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Victor, all the numbers I've been quoting you come straight from the HCDB-090128, which is the latest version of the database used in HCE, so you should probably use your variant b) from your last post. There are no decoys listed as available for PRC aircraft in the HCDB. (Actually, I'm not sure that anyone but the US has Decoy loadouts available.) An important note: The Kh-31 (both versions) is NOT a sea skimmer, it has a minimum altitude of Low. The radar horizon for a Low altitude target from a Medium ship (any of the listed escorts) is 51 nm, and 53 nm from a Large ship (the carrier). And finally, on a more personal note, I've got a busy weekend coming up. If I don't post anything for a day or two, it's nothing personal, I will be back ... [:D]

ORIGINAL: Warhorse64

Warhorse, I want to thank you again for providing the data.

Victor, all the numbers I've been quoting you come straight from the HCDB-090128, which is the latest version of the database used in HCE, so you should probably use your variant b) from your last post.

I have been kicking that one around in my head over the past couple of hours and I agree with you. We can let someone else worry about if the Harpoon game values are correct or not. However, because some of the numbers lead to a borderline failure, I will be raising some issues of interpretation (eg. minimum range) elsewhere in this thread.

There are no decoys listed as available for PRC aircraft in the HCDB. (Actually, I'm not sure that anyone but the US has Decoy loadouts available.)

Noted.

An important note: The Kh-31 (both versions) is NOT a sea skimmer, it has a minimum altitude of Low. The radar horizon for a Low altitude target from a Medium ship (any of the listed escorts) is 51 nm, and 53 nm from a Large ship (the carrier)

More useful numbers. I will use these particular ones for a crucial part of the next chunk of the calculation.

And finally, on a more personal note, I've got a busy weekend coming up. If I don't post anything for a day or two, it's nothing personal, I will be back ... 

By which time this calculation should be substantially completed. There are only about 3 chunks remaining.

Testing the Assumptions

A number of assumptions have been made in doing this calculation. We are working with the best data we can get, but some things are unknown, so we just have to use the best estimates available. Of course, some things are well known, and it's just me that doesn't know them.

Some of the assumptions are pretty crucial, and I am mentioning them here.


Missile Speeds

I have been using 600 kn for subsonic missiles and 1800 kn for fast missiles. These are not exactly assumptions; they are good working figures. 600 kn is very close to Mach .9, which is a standard published figure for subsonic missiles, and it is easy enough to work with that I can do the calculations in my head. 1800 kn is very close to Mach 2.7, which is close enough (within 10%) to any value between Mach 2.5 and Mach 3 that it gives good results, and it is easy enough to work with that I can do the calculations in my head.


SAM Minimum Range

The SM-2 has a minimum range of 3 nm. I am assuming that what this means is that the missile needs to travel 3 nm (5.5 s) before it can do anything useful. This means that it can only be used against an incoming missile if that missile is rather more than 3 nm away. For example, if the SM-2 is fired at a missile that travels at the same speed (1980 kn, Mach 3), then the target needs to be at least 6 nm away at that point. As the SM-2 moves out by 3 nm, the target moves in by 3 nm to the intercept point.

An alternative interpretation of the SAM minimum range is that the SM-2 cannot be launched at targets less than 3 nm away, but it will do fine as long as the target is further away at launch point, even if the SAM travels less than 3 nm to the intercept point. Of course, this interpretation gives a possibly crucial advantage to the SAM over the method I am using.


Missile Launcher Cycle Time

When SAMs get fired, a whole bunch of them leave the launcher simultaneously, in zero time. Is this reasonable? Viewed in isolation, the answer is probably not, but I think that the way to think about this is that Harpoon does not consider details at that scale. Rather, what it means is that, after the launcher has got a particular batch of SAMs in the air, then 30 s later, the launcher will have got the next batch in the air, the number in question being termed the Rate of Fire. Harpoon does not tell you the details of what that launcher was doing in the 30 s, only what the situation is at 2 instants of time separated by 30 s.


Radar

I have only a primitive understanding of the way radar systems work (not talking about radiation theory here !), and this could have a severe impact on the results of the calculation, so I will explain how I think it works and somebody can correct me if I'm wrong.

Early on, the American ships have their radars off, but their AWACS asset (or facsimile) is looking down on the region from some altitude. Due to its height, it can detect low-flying things far away. At some point, inbound Chinese aircraft and/or missiles are detected. The American ships now turn their radars on. However, because the ship radars are looking up, not down, they are limited by horizon effects and probably can't detect the Chinese yet. The radar horizon is about 25 nm to VL targets, 50 nm to low, and 100 nm to medium altitude. (I am using these values for my convenience. They are close enough to the "correct" values to give good results, and bear in mind that the radar horizon to any altitude range is actually a variable number depending on the exact target altitude.) I am assuming that the power of the ship radar (SPY-1B/D) is great enough that it is horizon-limited, not power-limited, at least out to 80 nm. (Someone could check this.) The fuel range of the SM-2s is 60 or 80 nm. I am assuming that the SAMs can only be fired at targets that the ship itself has detected, so that things could be safe within the SM-2 fuel range at low altitude and well within at VL. At some point the Chinese missiles (and maybe the aircraft) are close enough that the ship radars have line of sight to them. I am assuming that, at that point, the AWACS (or facsimile) "tells" the ships what it knows, so that the ships immediately get a certain detection and do not need to spend time in attempting to detect. Therefore I am assuming that the SAMs get launched as soon as the ships can establish the radar line of sight.

This set of assumptions could be entirely wrong; it might depend on whether the American air asset is an AWACS or just a scout helicopter; and it might depend on whether Aegis is available. In any case, unless corrected, that is the model that I am using.


AWACS

In this scenario, both sides are assumed to have AWACS or some facsimile in the area at the same time. The Chinese are using it to locate the ships exactly; the Americans are using it to tell them about missiles and stuff that the ship radars can't see. However, I think the Americans probably shouldn't be allowed any air assets. Given that the Americans, by definition of the scenario, have no fighters available, this means that the Chinese can safely park their fighters 55 nm from the ships at low altitude and hurl AAMs into the airspace above the ships. As long as the Chinese have AAMs with a range of 60 nm or so, no American aircraft would survive.

1-ship calculation for the CV:

The rate of fire for the Sea Sparrows (15) is so large compared to the number carried (16) that they all get used. They expect to kill 8 targets.

At 990 kn, the YJ-83 missiles need 18 s to move 5 nm. At 1940 kn, the Krypton ARMs need 9 s to move 5 nm. At 1320 kn, the RAMs need 13.5 s to move 5 nm and .8 s to move .3 nm. The first engagement with RAMs happens at 5 nm, and the launcher needs 16.5 s to shoot again, so the RAMs only get one shot at the ARMs. The YJ-83s are 1.5 s (.4 nm) out when the RAM launcher can shoot again, and they have moved a further .2 nm in the next .8 s, putting them inside the RAM minimum range, so the RAMs don't get a second shot at these missiles, either. The first batch of 15 RAMs expects to kill 8 targets.

Bear in mind that we are talking fractions of a second and a few yards here or there. These calculations should be taken with a grain of salt.

Ships defending other ships in the formation

If only one ship is attacked, most of the other ships will be able to fire their missiles in support.

a) If the attack comes in from the flank, then, when the YJ-83s are 15 nm from their target, the 3 other main US ships are about 25 nm away. The SAMs move twice as fast as the ASMs and will intercept them 16.7 nm out (6.7 nm from the target ship). So the other ships will get 1 or 2 SAM launches in defense of the target ship.

b) If the attack comes in from the front or back, then, when the YJ-83s are 10 nm from their target, the 3 other main US ships are about 20-25 nm away. The SAMs move twice as fast as the ASMs and will intercept them 16.7 nm or more out (1.7 nm or more from the target ship). So the other ships will get 1 or 2 SAM launches in defense of the target ship.

c) If the attack comes in from the flank, then, when the YJ-91s are 40 nm from their target, the 3 other main US ships are about 50 nm away. The SAMs move as fast as the ASMs and will intercept them 25 nm out (15 nm from the target ship). So the other ships will get at least 1 SAM launch in defense of the target ship.

d) If the attack comes in from the front or back, then, when the YJ-91s are 35 nm from their target, the 3 other main US ships are about 45-50 nm away. The SAMs move as fast as the ASMs and will intercept them 25 nm or more out (10 nm or more from the target ship). So the other ships will get at least 1 SAM launch in defense of the target ship.

e) If the attack comes in from the flank, then, when the YJ-91s are 25 nm from their target, the 3 other main US ships are about 35 nm away. The SAMs move as fast as the ASMs and will intercept them 17.5 nm out (7.5 nm from the target ship). So the other ships will get at least 1 SAM launch in defense of the target ship.

f) If the attack comes in from the front or back, then, when the YJ-91s are 25 nm from their target, the 3 other main US ships are about 35-40 nm away. The SAMs move as fast as the ASMs and will intercept them 20 nm or more out (5 nm or more from the target ship). So the other ships will get at least 1 SAM launch in defense of the target ship.

In all cases, all 4 main US ships can help to defend each other against any of the incoming missiles. The only exception is that, for the ARMs, only the ship directly on the other side of the formation will be able to properly respond. As discussed in my thread "outmaneuvering SAMs", the SM-2s that come in from the side will have difficulty solving the pursuit problem because they move at the same speed as the ARMs. However, against the YJ-83s, the SM-2s can simply catch them from behind.

Final version of the intercept calculations.

The YJ-83 missiles can be launched from outside of SM-2 range. There is no advantage to launching them from closer in (but see below). However, there may be an advantage to launching the YJ-91 ARMs from closer in than their maximum range. Because these missiles travel at low altitude or higher, the ships can see them further out than the sea-skimmers, and can fire at them more often.

There are 5 ways of getting the missiles in closer.

1) Just fly a whole bunch of planes in to short range at any altitude and ignore the casualties. With a billion or so people to call on, the Chinese can probably afford to throw away their pilots and maybe even their toys, and their military history over the past 100 years (or always) shows that this is exactly the approach they do use. I do not recommend this approach.

2) Fly the planes in at any altitude and launch missiles just before the planes are shot down. This is better than the first option, because it does have the effect of absorbing more SAMs than just the missiles alone would do. I do not recommend this approach.

3) Fly the planes in as close as safely possible at low altitude and launch from there. Warhorse has given us the radar horizon from low altitude to a medium ship as 51 nm. For my convenience, I will take the safe low altitude to be 50 nm (note 1). If a SM-2 shoots at such a plane, the plane will always be able to escape. This range does not give a significant advantage for the YJ-91 ARM (in fact, that is probably its practical launch range), and it gives no advantage at all for the YJ-83. It would give a definite advantage to long-range cruise missiles.

4) Fly the planes in as close as safely possible at VL altitude and launch from there. Warhorse has given us the radar horizon from VL altitude to a large ship as 26 nm. For my convenience, I will take the safe VL altitude to be 25 nm (note 1). If a SM-2 shoots at such a plane, the plane will always be able to escape. Most planes cannot safely fly at this altitude. For such planes, I do not recommend this approach, because even though it may save on missiles, I consider the cost in pilots, even if slight, to be far higher than any number of inanimate objects. (Ed: What about human lives on the other side?) However, Warhorse has pointed out that the Su-30MK2 DOES safely fly at VL. This is deadly, deadly stuff. These planes could just park at VL 25 nm from the ship; ordnance hurled from that range would give the ship very little time to respond, although how exactly such a plane would loiter at wavetop height is a bit of a mystery.

5) Fly in on an oblique approach. I have discussed this earlier and elsewhere. For example, see my post of 2/11/09 in my thread "outmaneuvering SAMs". I am not going to discuss this further here, because the present subject is already complex enough.

According to Brad's post 47, this thread ("I think the SM-2's could be launched on inertial/midcourse guidance before they enter the ship's own radar LOS, assuming that another asset (such as the E-2C Hawkeye with CEC) could "see" the targets."), the ship does not need to have radar LOS to the target to launch the SM-2s if a friendly unit can see the target, although the ship still needs to have the target on its own radar at the time of intercept. This means that the Chinese planes could safely park at 25 nm (VL) or 50 nm (low). On the other hand, for inbound missiles, which are definitely going to enter the ship's radar LOS, an intercept course would be calculated that would let the intercept happen at those ranges. (Exactly? I doubt it. How far in? I have no idea, so I am using the stated figures.) However, bear in mind that a missile does not actually exist as a target until after it has been launched, which means that a missile carried in by a plane has a definite advantage over one coming in on its own steam.

So then I have to raise the following issue the second time. Based on the data Warhorse has provided, Chinese fighters with 40-odd nm range AAMs are parked at VL 25 nm from the nearest US ship, and 30 nm from the center of the formation. Where can the Americans have air assets with radar to look down on the battle area? Nowhere, I think. Nevertheless, we are assuming that those assets exist.
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Finally, we get to the calculations.

a) YJ-83 cruise missile

This missile comes in at VL at 990 kn. I am using 1,000 kn in the calculation, which is only 1% different, makes my life a whole lot easier, and gives results that are as good as the exact calculation. (note 1) Since this missile is coming in from far away, it has been detected some time ago and an intercept has been calculated to take place at 25 nm. 1,000 kn means it travels 1 nm in 3.6 s and 25 nm in 90 s. The SM-2 moves twice as fast (2,000 kn) and travels 25 nm in 45 s. This means that the SAM launcher has cycled by the time the first intercept happens.

Intercept 2: The missiles are moving with a combined speed of 1,000 + 2,000 = 3,000 kn. It takes them 30 s to reach the intercept point, which is 25 nm x 2/3 = 16.7 nm out.

Intercept 3: There is no launcher delay. With only 2/3 of the distance to cover relative to the last step, the time to intercept is now 20 s. The intercept happens 25 nm x 2/3 x 2/3 = 100/9 = 11 nm out.

Intercept 4: There is a 10 s launcher delay. There are 30 s left on the clock, so the YJ-83s are 25 nm x 1/3 out. Time to intercept time is 10 s, and it happens 25 nm x 1/3 x 2/3 = 5.5 nm out. There is no minimum range problem.

No more intercepts are possible, because the SAM launcher will not cycle in time.

This is better for the US than what I calculated on 2/20 (post 39), because I am now utilizing the AWACS.

b) YJ-91 ARM launched from 50 nm, low

This missile comes in at 1,940 kn. I am using 2,000 kn in the calculation, which is only 3% different, makes my life a whole lot easier, and gives results that are as good as the exact calculation. (note 1) Since this missile has just been launched, the SAMs are still in their launcher. I am giving the ship an automatic detection, because there's a whole batch of incoming missiles, and the ship should be able to pick up some of them. 2,000 kn means the YJ-91 travels 1 nm in 1.8 s and 50 nm in 90 s. The SM-2 moves at the same speed (2,000 kn). Therefore the first intercept happens at 45 s and 25 nm out.

Intercept 2: There is no launcher delay. The intercept happens 22.5 s later and 12.5 nm out.

Intercept 3: There is a 7.5 s launcher delay. There are 15 s left on the clock, so the YJ-91s are 50 nm x 1/6 = 8.3 nm out. The intercept happens 7.5 s later and 4.1 nm out. There is no minimum range problem.

No more intercepts are possible, because the SAM launcher will not cycle in time.

The main reason the ship only gets 3 intercepts in this case is that the missile did not exist as a separate entity until it was 50 nm out.

c) YJ-91 ARM launched from 25 nm, VL

I initially thought that this was impossible, for the following reason. The YJ-91 has a minimum flight altitude of low, and I assumed that there was a reason for this, such as if the missile gets too close to the water, it crashes. So while the Flanker could safely be there, the missile couldn't.

I have since read something that confirms this. However, there is nothing to prevent the Flanker parking 25 nm out at VL, and then, when the time is ripe, popping up to low altitude to shoot, and dropping back below the horizon. This is deadly, deadly stuff.

The missile comes in at 1,940 kn. I am using 2,000 kn in the calculation, which is only 3% different, makes my life a whole lot easier, and gives results that are as good as the exact calculation. (note 1) Since this missile has just been launched, the SAMs are still in their launcher. I am giving the ship an automatic detection, because there's a whole batch of incoming missiles, and the ship should be able to pick up some of them. 2,000 kn means the YJ-91 travels 1 nm in 1.8 s and 25 nm in 45 s. The SM-2 moves at the same speed (2,000 kn). Therefore the first intercept happens at 22.5 s and 12.5 nm out.

Intercept 2: There is a 7.5 s launcher delay. There are 15 s left on the clock, so the YJ-91s are 25 nm x 1/3 = 8.3 nm out. The intercept happens 7.5 s later and 4.1 nm out. There is no minimum range problem.

No more intercepts are possible, because the SAM launcher will not cycle in time.

Only 2 intercepts! The ship can only fire 40 SAMs and destroy 20 ARMs. After that, the ARMs start hitting and disassembling the ship. It doesn't matter which US ship is under attack; none of them get to expend their full complement of SAMs.

d) combination attacks

If the ship radars are off, it is safe against ARMs. I don't know if the ship can protect itself against ARMs already in flight by turning its radars off. If it can, it will be necessary to keep enough of the bigger missiles coming in so that the ship can't reasonably turn its radars off. In that case, the ship will be able to use some of its SAMs against the slower missiles in addition to two batches against the ARMs.

e) Enhanced Sea Sparrows

Although the ESSMs are faster and have a smaller minimum range than the SM-2s, they do not provide a useful advantage (more intercepts) in this situation. Anyway, there aren't very many of them, and they have a rather short maximum range.

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Notes:

1) On the merits of calculations with approximate figures.

The first thing we need to realize is that, no matter how many significant figures our calculator gives us, no number in real life can be considered exact, as long as it relates to a measurement. Each number has some uncertainty associated with it, as well as a certain number of significant digits, and those 2 concepts are linked.

There are other reasons why stating an exact number may not be meaningful. For example, in Harpoon, the radar horizon from a medium ship to a low altitude plane is stated as the exact number 51 nm. But how high, exactly is the radar emitter/receiver? And, more importantly, exactly how high is the plane? The 51 nm could only be exactly correct for one specific plane height for each specific radar height, and the value of 51 nm is therefore only a representation of the actual radar horizon.

Another example is how Harpoon handles detection. Detection attempts only happen at discrete 30 s intervals, and those are fixed with respect to the game clock. Now, I am not arguing against that system. But it needs to be realized that this procedure guarantees that many results can only be approximate.

Indeed, the whole game is based on calculations done by a computer algorithm at discrete time intervals. All numerical analysis is inherently approximate, even though the results can be and often are stated to umpteen significant figures.

There are 3 main reasons I am working with approximate numbers in most of this thread.

a) Accuracy. The important thing I need to work out in this calculation is how many intercepts happen for a given ASM/SAM combination. Number of intercepts is an integer, generally a small integer less than 5. But the intermediate results are intercept locations or times, which are real numbers. Suppose that the exact calculation tells me that the number of intercepts is a number anywhere between 3.2 and 3.8. This means that exactly 3 intercepts can happen. (A fractional intercept means that the launcher hasn't cycled yet, or that the intercept is too close to the ship to actually work.) Now if the approximate calculation tells me that the number of intercepts is anywhere in that range, then I have obtained EXACTLY the same USEFUL answer as with the exact calculation. 3.5 + .3 = 3.5 + 10%. Generally, if my approximate calculation is within 10% of the exact calculation, there is no meaningful loss of accuracy. In some special cases, the result is borderline, so I might need to recalculate more carefully.

b) Transferability. Using a representative number such as 1800 kn for my calculations, I obtain a useful result for a 1940 kn missile. But I can take all the results and immediately apply them to any missile moving at speeds from 1600 to 2000 kn. There is no need to recalculate anything.

c) Convenience. Especially in this particular post, you will see that I have done the entire calculation in my head, without once ever needing to use a calculator. This saves me a great deal of time. For example, if a missile travels at 1800 kn, which is 1800 nm per 3600 s, then it needs exactly 2 s to go one mile, and that is a very easy figure to work with. If it is travelling at 2000 kn, which is 2000 nm per 3600 s, then it needs exactly 1.8 s to go one mile, which is a fairly easy figure to work with. But if I have to use 1940 kn, every single calculation requires me to punch a bunch of numbers into my calculator. Incidentally, 1940 kn is within 3% of 2000 kn and 8% of 1800 kn, so both approximate figures will give excellent results.

Did I mention round-off error? Calculators can easily introduce round-off error if you're not careful.

The points I have raised here are standard for any numerical analysis. It's not a matter of me being lazy or sloppy. Of course, any one who wants to can use the calculation steps I have shown and recalculate with the exact values. I think that you will get the same results as the approximate calculation.

ORIGINAL: VictorInThePacific
Missile Speeds
I have been using 600 kn for subsonic missiles and 1800 kn for fast missiles. These are not exactly assumptions; they are good working figures.

I am not sure why you would use anything other than the exact known figures for the missiles involved. But maybe you can rationalise this.

SAM Minimum Range
The SM-2 has a minimum range of 3 nm. I am assuming that what this means is that the missile needs to travel 3 nm (5.5 s) before it can do anything useful. This means that it can only be used against an incoming missile if that missile is rather more than 3 nm away.

I think this is the safer approach. The SM-2 is not pointed at the target immediately, as it needs to tip over from its vertical launch, and acquire the target through the ship's fire control system.

Missile Launcher Cycle Time
When SAMs get fired, a whole bunch of them leave the launcher simultaneously, in zero time. Is this reasonable?

Its not exactly zero time in real life, of course, but we are talking about a vertical launch system (VLS), where each missile has (in effect) its own "launcher". The VLS can essentially pump out missiles as fast as they can effectively clear the launcher without interfering with each other.

Radar
I am assuming that the SAMs can only be fired at targets that the ship itself has detected, so that things could be safe within the SM-2 fuel range at low altitude and well within at VL. At some point the Chinese missiles (and maybe the aircraft) are close enough that the ship radars have line of sight to them. I am assuming that, at that point, the AWACS (or facsimile) "tells" the ships what it knows, so that the ships immediately get a certain detection and do not need to spend time in attempting to detect. Therefore I am assuming that the SAMs get launched as soon as the ships can establish the radar line of sight.

I think the SM-2's could be launched on inertial/midcourse guidance before they enter the ship's own radar LOS, assuming that another asset (such as the E-2C Hawkeye with CEC) could "see" the targets.

The SPY-1B/D radar can, btw, detect Large/Medium targets at 175 nm, Small targets at 137 nm, Vsmall targets at 55 nm, and Stealthy targets at 16 nm.

AWACS
In this scenario, both sides are assumed to have AWACS or some facsimile in the area at the same time. The Chinese are using it to locate the ships exactly; the Americans are using it to tell them about missiles and stuff that the ship radars can't see. However, I think the Americans probably shouldn't be allowed any air assets. Given that the Americans, by definition of the scenario, have no fighters available, this means that the Chinese can safely park their fighters 55 nm from the ships at low altitude and hurl AAMs into the airspace above the ships. As long as the Chinese have AAMs with a range of 60 nm or so, no American aircraft would survive.

I'm unsure what any of this adds to the equation. Maybe the Chinese are detecting the CSG with their new over the horizon radar (OTH-B or OTH-SW). The E-2C Hawkeye is there to create the most powerful Aegis defense possible for the purposes of the analysis.

ORIGINAL: VictorInThePacific

AWACS

In this scenario, both sides are assumed to have AWACS or some facsimile in the area at the same time. The Chinese are using it to locate the ships exactly; the Americans are using it to tell them about missiles and stuff that the ship radars can't see. However, I think the Americans probably shouldn't be allowed any air assets. Given that the Americans, by definition of the scenario, have no fighters available, this means that the Chinese can safely park their fighters 55 nm from the ships at low altitude and hurl AAMs into the airspace above the ships. As long as the Chinese have AAMs with a range of 60 nm or so, no American aircraft would survive.

Well, actually, the longest-ranged AAM available to the PRC in-game is the AA-10 Alamo C (R-27ER), with a range of 47.6 nm, carried by the Su-30MK2 Flanker C. They'd probably have to come in to about 40 nm to get the shot off. On the other hand, these aircraft have the NoE flag, so they can fly at VLow for extended periods in reasonable safety. On the gripping hand, I'm not sure how or if the game models Cooperative Engagement, so they might be in danger from SAMs anyway.

ORIGINAL: VictorInThePacific
time. Based on the data Warhorse has provided, Chinese fighters with 40-odd nm range AAMs are parked at VL 25 nm from the nearest US ship, and 30 nm from the center of the formation. Where can the Americans have air assets with radar to look down on the battle area? Nowhere, I think. Nevertheless, we are assuming that those assets exist.

If it helps, the E-2C Hawkeye radar (lets say the APS-145 found aboard the most recent variants) can detect targets as follows: Large targets at 350 nm, Medium at 266 nm, Small at 190 nm, Vsmall at 76 nm, and Stealthy at 23 nm. It will probably be orbiting at High altitude. The Su-30, btw, is a Large target.

For a unit at High altitude, the radar horizon is some 232 nm distant, so there's lots of room for the Hawkeye to loiter out of harm's way.

ORIGINAL: Warhorse64
For a unit at High altitude, the radar horizon is some 232 nm distant, so there's lots of room for the Hawkeye to loiter out of harm's way.

Yup. [8D]

ORIGINAL: CV32

If it helps, the E-2C Hawkeye radar (lets say the APS-145 found aboard the most recent variants) can detect targets as follows: Large targets at 350 nm, Medium at 266 nm, Small at 190 nm, Vsmall at 76 nm, and Stealthy at 23 nm. It will probably be orbiting at High altitude. The Su-30, btw, is a Large target.

I think the main purpose of the Hawkeye for this particular situation (almost by definition) is to locate inbound missiles before the ship radars can. I think missiles are very small? So the Hawkeye would need to be pretty close by.

Highway ... to the ... danger zone!

Putting it all together.

What we have done so far:

1) We asked the question.

"What exact combination of anti-ship missiles, ARMs, jamming, and/or decoys is needed to defeat a modern day Aegis protected naval group?"

2) We chose the target group.

Carrier Strike Group Seven (CSG-7), including DESRON-7, and comprising:

Nimitz class carrier USS Ronald Reagan (CVN 76)
Ticonderoga (Bunker Hill) class Aegis cruiser USS Chancellorsville (CG 62)
Arleigh Burke Flight II class Aegis destroyer USS Decatur (DDG 73)
Arleigh Burke Flight IIA class Aegis destroyer USS Howard (DDG 83)
Arleigh Burke Flight IIA class Aegis destroyer USS Gridley (DDG 101)
Oliver Hazard Perry class guided missile frigate USS Thach (FFG 43)
Supply class fast combat support ship USNS Bridge (T-AOE 10)

The carrier has no aircraft of any sort available, except possibly one Hawkeye, so it is more proper to think of the US force as a surface action group that just happens to contain a large target that behaves in some ways like a CV, rather than a CVBG that is missing all its airplanes.

3) We chose the attacking force, which generally speaking, is PRC land-based air.

4) We set the ship formation. (post 25, this thread)

5) We selected the best available units from the Chinese arsenal.

Su-30MK2 (Soviet Flanker C), 4 * Kh-31P Krypton ARM (YJ-91), range 55 nm, speed 1940 kts.
H-6M (like Soviet Badger), 4 * C803 (YJ-83), range 135 nm, speed 990 kts.

The Flanker C can safely fly at VL, and we may use this fact. However, the YJ-91 cannot be launched from VL. The Flanker C definitely comes in as close as is safe at low altitude. The YJ-83 comes in at VL, and its launch platform never enters SAM range.

6) We calculated the number of intercepts the ships' SAMs would get against the 2 Chinese missiles being considered. (posts 44, 45, 46, this thread) This tells us how many attacking missiles to bring.

We almost have a definite answer. It's taken a lot of time and effort by a bunch of people. We explored some ideas that eventually were abandoned. It's not as though we knew the answer at the beginning, or even necessarily how best to approach the problem. There are a few details that remain to be worked out. Some of them are quite significant.
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In post 45, this thread, I showed that each of the 4 main ships can help to protect any other unit of the formation. So, in order to better overwhelm the defenses of any ship, all of these ships need to be engaged at the same time. The AO and the FF can be ignored by the main strike. The CV gets one or 2 shots, which can be used to defend itself or to defend another ship, but not both. For most purposes, the CV can be ignored in the main strike.

The purpose of the main strike is to sink or disable the 4 ships with effective SAM batteries. The follow-up strike can be done with cheapo bombs and will sink the remnants of the fleet.

The main strike will most likely attack the fleet with sufficient ASMs that it is not possible to shoot them all down, and most likely each of the 4 main ships will be attacked with enough ASMs to overwhelm their individual self-defense. Faced with such an attack, leaving each ship to its own devices means that they will all sink, whereas sacrificing some of them means that the rest can be saved.

Here's another question that a Harpoon expert will need to answer: Does the SAM computer intelligently try to save as many high-value ships as possible (for example ships that have not yet expended all their SAMs), or does it simply shoot at the closest or earliest-arriving ASMs?

The main strike will most likely involve some combination of ARMs and big missiles. The big missiles alone are easier to shoot down, but they do more damage. The ARMs go after particular systems. Here's another question that a Harpoon expert will need to answer: What sort of damage do the ARMs actually do? In particular, if an Arleigh Burke DD is hit with, say, 10 YJ-91 ARMs, what sort of damage are we looking at? Ship sunk? Weapons knocked out?

ORIGINAL: VictorInThePacific
I think the main purpose of the Hawkeye for this particular situation (almost by definition) is to locate inbound missiles before the ship radars can. I think missiles are very small? So the Hawkeye would need to be pretty close by.

Most missiles are Vsmall. Some are Small (like the big Russkies), some are stealthy. Fortunately for us, both the Kh-31A/P and the YJ-83 are Vsmall.

ORIGINAL: VictorInThePacific
Does the SAM computer intelligently try to save as many high-value ships as possible (for example ships that have not yet expended all their SAMs), or does it simply shoot at the closest or earliest-arriving ASMs?

Ships in the AAW ring of the formation will defend ships in the main body, but not because they have any particular value.

What sort of damage do the ARMs actually do? In particular, if an Arleigh Burke DD is hit with, say, 10 YJ-91 ARMs, what sort of damage are we looking at? Ship sunk? Weapons knocked out?

A YJ-91 (Kh-31P copy) carries 18 damage points (DP). (This will become 40 DP under newer annex data). The Arleigh Burke (depending on variant) anywhere from 226 up to 260 DP. (This will become 223 to 230 under newer annex data). Either way, 10x YJ-91 is going to inflict some serious, probably catastrophic, damage. Certainly it is very likely that sensors and weapons will be knocked out.

By the Numbers

Finally, I am going to provide the entire attack plan and results.

But first, there has been a bit of confusion about one of the concepts, so I will state it properly here and redo one part of the calculation.

In general, and I simplify, in real life a SAM launch system needs to be able to "see" the target both at the time of launch and at the time of intercept. In some cases, a friendly unit may assume that responsibility. In the case of the situation I am specifically concerned with, where an attack aircraft hiding below the horizon pops up to launch weapons and then hides below the horizon again, the aircraft is safe because the SAM launcher can't see it at the time the intercept might occur.

In Harpoon, the SAM launch system needs to be able to "see" the target only at the time of launch; after that, the missile will not lose tracking due to the target leaving the LOS of the launcher. However, if the aircraft pops up and hides again while the SAM launcher is engaged with something else, then the aircraft remains safe, and there is no observable difference between Harpoon and real life. A good time to pop up would be when the SAM launcher has just used up its ROF for that combat round; the aircraft will then be safe for about 30 s.

The calculations in posts 44, 45, this thread, remain correct. Calculation b) in post 46, this thread, will not be used. Calculation a), c) in post 46, this thread, are replaced with the following:

a) YJ-83 cruise missile

These missiles come in at VL at 990 kn. I am using 1,000 kn in the calculation, which is only 1% different, makes my life a whole lot easier, and gives results that are as good as the exact calculation. The missiles cross the radar horizon at 25 nm. I am giving the ship an automatic detection at this point, because there's a whole batch of incoming missiles, and the ship should be able to pick up some of them. 1,000 kn means they travel 1 nm in 3.6 s and 25 nm in 90 s. The SM-2s move twice as fast (2,000 kn) and travel 25 nm in 45 s.

Intercept 1: The missiles are moving with a combined speed of 1,000 + 2,000 = 3,000 kn. It takes them 30 s to reach the intercept point, which is 25 nm x 2/3 = 16.7 nm out.

Intercept 2: There is no launcher delay. With only 2/3 of the distance to cover relative to the last step, the time to intercept is now 20 s. The intercept happens 25 nm x 2/3 x 2/3 = 100/9 = 11 nm out.

Intercept 3: There is a 10 s launcher delay. There are 30 s left on the clock, so the YJ-83s are 25 nm x 1/3 out. Time to intercept time is 10 s, and it happens 25 nm x 1/3 x 2/3 = 5.5 nm out. There is no minimum range problem.

No more intercepts are possible, because the SAM launcher will not cycle in time.

b) YJ-91 ARM launched from 30 nm, low

Up to this point, I have been using 25 nm as the "safe" VL range. I have moved it out a little bit to give some room for error, because if something goes wrong, the plane can neither escape by distance nor by going below the radar horizon.

The missiles come in at 1,940 kn. I am using 2,000 kn in the calculation, which is only 3% different, makes my life a whole lot easier, and gives results that are as good as the exact calculation. Since the missiles have just been launched, the SAMs are still in their launcher. I am giving the ship an automatic detection, because there's a whole batch of incoming missiles, and the ship should be able to pick up some of them. 2,000 kn means the YJ-91s travel 1 nm in 1.8 s and 30nm in 54 s. The SM-2s move at the same speed (2,000 kn). Therefore the first intercept happens at 27 s and 15 nm out.

Intercept 2: There is a 3 s launcher delay. The YJ-91s move about 1.5 nm in that time. There are now 24 s left on the clock, and the YJ-91s are 13.5 nm out. The intercept happens 12 s later and about 7 nm out. There is no minimum range problem.

No more intercepts are possible, because the SAM launcher will not cycle in time.
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For the following attack plan, I will only say where the airplanes need to be at particular times and what they need to do there. It will be the reader's responsibility to calculate the various flight paths to the target for the different assets, including launch times, fuel expenditures, and so forth. Just kidding. We are going to ignore all those details and assume that someone else has done those calculations, and now the airplanes have arrived at the target on schedule. I include all these details when I say "park" your airplanes at such-and-such a location.

In order to prevent the 4 main ships from protecting each other, all of them will be attacked simultaneously. The carrier will be attacked at the same time as well. 10 planes and 40 missiles will be allocated against each of these ships in the main attack. It doesn't really matter how you allocate the planes, except that there should be at least 10 cruise missiles and 10 ARMs allocated to each target. The YJ-83s are actually quite small, and only do about twice as much damage as the ARMs. The ARMs are more likely to hit critical weapons and sensor systems. For example, let's choose 5 Su-30MK2s and 5 H-6Ms each. Park the SU-30 MK2s 30 nm from their targets at VL, in such a way that all the other ships are as far away from the attack lines as possible. Park the H-6Ms 50 nm out at low in the same way. Make sure that you know which of your plane groups is targeted on what, because the weapons allocation dialog box doesn't let you access the map, and the following depends on proper timing. Launch all 100 YJ-83s. As these missiles pass the 25 nm barrier, the ships start firing, and the countdown clock starts. There are 90 s until these missiles hit.

T - 90 s: Each of the 4 main ships launches 20 x SM-2s. The SAM launchers are now unavailable for the next 30 s. Bring all the SU-30MK2s up to low and launch all 100 YJ-91s. What happens next depends critically on how fast you can click the mouse. I am assuming that you can launch all these missiles in 15 s. After that, you still have 15 s to bring all your Su-30 MK2s down to VL. This may not be very important, because the SAM launchers will be busy. Very, very busy.

T - 75 s: The ARMs are at 30 nm.

T - 60 s: The ARMs are at 21.7 nm. They need 39 s to move this distance. At the time the next SAM launch will happen, the ARMS will be 9 s from impact (5 nm out). Since the SAMs move at the same speed, the intercept would happen 2.5 nm out, which is too close for the SAMs to be active. In fact, the ARMs would need to be 6 nm (10.8 s) out at T - 30 s for the third intercept to be able to target them, and since they need 43.2 s to travel the remaining 24 nm, they would have had to be launched at T - 72.2 s (17.8 s after the YJ-83 launch) OR LATER. So if you click the mouse rapidly when launching the ARMs, they can only be intercepted once, but if you are slow, they can be intercepted twice. Yes, we really are quibbling about fractions of a second here.

On the other hand, I have showed above that there will always be 3 intercepts against the YJ-83s, so let's assume that the second intercept is against the ARMs and the third one is against the YJ-83s.

Those ships with Enhanced Sea Sparrows could use them against ARMs in the third intercept, but there's not much point to that, because the ARMs have smaller warheads (unless the attack was ARM-heavy).

T - 40.5 s: The second intercept happens, 10.8 nm out. The launcher needs 10.5 s to cycle.

T - 30 s: The ARMs are 9 s from impact (5 nm out). The YJ-83s are 8.3 nm out. Time to intercept time is 10 s, and it happens 5.5 nm out.

T - 21 s: Any remaining ARMs impact.

T = 0: Any remaining YJ-83s impact.

Before any ASMs impact, the anti-missile decoys fire. Each of the ships in question has 4 at 15-20 %. I will allocate one soft kill per ship for these. Just before impact, the ship's guns will fire. I will allocate one hard kill per ship for these. I will assume that the first ASM misses, but that would actually be unlikely. So, with the assumptions and calculations above, this would be 7 ARM hits on each of the 4 main ships, and 20 ARM hits and 2 YJ-83 hits on the carrier.

That's 126 damage points plus ARM specials on each of the 4 main ships, which should wreck or sink them, and 426 damage points on the carrier.

Under these circumstances, an intelligent group defence would sacrifice some of the ships (make no attempt to defend them) in order to make sure that some of the ships survive. However, I think the way it actually works is that each ship will first attempt to defend itself, even if that means they all sink.

There will need to be a mopping-up strike to deal with the AO, FF, and whatever remains from the main strike. Second-rate airplanes could be used for this; the main strike used the very best in the Chinese arsenal.

On the assumption you've reached a conclusion(s), can we have a very brief summary of the minimum amount of force needed to obtain a hit(s) on the carrier?

Some Additional Comments

Tweaking the attack

What I considered in post 56, this thread, was 20 x YJ-83 plus 20 x YJ-91 against the 5 biggest ships. I will restate this here in terms of what will be necessary to defeat the SAMs; after this, all ASMs basically get free hits.

a) The threshold for the carrier is 17 missiles of either type. The first 8 would be stopped by the RAMs, and the next 8 would use up all the ESSMs, although the order in which the SAMs shoot is the other way around. Then the decoys will get one more. About 75% of all ASMs arriving within the next 30 s will hit the carrier.

b) I have showed that the 4 main ships get exactly 3 SAM launches against the YJ-83s. That is 3 batches of 20 SM-2s each, which will almost always shoot down exactly 30 ASMs. Then the guns and decoys will get one each, so the threshold for these ships is 32 ASMs (31 for the Gridley). About 75% of all ASMs arriving within the next 30 s will hit the ships. These ships will still have between 2 and 46 SAMs left at that point, so it is important to sink or wreck them quickly, before those SAMs can shoot again.

c) I have showed that the 4 main ships get a maximum of 3, and maybe as few as 1, SAM launches against the YJ-91s. However, in order to achieve the low number, the YJ-91s need to be launched as part of a specific coordinated attack. The SAM launchers will probably still get 3 shots, only not against the YJ-91s.

d) The main defence against the YJ-91s may be shutting down down the ships' radars, but that may not be useful after the ARMs have already been launched. To prevent this, a significant number of the ASMs need to be the YJ-83s. In fact, in order to get the best result with the YJ-91s, the first missiles in have to be the YJ-83s.

e) The FF can shoot down 2 ASMs.

KH-59M

I have not treated this ASM at all, but it should be mentioned. The Flanker C can carry 4 of them. It flies much more slowly than the 2 missiles I have considered, so it would be a lot easier to shoot down (the SAMs get to shoot more often). However, it has a much bigger warhead, doing twice as much damage as the YJ-83. It would be probably be the best choice for the mop-up attack, where the remaining ships have no significant defences.

ORIGINAL: CV32

a very brief summary of the minimum amount of force needed to obtain a hit(s) on the carrier?

17 missiles of either type - after that, about 75% of all ASMs arriving within the next 30 s will hit the carrier. (And the escorts need to be busy with other matters.)

ORIGINAL: CV32

Ships in the AAW ring of the formation will defend ships in the main body

Does this mean that in in all other situations, ships will defend only themselves?

ORIGINAL: VictorInThePacific
Does this mean that in in all other situations, ships will defend only themselves?

No, AAW capable ships in other rings of the formation will still shoot at incoming missiles. But ships in the AAW ring (or the main body) are in the best position to do so, since the further out a ship is in the formation, the more likely they'll be shooting at a crossing target.