055 DDG Large Destroyer Thread

[Jura The idiot]

I'm not sure why you think that if a first missile misses, so will a second, and vice versa. This is patently not true. The conditions under which the first engagement misses doesn't necessarily have anything to do with the conditions experienced during the second engagement, which will be further along the flight path of the bogey towards the 055, possibly significantly further. So local environmental conditions can be excluded. If you're talking about rain or other inclement weather, then decreased Pks would apply across the board to all missiles, both inbound and outbound, and not just to a single pair of HHQ-9s or DK-10As. You're probably thinking the engagements will be nearly simultaneous, whereas more likely they will be separated by a few seconds or even a few dozen seconds. Even if they were nearly simultaneous, the Pk for the engagement is being determined by the ability of the missile's on board sensor to correctly ascertain the position of the target, which instead of being binary ("I know where it or I don't know where it is"), is distributed on a curve ("it's most likely to be at X location with a Y degree of error"), the slope of which changes millisecond by millisecond as targeting information is updated. Each HHQ-9 will have a different curve and different slope, even ones separated by only a few seconds. They won't approach from exactly the same direction, they won't be flying at exactly the same speed, they won't be attacking from the same angle, etc. It is grossly unreasonable to think that if one missile misses, the other is also likely to miss. In reality both will definitely have independent individual Pks, even if they are same type of missile and attack the target only a few seconds apart.

Incidentally Pk can never be HIGHER than 1, so I'm not sure why you pointed out that Pk is lower than 1 in this discussion.

Today at 3:25 AM you used the conditional probability:

... vs vanilla missile Pks will be: HHQ-9 = 0.7, ...

... HHQ-9s would be doubly assigned to ...; of the 40 engaged by 2 HHQ-9s, 4 would likely get through (Pm = 0.3 x 0.3 = 0.09), ...

reformulating using your numbers now,

P_hit = 0.7 (of a single HHQ-9 against an unspecified incoming missile: obviously just examples here);

P_stop, the probability of stopping a bogey by firing a two-missile salvo, is

P_stop = 1.0 - (1.0 - P_hit)*(1.0 - P_hit)
P_stop = 1.0 - 0.3*0.3; P_stop = 0.91

(P_stop multiplied by your 40 incoming missiles is 36, meaning 4 leakers, as you said Today at 3:25 AM)

BUT some would tell you that a probability EACH missile hits is 0.7 no matter how big salvo you fired, in other words, P_stop = P_hit, and you would take down only 40*0.7=28 (not 36 as above) missiles;

what I had in mind Today at 7:22 AM was according to simulations I've heard of, in a salvo-firing of the type I talk about (which is sending two missiles against the same target), an increase in P_stop is SMALLER than obtained from
P_stop = 1.0 - (1.0 - P_hit)*(1.0 - P_hit)
(smaller by about 50%; in this case you would take down 32 missiles (neither the 36 or 28 above))

in plain words, quote, shooting the same stuff in a salvo doesn't that much increase your chance of hitting by such a salvo, end of quote

in real world,
How did a 30 year-old Su-22 defeat a modern AIM-9X?

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on June 18 over Syria, after a Super Hornet hadn't hit a Su-24 with a Sidewinder at close range (by the way I'm guessing the vendor declared Pk of almost one for such an encounter EDIT OK this is what vendors always do), the US pilot used a different TYPE of missile (an AMRAAM ... at close range) and didn't try again with yet another Sidewinder, then another, ...

just something for you to think about on the plane over the Pacific LOL just don't miss the view of the Jiangnan shipyard because of this, so that you tell us how many aft VLS cells you saw on the Type 055

 

[Iron Man]

BUT some would tell you that a probability EACH missile hits is 0.7 no matter how big salvo you fired, in other words, P_stop = P_hit, and you would take down only 40*0.7=28 (not 36 as above) missiles;

what I had in mind Today at 7:22 AM was according to simulations I've heard of, in a salvo-firing of the type I talk about (which is sending two missiles against the same target), an increase in P_stop is SMALLER than obtained from
P_stop = 1.0 - (1.0 - P_hit)*(1.0 - P_hit)
(smaller by about 50%; in this case you would take down 32 missiles (neither the 36 or 28 above))

in plain words, quote, shooting the same stuff in a salvo doesn't that much increase your chance of hitting by such a salvo, end of quote

Who are these "some" people, and did they take statistics in middle school or skip that course? Also, you are making two mutually incompatible claims here. One is that the gain in chance to hit the target with a salvo of 2 missiles is smaller than obtained from 1 - (1 - Pk) x (1 - Pk) by "about 50%" (sounds like a random percentage to me). The other is that there is NO gain at all in chance to hit the target with a pair of missiles. I have not only never heard of either of these claims before, but these claims are both exclusive of each other. Either there is some increase in chance to hit, or there is no increase in chance to hit. You can't claim both. Can you provide a source for these claims? I find it astounding that there are people who make these claims, especially since the math is based on basic logic/statistics that are accessible to anyone. I feel like people making these claims are about as hilariously wrong as claiming that 1 + 1 = 3.

 

[Jura The idiot]

... I feel like people making these claims are about as hilariously wrong as claiming that 1 + 1 = 3.

and I feel I should leave you to that (the point is if you don't look at what the first missile achieved, if it didn't go in the opposite direction for example, you don't increase so called redundancy much), with LOL

 

[jobjed]

what I had in mind Today at 7:22 AM was according to simulations I've heard of, in a salvo-firing of the type I talk about (which is sending two missiles against the same target), an increase in P_stop is SMALLER than obtained from
P_stop = 1.0 - (1.0 - P_hit)*(1.0 - P_hit)
(smaller by about 50%; in this case you would take down 32 missiles (neither the 36 or 28 above))

in plain words, quote, shooting the same stuff in a salvo doesn't that much increase your chance of hitting by such a salvo, end of quote

What sort of phenomenon are you talking about where consecutive launches will decrease the Pk of the second interceptor?

As long as the targeting apparatus on the ship isn't being overworked e.g. SPG-62 on Burkes not required to illuminate more than 3 targets, then Pk of all interceptors, consecutive or not, should be the same.

Terminal targeting calculations are done by the missiles themselves so there doesn't exist the problem of the ship running out of processing power because the missiles do all the processing; launch 1 missile - 1 processor active, launch 50 missiles - 50 processors active. So where would this "diminishing Pk of consecutive launches" come from?

 

[Jura The idiot]

What sort of phenomenon are you talking about where consecutive launches will decrease the Pk of the second interceptor?

As long as the targeting apparatus on the ship isn't being overworked e.g. SPG-62 on Burkes not required to illuminate more than 3 targets, then Pk of all interceptors, consecutive or not, should be the same.

Terminal targeting calculations are done by the missiles themselves so there doesn't exist the problem of the ship running out of processing power because the missiles do all the processing; launch 1 missile - 1 processor active, launch 50 missiles - 50 processors active. So where would this "diminishing Pk of consecutive launches" come from?

the assumptions (not mine, but from #4167 Iron Man, Yesterday at 3:25 AM )
again:

... vs vanilla missile Pks will be: HHQ-9 = 0.7, ...

... HHQ-9s would be doubly assigned to ...; of the 40 engaged by 2 HHQ-9s, 4 would likely get through (Pm = 0.3 x 0.3 = 0.09), ...

now, I'll use #1-missiles and #2-missiles to distinguish between "doubly assigned" missiles from the assumptions I just quoted:

40 of #1-missiles take down 0.7*40=28 bogeys
what 40 of #2-missile is going to achieve?

I can imagine 28 of #2-missiles would also hit the bogeys being hit by #1-missiles, and the remaining 12 #2-missiles would also miss the bogeys being missed by #1-missiles ... this way, P_stop would be 0.7, way below of 0.91 given by the chain rule Yesterday at 8:50 PM

if you guys still don't know what I mean, I quit, as it would take me to write a ridiculously long post, which is something I won't do LOL!

I think it comes down to this:
if you're convinced assigning FOUR missile in the present context would practically guaranteed ALL incoming missiles were taken down, 40*(1-0.3*0.3*0.3*0.3) is 40 when rounded to integer, then I'm telling you I'm not convinced

 

[Iron Man]

the assumptions (not mine, but from #4167 Iron Man, Yesterday at 3:25 AM )
again:

now, I'll use #1-missiles and #2-missiles to distinguish between "doubly assigned" missiles from the assumptions I just quoted:

40 of #1-missiles take down 0.7*40=28 bogeys
what 40 of #2-missile is going to achieve?

I can imagine 28 of #2-missiles would also hit the bogeys being hit by #1-missiles, and the remaining 12 #2-missiles would also miss the bogeys being missed by #1-missiles ... this way, P_stop would be 0.7, way below of 0.91 given by the chain rule Yesterday at 8:50 PM

if you guys still don't know what I mean, I quit, as it would take me to write a ridiculously long post, which is something I won't do LOL!

I think it comes down to this:
if you're convinced assigning FOUR missile in the present context would practically guaranteed ALL incoming missiles were taken down, 40*(1-0.3*0.3*0.3*0.3) is 40 when rounded to integer, then I'm telling you I'm not convinced

You don't need to be convinced of anything. It's just statistics and probability, something that is true regardless of whether you need convincing or not. The "1-0.3*0.3*0.3*0.3" is correct if you are calculating the probability that at least 1 out of the 4 interceptors will hit a target if each one has a Pk of 0.7. The "40*" is basically a rough estimate of the average number of missiles shot down if ONE HUNDRED AND SIXTY interceptors were used, which was not part of the scenario (who assigns FOUR interceptors per target???) Even a million missiles assigned to a single target won't make the probability that at least one of them will hit the target equal to 1, but it will sure get pretty damn close (look up "asymptote"). There is no such thing as a "guarantee", and the equation certainly never claims to provide one, nor do I claim to. This really isn't rocket science! Just ask yourself how likely is it to flip a coin 5 times and get all 5 of them to be tails? Do you really think that it's 50%? Seriously? Do the math, the probability of such an outcome is: 0.5*0.5*0.5*0.5*0.5 = 0.03125, i.e. the chance of this happening is 3.125%. Likewise, the probability that at least one of the coin flips turns up heads is 1 - 0.03125, or 96.875%. Now translate heads into "hit" and tails into "miss", and you will see the analogy is directly convertible into a missile engagement. 5 interceptors all assigned to one target, each with a 50% Pk, will result in a 96.875% probability of at least one of them hitting the target.

 

[Jura The idiot]

You don't need to be convinced of anything. It's just statistics and probability, something that is true regardless of whether you need convincing or not. The "1-0.3*0.3*0.3*0.3" is correct if you are calculating the probability that at least 1 out of the 4 interceptors will hit a target if each one has a Pk of 0.7. The "40*" is basically a rough estimate of the average number of missiles shot down if ONE HUNDRED AND SIXTY interceptors were used, which was not part of the scenario (who assigns FOUR interceptors per target???) Even a million missiles assigned to a single target won't make the probability that at least one of them will hit the target equal to 1, but it will sure get pretty damn close (look up "asymptote"). There is no such thing as a "guarantee", and the equation certainly never claims to provide one, nor do I claim to. This really isn't rocket science! Just ask yourself how likely is it to flip a coin 5 times and get all 5 of them to be tails? Do you really think that it's 50%? Seriously? Do the math, the probability of such an outcome is: 0.5*0.5*0.5*0.5*0.5 = 0.03125, i.e. the chance of this happening is 3.125%. Likewise, the probability that at least one of the coin flips turns up heads is 1 - 0.03125, or 96.875%. Now translate heads into "hit" and tails into "miss", and you will see the analogy is directly convertible into a missile engagement. 5 interceptors all assigned to one target, each with a 50% Pk, will result in a 96.875% probability of at least one of them hitting the target.

my final on this:

you're assuming the missiles in your salvo in the chunk of
#4167 Iron Man, Yesterday at 3:25 AM
I repeatedly quoted
are UNcorrelated: the first missile has P_hit of 0.7, so does the second, the second INdependently able to go after the bogey missed by the first missile (then the P_stop
is P_stop = 1.0 - (1.0 - P_hit)*(1.0 - P_hit) as I posted Yesterday at 8:50 PM)

in reality though, if you for example sent those two missiles in a completely wrong direction, both would miss (affecting the P_stop value, possibly in the way I described in the first part of Today at 10:12 AM post) ... now you may have the last word here (I'm not going to respond)

 

[jobjed]

you're assuming the missiles in your salvo
are UNcorrelated

Why do you think they're dependent events?

I'm pretty sure they're independent. One missile's Pk will not affect the next's.

in reality though, if you for example sent those two missiles in a completely wrong direction, both would miss

The missiles' Pks are still independent, they're just independently bad because you launched them in the wrong direction. If you launch a missile in the wrong direction, its Pk is no longer 0.7 but more like 0.07. Then combined chance of two missiles' taking down one target is (1- 0.93^2) = 13.5%.

Basically, why are you insisting that Pks of missiles are dependent? They're separate events and the chances of an occurrence for one of them doesn't affect the chances for other missiles.

 

[vesicles]

Why do you think they're dependent events?

I'm pretty sure they're independent. One missile's Pk will not affect the next's.




The missiles' Pks are still independent, they're just independently bad because you launched them in the wrong direction. If you launch a missile in the wrong direction, its Pk is no longer 0.7 but more like 0.07. Then combined chance of two missiles' taking down one target is (1- 0.93^2) = 13.5%.

Basically, why are you insisting that Pks of missiles are dependent? They're separate events and the chances of an occurrence for one of them doesn't affect the chances for other missiles.

I've been thinking exactly the same thing. I'm not a statistician, but the way the probability was calculated (assuming dependence among multiple targeting events) sounded very weird to me.

I would imagine the FCS would have treated each individual target separately. If the probability of "90%" is determined for individual targets, then it should be held equally among all individual target, no matter how many they are. In fact, that probability was most likely calculated by averaging accuracy among many attempts.

What about simultaneously targeting multiple targets? That would be a different mode for the FCS. And the way to calculate the probability would be different from the individual probability. And this value would most likely be independent of the individual probability.

 

[Jura The idiot]

I got quoted by another debater so one more post, will leave it to Readers after this:

Why do you think they're dependent events?

I'm pretty sure they're independent. One missile's Pk will not affect the next's.




The missiles' Pks are still independent, they're just independently bad because you launched them in the wrong direction. If you launch a missile in the wrong direction, its Pk is no longer 0.7 but more like 0.07. Then combined chance of two missiles' taking down one target is (1- 0.93^2) = 13.5%.

Basically, why are you insisting that Pks of missiles are dependent? They're separate events and the chances of an occurrence for one of them doesn't affect the chances for other missiles.

hitting by missiles of the same type fired in a salvo isn't "a process without memory", in the whole set of 80 missiles and 40 targets from Today at 10:12 AM, the outcome for the second missiles (= #2-missiles from Today at 10:12 AM) is not independent of what was the outcome for the first missiles (= #1-missiles from Today at 10:12 AM);
for example if one of the first missiles didn't make it to the kill box (and assuming two missiles are "double assigned" as in #4167 Iron Man, Yesterday at 3:25 AM), the probability the second missile from that salvo would make it to the kill box is lower than that 0.7 (EDIT which is a value from #4167 Iron Man, Yesterday at 3:25 AM I had been using as an example)

in other words,
you may ask yourself a question:
if you now shot multiple missiles from 1960s, each with some low Pk, in a salvo against for example latest-block Exocet, would you IN THIS WAY significantly increase the probability of taking down said Exocet? (or said probability would be still very close to said low Pk, huh?)
LOL!

another story is if two different types of missiles, one with for example IR homing and P_hit_IR,
another with for example active-radar-homing and P_hit_ARH, are fired in a salvo, then the chain rule should be a good approximation:
P_stop = 1.0 - (1.0 - P_hit_IR)*(1.0 - P_hit_ARH)

I'm not going to respond to anybody anymore on this topic