Key.Aero Network
Register Free

Page 3 of 5 FirstFirst 12345 LastLast
Results 61 to 90 of 132

Thread: TPY-2 can be radar OTH ?

  1. #61
    Join Date
    Aug 2011
    Posts
    4,370
    Quote Originally Posted by bring_it_on View Post
    As I had mentioned the US Government only awarded the contract to Lockheed to begin executing the Saudi order only in late December of 2015 so they are likely to begin getting deliveries this year and receive all missiles and launcher modifications by 2019. With US FMS sales it is easy to follow FMS notification date, or even vendor_customer contract finalization dates but what is the only 'real' completion metric to look at is the actual DOD contract award. For now their upgraded PAC-2s are the best they have and I don't think they have configuration 3+ either. UAE has the most advanced system in the region. Saudi's are likely to announce or finalize their THAAD and perhaps an IBCS order during Trump's visit there if everything could move along in the Congress. Informal, and back channel requests for IBCS and THAAD has been pending for a number of years as part of a broader GCC missile shield but much like Qatar's and UAE's request the previous administration sat on it longer than usual.

    https://www.google.com/url?sa=t&rct=...BD1Cx-MjVJ3tcA



    While it is easier to see missiles being fired, it is harder to deduce shot doctrine or tell how many targets they are actually attempting to defeat unless one has such a clear view that one can see tens of km in altitude, and downrange to deduce warheads, decoys or missile fragments.

    Few dozen interceptors at a time? A PAC-2 launcher can only carry 4 missiles so I'd love to see videos of them emptying 3-6 launchers in one go (Typically the Saudi's should be using 4-6 launchers per radar). 2 PAC-2's per Ballistic Missile target should be a good shot doctrine, but you could even allow 3 per target if you are protecting a very high value asset given the nature of this particular conflict (Saudi's have vastly more resources and are at home). Even PAC-3 shot doctrine calls for mostly a twin launch per TBM target. 3 Incoming Ballistic Missiles could warrant 6-8 PAC-2 interceptors depending upon the area they are protecting but unless there is a massive raid of say 6-10 incoming missiles you wouldn't need to launch 12, 24 or 36 interceptor missiles at them even if it were physically possible given the PAC-2 and its size.

    These are the raid sizes the PAC-3 with its 16 missile/launcher configuration is designed to tackle. PAC-2 not so much even if one gets past the fact that the PAC-3 exists primarily because there was a need for a considerably better TBM interceptor than the PAC-2 (even though the PAC-2 has been subsequently upgraded and improved).
    The problem with all these complex airdefence systems that they are very hard to upgrade and maintain and get obsolete soon against primitive systems like in Yemen. even F-16 so mass produced at many places but hardly 5% can be consider modern.

    http://www.defenseworld.net/news/191...e#.WQ7rcqllDqA
    Saudi Arabia Could Insist on Locally-made Spares For Imported Military Hardware

  2. #62
    Join Date
    Jun 2004
    Location
    Columbia, MD
    Posts
    11,530
    Man, finding the video again was a PIA:
    How do we even know that is a PAC-2 launch? There are a number of clips on youtube and particularly twitter that have over the last couple of years shown likely Astros ripple fires as Patriot missiles. PAC-2 launchers, part of a battery would be spread out (They would have 4-6 launchers per fire unit, and a couple of fire units protecting a high value asset. Since the Patriot uses a fixed, sectored sensor coverage doctrine, each radar and fire unit wold be covering a different threat sector), and you would not have couple of dozen missiles essentially following each other and it would be extremely weird to have emptied out pretty much all 5-6 of your launchers in just a few seconds. This to me looks like something totally different and not related to a Patriot interceptor launch.

    Below is a multi PAC-2 launch Saudi Video. You can see the radar and launcher in the shot as well, and as you can see the missiles (6+) are launched from different launchers, spread out over a distance, in relatively quick succession. It is impossible to gauge the shot doctrine from these videos since we really do not know how many incoming targets they are launching to engage but I would not find 6-8 missiles launched at 3-4 incoming missiles as unreasonable shot doctrine for their system. If they were US batteries using PAC-3's, the shot doctrine would have been 2 PAC-3's per TBM target and 1 for 1 for a cruise missile. An astute tactician would launch multiple ballistic missiles towards an area defended by the legacy patriot since he would know that the best strategy against a PAC-2 set up would be to saturate the defense system. A saturated TBM attack threat on troop or Command concentrations basically led to the PAC-3 program in the first place in an aim to quadruple the number of missiles per fire control unit. Therefore, I wouldn't expect one or two ballistic missiles being launched at a high value Saudi target that warrants Patriot coverage in the first place.

    In OIF, the shot doctrine with GEM's appeared to have been 3 Missile launches per TBM target likely due to the mismatch (plenty of missiles on station vs a limited threat) and the value of the assets they were protecting. There they engaged 9 TBM's with PAC-2 GEM's and PAC-3's (Mostly PAC-2s) and claim 8 TBM kills with one inconclusive but likely kill. Since the US is an expeditionary force and target concentrations are dense I would expect a slightly conservative shot doctrine from US units as opposed to a homeland defense mission for an export customer where there may be multiple targets they are protecting but similar economic, and resource mismatch exists with the Saudi vis-a-vis Yemen so I wouldnt be surprised that in the absence of a configuration 3+, and PAC-3 they are also using a conservative shot doctrine just to be safe.

    PAC-2 Launched from a firing unit (Radar, and well spaced PAC-2 launchers can be seen including an aircraft that gives a good hint of what this battery is protecting)





    Saudi Rocket Launch



    Here is another Saudi video of a rocket launch, erroneously titled Patriot interceptors. I could be wrong but I count 30 launches. This one is quite easy to classify as 'not a Patriot'.

    https://www.youtube.com/watch?v=6cT1aU-XG5A

    The U.S. has already signed contracts for the delivery of THAAD to the United Arab Emirates and Qatar. The fact that this was not done in the case of South Korea indicates that the true motives of deployment on the Korean Peninsula are far from declared.
    There is no pending THAAD request from South Korea unlike UAE, and Qatar that officially requested the purchase of the missile defense system. South Korea is slowly building its own BMD program but if they ever do exercise the option of buying an upper tier system their request for THAAD will likely be easily approved.

    As far as radar coverage and discrimination, this can be easily achieved by a number of ways including using ships designed for such a purpose so as to replicate the performance of the TPY-2 in FBM, a mode the US radars in South Korea cannot operate in without leaving the US Forces in South Korea, and the Korean mainland defenseless against the upper tier threat.

    There are South Korean Green Pine radars that they have acquired that can be used as stand alone Situational Awareness / Early Warning sensors and can depending upon their placement cover some of the same airspace the Chinese claims the TPY-2 can cover in FBM even though it is there to operate in support of the THAAD missile defense system and battery.
    Last edited by bring_it_on; 7th May 2017 at 20:26.
    Old radar types never die; they just phased array

  3. #63
    Join Date
    Dec 2015
    Posts
    933
    Quote Originally Posted by PeeD
    You are mistakenly simplifying the case. Soviet engineers should have known what they are doing with the gas system of the Iskander. You don't take into consideration the offset of the gas system to the CoG.The gas system of the Iskander is at a strong offset to the CoG.
    I couldn't care less if Russian engineers or US engineer do it. 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. You can't just ignore such fundamental physics law.
    And yes i did simplified it for you because the F variable in F= ma is resultant force not thrust, so F actually equal thrust minus drag. But that only mean the gas system need to produce even higher thrust to perform 30G maneuver as rumored.
    Quote Originally Posted by PeeD
    As for the maneuvering: it may travel at a wrong ballistic course and all the maneuvering during terminal phase will the result in the right target location
    You basically propose that the gas system on Iskander has enough fuel to produce the massive thrust ( more than 75% of LGM-30 Minuteman first stage ) not just once but many times.

    Quote Originally Posted by PeeD
    No precious kinetic energy is wasted. Under exo-atmospheric conditions there is no loss at all because of maneuvering as it causes no friction. In endo-atmospheric conditions speed is decreased due to maneuvering, however as the missile has to slow down anyway, the maneuvering does to some extend that what friction would have done. So you do a controlled amount of maneuvering to decease your speed to a level in which the energy is not wasted due to heatshield heat up. We have 3 parameters: Speed, heatshield temperature and gain of kinetic energy due to transformation of potential energy.
    Under the exo-atmospheric condition, the side force ( especially if very offset to CoG ) will mostly spin the missiles nose around rather than changing its direction of travel. Noiticed how the reaction control system on space shuttle or satellite operate?


    If the thruster is at the same location as the CoG similar configuration as the EKV, it still not "turn" the thing, but rather moving it perpendicular to direction of travel.




    In endo-atmospheric conditions, speed of missiles decelerate due to drag. The formula for drag is:


    when you turn the missiles, expose its side aspect to the air flow, both the reference area and the drag coefficient will increase. The side area of Iskander-M is at least several times the cross section of its nose, the Cd is much bigger too obviously. As a result, the drag will be several time bigger atleast, without main rocket operating the missiles will lose speed extremely quick, since the only things counter the drag is the weight of missiles. After the turn, you can gain some speed back through potential to kinetic energy convention but the acceleration due to gravity is rather small. In perfect condition (without any drag), the gravitational acceleration is only 9.8 m/s. When missiles entered the atmoshphere, the drag will be much higher, therefore acceleration rate is much slower and keep getting worse and worse as missiles fell down.
    Last edited by garryA; 7th May 2017 at 17:06.

  4. #64
    Join Date
    Dec 2003
    Posts
    108
    @garryA

    I couldn't care less if Russian engineers or US engineer do it. 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. You can't just ignore such fundamental physics law.
    And yes i did simplified it for you because the F variable in F= ma is resultant force not thrust, so F actually equal thrust minus drag. But that only mean the gas system need to produce even higher thrust to perform 30G maneuver as rumored.
    You might be a professional in one field and I an expert in another, however it would be too arrogant to think you or me can question the functioning of such a complex system such as the Iskander. If the Russians have implemented a gas system, we should expect that it's working and this also for a useful reason. We should rather try to interpret the details of those system in a way that makes sense.

    I agree that 30G maneuvers with the gas system is very unlikely, I just argumented on the numbers you brought up. It's rather likely that 30G would be possible via the fins insider layers of the atmosphere which allow 30g at mach 6 to be pulled.
    I claim that the exoatmospheric gas system could make re positioning in denser air layers necessary for the interceptor (loss of energy). I claim that at least its possible that the gas system might be used to evade a THAAD interceptor, whether via a maneuver. What we can be sure about is that the expertise of Soviets/Russians in that field and implementation a gas system is a clear indicator that it's supposed to counter exoatmospheric interceptors.

    You basically propose that the gas system on Iskander has enough fuel to produce the massive thrust ( more than 75% of LGM-30 Minuteman first stage ) not just once but many times.
    No I don't. Why would you use a "30g maneuver" or 75% Minuteman first stage thrust for a course correction? Any thrust will lead to a course change. A small continuous thrust will change the course and enable such position changes in order to bleed the interceptor.

    In endo-atmospheric conditions, speed of missiles decelerate due to drag. The formula for drag is:


    when you turn the missiles, expose its side aspect to the air flow, both the reference area and the drag coefficient will increase. The side area of Iskander-M is at least several times the cross section of its nose, the Cd is much bigger too obviously. As a result, the drag will be several time bigger atleast, without main rocket operating the missiles will lose speed extremely quick, since the only things counter the drag is the weight of missiles. After the turn, you can gain some speed back through potential to kinetic energy convention but the acceleration due to gravity is rather small. In perfect condition (without any drag), the gravitational acceleration is only 9.8 m/s. When missiles entered the atmoshphere, the drag will be much higher, therefore acceleration rate is much slower and keep getting worse and worse as missiles fell down.
    We talked about this. Let me put it simply: Because the velocity is not proportional to drag force, it's with ^2. If your velocity is lower our losses due to drag will be lower. The lower earth atmosphere is so dense that mach 3 is about the highest velocity you can reach with a conventional structure/materials. So the Iskander much be around mach 3 when impacting. It must loose 3 mach numbers from its speed outside the atmosphere. Hence it may enter the lower atmosphere at mach 5 or at mach 3,5.
    At mach 5 it may be at mach 3 at impact. --> 2 mach numbers loss due to drag
    At mach 3,5 it may be at mach 2,5 at impact. --> 1 mach number loss due to drag

    So the Iskander could bleed 1,5 mach numbers for endo-atmospheric maneuvering and end up by only 0,5 mach numbers below a non-maneuvering Iskander at impact.

  5. #65
    Join Date
    Dec 2015
    Posts
    933
    Quote Originally Posted by PeeD
    You might be a professional in one field and I an expert in another, however it would be too arrogant to think you or me can question the functioning of such a complex system such as the Iskander. If the Russians have implemented a gas system, we should expect that it's working and this also for a useful reason. We should rather try to interpret the details of those system in a way that makes sense.
    Iam not saying that Iskander can't change direction or can't fly to target, iam arguing that many numbers , details floating around the internet about Iskander are nonsense , just like the myth about plasma stealth that defeat everything or the jammer on Su-24 that shut down AEGIS ...etc. There are tons of BS information on internet, hardly a big surprise.

    Quote Originally Posted by PeeD
    I agree that 30G maneuvers with the gas system is very unlikely, I just argumented on the numbers you brought up. It's rather likely that 30G would be possible via the fins insider layers of the atmosphere which allow 30g at mach 6 to be pulled.
    I highly doubt that Iskander can pull 30G with its tiny fin and tube body either, especially consider high altitude. Air to air missiles can turn high G value only at sea level. I can't be bothered to do the calculation now but if you put the number in the lift equation, you probably end up with CL higher than subsonic airfoil for Iskander if it want to pull 30G at high altitude (pulling 30G at low altitude is kinda too late).


    Quote Originally Posted by PeeD
    implementation a gas system is a clear indicator that it's supposed to counter exoatmospheric interceptors.
    I don't know if Iskander has a mid body thruster or not ( since i can't find the holes like on the space shuttle). But gas system are not indicator that it is supposed to counter exoatmospheric interceptors. For example intercontinental ballistic missile generally all have gas system in their final stage.


    Quote Originally Posted by PeeD
    We talked about this. Let me put it simply: Because the velocity is not proportional to drag force, it's with ^2. If your velocity is lower our losses due to drag will be lower. The lower earth atmosphere is so dense that mach 3 is about the highest velocity you can reach with a conventional structure/materials. So the Iskander much be around mach 3 when impacting. It must loose 3 mach numbers from its speed outside the atmosphere. Hence it may enter the lower atmosphere at mach 5 or at mach 3,5.
    At mach 5 it may be at mach 3 at impact. --> 2 mach numbers loss due to drag
    At mach 3,5 it may be at mach 2,5 at impact. --> 1 mach number loss due to drag

    So the Iskander could bleed 1,5 mach numbers for endo-atmospheric maneuvering and end up by only 0,5 mach numbers below a non-maneuvering Iskander at impact.
    Mach 3 is not the fastest impact velocity for conventional material, structure either

    Rail gun for example intended to have impact velocity around Mach 5 and it has similar trajectory to most ballistic missiles


    Kh-15 reach terminal velocity of Mach 5 in a dive to target.
    Kh-22 reach terminal velocity of Mach 4.6 in a high altitude dive to target
    AQM-37 can reach terminal velocity of Mach 5 in dive too.
    http://www.designation-systems.net/dusrm/m-37.html

    As a theater ballistic missiles,without maneuver Iskander can probably reach around Mach 5-6 terminal velocity in impact. Consider the small fin, Iskander will need very high CL to pull high G, high CL mean either very high AoA or very thick , lift oriented air foil. From photos we can see, there is nothing lift oriented about Iskander fin. It has low aspect ratio, high wing sweep and not much thickness if there is any. The body itself is also a tube body. So the only other option is high AoA, and high AoA generate alot more drag. The missiles likely lose 2-3 Mach from each high AoA maneuver and then it has to maneuver back to still high the right target too so will lose even more speed. While Iskander can probably change course and direction, i don't think it can do what information on Wiki seem to suggest
    Last edited by garryA; 8th May 2017 at 22:57.

  6. #66
    Join Date
    Dec 2003
    Posts
    108
    @garryA

    I highly doubt that Iskander can pull 30G with its tiny fin and tube body either, especially consider high altitude. Air to air missiles can turn high G value only at sea level. I can't be bothered to do the calculation now but if you put the number in the lift equation, you probably end up with CL higher than subsonic airfoil for Iskander if it want to pull 30G at high altitude (pulling 30G at low altitude is kinda too late).
    High G's were witnessed by US MaRV programs too, partly due to de-acceleration and partly due to turns.
    Its the high speed in the equation with ^2 that makes it possible, aerodynamic maneuvers can be done at very high altitude due to the higher speed.
    The Iskander was rumored to fly a depressed trajectory and called a quasi-ballistic missile due to that. Shorter warning time due to later radar detection was one reason, but another reason is that at high speed it would remain maneuverable via it's fins. Other missiles would have almost no function on their fins at around 50km, but Iskander could due to its speed and resulting ram air pressure.

    I don't know if Iskander has a mid body thruster or not ( since i can't find the holes like on the space shuttle). But gas system are not indicator that it is supposed to counter exoatmospheric interceptors. For example intercontinental ballistic missile generally all have gas system in their final stage.
    They have because their post boost vehicle does maneuver to place the re-entry vehicles in trajectory. The Iskander would have no reason for a gas system, any possibly necessary course corrections would be done in terminal via it's fins. Hence the only reason would be to counter exo atmospheric interceptors or to change course to confuse/bleed endo-atmospheric interceptors. The gas system of the Iskander is at its TVC system.

    Mach 3 is not the fastest impact velocity for conventional material, structure either

    Rail gun for example intended to have impact velocity around Mach 5 and it has similar trajectory to most ballistic missiles
    Sounds exotic. For a system without any propulsion, just with a hardened GPS guidance and hardened fins a starting speed of mach 7,5 might be possible if the heat shielding is nearly all the projectiles weight. Still a very hard task and different than missile and their airframes.

    All other systems down to ICBM reentry vehicles need to de-accelerate down to around mach 3 at sea level to avoid disintegration due to thermal stress and drag forces. The Iskander is surely in that category and maybe mach 4 would be possible if it's hardened accordingly.

    Kh-15 reach terminal velocity of Mach 5 in a dive to target.
    Kh-22 reach terminal velocity of Mach 4.6 in a high altitude dive to target
    AQM-37 can reach terminal velocity of Mach 5 in dive too.
    http://www.designation-systems.net/dusrm/m-37.html
    Maybe at the start of the dive at ~10km altitude they could still have mach 5 but almost certainly not at impact, sea level.

    As a theater ballistic missiles,without maneuver Iskander can probably reach around Mach 5-6 terminal velocity in impact. Consider the small fin, Iskander will need very high CL to pull high G, high CL mean either very high AoA or very thick , lift oriented air foil. From photos we can see, there is nothing lift oriented about Iskander fin. It has low aspect ratio, high wing sweep and not much thickness if there is any. The body itself is also a tube body. So the only other option is high AoA, and high AoA generate alot more drag. The missiles likely lose 2-3 Mach from each high AoA maneuver and then it has to maneuver back to still high the right target too so will lose even more speed. While Iskander can probably change course and direction, i don't think it can do what information on Wiki seem to suggest
    I guess we have to first define high G in this scenario, in relation with THAAD.
    How many G's would the gas system of the THAAD be able to pull?
    How many G's can the Iskander pull if it fly a depressed trajectory in which its aerodynamic control via fins is still possible due to the high ram air speed?

    I haven't read the Wiki article about Iskander but in this scenario there is a good chance to outmaneuver the THAAD interceptor. There is also a good chance that at the depressed trajectory the Iskander would be outside the PAC-3 interception envelope.
    We have a fundamentally different understanding about how much speed is lost when doing a maneuver, you talk about 2-3 mach numbers being lost by each maneuver. You also ignore that speed lost due to maneuvering creates a more efficient drag economy.

    Hence let me make some claims too:
    -THAAD would only be able to do 10G maneuvers
    -Iskander would be able to dodge those maneuvers by a counter 10G maneuver at depressed trajectory 60km altitude via its fins
    - Each course changing 3G maneuver of the Iskander consumes mach 0,1 and each 10G evasive maneuver mach 0,3.
    This speculation is not really helpful.

    Btw. at speeds like mach 6 no high AoA maneuvers are necessary to create loads like 10 or 30G.

  7. #67
    Join Date
    Dec 2015
    Posts
    933
    Quote Originally Posted by PeeD
    High G's were witnessed by US MaRV programs too, partly due to de-acceleration and partly due to turns.
    Not what I heard of
    Quote Originally Posted by PeeD
    Its the high speed in the equation with ^2 that makes it possible, aerodynamic maneuvers can be done at very high altitude due to the higher speed.
    The Iskander was rumored to fly a depressed trajectory and called a quasi-ballistic missile due to that. Shorter warning time due to later radar detection was one reason, but another reason is that at high speed it would remain maneuverable via it's fins. Other missiles would have almost no function on their fins at around 50km, but Iskander could due to its speed and resulting ram air pressure.
    Not long ago, i been in similar discussion about AIM-120 turn performer, i don't mind doing it again
    Let start, here is a photo of Iskander


    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.

    Quote Originally Posted by PeeD
    The Iskander was rumored to fly a depressed trajectory and called a quasi-ballistic missile due to that. Shorter warning time due to later radar detection was one reason, but another reason is that at high speed it would remain maneuverable via it's fins. Other missiles would have almost no function on their fins at around 50km, but Iskander could due to its speed and resulting ram air pressure
    quasi-ballistic trajectory look like this

    I don't see how quasi-ballistic will shorter radar warning time, it help improve range but surely the warning time will be longer IMHO. Moreover, i don't see the relation with ram air pressure either since Iskander is a rocket, not a scramjet or ramjet missile.

    Quote Originally Posted by PeeD
    They have because their post boost vehicle does maneuver to place the re-entry vehicles in trajectory. The Iskander would have no reason for a gas system, any possibly necessary course corrections would be done in terminal via it's fins. Hence the only reason would be to counter exo atmospheric interceptors or to change course to confuse/bleed endo-atmospheric interceptors. The gas system of the Iskander is at its TVC system.
    On the otherhand, i believe that the gas system is for re-oriented the missiles nose before re-entry without the gas system the missiles will likely oriented wrong way and get burned in the atmostphere.The fins are most likely to turn missiles in atmosphere so that it follow quasi-ballistic path for extended range.

    Quote Originally Posted by PeeD
    Sounds exotic. For a system without any propulsion, just with a hardened GPS guidance and hardened fins a starting speed of mach 7,5 might be possible if the heat shielding is nearly all the projectiles weight. Still a very hard task and different than missile and their airframes.

    All other systems down to ICBM reentry vehicles need to de-accelerate down to around mach 3 at sea level to avoid disintegration due to thermal stress and drag forces. The Iskander is surely in that category and maybe mach 4 would be possible if it's hardened accordingly.
    The terminal velocity depending on forces vs drag relation, it is irrelevant of whether the missiles has heat shield or not. The heat shield only help prevent the air from destroying the missile but has nothing to do with the drag vs force equation.
    If the rail gun project can reach Mach 5 on impact, that mean at Mach 5 the resultant force of gravity minus drag is equal zero. Unless Iskander and others ICBM are significantly more draggy, there is no reason for their terminal velocity to be limited to Mach 3 at impact.
    Quote Originally Posted by PeeD
    I guess we have to first define high G in this scenario, in relation with THAAD.
    How many G's would the gas system of the THAAD be able to pull?
    How many G's can the Iskander pull if it fly a depressed trajectory in which its aerodynamic control via fins is still possible due to the high ram air speed?
    Consider the design of THAAD with no fin, seperate kill vehicle with divert gas system, you can see that is is mostly a interceptor intended to intercept missiles at very high altitude where the air is extremely thin and the fin doesn't work. So putting a high G value here for either missiles is simply nonsense



    Quote Originally Posted by PeeD
    Each course changing 3G maneuver of the Iskander consumes mach 0,1 and each 10G evasive maneuver mach 0,3.
    This speculation is not really helpful.
    Even aircraft with their high lift design will lose significant amount of speed if they attemp high G at altitude, there is no way a tube body with fin only lose 0.3 Mach when trying to do 10G maneuver at Mach 6
    Last edited by garryA; 9th May 2017 at 13:39.

  8. #68
    Join Date
    Jun 2004
    Location
    Columbia, MD
    Posts
    11,530
    Consider the design of THAAD with no fin, seperate kill vehicle with divert gas system, you can see that is is mostly a interceptor intended to intercept missiles at very high altitude where the air is extremely thin and the fin doesn't work. So putting a high G value here for either missiles is simply nonsense
    The broader consensus among the Missile defense experts is that THAAD has an envelope in the 50-150 km altitude with the Extended Range Missile increasing the upper envelope but leaving the lower one the same. There is a gap between the upper tier of the PAC-3/MSE, and the lower altitude of the THAAD that currently has a notional interceptor (Upper tier Patriot) as a defined gap for a future program.
    Old radar types never die; they just phased array

  9. #69
    Join Date
    Dec 2003
    Posts
    108
    @garryA

    Not what I heard of
    Then I recommend you to read the book lightning bolts about MARVs.

    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.
    I appreciate your effort. However I think your model is not applicable for this case. The reason is simply empirical knowledge.

    As describes the book lightning bolts is a good read about early US experiences with MARVs, the G numbers mentioned there go up to 50 IIRC.
    Then we have weapons like the AIM-54 which reached high speeds and had to perform at least two digit G maneuvers, at least even at 5 G your model would deliver impossible results.
    We have effects that have a huge impact:
    - Compared to high velocity objects, low velocity objects are exposed to much lower G loads for the same turn. Hence the vector change of a high velocity object is much smaller for the same G load.
    - At high velocities, the ram air effect on the aerodynamic surfaces is much higher, so the drag force. Hence a low velocity object can have the same drag force at low altitude as a high velocity object at high altitude.
    - Rotation axis position of the fins can be extreme for maximum ram air pressure --> turn capability.

    I don't see how quasi-ballistic will shorter radar warning time, it help improve range but surely the warning time will be longer IMHO. Moreover, i don't see the relation with ram air pressure either since Iskander is a rocket, not a scramjet or ramjet missile.
    A ballistic missile with a depressed trajectory will appear later on the LOS radar horizon due to curvature of the earth. Only inaccurate OTH radars won't have this problem, but they are useless for engagement.

    On the otherhand, i believe that the gas system is for re-oriented the missiles nose before re-entry without the gas system the missiles will likely oriented wrong way and get burned in the atmostphere.The fins are most likely to turn missiles in atmosphere so that it follow quasi-ballistic path for extended range.
    A ballistic missile does come down on it's own via the ballistic trajectory, no need for a gas system. The gas system has a different role as said.

    The terminal velocity depending on forces vs drag relation, it is irrelevant of whether the missiles has heat shield or not. The heat shield only help prevent the air from destroying the missile but has nothing to do with the drag vs force equation.
    If the rail gun project can reach Mach 5 on impact, that mean at Mach 5 the resultant force of gravity minus drag is equal zero. Unless Iskander and others ICBM are significantly more draggy, there is no reason for their terminal velocity to be limited to Mach 3 at impact.
    I put it simply: At sea level the airframe structure would have to be very robust to endure the drag forces of mach 4 and above. More or less no one is willing to build such a heavy airframe to survive mach 4 at sea level. This is the reason why TBMs like the Iskander up to a ICBM RV are designed to de-accelerate down to below mach 4 at ground impact is that this is the best weight-velocity trade-off (expect for not yet realized exotic stuff).

    Consider the design of THAAD with no fin, seperate kill vehicle with divert gas system, you can see that is is mostly a interceptor intended to intercept missiles at very high altitude where the air is extremely thin and the fin doesn't work. So putting a high G value here for either missiles is simply nonsense
    Good and this leads us to the starting point of the discussion. Iskander vs. THAAD.
    So if the kill vehicle of the THAAD has left useful atmosphere and the target, the Iskander, suddenly changes course via its gas system or still effective aerodynamic control: will the THAAD kill vehicle have enough fuel to chase the Iskander? This is the kinematic game that will leave one of the two as looser and these are among methods to counter a ABM system like THAAD.

    Even aircraft with their high lift design will lose significant amount of speed if they attemp high G at altitude, there is no way a tube body with fin only lose 0.3 Mach when trying to do 10G maneuver at Mach 6
    As said: The frontal cross section of a Iskander doing a 10G turn is not changing much because a small change of the vector will cause 10G load at mach 6. That for the tube body argumentation.
    Secondly the impulse of the Iskander is very high with it's mass and velocity. I think just because the speeds are magnitudes different, the comparison of aircrafts and BMs is leading you to wrong results.

  10. #70
    Join Date
    Dec 2015
    Posts
    933
    Quote Originally Posted by PeeD
    Then I recommend you to read the book lightning bolts about MARVs.
    I appreciate your effort. However I think your model is not applicable for this case. The reason is simply empirical knowledge.
    As describes the book lightning bolts is a good read about early US experiences with MARVs, the G numbers mentioned there go up to 50 IIRC.
    If you have the book, evidence, feel free to screenshot and upload it. I highly doubt that they mentioned a MARVs that can turn 50G but iam open to see the evidence. Btw, acceleration of speed due to booster at launch is not the same as acceleration of the turn, i think you may be confuse the two.

    Quote Originally Posted by PeeD
    Then we have weapons like the AIM-54 which reached high speeds and had to perform at least two digit G maneuvers, at least even at 5 G your model would deliver impossible results.
    AIM-54 (or any air to air missiles) cannot perform two digit G maneuver at high altitude like at 24 km (let alone some where like 50 km height). Moreover, the Phoenix is also said to have terrible agility after motor burned out.
    Another factor that you must consider is the ratio between fin area and weight of the AIM-54, it is considerably better than the Iskander so to perform the same maneuver the AIM-54 will need smaller CL.

    Quote Originally Posted by PeeD
    We have effects that have a huge impact:
    - Compared to high velocity objects, low velocity objects are exposed to much lower G loads for the same turn. Hence the vector change of a high velocity object is much smaller for the same G load.
    - At high velocities, the ram air effect on the aerodynamic surfaces is much higher, so the drag force. Hence a low velocity object can have the same drag force at low altitude as a high velocity object at high altitude.
    - Rotation axis position of the fins can be extreme for maximum ram air pressure --> turn capability
    Velocity, air density and fin area are all included in the equation. Moreover, CL of a plate can only increase up to a certain AoA , after that , you will have less CL and more Cd



    Quote Originally Posted by PeeD
    A ballistic missile with a depressed trajectory will appear later on the LOS radar horizon due to curvature of the earth. Only inaccurate OTH radars won't have this problem, but they are useless for engagement
    I don't think the quarsi ballistic trajectory can help reduce LoS since it still have the same top point and only extended the reentry length. Furthermore, let say the altitude of the depressed trajectory is 30 km, ground radar height is 15 meters, the radar horizon would already be 730 km which is longer than maximum range of Iskander already.

    Quote Originally Posted by PeeD
    A ballistic missile does come down on it's own via the ballistic trajectory, no need for a gas system. The gas system has a different role as said.
    It will eventually come down due to gravity, the gas system however is to oriented which part of it will point toward the earth, just like on a space shuttle

    Quote Originally Posted by PeeD
    I put it simply: At sea level the airframe structure would have to be very robust to endure the drag forces of mach 4 and above. More or less no one is willing to build such a heavy airframe to survive mach 4 at sea level. This is the reason why TBMs like the Iskander up to a ICBM RV are designed to de-accelerate down to below mach 4 at ground impact is that this is the best weight-velocity trade-off (expect for not yet realized exotic stuff).
    I can't find any source that stage ICBM are designed to decelerate to below Mach 4 before impact, moreover, the warheads of ballistic missiles doesn't cruise at Mach 5 at sea level, it merely comming down through the atmostphere, so likely spend a few seconds at sea level air density at most

    Quote Originally Posted by PeeD
    Good and this leads us to the starting point of the discussion. Iskander vs. THAAD.
    So if the kill vehicle of the THAAD has left useful atmosphere and the target, the Iskander, suddenly changes course via its gas system or still effective aerodynamic control: will the THAAD kill vehicle have enough fuel to chase the Iskander? This is the kinematic game that will leave one of the two as looser and these are among methods to counter a ABM system like THAAD.
    They both got gas system, however, THAAD is smaller and has a seperate stage for its kill vehicle, it also doesn't carry ECM or decoys.Which mean lower mass. As we already know, F=ma , so generally with same force and lower mass mean better acceleration .Thus, I find no reason to believe that the gas system on Iskander can change its direction quicker than the one on THAAD. Aerodynamic control is kinda useless above 22-25 km.

    Quote Originally Posted by PeeD
    As said: The frontal cross section of a Iskander doing a 10G turn is not changing much because a small change of the vector will cause 10G load at mach 6. That for the tube body argumentation.
    Secondly the impulse of the Iskander is very high with it's mass and velocity. I think just because the speeds are magnitudes different, the comparison of aircrafts and BMs is leading you to wrong results
    The AoA would pretty much depending on the air density, you will need very big AoA to turn 10 G at 50 km altitude ( if the fin even work at all ). Moreover, drag at high velocity and high AoA is much higher so the missiles will lose alot more speed than an aircraft making a turn

  11. #71
    Join Date
    Dec 2003
    Posts
    108
    @garryA

    If you have the book, evidence, feel free to screenshot and upload it. I highly doubt that they mentioned a MARVs that can turn 50G but iam open to see the evidence. Btw, acceleration of speed due to booster at launch is not the same as acceleration of the turn, i think you may be confuse the two.
    Unfortunately I have no access to it at the moment, but I recommend it. The 30-50G numbers were encountered during reentry, there of course a part of it is due to deceleration, I'm aware of that.

    AIM-54 (or any air to air missiles) cannot perform two digit G maneuver at high altitude like at 24 km (let alone some where like 50 km height). Moreover, the Phoenix is also said to have terrible agility after motor burned out.
    Another factor that you must consider is the ratio between fin area and weight of the AIM-54, it is considerably better than the Iskander so to perform the same maneuver the AIM-54 will need smaller CL.
    You misunderstood me: The AIM-54 was designed to operate at around 12km at speeds of mach 4+, there the fins still work. So that a G load would it encounter there, what do you think? Would that G load be consistent with whats possible via your lift formula? I'm quite sure the loads on the AIM-54 @ 12km and mach 4 would be 10+ and this because of the inertia of it at mach 4.
    To put it simply a 10G turn with a F-16 @ mach 0,9 would cause a e.g 45° direction change and expose it completely to the drag. A 10G turn with a AIM-54 @ mach 4 would just cause e.g a 10° direction change.

    There is one issue with the inertia effect at high speeds, which cause much higher G loads due to the velocity for the same change of direction.

    Then there is another issue I tried to explain. Your formula is as it seems for a wing aerofoil lift. Beside that there is a ram-air/drag aerodynamic control method, best explained by the T-50 and J-20 vertical stabilizers: These can turn at an angle which would cause stall with a wing aerofoil, but due to ram air pressure (i.e drag force at high velocity) they can offer high steering power if necessary.
    Your graph is good and shows this effect e.g @ 45° to some extend however I doubt it is also representative for high velocities (but low air density).
    I'm not a aerodynamic guy but I think this, "just lift" approach is the reason why your formula is not applicable in light of different sources talk about such high G numbers at those speeds.

    I don't think the quarsi ballistic trajectory can help reduce LoS since it still have the same top point and only extended the reentry length. Furthermore, let say the altitude of the depressed trajectory is 30 km, ground radar height is 15 meters, the radar horizon would already be 730 km which is longer than maximum range of Iskander already.
    Quasi as with you graph of the Indian TBM not, yes. I just mentioned this effect for ballistic missiles in general, agreed that it's not applicable for the relatively short ranged Iskander.

    It will eventually come down due to gravity, the gas system however is to oriented which part of it will point toward the earth, just like on a space shuttle
    The space shuttle is in a orbit, without gas system gravity and drag would decide that it land on earth in 4 or 5 years. A ballistic missile like the Iskander is on a ballistic trajectory that will take it back to earth in a accurately calculated time and place accurately to seconds not months or years. It's CoG will make sure that it will come down with the nose first.
    The gas system on the Iskander has almost certainly a anti-TBM purpose.

    I can't find any source that stage ICBM are designed to decelerate to below Mach 4 before impact, moreover, the warheads of ballistic missiles doesn't cruise at Mach 5 at sea level, it merely comming down through the atmostphere, so likely spend a few seconds at sea level air density at most
    Yes that's not easy to find. Here is a paper describing it:

    Name:  Fig-14-Altitude-and-cross-range-variation-with-down-range-for-different-reentry-vehicle.png
Views: 279
Size:  120.0 KB

    The Iskander would be the bi-conic conventional here and as you see this is for a mach 12 missile and its decelerated to less than mach 3 at impact. That delta of mach 9, minus some penalties could be used for maneuvering for this missile with the right kinematic management.

    They both got gas system, however, THAAD is smaller and has a seperate stage for its kill vehicle, it also doesn't carry ECM or decoys.Which mean lower mass. As we already know, F=ma , so generally with same force and lower mass mean better acceleration .Thus, I find no reason to believe that the gas system on Iskander can change its direction quicker than the one on THAAD. Aerodynamic control is kinda useless above 22-25 km.
    You assume same gas system power for Iskander and THAAD? We can't really find out which one has more endurance. I could as well say a F-16 has more endurance than a B-1 with that argumentation.
    This is not about quicker, it's about endurance and maybe endgame agility for a evasive maneuver. As said: Expect the Iskander to be detected, the THAAD interceptor launched and then it starts to deviated from the rendezvous point originally calculated and the THAAD has to follow. Now who wins that endurance game? Can the THAAD catch-up to the new impact trajectory that changes continuously? This is one, a exo-atmospheric anti-ABM scenario for Iskander vs. THAAD.

    The AoA would pretty much depending on the air density, you will need very big AoA to turn 10 G at 50 km altitude ( if the fin even work at all ). Moreover, drag at high velocity and high AoA is much higher so the missiles will lose alot more speed than an aircraft making a turn
    The velocities and inertia involved change that picture.
    The Iskander will experience much higher G loads for the same angular vector change due to its high speed --> high inertia, but this also means that the final position change is also much higher due to those velocities.
    The Iskanders fins will have a high drag force as reaction force beside the lift. Just due to the magnitudes higher speed compared to aircraft it will have a high steering capability even if the air density is much lower. If the air density would be much higher at 50km altitude the Iskander would just disintegrate with mach 6 drag forces.

  12. #72
    Join Date
    Dec 2015
    Posts
    933
    Quote Originally Posted by PeeD
    Unfortunately I have no access to it at the moment, but I recommend it. The 30-50G numbers were encountered during reentry, there of course a part of it is due to deceleration, I'm aware of that.
    Honestly, on one hand it is the lift equation and Su-27 manual, on the other hand, it is your memory. I have no choice but to go with the former here.
    Quote Originally Posted by PeeD
    You misunderstood me: The AIM-54 was designed to operate at around 12km at speeds of mach 4+, there the fins still work. So that a G load would it encounter there, what do you think? Would that G load be consistent with whats possible via your lift formula? I'm quite sure the loads on the AIM-54 @ 12km and mach 4 would be 10+ and this because of the inertia of it at mach 4.
    Let consider the AIM-54 then

    The length of front fin is around 0.4 times total missiles length (1.6 meters)
    The length of back fin is around 0.095 total missiles length ( 0.38 meters )
    Missiles diameter is 0.38 meters , wing span is 0.91 meters, so the height of each fin is (0.91 -0.38)/2 = 0.254 meters
    Front fin is practically a triangle so area of each front fin is 0.2 meters square
    Back fin is a rectangle so area of each back fin is 0.097 meters square
    => total lift area is (0.097 + 0.2 )*2 =0.594 meters square.
    AIM-54C loaded weight is 463 kg, using the same assumption that we had for Iskander ( 1/2 total weight is propellant ), the AIM-54C will be around 231 kg at burn out. To make 10G turn , the missiles will need to generate aerodynamic lift of 2,310 kg ( 22,638 Newton).Speed at burn out of AIM-54 is Mach 5
    Using the same lift equation: lift = CL* air density* 0.5*velocity^2 *wing area
    At altitude of 12 km , air density will be 0.31 kg/m3 , speed of sound is 295 meters/seconds ( so Mach 5 will be 1475 meters/seconds)
    So 22,638 = CL*0.31*0.5*1475^2 *0.549
    => 22,638 = CL*185,134.8
    => CL = 0.122
    In short, to turn 10 G at Mach 5 , altitude of 12 km, AIM-54 only need lift coefficient 1/10 times as good as a Su-27. As you see. The number make sense for AIM-54.

    Quote Originally Posted by PeeD
    To put it simply a 10G turn with a F-16 @ mach 0,9 would cause a e.g 45° direction change and expose it completely to the drag. A 10G turn with a AIM-54 @ mach 4 would just cause e.g a 10° direction change
    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.

    In other words, an AIM-54 turnning 10 degrees may expose more to drag than an F-16 turnning 45 degrees , if AIM-54 need higher AoA for the turn.

    Quote Originally Posted by PeeD
    There is one issue with the inertia effect at high speeds, which cause much higher G loads due to the velocity for the same change of direction.
    The force required to satisfy F=ma equation is the same regardless of speed. The turn rate and turn radius will change depending on speed but force required for turn acceleration is the same

    Quote Originally Posted by PeeD
    Then there is another issue I tried to explain. Your formula is as it seems for a wing aerofoil lift. Beside that there is a ram-air/drag aerodynamic control method, best explained by the T-50 and J-20 vertical stabilizers: These can turn at an angle which would cause stall with a wing aerofoil, but due to ram air pressure (i.e drag force at high velocity) they can offer high steering power if necessary.
    Firstly, lift equation can be applied to the whole aircraft not only the air foil. Secondly, lift is the way to call the force component that help you turn/steer aircraft toward desire direction, you can turn/ steer the fin past AoA where Cl is maximum but after that point the force that help you turn the aircraft/ missiles doesn't increase, they reduce. Instead what you increase is the drag. Btw, nose pointing is not the same as turning

    .


    Quote Originally Posted by PeeD
    The space shuttle is in a orbit, without gas system gravity and drag would decide that it land on earth in 4 or 5 years. A ballistic missile like the Iskander is on a ballistic trajectory that will take it back to earth in a accurately calculated time and place accurately to seconds not months or years. It's CoG will make sure that it will come down with the nose first.
    The gas system on the Iskander has almost certainly a anti-TBM purpose.
    ICBM are not in orbit and they still need a gas system. Moreover, without a gas system how can Iskander even follow a quasi ballistic path ? it require the missiles to be horizontal to the atmosphere instead of near vertical. Furthermore, if Iskander truly carry decoy, dropping them surely change center of gravity too.

    Quote Originally Posted by PeeD
    The Iskander would be the bi-conic conventional here and as you see this is for a mach 12 missile and its decelerated to less than mach 3 at impact. That delta of mach 9, minus some penalties could be used for maneuvering for this missile with the right kinematic management.
    I honestly don't understand what that graph trying to say. So all ballistic missiles has re-entry speed below Mach 2 when altitude smaller than 20 km ?
    Furthermore, that still not really justify wasting kinetic energy for the turn. While terminal velocity in the last few km may be similar , you still sacrificed considerable cruise velocity compared to missiles that doesn't turn

    Quote Originally Posted by PeeD
    You assume same gas system power for Iskander and THAAD? We can't really find out which one has more endurance. I could as well say a F-16 has more endurance than a B-1 with that argumentation.
    This is not about quicker, it's about endurance and maybe endgame agility for a evasive maneuver. As said: Expect the Iskander to be detected, the THAAD interceptor launched and then it starts to deviated from the rendezvous point originally calculated and the THAAD has to follow. Now who wins that endurance game? Can the THAAD catch-up to the new impact trajectory that changes continuously? This is one, a exo-atmospheric anti-ABM scenario for Iskander vs. THAAD.
    B-1 may has more endurance but it can't run aways from an F-16. Considering there is no ( very little) air in space, endurance would be the smallest problem. Moreover, there is no way Iskander know where is the interceptor so the best it could do is maneuver randomly, unlike THAAD will only maneuver when needed . So i think THAAD should win the endurance game. TBH, i honestly doubt that Iskander carry that much fuel for its gas system either, after all the alleged ECM , decoys and what not

    Quote Originally Posted by PeeD
    The velocities and inertia involved change that picture.
    The Iskander will experience much higher G loads for the same angular vector change due to its high speed --> high inertia, but this also means that the final position change is also much higher due to those velocities.
    The Iskanders fins will have a high drag force as reaction force beside the lift. Just due to the magnitudes higher speed compared to aircraft it will have a high steering capability even if the air density is much lower. If the air density would be much higher at 50km altitude the Iskander would just disintegrate with mach 6 drag forces.
    You are really confusing between lift and drag. Drag slow you down but not helping you change direction. The force component that help change direction is called lift
    Last edited by garryA; 11th May 2017 at 05:54.

  13. #73
    Join Date
    Dec 2003
    Posts
    108
    @garryA

    As said I think the problem here is that you consider aircraft aerodynamic at aircraft speeds.
    You talk about lift which does not enter stall region.
    I however think that the ram air/drag on fins at hypervelocity well within stall AoA would cause a huge moment on the center of pressure. This is not aerofoil lift anymore but a offset airbreak in stall region.

    Look at your chart:

    At 50° AoA the combined force of lift and drag are 2,5 while 2 for the non-stall region at 15° AoA. Now I doubt that this chart is applicable for our velocities and air pressures.



    In short, to turn 10 G at Mach 5 , altitude of 12 km, AIM-54 only need lift coefficient 1/10 times as good as a Su-27. As you see. The number make sense for AIM-54.
    Here it was my fault to request a calculation for 12km altitude while that of the Iskander was done at 24km. Indeed the airpressure at a conventional altitude of 12km is very much higher.
    But again, the empirical knowledge we have about early US MARV testing could be explained by applying the aerodynamic control like described above or the maneuvering took place at well below 20km. I go for the former taking into account the high deceleration G forces.

    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.
    Agreed. Hence in the context I don't expect high drag losses due to the Iskanders tube body as you suggested, more so at those speeds/inertia and low air pressures.

    In other words, an AIM-54 turnning 10 degrees may expose more to drag than an F-16 turnning 45 degrees , if AIM-54 need higher AoA for the turn.
    The turn is not so important, its the load factor experienced that counts. Iskander would try to do a maneuver and create a G load that the THAAD at similar speeds can't do and vice versa.

    The force required to satisfy F=ma equation is the same regardless of speed. The turn rate and turn radius will change depending on speed but force required for turn acceleration is the same
    Yes and here due to the high inertia/speed the Iskanders tube body will not create the huge drag losses you suggested in order to reach 10 or 30G.

    Firstly, lift equation can be applied to the whole aircraft not only the air foil. Secondly, lift is the way to call the force component that help you turn/steer aircraft toward desire direction, you can turn/ steer the fin past AoA where Cl is maximum but after that point the force that help you turn the aircraft/ missiles doesn't increase, they reduce. Instead what you increase is the drag. Btw, nose pointing is not the same as turning
    What you want is a reaction force that creates a moment on your center of pressure. Whether its efficient lift or brute force frag in stall region does not matter, you just need the reaction force to pull our G loads.

    ICBM are not in orbit and they still need a gas system. Moreover, without a gas system how can Iskander even follow a quasi ballistic path ? it require the missiles to be horizontal to the atmosphere instead of near vertical. Furthermore, if Iskander truly carry decoy, dropping them surely change center of gravity too.
    Neither that Indian TBM nor early generation ICBMs have a exo-atmospheric gas system. On a ballistic path, anything that has a heavier nose (CoG), will come down with the nose on a strict and well defined ballistic trajectory. The decoys and other potential penetration aids of the Iskander wont change the tip CoG.

    I honestly don't understand what that graph trying to say. So all ballistic missiles has re-entry speed below Mach 2 when altitude smaller than 20 km ?
    Furthermore, that still not really justify wasting kinetic energy for the turn. While terminal velocity in the last few km may be similar , you still sacrificed considerable cruise velocity compared to missiles that doesn't turn
    No, the graph says that the speed of a conventional tube RV starts to reduce at 20-25km and that so significantly from mach 12 at 25km to mach 3 at sea level, 9 mach numbers are lost due to drag/friction. It is a little confusing because both boost and terminal phase are displayed in one graph, but with some concentration on just the dot line you may understand it.
    In that case the RV airframe is exposed to dynamic pressure loads of well beyond 320kPa. Now the Iskander is a more fragile airframe than a RV and dynamic pressure loads below 320kPa are that engineering practice tells to be a sane trade-off between airframe weight and structural strenght.
    The Iskander seems to have a peak at mach 6 and needs to slow down to mach 3 at sea level to avoid plastification/disintegration of the airframe. A good kinematic management/profile would use 2 mach numbers early on for maneuvering via fins, leave one mach number for drag losses after 25km and enter the region below 25km at mach 4 instead of mach 6, in order to keep mach 3 at impact.

    B-1 may has more endurance but it can't run aways from an F-16. Considering there is no ( very little) air in space, endurance would be the smallest problem. Moreover, there is no way Iskander know where is the interceptor so the best it could do is maneuver randomly, unlike THAAD will only maneuver when needed . So i think THAAD should win the endurance game. TBH, i honestly doubt that Iskander carry that much fuel for its gas system either, after all the alleged ECM , decoys and what not
    A B-1 will almost always run away from a F-16 because of its larger fuel reserves/kinematics/endurance and this is just a example against your argumentation that the smaller THAAD KV can chase the Iskander due to its small size.

    The Iskander may maneuver randomly in a profile that bleeds the THAAD KV fuel in the best way and may have a RF proximity sensor for a last ditch maneuver. In the end such maneuvering can account for many dozens of km course changes. This is a numbers game, about margins, envelopes, times, we can just talk about the basic concepts and feasibility as we will never know for sure.

    You are really confusing between lift and drag. Drag slow you down but not helping you change direction. The force component that help change direction is called lift
    This is conventional aircraft, sub to low supersonic thinking. For pure kinematics I just need a reaction force that creates a moment to change course, if its hypervelocity drag instead of lift, so be it.

    Because you are a friend of conventional aircraft aerodynamics I have a interesting question for you: What would happen if (just) the left vertical "fin" (vertical stabilizer) of the T-50 would move to a 60° AoA position at mach 1,6 assuming that its structure would be strong enough not to disintegrate?

  14. #74
    Join Date
    Dec 2015
    Posts
    933
    Quote Originally Posted by PeeD
    As said I think the problem here is that you consider aircraft aerodynamic at aircraft speeds.
    You talk about lift which does not enter stall region.
    I however think that the ram air/drag on fins at hypervelocity well within stall AoA would cause a huge moment on the center of pressure. This is not aerofoil lift anymore but a offset airbreak in stall region.
    Look at your chart:
    At 50° AoA the combined force of lift and drag are 2,5 while 2 for the non-stall region at 15° AoA. Now I doubt that this chart is applicable for our velocities and air pressures
    The lift equation is applicable at all speed because at the core, lift is basically reaction force due to Newton law, they are used to estimate missiles agility too, not just aircraft. So your argument that i only consider aerodynamic at aircraft speed is fundamentally wrong. Moreover, you still fail to graps the different between nose pointing and turning (aka changing direction). So i will try one last time. If you still don't understand i just have to pass. Look at the photo below

    The first picture is when Iskander falling straight down ( it is never the case but for the sake of simplification, just imagine your missiles falling down with angle of 90 degrees), 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 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 )



    Quote Originally Posted by PeeD
    Here it was my fault to request a calculation for 12km altitude while that of the Iskander was done at 24km. Indeed the airpressure at a conventional altitude of 12km is very much higher.
    But again, the empirical knowledge we have about early US MARV testing could be explained by applying the aerodynamic control like described above or the maneuvering took place at well below 20km. I go for the former taking into account the high deceleration G forces.
    Even if i consider you emirical knowledge to be somewhat accurate
    1/ MIRV if come from ICBM can move at speed of around Mach 15-20, much faster than what Iskander capable of, they will have considerably more lift for maneuver
    2/ Iskander is a single stage missiles, so it is considerably heavier than MRIV that seperated from their carrier, thus require much more lift for the same maneuver
    3/ deceleration is not the same as turning
    The most practical possibility is that Iskander can perform high G at low altitude, like 5 km or something similar. So it is possibly defense against something like PAC-2, PAC-3

    Quote Originally Posted by PeeD
    Agreed. Hence in the context I don't expect high drag losses due to the Iskanders tube body as you suggested, more so at those speeds/inertia and low air pressures.
    Iskander has high speed and very high wing loading while Cl is low as well, so expect it to lose considerable amount of speed if attemped to maneuver.

    Quote Originally Posted by PeeD
    The turn is not so important, its the load factor experienced that counts. Iskander would try to do a maneuver and create a G load that the THAAD at similar speeds can't do and vice versa.
    G load isn't important, what important for a turn is ability to change direction

    Quote Originally Posted by PeeD
    Yes and here due to the high inertia/speed the Iskanders tube body will not create the huge drag losses you suggested in order to reach 10 or 30G
    inertia is irrelevance and high speed will mean more drag since the formular for drag is also propotional to speed



    Quote Originally Posted by PeeD
    What you want is a reaction force that creates a moment on your center of pressure. Whether its efficient lift or brute force frag in stall region does not matter, you just need the reaction force to pull our G loads.This is conventional aircraft, sub to low supersonic thinking. For pure kinematics I just need a reaction force that creates a moment to change course, if its hypervelocity drag instead of lift, so be it.
    Because you are a friend of conventional aircraft aerodynamics I have a interesting question for you: What would happen if (just) the left vertical "fin" (vertical stabilizer) of the T-50 would move to a 60° AoA position at mach 1,6 assuming that its structure would be strong enough not to disintegrate?
    See reply above about nose pointing vs turning.
    About your question: the air flow will collide with the vertical fin, thus point the aircraft to the same side ( if fin turn to the left then aircraft will point to the left). Once the aircraft point to same side, the air flow on the airframe and the engine thrust will make it change direction

    Quote Originally Posted by PeeD
    Neither that Indian TBM nor early generation ICBMs have a exo-atmospheric gas system. On a ballistic path, anything that has a heavier nose (CoG), will come down with the nose on a strict and well defined ballistic trajectory. The decoys and other potential penetration aids of the Iskander wont change the tip CoG.
    I don't follow Indian TBM program so i don't know abot that, but AFAIK, LGM-30 Minuteman has a gas system, so is Trident. Btw how can you know if decoys and ECM will change Iskander CoG or not when we don't even have a photos or diagram of the alleged system ?. Moreover, without gas system, Iskander won't be able to follow a quarsi ballistic path either.

    Quote Originally Posted by PeeD
    The Iskander seems to have a peak at mach 6 and needs to slow down to mach 3 at sea level to avoid plastification/disintegration of the airframe. A good kinematic management/profile would use 2 mach numbers early on for maneuvering via fins, leave one mach number for drag losses after 25km and enter the region below 25km at mach 4 instead of mach 6, in order to keep mach 3 at impact
    It more likely that Iskander if it indeed perform evasive maneuver will do so at low altitude since it clearly doesn't have enough lift for that at high altitude.

    Quote Originally Posted by PeeD
    A B-1 will almost always run away from a F-16 because of its larger fuel reserves/kinematics/endurance and this is just a example against your argumentation that the smaller THAAD KV can chase the Iskander due to its small size.

    The Iskander may maneuver randomly in a profile that bleeds the THAAD KV fuel in the best way and may have a RF proximity sensor for a last ditch maneuver. In the end such maneuvering can account for many dozens of km course changes. This is a numbers game, about margins, envelopes, times, we can just talk about the basic concepts and feasibility as we will never know for sure.
    I honestly, don't see how B-1 can run away from an F-16 unless it has a few hundred km head start. Back to THAAD vs Iskander case, i don't see how Iskander has more reserve either when it supposed to carry ECM, decoys and what not, or how is it even matter in exdo atmosphere when there is almost no air ( no drag) ?. The engagements time between a Mach 6 and Mach 8 missiles will be extremely short too so things like RF proximity sensor are unlikely to work ( and there isn't much time to drag out many maneuvers either).
    Last edited by garryA; 12th May 2017 at 08:09.

  15. #75
    Join Date
    Dec 2003
    Posts
    108
    @garryA

    The lift equation is applicable at all speed because at the core, lift is basically reaction force due to Newton law, they are used to estimate missiles agility too, not just aircraft. So your argument that i only consider aerodynamic at aircraft speed is fundamentally wrong.
    Ok, I try to explain why your lift equation is not applicable.
    Your lift is also applicable to high speeds, yes. Is it the only usable reaction force?
    Your aircraft aerodynamics where lift is everything because the use of aerofoils and never going to stall region and beyond, plus efficiency consideration is wrong for our hypervelocity missile case.

    I use your own graphs and sources to explain why: http://www.aerospaceweb.org/question...cs/q0194.shtml

    This is your lift and normal force diagram for a random case:



    At 15° AoA we see a stall. The max. lift value reached before stall is 1.

    Here is your aerofoil lift diagram with an additional normal force vector and components/resultants, defined by N and A (more explanation is given at the website):



    and here is the related diagram showing drag and axial force.



    For a reaction force to create a moment towards the center or pressure, the most effective situation is thus as following:

    Fin is at the AoA of 90°, drag = normal force. This is at a factor of 2 more effective than your aerofoil lift outside stall region.

    This error of a factor of two is due to your usage of the lift equation for maximum steering of a hypervelocity missile.

    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 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

    G load isn't important, what important for a turn is ability to change direction
    I understand that. Let's just keep the discussion within context.
    We have two objects with roughly the same velocity, Iskander and THAAD both do maneuvers of a certain G value.
    Both are in a environment with a very low air density.
    Due to the inertia of both, a relatively small change of AoA will cause a very high G load that could reach the structural limit of the airframe allowed. Hence aerodynamic losses due to the exposure of the airframe is lower at high speeds due to the smaller AoA.

    Result is that the one of the two that can do a higher G maneuver may win.

    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 )
    Correct, I should not have added the two. Above I detailed that and I hope to have shown you why your calculation is potentially at a factor of 2 wrong because you just consider lift.

    1/ MIRV if come from ICBM can move at speed of around Mach 15-20, much faster than what Iskander capable of, they will have considerably more lift for maneuver
    Yes, a ICBM, IRBM, MRBM, all have more potential for maneuvering, either via lift, dynamic pressure/drag or gas steering.

    2/ Iskander is a single stage missiles, so it is considerably heavier than MRIV that seperated from their carrier, thus require much more lift for the same maneuver
    Actually the Iskander appears to fly a depressed trajectory going into a glide phase at terminal. It will likely remain in a altitude region where due to it's speed a sufficient amount of dynamic pressure is available to react on its fins, as said, almost certainly outside non-stall lift region. The gas system is a secondary mean for maneuvering. Hence the lift or better said steering situation is likely quite good, just outside PAC-3 envelope.

    Iskander has high speed and very high wing loading while Cl is low as well, so expect it to lose considerable amount of speed if attemped to maneuver.
    It more likely that Iskander if it indeed perform evasive maneuver will do so at low altitude since it clearly doesn't have enough lift for that at high altitude.
    The Iskander will need small AoA changes for high G maneuvers at it's speed and it's operating altitude as well as secondary gas system will help it to keep aerodynamic losses low enough for useful anti-ABM maneuvering.


    inertia is irrelevance and high speed will mean more drag since the formular for drag is also propotional to speed
    Inertia is relevant because the higher it is the lower the AoA for direction change, hence the lower the losses due to exposure of airframe to drag. Interia --> G load on airframe becomes the limiting factor.

    In the formula for drag, speed is with ^2, not proportional.

    About your question: the air flow will collide with the vertical fin, thus point the aircraft to the same side ( if fin turn to the left then aircraft will point to the left). Once the aircraft point to same side, the air flow on the airframe and the engine thrust will make it change direction
    Would you describe that T-50 case as an effect of lift or a effect of drag/dynamic pressure/ram-air-pressure as I described above?

    I don't follow Indian TBM program so i don't know abot that, but AFAIK, LGM-30 Minuteman has a gas system, so is Trident.
    I doubt the Minueteman I had a steering gas system, only retro boosters.

    Btw how can you know if decoys and ECM will change Iskander CoG or not when we don't even have a photos or diagram of the alleged system ?. Moreover, without gas system, Iskander won't be able to follow a quarsi ballistic path either.
    I can know because Iskander works. I also know because the options for other re-entry points like aft or side can be excluded due to the burnt booster at aft and warhead at tip. The result will always be a re-entry with the top, also because ECM and decoys are certainly at the top half of the tube body. A quasi-ballistic path can be a depressed one with a following glide phase as proposed for the Iskander.

    Back to THAAD vs Iskander case, i don't see how Iskander has more reserve either when it supposed to carry ECM, decoys and what not, or how is it even matter in exdo atmosphere when there is almost no air ( no drag) ?. The engagements time between a Mach 6 and Mach 8 missiles will be extremely short too so things like RF proximity sensor are unlikely to work ( and there isn't much time to drag out many maneuvers either).
    I didn't say that Iskander has more fuel reserves for its gas system, which one can pull and endure more G and loads, if a RF proximity sensor is present and working.
    These are all detail design parameters, all we can speculate about is that these things are physically possible and hence must be taken into consideration when judging both systems against each other.

    I honestly, don't see how B-1 can run away from an F-16 unless it has a few hundred km head start.
    Just let them start at sea level and with a distance of 50km. The F-16 will be out of fuel at 1/3 the max. range of the B-1. This argument was against your claim that the THAAD KV has more endurance because it's smaller.

  16. #76
    Join Date
    Dec 2015
    Posts
    933
    It feels like talking to a wall trying to explain aerodynamic to you. I gave the equation, the calculation, the detail diagram explanation. But you just keep repeating your wrong idea again and again without even trying to understand where did you get it wrong. Ffs, how many times do you want me to repeat that lift equation is applicable at all speed and nose pointing is different from turning until you even consider it ???.
    That it, you are free to believe whatever fairy tales you want, this was a waste of time.

    Last edited by garryA; 12th May 2017 at 14:23.

  17. #77
    Join Date
    Oct 2013
    Posts
    1,873
    ^ I admire your patience

  18. #78
    Join Date
    Dec 2015
    Posts
    933
    Quote Originally Posted by mig-31bm
    I admire your patience
    Being patience doesn't work out quite well for me TBH.
    How does he even come up with " fin at 90 degrees equal most efficiency turning" is beyond me. It is ridiculous even in nose pointing case
    Last edited by garryA; 12th May 2017 at 15:24.

  19. #79
    Join Date
    Dec 2003
    Posts
    108
    I try to explain what has been witnessed in the real world with what you put against it.

    US MARV testing described in the book lighting bolts is more convincing for me than a single formula. I might have not enough knowledge about aerodynamic to explain it, but of course I will question your argumentation against something that actually happened during testing.

    So you should not be too offended and rather move to the other topics of this discussion if you think I'm wrong on this one out of some 10 points regarding the Iskander vs. THAAD discussion.

    This one point, the lift equation could well be applicable to our case yes but your comparison to the Su-27 is the mistake of yours that gives a wrong impression, a wrong argumentation. I'm not familiar with aerodynamic equations but if you are, I wonder why you don't get what I try to say? I just realized that the aerodynamic reaction force beyond stall region is higher than im effective lift region, your Cl value for Su-27 within effective lift region is thus not representative. Something didn't make sense and I tried to find out why.

    I don't expect from you to try to explain the high G loads of US MARV testing and its good for me that you try your best to bring up arguments against it. However you seems to want some kind of victory out of this discussion, so enjoy it if you feel so.

  20. #80
    Join Date
    Dec 2015
    Posts
    933
    Quote Originally Posted by PeeD
    I try to explain what has been witnessed in the real world with what you put against it.

    US MARV testing described in the book lighting bolts is more convincing for me than a single formula. I might have not enough knowledge about aerodynamic to explain it, but of course I will question your argumentation against something that actually happened during testing.

    So you should not be too offended and rather move to the other topics of this discussion if you think I'm wrong on this one out of some 10 points regarding the Iskander vs. THAAD discussion.

    This one point, the lift equation could well be applicable to our case yes but your comparison to the Su-27 is the mistake of yours that gives a wrong impression, a wrong argumentation. I'm not familiar with aerodynamic equations but if you are, I wonder why you don't get what I try to say? I just realized that the aerodynamic reaction force beyond stall region is higher than im effective lift region, your Cl value for Su-27 within effective lift region is thus not representative. Something didn't make sense and I tried to find out why.

    I don't expect from you to try to explain the high G loads of US MARV testing and its good for me that you try your best to bring up arguments against it. However you seems to want some kind of victory out of this discussion, so enjoy it if you feel so.
    I didn't get annoyed when you are wrong. I get annoyed when you repeat the same thing again and again even though said things have been elaborated in previous posts. 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, that why it is not considered. I gave you the diagram direction of the force. I even go as far as telling you the component of the force when missiles is at an AoA with the air flow. You can even draw a diagram yourself with actual force and angle value to see the acceleration direction. Adding up resultant forces with different direction is high school physics, you should be able to do it with else
    The fact that you said most efficient turning is at 90 degrees shown that you didn't even bothered to read anything i wrote at all, nor do you draw it on a paper to see the direction of the force when missiles or its fin is at 90 degrees with air flow.
    Even in MARV case, i told you before that it is very different situation from what we discussing here since MARV can potentially move at Mach 15-20, much faster than what Iskander capable of. They are also lighter because they are seperate stage. The chance for them to make high G turn is much higher because they can generate more lift (due to much high speed of several Mach) and has less required force (due to lighter weight) . Just like the comparision with AIM-54, the AIM-54 can generate more lift ( due to fin area and air density ) while has much smaller required force ( due to smaller G and lighter weight ) so the number make sense for AIM-54 while it doesn't force Iskander.
    The point is, if numbers , equation and diagram can't convince you then there is no reason for me to continues with others parts of discussion, it will just going in a circle like last time until i get bored and can't be bothered to reply.
    Last edited by garryA; 12th May 2017 at 16:44.

  21. #81
    Join Date
    Dec 2003
    Posts
    108
    You get annoyed because I repeat by objections. I took some time and checked your lift equation to understand it. I now know how to formulate my objections in the context of the lift formula.

    The key is Cl/Cd: The details like AoA are there. I objected that the AoA of the Su-27 Cl does not go beyond stall region while the fins of a missile might do to reach the maximum Cn value of your chart (at 90°). This changes Cl by a factor of 2.
    Hence I accept the lift equation for all speeds but not the direct quantification of it via Su-27 values.
    It's up to you whether you want to understand that somewhat complex detail or dismiss it and my understanding of the issue, as I doubt I can put it more understandable than this.

    This is one mean to explain high G values of MARV I found.

    I appreciate the two calculations for Iskander at 25km and AIM-54 at 12km altitude, it showed to me that altitudes of around 15km are more feasable for aerodynamic maneuvering. My goal is to conclude the Iskander vs. THAAD discussion we started.
    So let me state a new more refined scenario for the Iskander to defeat THAAD and other ABM:
    A depressed trajectory with an apogee of 50km (Iskander is kinematic strong enough to fly such a less efficient trajectory). At terminal phase it would just glide via body lift.
    If it would glide between 40 to 25km for the last 200km, nothing could intercept it until it dives for last terminal phase.

    I agree that Iskander on a high ballistic trajectory that would put it into the altitude envelope of the THAAD, has only its gas system working. Hence our discussion can be closed:
    1. The Iskander is said to fly a depressed trajectory, outside THAAD envelope.
    2. The specifics of the Iskander and THAAD gas system and endurance would be unknown to us even if the Iskander would for some reason fly a high ballistic trajectory. In such a gas system based high altitude encounter, drag and lift would play no role.

    Below 20km which is a more likely altitude for heavy evasive aerodynamic maneuvering PAC-2 and PAC-3 would become the opponents. There we could open up the lift discussion and your nose pointing discussion. However I don't feel a need for that anymore. As discussed it will use its 2-3 mach numbers kinematic reserve for aerodynamic maneuvering, you think its a small amount of maneuvering, I think its more.

    The fact that you said most efficient turning is at 90 degrees shown that you didn't even bothered to read anything i wrote at all, nor do you draw it on a paper to see the direction of the force when missiles or its fin is at 90 degrees with air flow.
    This is a good example for how you either don't understand what I say or get it wrong for some other reason, this is what I said in fact:

    "Fin is at the AoA of 90°, drag = normal force. This is at a factor of 2 more effective than your aerofoil lift outside stall region."

    I hope I know the difference between effective and efficient and I don't know how you got drag/normal force (reaction force) for turn.
    I wonder how you can save the information of my text in such a wrong way in your memory.

  22. #82
    Join Date
    Dec 2015
    Posts
    933
    Quote Originally Posted by PeeD
    The key is Cl/Cd: The details like AoA are there. I objected that the AoA of the Su-27 Cl does not go beyond stall region while the fins of a missile might do to reach the maximum Cn value of your chart (at 90°). This changes Cl by a factor of 2.
    Hence I accept the lift equation for all speeds but not the direct quantification of it via Su-27 values.
    It's up to you whether you want to understand that somewhat complex detail or dismiss it and my understanding of the issue, as I doubt I can put it more understandable than this.
    ........................sigh

    Maximum CN is not important, because it represents the total of lift (Cl) and drag (Cd) . To turn, you want high lift not high drag. The only thing high drag would do is slow missiles down and that is not what you want. To turn better, you want to maximize the force toward desired direction of travel.
    Look at the picture i gave you, why can Cd get so high at 90 degrees while there is no AoA where Cl can get that high ?. No, not because of anything to do with stalling. It is because at 90 degree, the fin is perpendicular to the air flow.At which point 100% of normal force is drag.On the otherhand, there is no situation where 100% normal force is lift (no, there is no opposite situation where drag is 0% and lift is 100%). Furthermore, lift cannot reach value as high as drag, because that it will require the air flow perpendicular to the airframe (which will not happen)


    Take out a piece of paper and draw the force distribution on it and you will see.

    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.
    Btw when i talked about Su-27'CL at 18 degree AoA. It is not the horrizontal tail of Su-27 at 18 degrees with airflow but its whole airframe at AoA of 18 degrees. Some what like this

    Now try to imagine the Iskander fly at similar AoA or higher at Mach 5

    Quote Originally Posted by PeeD
    I appreciate the two calculations for Iskander at 25km and AIM-54 at 12km altitude, it showed to me that altitudes of around 15km are more feasable for aerodynamic maneuvering
    I haven't done the calculation for 15 km yet, but remember the AIM-54 is alot lighter, has very low wing loading compared to Iskander and was only done a 10 G maneuver in the 12 km case.

    Quote Originally Posted by PeeD
    If it would glide between 40 to 25km for the last 200km, nothing could intercept it until it dives for last terminal phase.
    1. The Iskander is said to fly a depressed trajectory, outside THAAD envelope.
    what make you think not things can intercept Iskander before it dive for terminal phase ? there are still SM-2, SM-6 and PAC-3MSE

    Quote Originally Posted by PeeD
    "Fin is at the AoA of 90°, drag = normal force. This is at a factor of 2 more effective than your aerofoil lift outside stall region."

    I hope I know the difference between effective and efficient and I don't know how you got drag/normal force (reaction force) for turn.
    I wonder how you can save the information of my text in such a wrong way in your memory.
    It is not 2 times more effective because drag component of normal force doesnot contribute to the force required for your turn.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.Drag force doesn't count because it is not the force toward desired direction of travel. I don't know how i can make this anymore simple.
    Last edited by garryA; 13th May 2017 at 06:09.

  23. #83
    Join Date
    May 2013
    Posts
    218
    How the THAAD can against DF-26 ?

    China says it has successfully tested new type of missile

    http://www.foxnews.com/world/2017/05...e-missile.html

    The DF-26 has a range of> 2000 km, Mach 10 speed (as it descends from> 800 km at high-altitude)

    THAAD has a range of 200 km, altitude 150km. The TPY-2 radar has a range of 1000km with unknown altitude ?!
    Last edited by blackadam; 13th May 2017 at 04:32.

  24. #84
    Join Date
    Jun 2004
    Location
    Columbia, MD
    Posts
    11,530
    None of those questions are that simple. Ballistic Missile radar performance much like any other radar performance is mode dependent and ABM capability depends upon a lot of variables. THAAD has been designed to defeat the SRBM, MRBM, and IRBM ranged missiles and has so far been tested against the SRBM and MRBM ranged targets with IRBM testing scheduled in the near term.
    Old radar types never die; they just phased array

  25. #85
    Join Date
    Aug 2011
    Posts
    4,370
    what make you think not things can intercept Iskander before it dive for terminal phase ? there are still SM-2, SM-6 and PAC-3MSE
    Are all those systems located in same area. its very expensive way of defense. infact these can be saturated with smaller cruise missiles like Kh-35 and let Iskander fly through after portable missile strike.

  26. #86
    Join Date
    Dec 2003
    Posts
    108
    @garryA

    I realize why we have to repeat ourselves so many times....

    You talked about highschool physics: So in your physics lift is a usable reaction force but drag not. Your lift equation with the fin area creates a Cl that can be used for generating a moment in a direction on the missile but Cn can't be used for that. Lift force can be used as reaction force to create a moment towards the center of pressure but the higher quantity normal-force/drag is for some reason not usable...

    Lets close the case and say we have fundamental different understandings of mechanics/physic. More so as I already agreed that air is apparently too thin at 25km for high G aerodynamic maneuvers (fortunately for the Iskander this is also true for the ABM interceptors with aerodynamic steering/hybrids).

    what make you think not things can intercept Iskander before it dive for terminal phase ? there are still SM-2, SM-6 and PAC-3MSE
    So SM-2 can work in exo-atmospheric conditions above ~25km?
    SM-6/SM-3 has no kill vehicle altitude limitation like THAAD and can engage at ~30km?
    PAC-3MSE improved the interception altitude to above ~30km?

    The argumentation here is that Iskander would fly in a altitude-band where ABM interceptors have engagement problems and dive down at ~90° in the very last part of the flight.

  27. #87
    Join Date
    Dec 2015
    Posts
    933
    Quote Originally Posted by PeeD
    I realize why we have to repeat ourselves so many times....

    You talked about highschool physics: So in your physics lift is a usable reaction force but drag not. Your lift equation with the fin area creates a Cl that can be used for generating a moment in a direction on the missile but Cn can't be used for that. Lift force can be used as reaction force to create a moment towards the center of pressure but the higher quantity normal-force/drag is for some reason not usable...
    No, there is no mine or your physics here. Physics laws are not like religions, it doesn't change from person to person
    When i mentioned high school, i mean you should know how to add up resultant force when direction is not the same because that stuff are taught at high school

    Lift equation are not taught at school but if you know how to add up resultant force from 2 forces with different direction you will understand what is Cn when you look at the diagram and why it is not used to measure turn acceleration.
    The reason why drag force and normal force cannot be used when measure G load is simple, normal force is simply the resultant force of both lift and drag, drag force is just like its name, literally opposite the air flow and slow down the missiles. Both drag force and normal force (normal is in fact already included drag force) doesn't point toward acceleration direction of the turn. As a result, they doesn't contribute toward how many G the aircraft or missiles could turn.
    Quote Originally Posted by PeeD
    More so as I already agreed that air is apparently too thin at 25km for high G aerodynamic maneuvers
    (fortunately for the Iskander this is also true for the ABM interceptors with aerodynamic steering/hybrids).
    So SM-2 can work in exo-atmospheric conditions above ~25km?
    SM-6/SM-3 has no kill vehicle altitude limitation like THAAD and can engage at ~30km?
    PAC-3MSE improved the interception altitude to above ~30km?
    The argumentation here is that Iskander would fly in a altitude-band where ABM interceptors have engagement problems and dive down at ~90° in the very last part of the flight.
    Interceptor are launched from sea level to high altitude, as such they have plenty of time to correct their course in high density air. Moreover, system such as SM-6, SM-2 has very high ratio of wing area vs mass compared to something like Iskander or any ballistic missiles .Moreover, in reality, anti ballistic missiles missiles are launched to intercept instead of chase, so they are guided by radar to reach a specific coordinate at a specific point in time. You can imagine it like thowing a trap toward the road just so car run over it.
    Regarding those interceptors listed above


    SM-3 with seperate kill vehicle is for very high altitude, exo-atmospheric condition


    THAAD with moveable nozzle and mid body jet system is for high altitude intercept ,mostly exo-atmospheric condition between 40-150 km


    PAC-3MSE has both mid body jet system and aerodynamic fin can operate both edo and exo atmoshpheric condition


    Both SM-2 and SM-6 can be used against target descend inside atmosphere since they both have aerodynamic fin
    Last edited by garryA; 13th May 2017 at 07:35.

  28. #88
    Join Date
    Dec 2003
    Posts
    108
    @garryA

    No, there is no mine or your physics here. Physics laws are not like religions, it doesn't change from person to person
    Exactly. Because you does not accept Cn or drag as a usable reaction force I don't accept your understanding of physics.
    To be more direct, you are fundamentally wrong and I don't want to argue about it with you. As you put it well, for me its also like talking against a wall.

    Interceptor are launched from sea level to high altitude, as such they have plenty of time to correct their course in high density air.
    Aha. So lets say an Iskander manages to do a continuous 0,5G maneuver at 40km altitude via aerodynamic control. How do you think the necessary course corrections of the ABM interceptor kill its kinematic potential in that dense air?

    Moreover, system such as SM-6, SM-2 has very high ratio of wing area vs mass compared to something like Iskander or any ballistic missiles
    A bold claim. How much more fin area - mass ratio does a SM-2 has over a Iskander? There is a general tendency yes but not "very high".

    Moreover, in reality, anti ballistic missiles missiles are launched to intercept instead of chase, so they are guided by radar to reach a specific coordinate at a specific point in time. You can imagine it like thowing a trap toward the road just so car run over it.
    Regarding those interceptors listed above
    I'm aware of that... As above said, continuous maneuvers by Iskander (via fins or gas system) make course corrections of the interceptor necessary, just that the interceptor is in much denser air.

    Thanks for naming all the interceptors.

  29. #89
    Join Date
    Jun 2004
    Location
    Columbia, MD
    Posts
    11,530
    PAC-3MSE has both mid body jet system and aerodynamic fin can operate both edo and exo atmoshpheric condition
    The PAC-3 MSE does not loft that high. There is a fairly decent gap between its upper limit and the lower limit of the THAAD interceptor. Enough of a gap for OEM's to think about a filler upper tier interceptor.
    Old radar types never die; they just phased array

  30. #90
    Join Date
    Dec 2015
    Posts
    933
    Quote Originally Posted by PeeD
    Exactly. Because you does not accept Cn or drag as a usable reaction force I don't accept your understanding of physics.
    To be more direct, you are fundamentally wrong and I don't want to argue about it with you. As you put it well, for me its also like talking against a wall.
    Isn't it funny, how my "fundamentally wrong" physics are so widely accepted by engineers?

    And it not that i don't accept drag and normal force as real force, they are real and used to pivot the AoA, however, the force component that change missiles direction of travel is lift
    Quote Originally Posted by PeeD
    Aha. So lets say an Iskander manages to do a continuous 0,5G maneuver at 40km altitude via aerodynamic control. How do you think the necessary course corrections of the ABM interceptor kill its kinematic potential in that dense air?
    Not much if anything at all, missiles can sustain more than 0.5G

    Quote Originally Posted by PeeD
    A bold claim. How much more fin area - mass ratio does a SM-2 has over a Iskander? There is a general tendency yes but not "very high".
    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)

    Regarding fins area, this is an SM-6 (same airframe as SM-2 but different seeker)



    This is an Iskander


    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
    Last edited by garryA; 13th May 2017 at 17:28.

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