Key.Aero Network
Register Free

Page 4 of 5 FirstFirst 12345 LastLast
Results 91 to 120 of 132

Thread: TPY-2 can be radar OTH ?

  1. #91
    Join Date
    Dec 2003
    Posts
    98
    @garryA

    Isn't it funny, how my "fundamentally wrong" physics are so widely accepted by engineers?
    In your physics world 2 neighboring out of 4 fins of the Iskander @ 90° AoA where we have as you said "pure drag", would not create a moment on the CoP/CoG. So something in your world is fundamentally wrong.
    In your world, a T-50 with one malfunctioning vertical fin/stabilizer @ 90° AoA would not create a moment, but just slow it down.

    As for your textbook rules: Yes good for a missile that has enough Cl to remain in the efficient non-stall lift region. But if you have high G requirements, you are not picky on which reaction forces you use, it's in stall region? pure drag? So be it.
    Heh, I don't even say that this is the case for Iskander, but your lack to understand it makes me wonder.

    Not much if anything at all, missiles can sustain more than 0.5G
    It's about the loss of kinetic energy due to drag in dense air. Iskander maneuvers at 0,5 at 30km altitude and the interceptor needs to do the necessary course corrections in dense air.

    This is a basic situation between ABM systems and BMs.
    Another basic situation that can be discussed is when the Iskander releases sub munitions before entering altitudes where endo-atmospheric interceptors can work. What does the PAC-3 want to do against 6 submunitions? These questions need all to be answered by the ABM side.

    An Iskander-M weight 4,615 kg, and it is single stage( meaning no discard parts of airframe to get lighter )
    An SM-2 or SM-6 weight 1,500 kg , and they are 2 stages missiles ( meaning at burn out they discard parts of their airframe to get lighter, and the boost stage of SM2, SM-6 is around 25-30% of their total length with bigger diameter)
    SM-2 and -6 have a start booster which they loose after a few seconds. The rest of the large missile remains the same until impact. The missile you see with its fins at the rear is what the mass-fin ratio is.
    The older Tochka and the Oka TBM both had high lift waffle fins. I would ask my self why they went back to the lower lift fins? Are you and me smarter? Or did they do calculations that lead them to the conclusion that based on their experience with Tochka and Oka, for required anti-ABM operations and tactics, conventional fins would be sufficient? So pardon me that I try to make sense of what we got.

    The standrad series has any where between 7-8 times more wing area and around 1/3 the weight, if that doesn't translate into much better wing loading then i don't know what will
    You are aware that the long wings of SM-2 and SM-6 are static? The fins at rear move.

  2. #92
    Join Date
    Dec 2015
    Posts
    920
    Quote Originally Posted by PeeD
    In your physics world 2 neighboring out of 4 fins of the Iskander @ 90° AoA where we have as you said "pure drag", would not create a moment on the CoP/CoG. So something in your world is fundamentally wrong.
    In your world, a T-50 with one malfunctioning vertical fin/stabilizer @ 90° AoA would not create a moment, but just slow it down.
    Sigh.... once again, you repeat exactly what you said before, it likes you didn't spend any time looking at the diagrams
    Read here: http://forum.keypublishing.com/showt...58#post2389958
    I explicitly stated that a turn has 2 parts:
    The first part: changing tail fin AoA, when the AoA of tail fin change there will be a moment on CoG/CoP of the aircraft or missiles that will pivot (pitch) the AoA of their whole airframe. For example : for an aircraft when the tail turn up , its nose will pitched up. If the tail is down , its nose will pitch down.This action is called the nose pointing


    The second part: once the whole airframe of aircraft/missiles is at an AoA with its current direction of travel , there will be more force on the whole airframe ( simply due to higher sectional area). This force has 2 components, the first is lift, the second is drag. Drag slow aircraft/missiles down while, Lift component will pull the missiles or aircraft off its current flight path. To sum up, what "turned" an aircraft or missiles is their whole airframe interrracting with airflow when at positive or negative AoA


    The fundamental different between a turn and nose pointing is the change in direction of travel.Nose point only change AoA..
    In your example, the fin at 90 degrees will create a pivot to turn AoA of missiles, but whether the direction of travel of the missiles change or not depending on how the whole airframe interact with the airflow.

    Nose pointing (pitch) without turning look like this:



    Another great example of nose pointing without changing direction of travel is the famous cobra maneuver by Su-27. The nose direction change by 90 degrees but the general direction of travel remain the same


    On the otherhand, turning looks like this


    So, why did i said a flat plate at 90 degrees provide no lift. It because it can only provide a pivot to change missiles AoA not provide lift to change direction of travel.Changing AoA and turning is vastly different.

    Quote Originally Posted by PeeD
    It's about the loss of kinetic energy due to drag in dense air. Iskander maneuvers at 0,5 at 30km altitude and the interceptor needs to do the necessary course corrections in dense air.
    I don't know what make you change your stand from "Iskander can turn 30G at altitude higher than 50 km" to Interceptor can't sustain barely 1.5 G at low altitude. But both are too extreme, if interceptor can't sustain merely 1.5G at low altitude, they won't be able to even point toward threat direction (consider straight up launch system). Moreover, interceptor has thrust, very high thrust while a ballistic missiles falling down only has gravity working for it

    Quote Originally Posted by PeeD
    Another basic situation that can be discussed is when the Iskander releases sub munitions before entering altitudes where endo-atmospheric interceptors can work. What does the PAC-3 want to do against 6 submunitions? These questions need all to be answered by the ABM side.
    I don't think Iskander can carry 6 submutions, the much bigger missiles likes Trident and R-36 only carry 8-12 submutions. Nevertheless, the most common solution is either to destroy target before they released submutions or launch more interceptors. 1 launch vehicle of Iskander has 2 missiles. 1 launch vehicles of PAC-3 has 16 missiles


    Quote Originally Posted by PeeD
    SM-2 and -6 have a start booster which they loose after a few seconds. The rest of the large missile remains the same until impact. The missile you see with its fins at the rear is what the mass-fin ratio is.
    You are aware that the long wings of SM-2 and SM-6 are static? The fins at rear move.
    Yes iam aware that the long fin are statics and only the rear fin move. But when the missiles is at an AoA with the air flow, the long fin will interract with the air flow to turn (change direction of travel) the missiles. Think about aircraft, why does aircraft with lower wing loading often advertised as better for maneuver eventhough the main wing cann't move ( except for the really small parts of the slat and flap ) ?. Wing loading never accounted for tail fin area either.

    Quote Originally Posted by PeeD
    The older Tochka and the Oka TBM both had high lift waffle fins. I would ask my self why they went back to the lower lift fins? Are you and me smarter? Or did they do calculations that lead them to the conclusion that based on their experience with Tochka and Oka, for required anti-ABM operations and tactics, conventional fins would be sufficient? So pardon me that I try to make sense of what we got.
    There could be various reason to change from grid fin to planar fin, for example grid fins can have considerably higher radar cross section due to their extremely high number of coner reflectors (those grids).
    Last edited by garryA; 13th May 2017 at 19:48.

  3. #93
    Join Date
    Dec 2015
    Posts
    920
    Iskander airframe at 90 degrees AoA looks like below, do you not see the ridiculous here ?
    Last edited by garryA; 13th May 2017 at 19:30.

  4. #94
    Join Date
    Dec 2003
    Posts
    98
    @garryA

    So now, once against, you repeat exactly what you said before, it likes you didn't spend any time looking at the explaination or diagrams

    ...It because it can only provide a pivot to change missiles AoA not provide lift to change direction of travel.Changing AoA and turning is vastly different.
    I did and I checked your calculations with the lift formula where you estimated the Iskander fin area and used that in the lift equation. Something does not add up if you used just the fin area of the Iskander, compared it to the whole lift of the Su-27 and dismissed the Iskander tube body lift in the calculation? Is there anything else than the velocity, altitude and fin area of the Iskander in your lift calculation that I missed?
    Your calculation was for that pivot force you talk about and I'm sticking to that.

    I don't know what make you change your stand from "Iskander can turn 30G at altitude higher than 50 km" to Interceptor can't sustain barely 1.5 G at low altitude. But both are too extreme, if interceptor can't sustain merely 1.5G at low altitude, they won't be able to even point toward threat direction (consider straight up launch system)
    There is a difference between a high G evasive maneuver and continuous low G maneuver to bleed a interceptor kinematics. I took your 30G number and defended it in light of the experiences with US MARV testing. Now you have rised some doubts about this with your calculations. However I see you evolving, now you seperate between the pivot force and the airframe drag. Your calculation to disprove 30G at 50km with the Iskander was just for the fins, the pivot force. You in the role of the devils advocate helps to get closer to the real deal.

    It seems you don't realize that a continuous say 1,5G maneuver could cause equal aerodynamic losses to a 30G maneuver if it's in denser air and over the course of many seconds instead of 1 second of an 30G evasive maneuver. This is a game to decrease/bleed the kinematic performance and envelope of the interceptor.

    I don't think Iskander can carry 6 submutions, the much bigger missiles likes Trident and R-36 only carry 8-12 submutions. Nevertheless, the most common solution is either to destroy target before they released submutions or launch more interceptors. 1 launch vehicle of Iskander has 2 missiles. 1 launch vehicles of PAC-3 has 16 missiles
    It certainly can, that Indian TBM of your graph does too:

    Name:  prithvi+missile+warheads.jpg
Views: 192
Size:  47.1 KB

    It's not about MIRVs. Intercepting 6 spin stabilized low cost submunitions with 6 or more PAC-3 would be one of the worst saturation scenarios imaginable.

    Yes iam aware that the long fin are statics and only the rear fin move. But when the missiles is at an AoA with the air flow, the long fin will interract with the air flow to turn (change direction of travel) of the missiles.
    They interact and contribute somewhat. If you want to count them and the fins with a 8x larger fin aera - mass ratio compared to the Iskander in your lift calculation you are doing a huge mistake.

  5. #95
    Join Date
    Dec 2003
    Posts
    98
    @garryA

    Iskander airframe at 90 degrees AoA looks like below, do you not see the ridiculous here ?
    Please, this seems to get confusing for you. I always said fins (2) to be at 90° AoA, never the missile... Your whole argumentation stated with your calculations where you only considered the fins. Your goal was to prove that they are too small to created a strong enough pivot force to turn the airframe into a AoA which then would lead to a change of direction with G loads...

  6. #96
    Join Date
    May 2013
    Posts
    215
    @garryA

    falling missile remain at 1 g (g = 9.8 m/s2) and speed constant? is right ? Or vary according to its corner fall, altitude and max speed?

    For example, the DF-21D missile has a cruising speed of Mach 10 at a height of 100km, a length of 14 meters and a diameter of 1 m, range 1500km, with a weight of 14 tons. So when it falls it can reach speeds on more than Mach 10? And force g will change!

    Do we consider the resistance of the air and the weight of the warhead ?

    Last edited by blackadam; 14th May 2017 at 00:24.

  7. #97
    Join Date
    Dec 2015
    Posts
    920
    Quote Originally Posted by PeeD
    I did and I checked your calculations with the lift formula where you estimated the Iskander fin area and used that in the lift equation. Something does not add up if you used just the fin area of the Iskander, compared it to the whole lift of the Su-27 and dismissed the Iskander tube body lift in the calculation? Is there anything else than the velocity, altitude and fin area of the Iskander in your lift calculation that I missed?
    Body lift, negative stability, high lift device..etc are included in lift coefficients. That why they are there. In other words, the fin area is a references, and how good the combination of fin and body generated lift is represented by the lift coefficient.


    The bigger your wing area relative to mass , the smaller CL value you will need to perform a certain maneuver and vice versa. When i said the Iskander fin is too small, i mean that it will need very big CL to compensate, which i find highly unlikely with its tube body configuration. Most missiles don't have CL very high, even air to air missiles


    Quote Originally Posted by PeeD
    Your calculation was for that pivot force you talk about and I'm sticking to that. Your whole argumentation stated with your calculations where you only considered the fins. Your goal was to prove that they are too small to created a strong enough pivot force to turn the airframe into a AoA which then would lead to a change of direction with G loads...
    No, i never once said i was calculating pivot force
    in #63 the first post where i talked about G-load , i explicitly stated that
    The location of the gas system relative to the CoG only matter as a pivot for nose pointing, but irrelevance to how many G the turn make.The acceleration of the turn itself have to obey F= ma equation.
    See turn acceleration and force? i even go as far as saying pivot for nose pointing is irrelevance of how many G the aircraft can make




    then in #72 , i explicitly stated the following
    You mistaken between nose pointing and turning again. Drag caused by the air flow and it is always opposite direction of travel. An aircraft turning 45 degrees will not expose its body to the air flow by 45 degrees, it will only expose its body as high as the angle of attack it used to generate lift for the turn.
    Firstly, lift equation can be applied to the whole aircraft not only the air foil ......Btw, nose pointing is not the same as turning


    Then in #74 , i posted the photo below and the description of how a turn is made, carefully stated that there are 2 parts for a turn

    let say you want to change direction that missiles is traveling. That bring us to the second picture below it. To change direction of an aircraft or missiles, or anything, you need to apply a force on it. So how do you apply this force ?.
    The first step changing the AoA of the fins ( stabilizer in case of aircraft).When the AoA of the fin changed there will be a force acted on it due to air flow.This may be slightly confusing but if you pointed the fin up then there will be a down force on the fin and vice versa. This force will change the nose position of the missiles or aircraft. Then it is simple, very much like a pivots, down force on the tail fin will pull missiles nose up and up force on the tail fin will pull missiles nose down. You can see how it behave in following photo:
    ....
    The problem here is you mistakenly think that the turn (chaging direction) is done once the nose of aircraft or missile is turned toward desired direction. But that is wrong. Changing nose direction without turning will look somewhat like this

    .If you think carefully about it, pointing the nose toward a certain direction mean the airflow has to applied an opposite force on the tail of the missiles, hence move it to opposite direction.So how it is possible that missiles moving toward something and still has their nose toward the same target?. Here is the stuff you missing. Which is step 2, look at the photo again.In the example, once the missiles nose pull up, its whole body will be at a certain AoA with the air flow. Thus, the air flow will then applied a force on the whole missiles, with much higher force than if missiles fly straight with zero AoA.This force can be thought of as 2 components :the first one is called lift, which is what make missiles changes direction toward the desired target, the second component is drag which is what slowed the missiles, aircraft down. When you keep increasing AoA of missiles the lift will keep increase up until a certain point, where it start to decrease and drag will increase. In other words, extreme AoA will not help missiles turn quicker but rather slow it down quicker
    When i talked about lift and AoA, i mean the force that actually help missiles change direction not the one that slow it down, that why you can't add CL and CN together. CN is practically the combined resultant coefficient of Cl (lift ) and Cd ( drag )



    Then in #80 where i get annoyed , i said this
    . I already told you the different between reaction force required to pivot the missiles and reaction force required to turn the missiles. I already told you that drag force doesn't change missiles direction of travel but only slow it down and used to pivot nose AoA
    I repeat the same thing again in #82
    Moreover, it is not the reaction force on the fins that change the missiles direction of travel, it is the aerodynamic lift (reaction force) on the whole missile after the fins pivot the AoA that will change its direction of travel. This may sound super confusing, but that why i have to draw a diagram for you.
    Think of any turn as acceleration toward a new direction. This acceleration is measure by G. Since acceleration is equal to force divided by mass. The minumum force you need to applied on missile body to make a 30G turn will be equal to missiles mass*G-load*9.8 .This force has to direct toward the new direction of travel
    Then there is #92 right above where i elaborate turning again

    If i only said it once or twice then admittely there could be miscommunication, but i emphasize that i was calculating lift not pivot force not just once but multiple times. I even said that pivots force has nothing to do with how many G the missiles can make and the G-load has to obey F=ma. I don't know how much clearer can i be.





    Quote Originally Posted by PeeD
    However I see you evolving, now you seperate between the pivot force and the airframe drag
    I have always seperate pivot force and airframe drag, did you not see my photo where i draw a massive arrow for drag and lift and a tiny arrow for pivot force. Do you not think it was intentional ????





    Quote Originally Posted by PeeD
    It seems you don't realize that a continuous say 1,5G maneuver could cause equal aerodynamic losses to a 30G maneuver if it's in denser air and over the course of many seconds instead of 1 second of an 30G evasive maneuver. This is a game to decrease/bleed the kinematic performance and envelope of the interceptor
    No , missiles and aircraft has thrust so there are sustained and intantaneous turn. A sustained turn is the condition where you can conserved your speed and altitude ( basically they have excess thrust for a turn, sustained turn is where your thrust can balance the extra drag ). Even though air density is lower at high altitude, it is much easier to sustain high G at low altitude. That because, due to lower air density, you will need to either fly much faster or at higher AoA at high altitude to generate the same amount of lift.
    For example: an F-16 at 30k feet can pull maximum of 9G ( which is what its CL allow it to do ) but it can barely sustain 3.9G
    .

    On the otherhand, an F-16 at sea level can easily pull and sustain 9G



    Quote Originally Posted by PeeD
    It's not about MIRVs. Intercepting 6 spin stabilized low cost submunitions with 6 or more PAC-3 would be one of the worst saturation scenarios imaginable.
    if it is not nuclear equipped warhead but simply bomlets then they can be intercepts by Centurion C-RAM, RIM-116, Iron dome or similar stuff, they unlikely to cause much damages given their limited size
    Heck, the MGM-140 carry over 950 M74 bomblets and no one bat an eye
    http://www.designation-systems.net/dusrm/m-140.html
    The MBDA Apache and JSOW used to carry loads of submutions too





    Quote Originally Posted by PeeD
    They interact and contribute somewhat. If you want to count them and the fins with a 8x larger fin aera - mass ratio compared to the Iskander in your lift calculation you are doing a huge mistake
    On the otherhand, they are the main contributors, as previously elaborated , it is the interraction between the airflow and the whole airframe that change the direction of travel, the tail fin acts as initial pivots force. Alternatively, you can think about why wing loading is important for aircraft agility but not tail fin area.

    P/s :To be honest, I don't think we can ever reach any agreement on anything, the last post just demonstrated that not even my intention was understood. Moreover, iam so bored of talking about lift constantly, and i still have a blog post to finish. Hence this will be my last reply.
    Last edited by garryA; 14th May 2017 at 05:33.

  8. #98
    Join Date
    Dec 2015
    Posts
    920
    Quote Originally Posted by blackadam
    @garryA

    falling missile remain at 1 g (g = 9.8 m/s2) and speed constant? is right ? Or vary according to its corner fall, altitude and max speed?

    For example, the DF-21D missile has a cruising speed of Mach 10 at a height of 100km, a length of 14 meters and a diameter of 1 m, range 1500km, with a weight of 14 tons. So when it falls it can reach speeds on more than Mach 10? And force g will change!

    Do we consider the resistance of the air and the weight of the warhead ?
    Object reached terminal velocity when resutant force equal zero. In other words, when thrust - drag = 0. A ballistic missiles doesn't have thrust but it is accelerated by the force of gravity acting on it. Since F = ma , the maximum acceleration due to gravity will be 9.8 m/s2 (a free fall). However, this will only happen in space. In atmosphere, there is drag, the faster you go the more drag you got.This drag will keep increasing more and more as you go faster, up until a point where drag equal gravity. So to answer your question, missiles doesnot accelerate at constant g = 9.8 m/s2 when they falling down. The acceleration will get smaller as time go on. The missiles will even decelerate.
    Last edited by garryA; 14th May 2017 at 04:49.

  9. #99
    Join Date
    May 2013
    Posts
    215
    @garryA

    I don;t think so ! suppose you let a rock fall from the top of a really tall building. It's speed will change as follows:

    after 1 sec - 9.81 m/s
    after 2 secs - 19.62 m/s
    after 3 secs - 29.43 m/s

    And can you tell me which formula you are using? I use this formula:V2-V02=2as

  10. #100
    Join Date
    Dec 2015
    Posts
    920
    Quote Originally Posted by blackadam
    I don;t think so ! suppose you let a rock fall from the top of a really tall building. It's speed will change as follows:

    after 1 sec - 9.81 m/s
    after 2 secs - 19.62 m/s
    after 3 secs - 29.43 m/s
    Iam talking about acceleration rate, you are talking about speed. Those are different quantity. Iam too lazy to type down now but you can check out this video

    I used this formular
    Last edited by garryA; 14th May 2017 at 06:10.

  11. #101
    Join Date
    Dec 2003
    Posts
    98
    Well garryA, now I understand...

    You thought you only need to consider the fin area in you Iskander Cl calculation because its body lift is negligible?

    Now for dozens of posts I talk about the fins only, because your calculation is for that only and first now I have to hear from you that you considered bodylift negligible?

    At least state such thing or better realize why I'm talking about the fins only if you are aware that you have done such a huge simplification...


    I neither agree with you that a tube body with the l/d ratio of the Iskander has negligible body lift, nor do I think it is necessary to expand the discussion beyond the fins/pivot force.

    No , missiles and aircraft has thrust so there are sustained and intantaneous turn. A sustained turn is the condition where you can conserved your speed and altitude ( basically they have excess thrust for a turn, sustained turn is where your thrust can balance the extra drag ). Even though air density is lower at high altitude, it is much easier to sustain high G at low altitude. That because, due to lower air density, you will need to either fly much faster or at higher AoA at high altitude to generate the same amount of lift.
    For example: an F-16 at 30k feet can pull maximum of 9G ( which is what its CL allow it to do ) but it can barely sustain 3.9G
    So my argument is wrong and the drag losses due to a sustained 1,5G turn won't have bigger impact on the lower altitude interceptor than the Iskander in terms of kinematic performance?

    if it is not nuclear equipped warhead but simply bomlets then they can be intercepts by Centurion C-RAM, RIM-116, Iron dome or similar stuff, they unlikely to cause much damages given their limited size
    Heck, the MGM-140 carry over 950 M74 bomblets and no one bat an eye
    http://www.designation-systems.net/dusrm/m-140.html
    The MBDA Apache and JSOW used to carry loads of submutions too
    You want to intercept six small mach 3 submunitions with C-RAM? For hat velocities is the iron dome cleared? Then you think that a single warhead produce more frag = radar (TPY-2) killing damage than 6 submunitions? They will have a wider CEP but also larger damage area and well if the damage is not enough send a second Iskander.

    There could be various reason to change from grid fin to planar fin, for example grid fins can have considerably higher radar cross section due to their extremely high number of coner reflectors (those grids).
    OK. So they changed the grid fins from Oka to conventional ones in Iskander due to RCS issues, sacrificing anti-ABM maneuvering capability for it? Possible if the Russians believe they can lower the RCS of the Iskander in a amount that could make it stealth against a high power radar like the TPY-2, maybe they made giant leaps in temperature and drag resistant RAM. I however think Russians do not of such a RAM technology nor would they do this sacrifice, I rather think for some reason the fins of the Iskander are sufficient for its anti-ABM strategy.

    On the otherhand, they are the main contributors, as previously elaborated , it is the interraction between the airflow and the whole airframe that change the direction of travel, the tail fin acts as initial pivots force. Alternatively, you can think about why wing loading is important for aircraft agility but not tail fin area.
    As said, yes they contribute, however your simplification to take them together with the fins into a lift calculation against the Iskanders fins 1:1, is a non-representative simplification.
    Last edited by PeeD; 14th May 2017 at 06:09.

  12. #102
    Join Date
    Dec 2015
    Posts
    920
    Quote Originally Posted by PeeD
    You thought you only need to consider the fin area in you Iskander Cl calculation because its body lift is negligible?
    Now for dozens of posts I talk about the fins only, because your calculation is for that only and first now I have to hear from you that you considered bodylift negligible?


    .........................................no
    I never said i don't consider body lift
    This is why it is so frustrating talking to you, i literally said in the first line of #97 that
    Body lift, negative stability, high lift device..etc are included in lift coefficients. That why they are there. In other words, the fin area is a references, and how good the combination of fin and body generated lift is represented by the lift coefficient
    In other words instead of having different equation every time you calculate for different aircraft , airfoil, missiles. The lift equation was invented so you can have similar references factors which is the fin/wing area. The CL represent how effective the whole airframe generate lift.Which include body lift, airfoil thickness , negative stability ...etc. Do you think the CL graph for Saparrow is only for the fin ? or do you think the CL chart for Su-27 is only for the wing ?
    Don't know why i even bothered when you keep twisting my words......
    Last edited by garryA; 14th May 2017 at 07:41.

  13. #103
    Join Date
    Dec 2003
    Posts
    98
    The mess started with your post 67:

    As far as we know, Iskander is 7.3 meters in length and 0.92 meters in body diameter, it has several trapezoid fins. If you use the ruler scales in paints or pts, you can estimate the inner length of the fin is 1/10 of missiles length (0.73 meters), the outter length of the fin is 1/27 of missiles length (0.27 meters), the heigh of the fin is 1/4 of missiles body diameter (0.225 meters).Iam not saying the estimation is 100% accurate, but it surely close enough ( if you see the result later, you will find that even if the fins are several times bigger than i estimated, it still doesn't really matter). From the photo we can see that Iskander's fins has trapezoid shape ,so with values given earlier we can calculate area of the fins to be around 0.11 square meters
    Lift as cited earlier is CL* air density* 0.5*velocity^2 *wing area
    By this http://www.hochwarth.com/misc/AviationCalculator.html
    let say altitude is 24 km (which is half of what you propose so that we can have some what thicker air for the missiles to turn), the air density will be 0.046 kg/m3, speed of sound at that altitude will be 297 m/sec ( so Mach 6 will be 1782 meters/ second)
    Wing area is 0.11 meters squares as calculated earlier.
    To make 30G turn , the missiles will need to generate aerodynamic lift of 69,225 kg ( or 678,405 Newtons)
    => CL*0.046*0.5*1782^2*0.11 = 678,405
    => CL*8,034 =678,405
    => CL = 84.44
    For comparision, the flanker airframe ( with LERX, blended body, negative stability and what not) has CL of 1.2 at Mach 1 and AoA of 18 degrees

    In short, the Iskander will need the lift coefficient around 70 times bigger than Su-27 for it to be able to pull 30G at Mach 6 and altitude of merely 24 km .No chance.
    Before, i may have a slight doubt but now iam 99% certain that the G-load of Iskander on Wiki is BS.
    Is the statement in bold is correct?

    No. And there can be no other answer.

  14. #104
    Join Date
    May 2013
    Posts
    215
    @garryA

    You forgot that the motorized missile, when it falling the engine still working, it pushed the speed + the gravitational force of the earth, making the missile faster

  15. #105
    Join Date
    Dec 2015
    Posts
    920
    Quote Originally Posted by blackadam
    You forgot that the motorized missile, when it falling the engine still working, it pushed the speed + the gravitational force of the earth, making the missile faster
    If the motor still working all the way down to sea level then yes, but motors of ballistic missiles are generally empty when they falling down. Rocket are unlike ramjet, tuborjet or tuborfan, they burned through their fuel very quick.


    Quote Originally Posted by PeeD
    The mess started with your post 67:Is the statement in bold is correct?

    No. And there can be no other answer.
    My statement is correct actually, CL is not only an airfoil quality in the case of specific airfoil but also an airframe qualiity in case of specific airframe. CL of Su-27 airframe will include both its body lift and wing efficiency. Similarly, CL of Iskander airframe will include both its body lift and fin efficiency. The fin/ wing area is a reference area in equation (regardless whether you calculate lift for specific airfoil or specific airframe, this is the base reference).
    For example: this is CL of F-16 and YF-16 with their wing area as reference area (obviously if you use non SI unit for wing area then velocity and altitude have to be non SI too)

    http://www.f-16.net/forum/viewtopic....52510&start=45
    Last edited by garryA; 14th May 2017 at 15:26.

  16. #106
    Join Date
    Dec 2003
    Posts
    98
    @garryA

    My statement is correct actually, CL is not only an airfoil quality in the case of specific airfoil but also an airframe qualiity in case of specific airframe. CL of Su-27 airframe will include both its body lift and wing efficiency. Similarly, CL of Iskander airframe will include both its body lift and fin efficiency. The fin/ wing area is a reference area in equation (regardless whether you calculate lift for specific airfoil or specific airframe, this is the base reference).
    For example: this is CL of F-16 and YF-16 with their wing area as reference area (obviously if you use non SI unit for wing area then velocity and altitude have to be non SI too)
    Aha, so the Cl for the Su-27 is like that of the F-16 based on the wing area only? Not a experimentally gained value that can be confidently written in a manual? We might should do the test and calculate Cl via wings only and compare it with the manual value...

    An equation result is only as good as it's inputs. Your input for the Iskander are its fins and hence you only get the Cl value for the fins.
    This is pure logic, you cant get anything about the airframe/body lift with that equation, no input about it, just impossible.

    I neglected body lift in our discussion because I knew your Iskander Cl value is just for the fins. For our discussion it would have been sufficient to just compare the Cl value of the fins, the pivot force, the force that creates the moment. No need to extend it into body lift and comparison with Su-27...

    I appreciated your effort to contribute to the discussion. Now that I understand the lift equation, got interested and rechecked your calculation. I discovered that you took just one fin out of four for the Iskander.
    My own calculation shows a Cl for Iskander fins at 30G at 24km a Cl value of 6,4. This value is 13 times lower than what you calculated. That mig-31bm guy might have cheered you up so that you start to post funny images but try to remain sober and get your facts straight.

    Of course I won't compare that Iskander Cl of 6,4 to the Cl of Su-27 because its a experimental value with body lift included and conventional outside stall region.

    In total I dismiss your value for Iskander fin Cl and what you want to compare and quantify it.

  17. #107
    Join Date
    Dec 2015
    Posts
    920
    Quote Originally Posted by PeeD
    Aha, so the Cl for the Su-27 is like that of the F-16 based on the wing area only? Not a experimentally gained value that can be confidently written in a manual? We might should do the test and calculate Cl via wings only and compare it with the manual value.

    Where did i said the CL is only for the wing ?
    I explicitly said that

    CL of Su-27 airframe will include both its body lift and wing efficiency. Similarly, CL of Iskander airframe will include both its body lift and fin efficiency
    and
    Body lift, negative stability, high lift device..etc are included in lift coefficients. That why they are there. In other words, the fin area is a references, and how good the combination of fin and body generated lift is represented by the lift coefficient.
    what parts of that is so hard to understand ? you free to do the test to compare it with the F-16 manual charts i posted above. But remember

    and FYI neither F-16 or Su-27 CL was for their wing only, it was for their whole airframe with wing area as reference area, the CL (lift coefficient) of NACA 64-206 airfoil on F-16 is the one below:


    Quote Originally Posted by PeeD
    An equation result is only as good as it's inputs. Your input for the Iskander are its fins and hence you only get the Cl value for the fins.
    This is pure logic, you cant get anything about the airframe/body lift with that equation, no input about it, just impossible
    Sigh................
    Let put it like this
    Imagine you are an aerodynamic engineer, when you have a totally new airframe, you want to know how much lift it can create in a certain condition ( or in other words, how many G it can pull).
    Then you have to do actual experiment like wind tunnel or what not..etc.
    From the lift value (actual force) that you measured in experiments, you can put it in the lift equation to measure the Lift coefficient (CL).
    This CL values can be understood as how efficient your aircraft/missiles is at generating lift.Either with very efficient air foil or very efficient body lift or both.
    After you got many value for the total lift force at different AoA, you will also get many value for the CL for different AoA. As a result from that you got the CL-AoA curve.
    With the CL-AoA curves, anyone that go after you, can easily calculate how much lift your missiles or aircraft can generate for any given AoA represent on the curve. AoA is limited by speed of course
    The only thing they have to do is, to take the reference wing area and multiply it with the given CL.
    The whole purpose of the CL value and lift equation are to make it easier for engineers to calculate lift of various vehicles or same aircraft in various situation.


    On the otherhand, if you already know how much lift is generated in a given condition, you know the reference wing area. You will be able to calculate CL. In the Iskander case, you know how many G it have to make, you know the reference fin area, hence, you will be able to calculate the required CL value for it to make the turn.

    In other words, because
    Lift = CL* air density*0.5 V^2 * reference wing area
    If you know how much lift generated, you will be able to calculate CL values required
    If you know the CL values then you will be able to calculate lift.


    In short, to create a high amount of lift you either need very big wing area or extremely efficient wing (very thick, low sweep, slat.. etc) + high body lift (lerx, flat body.. etc)}. For example: there are 2 aircraft, the first has wing area of X square meter, the second has wing area of 2X square meter, however the first has CL of 2 while the second has CL of 1. They will ended up generate the same amount of lift.

    Without the lift equation, and the same point of reference (wing area) , you will have to test every situation possible. Imagine, you tested the original aircraft and the data shown that at 100% fuel it can pull 9G at 10k feet. Let say, if the aircraft is at 15k feet and 27% fuel, how many G it can pull ? Do you as an engineer decided to make another test for that exact data point ?

    Quote Originally Posted by PeeD
    I neglected body lift in our discussion because I knew your Iskander Cl value is just for the fins. For our discussion it would have been sufficient to just compare the Cl value of the fins, the pivot force, the force that creates the moment. No need to extend it into body lift and comparison with Su-27..
    The discussion was never about the pivot force in the first place. Why do you think i have to repeatedly elaborate how a turn is made ?
    Body lift, wing effeciency are all parts of lift coefficient, they are not extended parts



    Quote Originally Posted by PeeD
    I appreciated your effort to contribute to the discussion. Now that I understand the lift equation, got interested and rechecked your calculation. I discovered that you took just one fin out of four for the Iskander.
    My own calculation shows a Cl for Iskander fins at 30G at 24km a Cl value of 6,4. This value is 13 times lower than what you calculated.
    Of course I won't compare that Iskander Cl of 6,4 to the Cl of Su-27 because its a experimental value with body lift included and conventional outside stall region.
    Actually, yes i forgot and only took 1 reference fin, but you should only took 2 fins as reference not 4 fin. The lift due to + and X configuration will be the same due to force distribution. I only take 4 fin in AIM-54 calculation because it actually has 8 fin in total). At least, you finally say something right

    Nevertheless, even 6.4 is considerably bigger than CL of even air to air missiles

    Btw, do you know what "stall" mean?

    P/S: how is your calculation get value 13 times smaller if area is only 4 times bigger? mind writing it down?
    Last edited by garryA; 15th May 2017 at 06:52.

  18. #108
    Join Date
    Dec 2003
    Posts
    98
    Well garryA,

    You are a genius. You managed to calculate the Cl of the Iskander by just inputting the fin area. You basically found a method to analytically calculate a airframes Cl via the fin area and the lift that is necessary to pull the required Gs, and it which also includes the body lift. You did it with the Iskander and compared it to experimentally gained Cl data of the Su-27. Your novel method will make complex geometric area measurements and experimental testing obsolete... fin area --> necessary lift = total airframe Cl... amazing

    I appreciate your efforts to show me:
    - Importance of body lift
    - Nosepointing vs. flight direction
    - resulting vectors
    - Usefulness of Cl in airframe design
    etc.
    Thanks but I was aware of all of them...

    As for the other points:

    Take 4 fins for the Iskander, X-tail =/= +-tail

    My 13x value... well lets say my dead mass was lower, my fin area larger and of course the factor 4 for # fins.
    However its useless and a waste of time to elaborate. You were brave to think you can just calculate this complex situation with the Iskander, but it has just lead us to some fin Cl values for which we lack a comparison to quantify and judge. The discussion made me just aware that high G's via fins become feasible at below 20km altitude, I falsely expected more atmosphere at 50km...

  19. #109
    Join Date
    Dec 2015
    Posts
    920
    Quote Originally Posted by PeeD
    Your novel method will make complex geometric area measurements and experimental testing obsolete... fin area --> necessary lift = total airframe Cl... amazing
    I know you are being sacastic, but i never said my methods can replace complex experiments and testing, the reason is that for a new airframe, without testing, you will not know how much lift it can generate (or in other words how many G it can pull) . Thus you have nothing to put in lift equation to find CL (lift coefficients).
    On the otherhand, here, you have the Iskander with alleged 30G at 50 km, Mach 5. If you have those value, you can easily calculate the lift required to make the turn at that condition because F=ma. From that you can calculate the CL required to make the lift.

    If you paid attention, you will noiticed that i didn't calculate the CL of actual Iskander. I calculated the CL that it will need to perform specific maneuver that you proposed. And because the CL would be so ridiculously big, the conclusion is that such maneuver in such condition isn't possible.
    Quote Originally Posted by PeeD
    Take 4 fins for the Iskander, X-tail =/= +-tail
    lead us to some fin Cl
    It is the same thing since force at 45 degrees surface will be divided into horizontal and vertical components ( the same ways you can calculate force when banking).Nevertheless, it more about using same form of reference area actually. Technically you can even use wetted area to compare CL, but then both has to be wetted area. As a general rule of thumb, they go with wing area as reference area because it is the most simple to calculate
    Btw ,it isn't fin CL but airframe CL since calculation based on total lift required to turn 30G
    Last edited by garryA; 15th May 2017 at 10:48.

  20. #110
    Join Date
    May 2013
    Posts
    215
    The ABM radar system is primarily for detecting and tracking end-stage targets, as it is slowed by air resistance. The radar TPY-2 and SBX-1 or SPY-1D can tracking escape velocity of missile ? like DF-26 ?!

    A rocket launcher will have a velocity of 7 km / s to escape from the atmosphere (escape velocity)

    Recent Korean and Chinese missiles recently tested, the TPY-2 system has not yet announced any specific detection or tracking capabilities!


    Last edited by blackadam; 16th May 2017 at 10:45.

  21. #111
    Join Date
    Jun 2004
    Location
    Columbia, MD
    Posts
    11,305
    TPY-2 in TBM tracks end stage targets to provide cues to the THAAD battery and its associated missile. In FBM, it provides early warning, and discrimination data to be used for mid-course SM3 intercepts using LOR.
    Old radar types never die; they just phased array

  22. #112
    Join Date
    Dec 2003
    Posts
    98
    The forums own stealthflanker did this exercise and you garryA even discussed with him. How unfortunate that you didn't remember the discussion, his calculation model is much better and tells us the actual G numbers:

    http://forum.keypublishing.com/showt...urn-rate/page2

    Instead of arguing about your proposed model let's do everyone a favor and use stealthflankers spreadsheet which is based on a book and looks correct as far as I checked.

    Here are the results I get for Iskander aerodynamic maneuvering (excluding gas system) at different relevant altitudes:

    @ 45km, the range in which it starts to come withing the slant range of the THAAD (with a assumed apogee of 50km), we get the following result for Gs pulled:
    1,14G for the fins @63° AoA considering newtonian impact theory (now I know the name for what I described).
    2,2G for missile body at 10° AoA consistent with a shallow dive of a depressed trajectory
    TOTAL: 3,4G (This number is for a contentious maneuvering in order to reduce the kinematic power of the THAAD)

    @ 37km, the range in which the THAAD KV (if modified to operate at this altitude) could reach the Iskander (still outside the envelope of PAC-2, PAC-3, SM-2), we get the following result for Gs pulled:
    3,3G for the fins @63° AoA considering newtonian impact theory.
    13,1G for missile body at 15° AoA consistent with a shallow dive of a later stage depressed trajectory
    TOTAL: 16,5G (This number is for evasive maneuvering for an exo-atmospheric interceptor)


    @ 28km, the range in which large endo-atmospheric interceptors become feasable (PAC-2, PAC-3SME if somehow possible, SM-2), we get the following result for Gs pulled:
    12,35G for the fins @63° AoA considering newtonian impact theory.
    81,8G for missile body at 20° AoA consistent with a shallow dive of a terminal stage depressed trajectory
    TOTAL: 94,2G (This number is for evasive maneuvering for an endo-atmospheric interceptor, it already exceeds the air frame structural capability by the factor of at least 3)

    Lower altitudes doesn't need to be considered due to even higher G load capabilities. At higher altitudes the gas system would contribute for higher G numbers if required.

    So much for this discussion, nothing better than hard numbers.

  23. #113
    Join Date
    Dec 2003
    Posts
    98
    Note:
    That the spreadsheet and thus also the discussion about AIM-120 too just seem to count 1 control surface/wing into the calculation instead of 2 described in the source (book Tactical missile design) (calculation 4 for x-configuration is too complex and the difference indeed negligible).
    Hence also the values for fins I posted have to be two times higher:
    2,28G @45km
    6,6G @37km
    24,7G @28km

    Further notes: The book Tactical missile design proposes a 3 times higher G manouver for a interceptor for robust interception performance, This is to some extend applicable to the THAAD and makes the scenario very adverse for it.
    Note that my fin AoA values work for the equation of the book and spreadsheed but they would be in stall region. I assumed that at hyper velocity, newtonian impact theory is the strongest factor for Cn and hence enable post-stall operation.

  24. #114
    Join Date
    Dec 2015
    Posts
    920
    Quote Originally Posted by PeeD
    Instead of arguing about your proposed model
    Lift equation, and reference area are not my proposed model.It never was, it is official model used to measure CL. Iam not the first nor the last person ever used the equation


    Quote Originally Posted by PeeD
    use stealthflankers spreadsheet which is based on a book and looks correct as far as I checked.
    Do you really checked and understood it ? or do you only use it because it sound like it support your theory ?

    Quote Originally Posted by PeeD
    @ 45km, the range in which it starts to come withing the slant range of the THAAD (with a assumed apogee of 50km), we get the following result for Gs pulled:
    1,14G for the fins @63° AoA considering newtonian impact theory (now I know the name for what I described).
    2,2G for missile body at 10° AoA consistent with a shallow dive of a depressed trajectory
    TOTAL: 3,4G (This number is for a contentious maneuvering in order to reduce the kinematic power of the THAAD)
    So not only the amount of G Iskander is about 10 times smaller than your proposed values, you also intentionally forget that THAAD will have thrust to maintain speed. How exactly does this supposed to "prove" that iam wrong ?

    Quote Originally Posted by PeeD
    @ 37km, the range in which the THAAD KV (if modified to operate at this altitude) could reach the Iskander (still outside the envelope of PAC-2, PAC-3, SM-2), we get the following result for Gs pulled:
    3,3G for the fins @63° AoA considering newtonian impact theory.
    13,1G for missile body at 15° AoA consistent with a shallow dive of a later stage depressed trajectory
    TOTAL: 16,5G (This number is for evasive maneuvering for an exo-atmospheric interceptor)


    @ 28km, the range in which large endo-atmospheric interceptors become feasable (PAC-2, PAC-3SME if somehow possible, SM-2), we get the following result for Gs pulled:
    12,35G for the fins @63° AoA considering newtonian impact theory.
    81,8G for missile body at 20° AoA consistent with a shallow dive of a terminal stage depressed trajectory
    TOTAL: 94,2G (This number is for evasive maneuvering for an endo-atmospheric interceptor, it already exceeds the air frame structural capability by the factor of at least 3)
    How exactly do you want people to double check your calculation if you don't tell them what exact value you used for wing area, mass, speed and air density ???
    Instead of trying to omitted PAC-2, PAC-3MSE and SM-2 , SM-6 from the calculation saying it is out of their evelope, why don't you tried to put their number in and estimate how many G they could turn according to that spread sheet ? You said you checked it to be correct, yet another member came up with this
    I put air density of 0.12 kg/m3 (59k feet) and missiles speed of mach 4 into the file , and it show Aim-120 can turn 78g , and able to deal with target making 16g turn ,
    then I change the air density to 1.225 kg/m2 ( sea level ) and missiles speed to mach 2 into the file and it show aim-120 can turn 303g and intercept target making 61g turn
    how did any aircraft evade AIM-120 if that was the case ?
    Last edited by garryA; 16th May 2017 at 18:39.

  25. #115
    Join Date
    Dec 2015
    Posts
    920
    Quote Originally Posted by PeeD
    Note that my fin AoA values work for the equation of the book and spreadsheed but they would be in stall region. I assumed that at hyper velocity, newtonian impact theory is the strongest factor for Cn and hence enable post-stall operation.
    Do you understand the meaning of "stall" ?
    Rocket obvious has the highest CN when it is perpendicular to the airflow. So that mean, if we used your theory, missiles can turn highest G value when its airframe is at 90 degrees to the airflow ?
    Last edited by garryA; 16th May 2017 at 18:57.

  26. #116
    Join Date
    Dec 2003
    Posts
    98
    @garryA

    Lift equation, and reference area are not my proposed model.It never was, it is official model used to measure CL. Iam not the first nor the last person ever used the equation
    You thought you were smart enough to start the numbers game. However, your simplifications were so much that they had no representation of the reality. The spreadsheet of stealthflanker on the other hand used a book in which a smart guy defined the necessary model and it's results make sense. I give you credit to have been brave enough to try it, that awaked by interest and lead to this useful results.

    Do you really checked and understood it ? or do you only use it because it sound like it support your theory ?
    Sure I did otherwise I would not have found the error with the control surface area. Surely a book about missile design also is a good reason to take it over a model from someone with potential no aerospace, engineering or physics background at all.

    So not only the amount of G Iskander is about 10 times smaller than your proposed values, you also intentionally forget that THAAD will have thrust to maintain speed. How exactly does this supposed to "prove" that iam wrong ?
    You proposed the numbers and I took them as they made sense to me. It's still possible that 30G is possible via contributions from the gas system or more depressed trajectory.
    You are wrong because you quite confidently presented a useless model to quantify the case, maybe intentionally because you might have remembered the discussion with stealthflanker and his spreadsheet.

    How exactly do you want people to double check your calculation if you don't tell them what exact value you used for wing area, mass, speed and air density ???
    People can put their values into the spreadsheet and get an idea, the results will be similar. (air density is included in the spreadsheet thanks to stealthflankers efforts)

    Instead of trying to omitted PAC-2, PAC-3MSE and SM-2 , SM-6 from the calculation saying it is out of their evelope, why don't you tried to put their number in and estimate how many G they could turn according to that spread sheet ?
    They are not cleared for those altitudes? Why should I do a theoretical training?

    You said you checked it to be correct, yet another member came up with this

    I put air density of 0.12 kg/m3 (59k feet) and missiles speed of mach 4 into the file , and it show Aim-120 can turn 78g , and able to deal with target making 16g turn ,
    then I change the air density to 1.225 kg/m2 ( sea level ) and missiles speed to mach 2 into the file and it show aim-120 can turn 303g and intercept target making 61g turn
    how did any aircraft evade AIM-120 if that was the case ?
    I think you have no kind of engineering approach but want to talk about it. The AIM-120 airframe is not capable to do very high G's, its systems will fail and it's structure disintegrate. It might be designed to do overloads from 30-50Gs, that's it. The airframe is thus the limiting factor not the aerodynamic capability to pull G's. Stealthflankers spreadsheet is not failproof but as it uses the equations of a good book, I at least have not found any errors beyond that with the wing/fin area.

    As for the stall thing.... just ignore it, I have no fun to try to explain such basics

  27. #117
    Join Date
    Dec 2015
    Posts
    920
    Quote Originally Posted by PeeD
    You thought you were smart enough to start the numbers game. However, your simplifications were so much that they had no representation of the reality
    No representation of reality yet even NASA used the equation ?

    Quote Originally Posted by PeeD
    The spreadsheet of stealthflanker on the other hand used a book in which a smart guy defined the necessary model and it's results make sense
    I think you have no kind of engineering approach but want to talk about it. The AIM-120 airframe is not capable to do very high G's, its systems will fail and it's structure disintegrate. It might be designed to do overloads from 30-50Gs, that's it. The airframe is thus the limiting factor not the aerodynamic capability to pull G's. Stealthflankers spreadsheet is not failproof but as it uses the equations of a good book, I at least have not found any errors beyond that with the wing/fin area.
    According to you the reason for AIM-120 not able to make those turn is that its structure is not strong enough. Fine then. How about you put values of an F-16 in the spread sheet ?
    From the sheet, F-16 at 50.000 feet , mach 1.85 will be able to make a turn over 28 G with AoA of merely 5 degrees, yet in reality this is what it can do

    So tell me then, what is the limiting factors here ? F-16 airframe was designed to sustain over 9G, its max controlable AoA is also far higher than 5 degrees. Let see what excuse you come up this time ?




    Quote Originally Posted by PeeD
    People can put their values into the spreadsheet and get an idea, the results will be similar
    And how do they know the values for dead mass and fin area that you used ?

    Quote Originally Posted by PeeD
    You are wrong because you quite confidently presented a useless model to quantify the case
    Oh, so now the lift equation is wrong ?


    Quote Originally Posted by PeeD
    They are not cleared for those altitudes? Why should I do a theoretical training?
    Are they not or you don't want to do the calculation because it will result in much higher G values ?

    Quote Originally Posted by PeeD
    As for the stall thing.... just ignore it, I have no fun to try to explain such basics
    You have no fun explain it or you don't understand what it mean ?. Given how you keep talking about it clearly indicate the later one
    Last edited by garryA; 16th May 2017 at 20:36.

  28. #118
    Join Date
    Dec 2003
    Posts
    98
    @garryA

    No representation of reality yet even NASA used the equation ?
    Only useful in it's context. The book Tactical missile design provides a much better quantifiable and more accurate calculation model.

    According to you the reason for AIM-120 not able to make those turn is that its, structure is not strong enough. Fine then. How about you put values of an F-16 in the spread sheet ?
    From the sheet, F-16 at 50.000 feet , mach 1.85 will be able to make a turn over 21G with AoA of merely 5 degrees, yet in reality this is what it can do

    So tell me then, what is the limiting factors here ? F-16 airframe was designed to sustain over 9G, its max controlable AoA is also far higher than 5 degrees. Let see what excuse you come up this time ?
    The calculation model is only accurately applicable for missiles. There is no mean to calculate a complex geometry like the F-16. Only a tube body with fins.
    You and your friends have problems to judge context and if models are applicable...

    If the F-16 has a good FBW system it will limit the AoA to avoid catastrophic overload... totally off-topic again...

    And how do they know the values for dead mass and fin area that you used ?
    By best practice advise: take something between 0,5-0,3 * total mass. Fin area? Get some good photos and compare it with the length or diameter...

    Oh, so now the lift equation is wrong ?
    No. Your application of it to this problem and comparison with values like Su-27 is.

    Are they not or you don't want to do the calculation because it will result in much higher G values ?
    Do it yourself... The fire control system will not let missiles operate above their max. altitude (25km), they might selfdistruct.
    It would be interesting nonetheless, but I doubt your ability to use such a spreadsheet.
    As for G numbers? The book Tactical missile design assumes that the interceptor must have a G capability of 3 times that of the target to perform a high PK interception. Now as the Iskander can easily pull G's at 28km up to its airframe limit of lets say 20G, it would be interesting to see if 60G could be possible by ABM/SAM airframes.

    You have no fun explain it or you don't understand what it mean ?
    You don't understand control surface deflection angle/AoA...

  29. #119
    Join Date
    Dec 2015
    Posts
    920
    Quote Originally Posted by PeeD
    Only useful in it's context. The book Tactical missile design provides a much better quantifiable and more accurate calculation model.
    It gives 28.4G for F-16 at 50.000 feet, as opposed to real life value of 4G. So accurate.Mind blow



    Quote Originally Posted by PeeD
    The calculation model is only accurately applicable for missiles. There is no mean to calculate a complex geometry like the F-16. Only a tube body with fins.
    You and your friends have problems to judge context and if models are applicable
    So F-16 will be able to pull 28.4G at 50K feet, Mach 2 if it was a tube body with fin ?. Or the complex body lift design reduced F-16 ability to generate lift by over 7 times ? Which one is that ?


    Quote Originally Posted by PeeD
    If the F-16 has a good FBW system it will limit the AoA to avoid catastrophic overload... totally off-topic again
    In CAT I configuration F-16 maximum AoA limits is 25 degrees at 1G and 15 degrees at 9G. 5 degrees AoA is far from the AoA limit. Wrong excuse, try different one.



    Quote Originally Posted by PeeD
    By best practice advise: take something between 0,5-0,3 * total mass. Fin area? Get some good photos and compare it with the length or diameter
    Yet, you gave no values of your own so others can dobble check.

    Quote Originally Posted by PeeD
    Do it yourself... The fire control system will not let missiles operate above their max. altitude (25km), they might selfdistruct.
    Where do you find the information that PAC-3MSE, SM-6, SM-2 will self destruct once they fly above 25 km ? Care to cite ?

    Quote Originally Posted by PeeD
    The book Tactical missile design assumes that the interceptor must have a G capability of 3 times that of the target to perform a high PK interception
    Have that take into account the fact that interceptors can still retain their speed due to thrust while their targets will lose all speed ?

    Quote Originally Posted by PeeD
    You don't understand control surface deflection angle/AoA
    Do I not ?or you don't want to explain in case you got it wrong ?
    Last edited by garryA; 16th May 2017 at 20:58.

  30. #120
    Join Date
    Dec 2003
    Posts
    98
    @garryA

    It gives 28.4G for F-16 at 50.000 feet, as opposed to real life value of 4G. So accurate.Mind blow
    So F-16 will be able to pull 28.4G at 50K feet, Mach 2 if it was a tube body with fin ? Great.
    Good luck with applying specialized formulas for completely different problems... I take what "Tactical missile design"s author tells over your models any day.

    In CAT I configuration F-16 maximum AoA limits is 25 degrees at 1G and 15 degrees at 9G. 5 degrees AoA is far from the AoA limit. Wrong excuse, try different one.
    Good and I hope you have now understood that airframe G's and aerodynamic G's are not the same.

    Yet, you gave no values of your own so others can dobble check.
    dia 0,914
    lenght 7,28
    weight 1254kg
    missile AoA 10°
    control surface deflection 63°
    control surface area 0,16m² *2
    control surface aspect ratio 0,66
    speed mach 6,5

    Where do you find the information that PAC-3MSE, SM-6, SM-2 will self destruct once they fly above 25 km ? Care to cite ?
    Didn't talk about SM-6. For the others, check Wiki für max. altitude...

    Have that take into account the fact that interceptors can still retain their speed due to thrust while their targets will lose all speed ?
    It's not about speed, at around 32km the Iskander would reach its airframe load limit with 30G. How intercept a 30G target with reasonable PK? With luck the interceptors airframe might survive 50G.

Thread Information

Users Browsing this Thread

There are currently 1 users browsing this thread. (0 members and 1 guests)

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  

- Part of the    Network -

KEY AERO AVIATION NEWS

MAGAZINES

AVIATION FORUM

SHOP

 

WEBSITES