5 ms·

> missed their target ... A kilometer is a pretty normal miss rate for a Scud Assume a target of 0.1km x 0.1km, but the missile may hit anywhere in an area of

> missed their target ... A kilometer is a pretty normal miss rate for a Scud Assume a target of 0.1km x 0.1km, but the missile may hit anywhere in an area of 1km x 1km around the target. That means the area of the target is 1/100 the size of the overall area. They'd have to launch at least 50 Scuds to have even a 50/50 chance of hitting the target. Can someone tell me if that calculation is way off? It sounds like this attack, other than spreading fear and panic, was very unlikely to work.

matt4077 9y ago
I did a rough outline of what I guessed to be the "valuable target area" of the airport, and it came to 0.15m^2: https://imgur.com/a/t7vhd https://imgur.com/a/t7vhd. Such a target would only need 5 Scuds for even odds. But, yes: The Scud is terribly (wonderfully) inaccurate.
- stats300 9y ago
  I think you meant 0.15km^2, not m^2. Really good point you make; the "valuable target area" is quite a bit larger than I assumed.
virtualized 9y ago
The chance of a hit is even lower: 1 - (0.99^50) ~= 40%
- gregshap 9y ago
  Assuming the chance of each hit is independent, why do you think they wouldn't be additive?
  - lostmsu 9y ago
    And potentially add to >100%?
    - chroem- 9y ago
      Yes, that would correspond to multiple missile strikes which is entirely reasonable.
      - wlesieutre 9y ago
        There is no number of missiles you could fire that would guarantee 100% chance of hitting the target (even once, let alone multiple times). Think of it like flipping coins. For a given coin flip, 50% odds of heads and 50% odds of tails. But if you flip the coin 100000000 times there's still a tiny outside chance of tails coming up every time. As you flip more coins, the odds of getting a heads on at least one of them will asymptotically approach 100%, but it will never get there.
  - matt4077 9y ago
    Indendence doesn't play into it. Simple example: flipping two coins doesn't give you a (0.5 + 0.5) = 100% chance of getting heads. The easiest way to solve "at least once"-type questions is invariably by subtracting P(!"at least once") = P("zero hits") from 1. To do it with addition, you'd have to list all possible outcomes completely. Those would, individually, be far lower probabilities. Example: P("at least one "six" when rolling two dice") = P("first not six, second six") + P("first six, second not six") + P("first six, second also six") = (1/6 * 5/6) + (5/6 * 1/6) + (1/6 * 1/6) = 11/36 Shorter: 1 - P("never six") = 1 - P("first not six, second not six") = 1 - (5/6 * 5/6) = 36/36 - 25/36 = 11/36 The number of possible outcomes you have to list and add together is 2^<number of trials> - 1. For two dices, that's a manageable 2^2 - 1 = 1. For 100 rockets it's 2^100 - 1 = 1.2676506E30 (- 1).
  - mnx 9y ago
    They are independent, and that's exactly why they aren't additive. Flipping a coin two times doesn't give you a 100% chance of getting heads. To expand on the intuition a little - they may well hit the same non-target spot multiple times before they hit the target at least once, in fact that is to be expected.
  - elsherbini 9y ago
    Intuitively it might make sense to add the probabilities, but it doesn't actually work out that way[0]. The parent used the law of total probability to solve for the probability of hitting at least once because it's simpler: P(hits >= 1) = 1 - P(hits = 0). Solving for the left side of the equation directly gets trickier: P(hits >= 1) = P(hits = 1) + P(hits = 2) + ... + P(hits = k) P(hits = 1) = 100 chose 1 * 0.01^1 * 0.99^99 ... p(hits = n) = 100 chose n * 0.01^n *0.99^(100-n) Taking that sum where n=100 yields the same as 1 - P(hits = 0) = 1 - (100 chose 0 *0.1^0 *0.99^100) = 1 - 0.99^100 [0] https://en.wikipedia.org/wiki/Binomial_distribution https://en.wikipedia.org/wiki/Binomial_distribution
  - deleted 9y ago
    [deleted]
  - gregshap 9y ago
    Great explanations, makes perfect sense with a little more thought.
  - deleted 9y ago
    [deleted]
ddalex 9y ago
I got the feeling that SCUD was always intended to deliver wire-area weapons, such as biological or chemical, and not small explosive warheads. If you're firing area-denial weapons, precision doesn't matter much.
- varjag 9y ago
  The very first SCUD warhead accepted into service was 50KT nuclear.
vilhelm_s 9y ago
Yes, I think they are not useful against point targets. In practice, it seems scud missiles have mostly been used against area targets, particularly terror bombing of cities. To get military effects you probably have to fire a lot of them, e.g. Wikipedia mentions > During the mujahideen attack against Jalalabad, between March and June 1989, three firing batteries manned by Afghan crews advised by Soviets fired approximately 438 missiles in defense of the embattled garrison. https://en.wikipedia.org/wiki/Scud#Operational_use https://en.wikipedia.org/wiki/Scud#Operational_use
Someone 9y ago
Apart from the error made by adding probabilities, you miss a factor 4 because your max error is only 500m, and a factor of (about (1)) π/4 because you are using a square hit region instead of a circular one, but ballpark, none of them matter. That airport terminal is quite a bit larger, though, and the size of the blast radius of the explosion will increase the likelihood of doing some damage (significantly so for large blast radiuses. If this warhead carried a nuclear device, a miss by a few hundred meters wouldn’t really be a miss) (1) I guess proper modeling of the error distribution will make that shape something else than a circle, but I also guess a circle is a much better first approximation than a square. Finally, that attack may have made economic sense. That scud cost a lot less than the rockets fired at it, and showing the capability of firing at targets that far away will pin down quite a few expensive forces around targets in Saudi Arabia.
bertil 9y ago
I believe the point is to render the airport unsafe to operate commercially. If you manage to hit nearby twice, it's unlikely civilians will want to hang around their gate for 45 minutes.

matt4077 9y ago

I did a rough outline of what I guessed to be the "valuable target area" of the airport, and it came to 0.15m^2: https://imgur.com/a/t7vhd https://imgur.com/a/t7vhd. Such a target would only need 5 Scuds for even odds. But, yes: The Scud is terribly (wonderfully) inaccurate.

stats300 9y ago
I think you meant 0.15km^2, not m^2. Really good point you make; the "valuable target area" is quite a bit larger than I assumed.

virtualized 9y ago

The chance of a hit is even lower: 1 - (0.99^50) ~= 40%

gregshap 9y ago
Assuming the chance of each hit is independent, why do you think they wouldn't be additive?
- lostmsu 9y ago
  And potentially add to >100%?
  - chroem- 9y ago
    Yes, that would correspond to multiple missile strikes which is entirely reasonable.
    - wlesieutre 9y ago
      There is no number of missiles you could fire that would guarantee 100% chance of hitting the target (even once, let alone multiple times). Think of it like flipping coins. For a given coin flip, 50% odds of heads and 50% odds of tails. But if you flip the coin 100000000 times there's still a tiny outside chance of tails coming up every time. As you flip more coins, the odds of getting a heads on at least one of them will asymptotically approach 100%, but it will never get there.
- matt4077 9y ago
  Indendence doesn't play into it. Simple example: flipping two coins doesn't give you a (0.5 + 0.5) = 100% chance of getting heads. The easiest way to solve "at least once"-type questions is invariably by subtracting P(!"at least once") = P("zero hits") from 1. To do it with addition, you'd have to list all possible outcomes completely. Those would, individually, be far lower probabilities. Example: P("at least one "six" when rolling two dice") = P("first not six, second six") + P("first six, second not six") + P("first six, second also six") = (1/6 * 5/6) + (5/6 * 1/6) + (1/6 * 1/6) = 11/36 Shorter: 1 - P("never six") = 1 - P("first not six, second not six") = 1 - (5/6 * 5/6) = 36/36 - 25/36 = 11/36 The number of possible outcomes you have to list and add together is 2^<number of trials> - 1. For two dices, that's a manageable 2^2 - 1 = 1. For 100 rockets it's 2^100 - 1 = 1.2676506E30 (- 1).
- mnx 9y ago
  They are independent, and that's exactly why they aren't additive. Flipping a coin two times doesn't give you a 100% chance of getting heads. To expand on the intuition a little - they may well hit the same non-target spot multiple times before they hit the target at least once, in fact that is to be expected.
- elsherbini 9y ago
  Intuitively it might make sense to add the probabilities, but it doesn't actually work out that way[0]. The parent used the law of total probability to solve for the probability of hitting at least once because it's simpler: P(hits >= 1) = 1 - P(hits = 0). Solving for the left side of the equation directly gets trickier: P(hits >= 1) = P(hits = 1) + P(hits = 2) + ... + P(hits = k) P(hits = 1) = 100 chose 1 * 0.01^1 * 0.99^99 ... p(hits = n) = 100 chose n * 0.01^n *0.99^(100-n) Taking that sum where n=100 yields the same as 1 - P(hits = 0) = 1 - (100 chose 0 *0.1^0 *0.99^100) = 1 - 0.99^100 [0] https://en.wikipedia.org/wiki/Binomial_distribution https://en.wikipedia.org/wiki/Binomial_distribution
- deleted 9y ago
  [deleted]
- gregshap 9y ago
  Great explanations, makes perfect sense with a little more thought.
- deleted 9y ago
  [deleted]

ddalex 9y ago

I got the feeling that SCUD was always intended to deliver wire-area weapons, such as biological or chemical, and not small explosive warheads. If you're firing area-denial weapons, precision doesn't matter much.

varjag 9y ago
The very first SCUD warhead accepted into service was 50KT nuclear.

vilhelm_s 9y ago

Yes, I think they are not useful against point targets. In practice, it seems scud missiles have mostly been used against area targets, particularly terror bombing of cities. To get military effects you probably have to fire a lot of them, e.g. Wikipedia mentions > During the mujahideen attack against Jalalabad, between March and June 1989, three firing batteries manned by Afghan crews advised by Soviets fired approximately 438 missiles in defense of the embattled garrison. https://en.wikipedia.org/wiki/Scud#Operational_use https://en.wikipedia.org/wiki/Scud#Operational_use

Someone 9y ago

Apart from the error made by adding probabilities, you miss a factor 4 because your max error is only 500m, and a factor of (about (1)) π/4 because you are using a square hit region instead of a circular one, but ballpark, none of them matter. That airport terminal is quite a bit larger, though, and the size of the blast radius of the explosion will increase the likelihood of doing some damage (significantly so for large blast radiuses. If this warhead carried a nuclear device, a miss by a few hundred meters wouldn’t really be a miss) (1) I guess proper modeling of the error distribution will make that shape something else than a circle, but I also guess a circle is a much better first approximation than a square. Finally, that attack may have made economic sense. That scud cost a lot less than the rockets fired at it, and showing the capability of firing at targets that far away will pin down quite a few expensive forces around targets in Saudi Arabia.

bertil 9y ago

I believe the point is to render the airport unsafe to operate commercially. If you manage to hit nearby twice, it's unlikely civilians will want to hang around their gate for 45 minutes.