http://bmpd.livejournal.com/1870660.html

Myanmar will get its first 3 Yak-130s this year.

EDIT: Nobody wants to comment on the R-33s on the last page?

tell us if that R-33 with different color has ARH seeker. Or give R-37 instead 😀

Now i wonder if Su-28 AEW/C concept ever make it to prototype like Su-30K2 :3

I wonder if there’s any plan in the future to use AL-31FM-1 or even 117S on Su-30SM.

Hmm as far as i know, the Su-30MKI family use technologies demonstrated at Su-37 program. The Su-37 itself however were not supposed to use Al-31FP. It was selected as interim because the intended engine, the AL-31FM with TVC was not ready for flight atm. Thus it might be bit under powered.

G

Ok, it’s clear now, thanks for the explanation. Why then does the director of fighter business development for NG say it can cause physical damage to a system? This guy certainly knows what he is talking about, and I don’t see why he would say that if it had no practical utility.

Yes, but as previously demonstrated by calculations above.. does it really practical ?

The NG’s director clearly know what he’s talking about and select one which looks promising and not telling the whole picture.

I don’t see where he takes into account that it is a beam. There should be another parameter in the equation for the narrowness of the beam. ( instead of 4pi^2 I think )

High gain equals Narrow beam. If you interested on how narrow a beam generated by 38 Db antenna.. then Let’s visit some simple equations by Barton.

Gain=30000/horizontal beamwidth*vertical beamwidth.

Gain is in dimensionless power ratio not Db.
beamwidth is in degrees.

We can assume that vertical and horizontal beamwidth of the radar would be the same a.k.a pencil beamwidth which is the case for typical fighter radar and HPM. The equations can then be written as follows

Gain=30000/beamwidth^2

Now we can arrange it to find the beamwidth of the antenna if the gain is known.

beamwidth=SQRT(30000/Gain)

Our HPM have gain of 38 Db.. Thus converting it to power ratio 10^(38/10) = 6309.573

The beamwidth would be :

beamwidth=SQRT(30000/6309.573)
beamwidth=2.1 degrees.

Does that looks narrow enough to you ?

Additionally. If you know the dimension of the antenna and its operating frequency, you can calculate the beamwidth with following equations :

Beamwidth in radians=1.25*lambda/d

Where d is the dimension of the antenna (width or height) If the antenna is not axis-symmetric then you’ll have to repeat the calculation using each dimension of the antenna before plugging in to the Barton’s equation for gain. Oh and be sure to multiple it by 57.3 first or simply use excel function DEGREES() to convert it to degrees. Lambda is of course the operational wavelength.

If one ever ask where factor 1.25 came from.. It’s antenna weighting factor.. correspond to Taylor weighting scheme. Modern slotted array radar today are arranged to have certain weighting scheme. Why such scheme needed ? Sidelobe reduction which would give numerous advantages. like better clutter filtering, less susceptibility to ESM. There are many other weighting scheme available today.. ESA Radar have advantage that it can actually change its weighting factor.

What exactly the difference between disruptive and destructive effect here? i first thought disruptive mean render the radar unable to detect targets but that would be jamming, and iam sure the radar can be jammed from much further distance
Anyway, it seem that create destructive effect on AAM and SAM will be much easier, seem possible

Disruptive effect is sort of computer bug or when your radar computer start to slow down. and destructive effect is when there is enough voltage or power to cause actual physical damage to your radar (burned computer chips, or melted gimbal actuator.)

and yes AAM, SAM and precision munitions are targeted application for HPM Like Ranets E and Vigilant eagle. But those system use either pulsed power source like super reltron tube or very large number of TR modules in megawatt range. Clearly not something fighter radar can afford for now and maybe later.

Where do you take into account that the AESA emits like a beam? The energy receive might be 100 or 10000 times larger, hard to say given the fact that we don’t know how narrow the beam is.

The gain tells everything. The narrower the beam.. the higher your gain.

Saw old CG’s of air launched kalibr from MiG-29. I wonder how far the program now.

The APG-77 and 81 can make EAs. There’s supposed to be able to damage enemy aircraft electronics. But the enemy aircraft might absorb a lot of energy through its antenna.

No. The EA Mode in both radars are basically jamming, probably deception one. Not HPM.

As for the range, that’s what I am wondering. If the stealth aircraft comes in at very low altitude using the terrain and suddenly pops up above a enemy column of vehicles. What would happened if it did a fast EA from close? Could it rapidly Attack all the vehicles? It could damage the enemy AD systems at the same time.

Of it’s not stealthy it’s pretty much the same in fact.

The answer is no.

For range however you can make basic calculations based on power density. to determine how much power received by target at range. You can look at Filippo Neri’s “Introduction to Electronic Defense” 2nd Edition. for the calculation method.