WARNING: This post contains a bit of basic math, formulas and pictures. I tried to make it somewhat readable, but feel free to point out if its shows up weird on your screen and I will try to change it.
Looks like I didn’t forget anything. It is explicitly excluded with a stipulation as to why.
Yes. Exactly as I did in the first place with tracking speed.
So can we put this part of the discussion to rest by saying that tracking speed has no implication on the comparison between final fire and area defence if (and only if) both systems use the same tracking speed (which they should anyway).
Looks like you were missing the "factor of 2" as well. ... Your still not addressing the possibility of a weapon that both has a range greater than a missiles movement in a single impulse and has the rate of fire to take advantage of it. This is actually minor and not relevant to the weapon in this discussion. It becomes relevent if the evaluation expands to include longer ranged weapons. This is handled in my formula by incrementing the number of impulses a salvo will need to cross the beam range (or the maximum range the fire control is set fire point defense).
No, actually its all in there, but maybe the whole issue becomes clearer when I provide a full derivation. The way we will do this is consider each possible interception separately, and then sum up the results in the end. So we will go through this shot by shot and compute the expected hits.
Lets consider the usual setup: we are engaging a large salvo of missiles that flies exactly past the ship with a relative velocity
v. We are assuming the ship can fire every 5 seconds.
Let
R be the maximum range of the system (i.e. the smaller of the maximum range of the firecontroll, the laser or the sensor).
Let
d be the distance that a missile can cover in 5 seconds (i.e. d=5s*v).
The first shot:At some point the missile will enter our engagement range. At the end of the 5 second pulse we will get the first shot at the missile. Let us denote the distance of the missile from the ship with
r. So the first time that we shoot at the missile it will be anywhere between r=R-d and r=R. It cant be further than R, because then we could not shoot at it this interval and would have to wait for the first shot. It cant be closer than r=R-d either, because d is the maximum distance that the missile will cover relative to the ship, i.e. if it was closer than R-d it would have been within engagement range earlier and this would not be the first shot.
Figure 1: situation during the first shot
The ship is positioned in the centre at 0. During the first shot the enemy missile
might be anywhere indicated by the blue arrow. (Not! necessarily at the tip) Figure 1 illustrates the point, where the full length of the blue arrow indicates the possible positions of the enemy missile when we first open fire at it. Let us first determine the expected probability that we score a hit during the first interval. We can write the probability that we hit a missile at any particular distance
r as
p(r). This is not to say that the probability depends only on r (it doesnt), but its the only thing we are going to mention explicitly (for now). Just to be clear: This probability includes many factors and the symbol stands for the full effect.
Now we can computed the
expected hits for this first shot. This is done by multiplying i) the chance to score a hit at any point in the range R-d < r < R time ii) the probability that the missile is at this particular point. This is done for all points, then the sum is taken. This can be written as an integral over the full range of
R-d<r<R as an integral of the probability
p(r) at any particular point
r times the probability
rho density to find the missile at this spot during this interval:
%5Ccdot%20p(r)%5Ccdot%20dr)
Now of course we could calculate that already, but lets wait a moment.
What will happen during the
second shot? Well the missile will move further by
d:
Figure 2: situation during the second shot
The salvo will have moved further by d (the distance it travels in 5s).
It is thus between R-2d < r < R -d Now for this bit we can analogously compute the expected probability of a hit during the second interval:
%5Ccdot%20p(r)%5Ccdot%20dr)
There is a nice property of integrals which allows us to write the sum of p
1 and p
2 as follows (note the change in the limits of the integral):
%5Ccdot%20p(r)%5Ccdot%20dr)
So at this point we have already taken care of two possible interceptions. But there could be more! And every time we add one more possible interception, the lower limit of the integral in the last equation shifts down by d. Now how long do we have to do that? Easy, lets say we have to do it
n times:
Figure 3: situation during the first shot
The last time that a salvo could be shot at. Notice that with a certain probability
the missile might have moved out the the engagement range already. Now for this last possible interception we can write:
d}^{R-nd}%5Cvarrho(r)%5Ccdot%20p(r)%5Ccdot%20dr)
However we know that p(r<-R) must be 0, because we have defined R as the maximum range at which a hit could be scored at all. With this knowledge we can re-write the last term a bit (note the change in the lower limit):
%5Ccdot%20p(r)%5Ccdot%20dr)
And now we can collect all terms from
all shots put together. The full expected hits can be written as the sum during each interval:
%5Ccdot%20p(r)%5Ccdot%20dr)
Lets simplify that a bit by noting that the probability density rho does not depend on r, so we can take it out of the integral:
%5Ccdot%20p(r)%5Ccdot%20dr=%5Cvarrho%5Ccdot%5Cint_{-R}^{R}%20p(r)%5Ccdot%20dr)
Moreover, we can write
=\widetilde{p}(r)\cdot\widetilde{p}_{other})
, that is we can factor out the range-dependent bit of the hitchance from all other factors. Dont worry, all other factors are still there, and we have not dropped them at all, we just merely separated the term:
%5Ccdot%20dr)
At this point however we will make one more assumption, which is that the tracking bonus is maxed out before the missile enters engagement range. Firstly this means that the range-dependent bit of the hitchance is symmetric around 0, i.e. we can can substitute the lower limit of the remaining integral with 0 and get a factor of 2 in front of the whole thing (that is effectively the factor we had been talking about earlier). Also we can also write
=(1-r/r_{FC}))
, where r
FC is the range of the firecontroll. Thus the equation becomes:
dr)
Solving the integral:
^2%20\right%20]_{0}^{R})
Factoring out the fraction and inserting the limits:
^2%20-%201\right])
Writing out the quadratic term in the the brackets and canceling the factors of 2 and 1/2:
^2%20-%201\right])
Consolidating the bit in the brackets of 1 and -1 and factoring out the minus:
^2%20\right])
Factoring out the fraction of R/r
FC, where the r
FC cancels out:
)
Substituting the probability density rho with its known value of 1/d=1/(5*v)
)
Bingo! That is exactly the formula I posted initially (only the names are switched a bit, for which I apologize). And its all there: Multiple interceptions, engaging missile both inbound and outbound, and all other factors but range included in the "other" factor. And that includes the tracking factors. Also note that because I considered v to be the
relative speed of the missile versus the ship (and did so from the beginning), so the movement of the ship is actually already figured in.
