Forum: F-16 versus XYZ

hansundfranz wrote:

This is called STT and can be done with conventional antennas just as easily.

BTW I don´ t think any modern radar would do this to avoid alerting the emenys RWR

I know conventional radars can in STT but not in TWS. AESA allows you to independently STT mulitple targets simultaneously, like a multiple target track. This of course will betray the raptors presence to the vipers but an AESA can scan and track multiple targets faster than a mechanical radar can in TWS. All data from every sensor in the raptor is compiled and fused together thru its common integrated processors. So once precise bearing, altitude and range data for the targets that was lacking from the alr-94 is obtained, the radar no longer needs to be emitting since the alr-94 now has precise data to track the targets from their radar emissions. the vipers will detect the raptors radar emissions but not for very long. the alr-94 on the raptor can now track the vipers passively. this of course depends on having the vipers flying with their radars on.

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Keep in mind that use of radar varies. The F/A-22 will first get its primary targeting information from an off-board source (AWACS, other F/A-22, etc). Second, it only has to 'fire' short bursts of the radar to confirm the location of the target, which are designed to not be easy to detect. Once a missile is in the air, the F-16s become the prey and will have to evade that. Tracking the F/A-22 at the same time would be difficult. With proper skills the F/A-22 pilot will likely have the advantage over the -16s to some degree. Would it get all four? I can't tell from published sources, but I'd bet it has a good shot at it.

Kevin
The Ejection Site

From the previous link I posted, the last paragraphs of page 5 should be good reading. However, it does allude to staitionary targets which is it is not limited to.

Lamoey. GPS works different.
The satellites contain 4 very precise clocks.
A satellite sends a timestamp and information about his orbit.
By solving a system of equitations the gps reciver can calculate the 3 coordinates and the time (time must be calculated as it is kind a hard to carry a caesium clock around)
Your idea, compareing where on the waveform you are for differenct recivers is exatly what interferomerty means.

Pat1, your link explains to use 2 pairs of dual recivers to calculate an angle for the emitting source and finally use trigonometry to get a range out of these 2 angles.

Now. lets assume the triple revivers spaced 25 ft appart and perpendicular to the emitter. (the ideal setup)

Again our emitting souce is 35 miles away.

The difference between the distance to the center reciever and to one of the outer recivers will be
sqrt (25^2 + 210000^2)-210000 = 0.001488 ft = 0.045mm

OK, lets say that our jet where you mount the sensors is bending, changeing size and vibrating in flight. Despite these problems (and with inetrpolating sverela measurements to average out the vibrations) you manage to position your reciver with a precission of 0.2 mm
0.2 mm = 0.0065 ft

The measured ange would be cosx = 210,0000 / 210,000.001488
=> X = 0.0068° real value
or
cos x = 210,000 / 210,000.007988
=> x = 0.0158°
With that we would reach a precission of about 0.01° which is a a good bit better then the MOA precission I assumed before but

A) I find it pretty unliekey that it is possible to reach such precission in mounting the recievers
B) it won t cut the bacon
If I´d repeat my initial calculations of the trinagulation with an error or 0.01° instead of 0.017° it will still be not precise enough to speak of "known range" And of course now our 2 points for the trinagulation are only 25 ft appart not 50 as initially assumed.

BTW ths webpage http://www.st-andrews.ac.uk/~www_pa/Sco ... intro.html
is agood find. A pity that a few chapters are missing.

I find Runningmans idea pretty interesting to use RWR data to keep updating positions, with ever increasing uncertanty but better then nothing) of targets that have been painted (and by that range measued with the radar before)

hansundfranz wrote:

Pat1, your link explains to use 2 pairs of dual recivers to calculate an angle for the emitting source and finally use trigonometry to get a range out of these 2 angles.

Now. lets assume the triple revivers spaced 25 ft appart and perpendicular to the emitter. (the ideal setup)....

Hi Hansundfranz, not sure what you mean with the above. You keep on assuming and you fail to see how an interferometer works. You are still considering “eyeball” sensors. Again, interferometers were introduced to replace these systems providing orders of magnitude in improvements over slit instruments. But don’t take my word for it, the website says: "THz systems can provide spatial accuracies of the order of metres in range and millimetres in offset at ranges of the order of 10 km. Hence they can compete with radar in terms of mapping resolution/accuracy."
It’s OK by me if you still don’t want to believe it. Please, I beg, leave me out of further discussion in this thread.

I'm with habu2 and Pat_1 on this. Adios!

I just want to go back to the original question.

Read this from the latest issue of Code One:
Five Eagles In Three Minutes

A single Raptor pilot from the Combined Test Force at Edwards AFB, California, launched four guided AIM-120 missiles against four separate targets on 24 July. The Raptor's integrated weapons system successfully identified, tracked, and linked each target's data to the AIM-120s and each missile passed within lethal range of its target. During a press conference at Andrews AFB, Maryland, in May, Secretary of the Air Force Dr. James G. Roche told reporters that in one test "We had five F-15 Eagles against one Raptor and the engagement was over in three minutes. None of the F-15s even saw the Raptor. The Raptor simply went down the line and, in simulation, took out all five of the F-15s."

Now, the F-15 has a radar with a longer range than the F-16's. So I do not think the F-16 will have a better chance, despite it's smaller size. But who knows if an F-16 pilot with great strategy skills could come up with a way to defeat the Raptor.

Pat1, come on, do the calculation in here and calculate the precission of range measurement, 3 sensors 25ft spacing between each other.
The only uncertantiy is that you do not know if they are truely on a line and the position error perpendicular to the ideal line can be 0.2 mm.

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THz systems can provide spatial accuracies of the order of metres in range and millimetres in offset at ranges of the order of 10 km. Hence they can compete with radar in terms of mapping resolution/accuracy."

1. How do you know the author is talking about an airborne system and not something like the colchuga (recivers on different trucks, several hundred meters apart from each other).
2.10 km are only 5 miles which is short range,

hansundfranz, drop the attitude and open your mind. The discriminator here is the wavelength of the signal being detected, not the sensor spacing. That is the definition of interferometry and is what determines the angular resolution.

h2 out.

Thats it. I am getting that Tricorder I saw on E-Bay. Cool

hansundfranz wrote:

Pat1, come on, do the calculation in here and calculate the precission of range measurement, 3 sensors 25ft spacing between each other.
The only uncertantiy is that you do not know if they are truely on a line and the position error perpendicular to the ideal line can be 0.2 mm.

I really don't feel like solving a system of non-linear partial differential equations that lack any form of closed solution. This is why we use computers, so people can spare lifetimes of insanity. I don't even know what algorithms I should use!

Quote:

1. How do you know the author is talking about an airborne system and not something like the colchuga (recivers on different trucks, several hundred meters apart from each other).

I don't know. Read our posts carefully. This subject veered so far off course that this doesn't even matter. I proposed a solution in principle and I've only tried to back of my claims because of your denial. Why should the platform even matter, after all, I've been warning about the practical limitations.

Quote:

2.10 km are only 5 miles which is short range,

Order of 10 Km means order of magnitude of ten kilometers (0-100Km). Again this is a practical limitation. Technology can mature...

hansundfranz, you are correct for position determination, but not for attitude determination, as antennas in relative close proximity are used, and not only the content of the GPS signal is used, but also its carrier frequency to make wavelength comparisons and calculations to get the millimeter accuracy needed. If an airplane had a GPS antenna at each wingtip and add one on the tail it would almost be as accurate as a gyro to determine attitude. This is obviously of no use on a fighter than loops and roles at high rates, but would probably do well on an airliner. Incidentally, it is used in maritime applications to measure role and pitch movements of the GPS antenna of a high accuracy survey vessel.

Sorry people for being such a pain in the a$$. I just want to clear this up for people following the thread. I got carried away and I didn?t explain how this system works (mostly because I don?t know that much about this particular application).

This is based on the link I listed since these people seem to be aware on the practical limitation of 3-port interferometric ranging.

In principle, such a system does the same thing as a basic ?triangulation? sensor. The application of such systems in to improve accuracy over its basic counterpart. Remember, error is due to the instruments capabilities. Interferometers have inherently a small error due to the size of the signal produced, the interferogram, which in theory could be infinite. However, the usefulness of the information falls extremely rapidly. On top of this, this source of errors can be accounted for in the software (its very complex and tedious) to a very high degree. In practice, they can provide orders of magnitude in accuracy over the other more obvious systems.

This system does not triangulate in the basic sense, hence trigonometry is not used to calculate the range or position of the emitter. This system measures the receiving radiations' curvature. There is an obvious relationship between the wavefront curvature and range (see bottom figure of page 4 on the link, remember real curvature is 3D). The trig is used to ?see how? but not how the measurement is done.

Technically, I don?t think you need interferometers to measure this, but this is about accuracy and there are associated problems with measuring this directly. Considering a 2-port system, this system will allow you to draw a sphere determining how far away the emitter is. Considering the error in range, it will draw two concentric spheres centered at the detector where the closest distance between the two shells is the error (a few meters according to the link). Draw yourself a mental picture? The emitter is somewhere sandwiched between these. With the 3rd port added, you can determine not only how far, but where the target is with milimetric precision. Accordingly, I interpret:

Quote:

THz systems can provide spatial accuracies of the order of metres in range

(radial ?in-and-out? shift)

Quote:

and millimetres in offset

(left-right and up-down shift)

Quote:

at ranges of the order of 10 km.

Interferometry means to measure the shift between the same signal measured on 2 sepreate recievers.

Quote:

This system does not triangulate in the basic sense, hence trigonometry is not used to calculate the range or position of the emitter. This system measures the receiving radiations' curvature.

If you use the 2 angles deterimend by each pair or recievers (the middle one is used 2 times) and then use trigonemtry or if you calculate the wafefronts radius directly does not change your precission.

Quote:

Considering a 2-port system, this system will allow you to draw a sphere determining how far away the emitter is.

Wrong, there is a infinite number of circles, touching to points in space. The circles radius ranges from: half the distance between your 2 recievers (distance zero) to infinity.

You need the 3rd port to determine the radius.
with 3 sensors on line and spaced equally you know that the wavefront will arrive at the average time of the 2 outer sensors ,if the emitter is a infinite distance anway, if the wafefront is not curved but straight.

Now if the emitter is not an infiinte distance away the wavefromt will be curved. The distance to the middle sensor will be slightly shorther then the average distance to the 2 outer sensors.

All this is only for 2d for 3d you need at least 4 recievers.

The problem is, the wavefront emitted from a point source 35 miles away will be pretty damm straight.

Here´s the calculation again)

Quote:

Now. lets assume the triple revivers spaced 25 ft appart and perpendicular to the emitter. (the ideal setup)

Again our emitting souce is 35 miles away.

The difference between the distance to the center reciever and to one of the outer recivers will be
sqrt (25^2 + 210000^2)-210000 = 0.001488 ft = 0.045mm

That means you have to know the position of your recievers with precission in the same magnitude to make valuable measurements.

And this sould be next to impossible on a flying high performance jet.
It bends under load, it changes size due to temperature changes caused by speed and altitude changes.

hansundfranz wrote:

Wrong, there is a infinite number of circles, touching to points in space. The circles radius ranges from: half the distance between your 2 recievers (distance zero) to infinity.

First, forget circles!!! Second, there is only ONE sphere with that particular curvature. Curvature, got it? This is a time dependant measurement, not a static one were you draw circles with a pencil passing through two points

Quote:

Interferometry means to measure the shift between the same signal measured on 2 sepreate recievers.

I'm glad you bought a manual, now lets get pass the preface

Quote:

You need the 3rd port to determine the radius.
with 3 sensors on line and spaced equally you know that the wavefront will arrive at the average time of the 2 outer sensors ,if the emitter is a infinite distance anway, if the wafefront is not curved but straight.

Quote:

All this is only for 2d for 3d you need at least 4 recievers.

I guess the university faculty and other couple hundred PhD who work on this are wrong. You caught us! The link is a hoax... Rolling Eyes

Thanks, but forgive me if I disregard unfunded statements

Quote:

Now if the emitter is not an infiinte distance away the wavefromt will be curved. The distance to the middle sensor will be slightly shorther then the average distance to the 2 outer sensors.

The problem is, the wavefront emitted from a point source 35 miles away will be pretty damm straight.

Here´s the calculation again)

Quote:

Now. lets assume the triple revivers spaced 25 ft appart and perpendicular to the emitter. (the ideal setup)

Again our emitting souce is 35 miles away.

The difference between the distance to the center reciever and to one of the outer recivers will be
sqrt (25^2 + 210000^2)-210000 = 0.001488 ft = 0.045mm

That means you have to know the position of your recievers with precission in the same magnitude to make valuable measurements.

And this sould be next to impossible on a flying high performance jet.
It bends under load, it changes size due to temperature changes caused by speed and altitude changes.

For the millionth time, what is this calculation for?? READDDDD!!!
I been gathering that you just want to prove yourself. In that case....

CONGRATULATIONS!!! By the power the Pokemon's have invested in me, I declare you proficient in high school trigonometry. Go ahead... print this, frame it and hang it in your bedroom.

And to end on a good note, I'll be happy to clear this out for anybody else. As for this nonsense, I will no longer post a reply to Hansundfranz

Forum: F-16 versus XYZ

Four F-16s vs one F/A-22