, or the frequency decreases.

It is taken from here:
http://www.globalsecurity.org/military/systems/aircraft/f-22-stealth.htm
ratio approaches zero - 0.0000005 m - length of optical wave - is almost 0 compared to that cut (see above)
But how about 0.03 m length of X-band wave?

There is NO alternative physics for reflection. Just Huygens–Fresnel principle I've mentioned previously.

[archangel117]

Let me get this straight, you are trying to cut a flat plane out of an object designed with continuous curvature? Seems to me it would be obvious that you would get ...

Sure!
And I do it with success, like all mathematicians in the world.

surface.png

You must be surprized that delta Si at the picture is a flat square.
I just cut this:

canopy_cut.png

This is flat enough for 3-cm wave dont you think?
Or do you think 1-mm inaccuracy (difference between flat and curve) may result in 10 or even 100 times in reflection?

Let me get this straight too. You are trying to apply optical propagation laws to X-band radiowave in spite of incomparable wavelength/object size ratio? Seems to me it would be obvious that you would get a much lower RCS just like radar signal is a laser beam and canopy is usual mirror.

The radiated energy, in fact, takes a pattern like a typical reflected wave structure. The width of the main forward scattered spike is proportional to the ratio of the wavelength to the dimension of the reradiating surface, as are the magnitudes of the secondary and tertiary spikes. The classical optical approximation applies when this ratio approaches zero. Thus, the backscatter - the energy radiated directly back to the transmitter increases as the wavelength goes up, or the frequency decreases.

It is taken from here:
http://www.globalsecurity.org/military/systems/aircraft/f-22-stealth.htm
ratio approaches zero - 0.0000005 m - length of optical wave - is almost 0 compared to that cut (see above)
But how about 0.03 m length of X-band wave?

There is NO alternative physics for reflection. Just Huygens–Fresnel principle I've mentioned previously.

Mityan, respecfully, you seem to have a skewed view on how this works. In the world of stealth even the smallest features can radically change the RCS of an object, yes even a Milimeter, this is why the F-22 and F-35 have borderline ludicrous tolerances.
You seem on the right track with some of your math but still off the mark with the final numbers.
Here: http://www.academia.edu/5672531/Radar_C ... _UHF_Bands
We can see that the Frontal RCS of the F-22 as simulated in the X band is roughly -18 dBsm. This is without R.A.M or R.A.S and is admitted by them to be a imperfect model of the craft. Yet if we apply the values for the R.A.M depicted in figure 7 here: http://www.freepatentsonline.com/5164242.pdf it can easily be seen how the RCS in the X-band can be -40 dBsm In the frontal aspect.

This is the physics as done by a professional software setup by professional people and using data from a patent you found. It can be seen that your method of calculation needs some work as your values do not match the simulations.

[mityan]

In the world of stealth even the smallest features can radically change the RCS of an object, yes even a Milimeter

Very doubtful statement. According to conventional physics 1 mm should not have a significant influence on 30-mm wave.
Where is that world of stealth - in alternative universe?

Here: http://www.academia.edu/5672531/Radar_C ... _UHF_Bands
We can see that the Frontal RCS of the F-22 as simulated in the X band is roughly -18 dBsm.

First of all - not -18 but -9 dBsm:

rcs2.png (36.89 KiB) Viewed 114560 times

This is without R.A.M or R.A.S and is admitted by them to be a imperfect model of the craft. Yet if we apply the values for the R.A.M depicted in figure 7 here: http://www.freepatentsonline.com/5164242.pdf it can easily be seen how the RCS in the X-band can be -40 dBsm In the frontal aspect.

I dont know the CST program, but ADS by Agilent and Microwave office by AWR are similar to it (designed for the same tasks). I simulated just a small phased array antenna (near 30 elements) and the process took a significant part of my 64 GB RAM (compare to 500 MB in the article) and lasted also several hours. And very doubtful that they have a 3D model of Raptor with 1-mm precision. But the improvement in model precision may lead to a better or a worse result with an equal probability.
Go ahead. Take this picture:

rcs.png (49.53 KiB) Viewed 114560 times

and this:

ram.png (60.59 KiB) Viewed 114560 times

You see that we can get a value near -13 dBsm (0.05 sq.m.) at 10 GHz.
And what I calculated earlier - a piece of canopy, it is not covered with RAM so the laue I got - 0.00058 - is still a part of overall RCS, even if the RAM attenuation will be applied to a model.

I want to say again. I DO NOT reject the value of -40 dBsm at all.
I believe that it is true, but with some more specification of its applicability. I've told about its narrow angle localization, and here is frequency dependence too.
So this is not a figure we can rely on upon estimating tracking range etc.

What about USAF officials - they wanted to compliment themselves.
And they wanted to get more money for further Raptor production.
Just businness, nothing wrong with it.