it is about what it displays not that it is displaying an imgae
So, it got 2km close and can display F-22 here? well…
it was cheaper to upgrade existing aim9m to 9xs, especially since the US has to field a lot more missiles for a lot more aircraft. Also thanks to some geopolitics, the US wasn’t allowed to develop any new ir missile, only upgrade existing one. (I could have some parts of this wrong since i forgot the full details of it but it’s off topic to begin with so I’ll leave it at that)
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i know that but that dosent make the 9x any less bad compared to its equivalents
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Can you provide a source?
restricted manual
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In-game yeah it’s awful. Better off using a Wii controller as an IRST than PIRATE in-game. But IRL PIRATE is a beast. It’s leagues above the Russian IRST’s if what we know about it is true. Just a shame the devs can’t be bothered to model it correctly or fix any of the bugs with it.
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Maintenance and compatibility with existing parts for reduced costs, performance was good too.
Eurofighter came into service later with a mech scan (behind peers, not ahead), and still hasn’t seen widespread use of the radar upgrade.
It was 2024 and Britain still hadn’t equipped new radar right?
The F-14A’s modified for use with AMRAAM could certainly have been equipped with them. They needed only software updates and the pylons iirc.
Can’t be letting the enemy know their vulnerabilities. Hard to hide them when it’s so obvious, tho.
To be fair, that’s the best thermals I’ve seen on anything Russian and it’s almost brand new. We had better thermals on our tanks since 1979.
The F-22 in the image seen has the length of the diagonal of a rectangle measuring 113px*85px.
solving pythagroas therum for the diagonal we get:
Image used to get pixel measurements
x=sqrt(113^2 + 85^2)
x=sqrt(12769 + 7225)
x=sqrt(19994)
x=141.4px
The F-22 on the picture ha a length of 141.4 pixels which equates to 62’1" or 18.923 meters
To calculate he distance we need the angular size of the F-22 which we can get by dividing the earlier calculated 141.4px by the width of the image itself and the multiply it by the FOV of the sensor/ camera
The FOV of the OLS-35 (Irst of the Su-35) is 10*7.5° or 75°
The Image is 873px in width, with that we can calculate the angular size of the raptor on the image:
x=(141.4px / 873px) * 75°
x=12.147766323° ~ 12.15°
To get the distance we can use the following formular:
Distance = actual size / 2tan( radiant / 2 )
to get the radiant we need to take the angular size and multiply it by PI/180
d= 18.923m / 2 * tan( ( 12.15° * PI / 180 )/2)
d= 18.923m / 2 * ~0.002 | the ~0.002 is a really long irrational number, which i used in the further calculation, it was just a bit to long for my taste ^^
d=4,969.637m
the result is a distance of ~5000m and that is fully ignoring any zoom that the camara has
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the widespread change to CAPTOR-E hasn’t been done because CAPTOR-M is Superior to AESA Based Radars of the 1990s and Early 2000s
No it isnt??? What the hell are you talking about? AESA radars, even earlier ones, are still much better than mechanical radars.
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better at what?
give me something technical so I can pick it apart please
wow… 5 km. insane!
the antenna type itself is Inferior as proven by US made Radars of WWII which were superior to German “PESA” Radars by using well… Mechanically scanned focussed beam antennas instead of electronically timed Dipole Arrays
Js use binoculars if it’s 5km lolll
ignoring zoom
the image is zoomed in the su35 is even further away
you guys cant read, can you?
the 5km is ignoring the zoom of the camera
with the sensor resolution we could correct the image but I doubt we’ll ever get that
Higher resolution, better detection and tracking, higher resitance to jamming / better jamming capabilities…
Bro is comparing WW2 radar to end of cold war
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