F-22 and F-35 RCS revealed by USAF

[Roscoe]

Gums, you surprise me...surely you of all people know that it is really 69 milliseconds...

[elp]

Is that like some sick preverted $ex joke Roscoe? That fighter pilots only use numbers in their briefings that are a multiple of 69 ?

[Roscoe]

Of course not...

[2sBlind]

Yeah, what do you think we have our craniums in the gutter or something?

[JR007]

"So to speak"....

[avon1944]

F-22 = metal marble
F-35 = metal golf ball (slightly less than B-2).

now, this metal marble and metal golf ball is detectible at what range? How far away is the signal being transmitted?

Adrian

[parrothead]

F-22 = metal marble
F-35 = metal golf ball (slightly less than B-2).

now, this metal marble and metal golf ball is detectible at what range? How far away is the signal being transmitted?

Adrian

Adrian,

Detection range is completely dependant on the radar being used. Think of it this way - the better the radar's "eyesight" the longer the detection range. It's like an eye chart for people. Someone with superior eyesight (say 20/10 for example) can read the smaller characters at a greater distance than someone with poor eyesight. The characters on the chart are the same size.

The question here is how close does an enemy aircraft have to be to see something the size of a marble or a golf ball? That will vary depending on the radar in question that's trying to detect the object in question .

[mityan]

Excuse me guys for bringing to top such an old topic, but there is one little thing that annoys me. I just cant get it out of my head.

I've heard that Raptor cockpit glass has a metallization layer.

I'm sure that overall reflections comrises of that ones from cockpit and from the rest of the airframe.

So how could the reflections from metallized cockpit glass (such a big metal plate - 45х27 inches front view) be less than from a small marble (1/2 inch wide)? Especially in broad range of angles.

Does anyone have a guess?

[hornetfinn]

It's because the cockpit glass is such a shape that it redirects the reflections away from the radar. The glass is shaped in such a way that it does have continuously changing curves in three dimensions, which does this effect (doubly curved surface). A sphere has the same radii all around and it does reflect some of the radar energy back to the radar. A max RCS of a sphere is the area of the sphere as if it was a circle and it depends on the radar frequency. 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. The amount of overall reflected energy is not reduced much by the cockpit glass but reflecting it away from the radar does effectively the same thing.

[sprstdlyscottsmn]

That is the same reason why the "bumps" on the bottom of the F-35 that result from the skin being "shrink wrapped" around the munitions does not impact RCS. The location of said bumps also looks to me to be effective area ruling, but that is another discussion for another time.

[smsgtmac]

Wow. Not only was this a dead thread, but its since been acknowledged that the marble-golfball [phone spellchecker? hah!] comparison was wrong
Shape is an important element, but there's other considerations.
It helps to first think of the canopy as a "transparency" composed of multiple layers. Then recognize that whether a part of the plane you are dealing with will benefit the total system LO by being a conductor or insulator. From a physics POV there is no 'teflection' per se, but conversion to heat or re-radiation. The direction of the re-radiation is controlled in the design. GKNs website has a little "transparency" slide show illustrating how the canopy handles EM energy of a different sort. Handling energy from distant RF sources is trivial in comparison.
This is a particularly important design criteria for the F-35, because of the pyros installed in the transparency stack up. Sorry for any typos, but this was sent via a tiny phone screen and keyboard (fixed from the PC the same evening).

[mityan]

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:

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

canopy.png (30.16 KiB) Viewed 60242 times

For tilted flat plane here is the RCS formula:

formula.png (5.04 KiB) Viewed 60242 times

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

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.

[mityan]

From a physics POV there is no 'teflection' per se, but conversion to heat or re-radiation. The direction of the re-radiation is controlled in the design.

Reflection and reradiation are the same.
What about heat?
For consuming the electromagnetic wave and revert it to heating the matching is required. The impedance of the canopy's metal plate should be equal to impedance of a free space - 377 Ohm in broad range of frequencies.
As for me, I think that's a bullshit, but maybe this is due to a lack of education.
Consider the matching is implemented. For very good high frequency design the voltage standing wave ratio (VSWR) of near 1.5 is an adequate value. It means that only 20% of energy is reflected back to space.
So divide the estimated value above by 5 and you get the rcs still greater than reported.
There is no magic.

And what I think also.
Nobody could think of russians as stupid dumbass.
And if the CEO of OAK (united aviation corporation of russia) Pogosyan tells Putin that PAK-FA RCS is about 0.3 sq.m. and this is close to Raptor, I think he's got his reasons. There are too many candidates for his chair to fell free in lying.

[smsgtmac]

From a physics POV there is no 'teflection' per se, but conversion to heat or re-radiation. The direction of the re-radiation is controlled in the design.

Reflection and reradiation are the same....

Not exactly. The terms are used interchangeably as an irritatingly sloppy shorthand even by CEM guys, but pure reflection involves no absorption whereas re-radiation involves some increase in potential followed by the discharge/shedding or conversion of same.

[KamenRiderBlade]

From a physics POV there is no 'teflection' per se, but conversion to heat or re-radiation. The direction of the re-radiation is controlled in the design.

Reflection and reradiation are the same....

Not exactly. The terms are used interchangeably as an irritatingly sloppy shorthand even by CEM guys, but pure reflection involves no absorption whereas re-radiation involves some increase in potential followed by the discharge/shedding or conversion of same.

The thing is, there will be some absorption, some reflection back to source, some radiating away from source when it comes to EM waves.

The question is how much of each will happen on the surface of the F-35 / F-22.

That's an answer that we'll probably never know until Stealth coatings become obsolete by some other tech and the US government reveals that fact.

In reality, there will probably be very little reflection back, a lot more radiating away via the wave hitting the surface at a non perpendicular angle, some absorption into heat depending on what type of RAM coating is used.