@garryA

It gives 28.4G for F-16 at 50.000 feet, as opposed to real life value of 4G. So accurate.Mind blow
So F-16 will be able to pull 28.4G at 50K feet, Mach 2 if it was a tube body with fin ? Great.

Good luck with applying specialized formulas for completely different problems… I take what “Tactical missile design”s author tells over your models any day.

In CAT I configuration F-16 maximum AoA limits is 25 degrees at 1G and 15 degrees at 9G. 5 degrees AoA is far from the AoA limit. Wrong excuse, try different one.

Good and I hope you have now understood that airframe G’s and aerodynamic G’s are not the same.

Yet, you gave no values of your own so others can dobble check.

dia 0,914
lenght 7,28
weight 1254kg
missile AoA 10°
control surface deflection 63°
control surface area 0,16m² *2
control surface aspect ratio 0,66
speed mach 6,5

Where do you find the information that PAC-3MSE, SM-6, SM-2 will self destruct once they fly above 25 km ? Care to cite ?

Didn’t talk about SM-6. For the others, check Wiki für max. altitude…

Have that take into account the fact that interceptors can still retain their speed due to thrust while their targets will lose all speed ?

It’s not about speed, at around 32km the Iskander would reach its airframe load limit with 30G. How intercept a 30G target with reasonable PK? With luck the interceptors airframe might survive 50G.

@garryA

Lift equation, and reference area are not my proposed model.It never was, it is official model used to measure CL. Iam not the first nor the last person ever used the equation

You thought you were smart enough to start the numbers game. However, your simplifications were so much that they had no representation of the reality. The spreadsheet of stealthflanker on the other hand used a book in which a smart guy defined the necessary model and it’s results make sense. I give you credit to have been brave enough to try it, that awaked by interest and lead to this useful results.

Do you really checked and understood it ? or do you only use it because it sound like it support your theory ?

Sure I did otherwise I would not have found the error with the control surface area. Surely a book about missile design also is a good reason to take it over a model from someone with potential no aerospace, engineering or physics background at all.

So not only the amount of G Iskander is about 10 times smaller than your proposed values, you also intentionally forget that THAAD will have thrust to maintain speed. How exactly does this supposed to “prove” that iam wrong ?

You proposed the numbers and I took them as they made sense to me. It’s still possible that 30G is possible via contributions from the gas system or more depressed trajectory.
You are wrong because you quite confidently presented a useless model to quantify the case, maybe intentionally because you might have remembered the discussion with stealthflanker and his spreadsheet.

How exactly do you want people to double check your calculation if you don’t tell them what exact value you used for wing area, mass, speed and air density ???

People can put their values into the spreadsheet and get an idea, the results will be similar. (air density is included in the spreadsheet thanks to stealthflankers efforts)

Instead of trying to omitted PAC-2, PAC-3MSE and SM-2 , SM-6 from the calculation saying it is out of their evelope, why don’t you tried to put their number in and estimate how many G they could turn according to that spread sheet ?

They are not cleared for those altitudes? Why should I do a theoretical training?

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

I put air density of 0.12 kg/m3 (59k feet) and missiles speed of mach 4 into the file , and it show Aim-120 can turn 78g , and able to deal with target making 16g turn ,
then I change the air density to 1.225 kg/m2 ( sea level ) and missiles speed to mach 2 into the file and it show aim-120 can turn 303g and intercept target making 61g turn

how did any aircraft evade AIM-120 if that was the case ?

I think you have no kind of engineering approach but want to talk about it. The AIM-120 airframe is not capable to do very high G’s, its systems will fail and it’s structure disintegrate. It might be designed to do overloads from 30-50Gs, that’s it. The airframe is thus the limiting factor not the aerodynamic capability to pull G’s. Stealthflankers spreadsheet is not failproof but as it uses the equations of a good book, I at least have not found any errors beyond that with the wing/fin area.

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

The forums own stealthflanker did this exercise and you garryA even discussed with him. How unfortunate that you didn’t remember the discussion, his calculation model is much better and tells us the actual G numbers:

http://forum.keypublishing.com/showthread.php?137398-What-factors-affecting-aircraft-missiles-sustained-turn-rate/page2

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

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

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

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

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

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

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

@garryA

My statement is correct actually, CL is not only an airfoil quality in the case of specific airfoil but also an airframe qualiity in case of specific airframe. CL of Su-27 airframe will include both its body lift and wing efficiency. Similarly, CL of Iskander airframe will include both its body lift and fin efficiency. The fin/ wing area is a reference area in equation (regardless whether you calculate lift for specific airfoil or specific airframe, this is the base reference).
For example: this is CL of F-16 and YF-16 with their wing area as reference area (obviously if you use non SI unit for wing area then velocity and altitude have to be non SI too)

Aha, so the Cl for the Su-27 is like that of the F-16 based on the wing area only? Not a experimentally gained value that can be confidently written in a manual? We might should do the test and calculate Cl via wings only and compare it with the manual value…

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

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

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

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

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

Well garryA, now I understand…

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

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

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

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

No , missiles and aircraft has thrust so there are sustained and intantaneous turn. A sustained turn is the condition where you can conserved your speed and altitude ( basically they have excess thrust for a turn, sustained turn is where your thrust can balance the extra drag ). Even though air density is lower at high altitude, it is much easier to sustain high G at low altitude. That because, due to lower air density, you will need to either fly much faster or at higher AoA at high altitude to generate the same amount of lift.
For example: an F-16 at 30k feet can pull maximum of 9G ( which is what its CL allow it to do ) but it can barely sustain 3.9G

So my argument is wrong and the drag losses due to a sustained 1,5G turn won’t have bigger impact on the lower altitude interceptor than the Iskander in terms of kinematic performance?

if it is not nuclear equipped warhead but simply bomlets then they can be intercepts by Centurion C-RAM, RIM-116, Iron dome or similar stuff, they unlikely to cause much damages given their limited size
Heck, the MGM-140 carry over 950 M74 bomblets and no one bat an eye
http://www.designation-systems.net/dusrm/m-140.html
The MBDA Apache and JSOW used to carry loads of submutions too

You want to intercept six small mach 3 submunitions with C-RAM? For hat velocities is the iron dome cleared? Then you think that a single warhead produce more frag = radar (TPY-2) killing damage than 6 submunitions? They will have a wider CEP but also larger damage area and well if the damage is not enough send a second Iskander.

There could be various reason to change from grid fin to planar fin, for example grid fins can have considerably higher radar cross section due to their extremely high number of coner reflectors (those grids).

OK. So they changed the grid fins from Oka to conventional ones in Iskander due to RCS issues, sacrificing anti-ABM maneuvering capability for it? Possible if the Russians believe they can lower the RCS of the Iskander in a amount that could make it stealth against a high power radar like the TPY-2, maybe they made giant leaps in temperature and drag resistant RAM. I however think Russians do not of such a RAM technology nor would they do this sacrifice, I rather think for some reason the fins of the Iskander are sufficient for its anti-ABM strategy.

On the otherhand, they are the main contributors, as previously elaborated , it is the interraction between the airflow and the whole airframe that change the direction of travel, the tail fin acts as initial pivots force. Alternatively, you can think about why wing loading is important for aircraft agility but not tail fin area.

As said, yes they contribute, however your simplification to take them together with the fins into a lift calculation against the Iskanders fins 1:1, is a non-representative simplification.

@garryA

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

…It because it can only provide a pivot to change missiles AoA not provide lift to change direction of travel.Changing AoA and turning is vastly different.

I did and I checked your calculations with the lift formula where you estimated the Iskander fin area and used that in the lift equation. Something does not add up if you used just the fin area of the Iskander, compared it to the whole lift of the Su-27 and dismissed the Iskander tube body lift in the calculation? Is there anything else than the velocity, altitude and fin area of the Iskander in your lift calculation that I missed?
Your calculation was for that pivot force you talk about and I’m sticking to that.

I don’t know what make you change your stand from “Iskander can turn 30G at altitude higher than 50 km” to Interceptor can’t sustain barely 1.5 G at low altitude. But both are too extreme, if interceptor can’t sustain merely 1.5G at low altitude, they won’t be able to even point toward threat direction (consider straight up launch system)

There is a difference between a high G evasive maneuver and continuous low G maneuver to bleed a interceptor kinematics. I took your 30G number and defended it in light of the experiences with US MARV testing. Now you have rised some doubts about this with your calculations. However I see you evolving, now you seperate between the pivot force and the airframe drag. Your calculation to disprove 30G at 50km with the Iskander was just for the fins, the pivot force. You in the role of the devils advocate helps to get closer to the real deal.

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

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

It certainly can, that Indian TBM of your graph does too:

[ATTACH=CONFIG]253382[/ATTACH]

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

Yes iam aware that the long fin are statics and only the rear fin move. But when the missiles is at an AoA with the air flow, the long fin will interract with the air flow to turn (change direction of travel) of the missiles.

They interact and contribute somewhat. If you want to count them and the fins with a 8x larger fin aera – mass ratio compared to the Iskander in your lift calculation you are doing a huge mistake.

@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.

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.