Formations

[Bgreman]

You guys are arguing way too strenuously over methodology when your results both indicate that final fire is more effective.

[Nathan_]

I have noticed that not every gun will fire in final defensive fire if I stack too many of them, though that might be me not using the appropriate PD options, so I'm experimenting with area defense and longer ranged(even main battery grade) weaponry.

[Charlie Beeler]

You also argue that since tracking speed and target speed (sV/tV) are fixed data points and not variable they can be functionally ignored.  I stipulate that you are wrong.

Hah! And I stipulate that you are wrong! _{/Tongue-in-cheeck}

Ok, seriously though, which bit do you disagree with?
a) Target speed and tracking speed are the same for the final-fire and area-defence variant. The tracking bonus is likely to be the same, or only very marginally different.
b) When all these three input-variables have the same respective value in both scenarios, then the factor computed from these inputs also takes the same value in both scenarios.
b) When the factor is a constant multiplier it does not bear influence on the comparison between the two systems.

This actually goes all the way back to your first post on this topic.

Let’s plug in the numbers from the Wiki for an early beam ship: r=90k km, R = 192k km (i.e. we will want a very long range and pick the longest-reaching FC for the area defence version), L=48k km. v=24k km/s (pretty conservative). Let us say we have 1,400t worth of payload that we could use as beam defence. I think that is quite high, because with engines, fuel etc that would make a ship of about 2,500t and that is probably on the relatively large side for an advance picket.

I started my evaluations of your figures based on the Wiki example ship as published.

Code: [Select]

Ark Royal class Cruiser    6200 tons     611 Crew     639 BP      TCS 124  TH 250  EM 0
2016 km/s     Armour 3-30     Shields 0-0     Sensors 10/10/0/0     Damage Control Rating 3     PPV 26
Annual Failure Rate: 102%    IFR: 1.4%    Maintenance Capacity 193 MSP    Max Repair 45 MSP

Nuclear Thermal Engine E10 (10)    Power 25    Fuel Use 100%    Signature 25    Armour 0    Exp 5%
Fuel Capacity 100,000 Litres    Range 29.0 billion km   (166 days at full power)

Twin 10cm C3 Near Ultraviolet Laser Turret (2x2)    Range 90,000km     TS: 15000 km/s     Power 6-6     RM 3    ROF 5       
3 3 3 2 1 1 1 1 1 0
15cm C3 Near Ultraviolet Laser (2)    Range 180,000km     TS: 3000 km/s     Power 6-3     RM 3    ROF 10       
6 6 6 4 3 3 2 2 2 1
Fire Control S04 96-3000 (1)    Max Range: 192,000 km   TS: 3000 km/s     
95 90 84 79 74 69 64 58 53 48
Fire Control S04 24-12000 (1)    Max Range: 48,000 km   TS: 12000 km/s     
79 58 38 17 0 0 0 0 0 0
Pebble Bed Reactor (6)     Total Power Output 18    Armour 0    Exp 5%

Active Search Sensor S10-R1 (1)     GPS 10     Range 100k km    Resolution 1
Active Search Sensor S20-R100 (1)     GPS 2000     Range 20.0m km    Resolution 100
Thermal Sensor TH2-10 (1)     Sensitivity 10     Detect Sig Strength 1000:  10m km
EM Detection Sensor EM2-10 (1)     Sensitivity 10     Detect Sig Strength 1000:  10m km

This design is classed as a military vessel for maintenance purposes

Your statement is that your taking the values from the Wiki ship but when you go to plug in the number you state

So for the area-defence version we would have to spend 800t on the firecontroll (size 16, 4x range, 4x tracking)

Which is clearly not from the Wiki ship.  Once I realized that you then changed the longer ranged BFC’s tracking speed I changed the comparison formulas and should have altered my statement above as well. 

It should have been; as long as the tracking speeds are the same between the fire controls the calculation values are the same.  But that they should not be excluded since they are known values and fundamental to missile intercept calculations.  By excluding the tracking speed modifier you don't have a way to evaluate the impact of changes in fire controls and/or turreted systems.

On this we do not agree.

Maybe the last bit is the most controversial here, because you write:

But since we’re actually talking about a modifier that can substantially reduce the hit probability it must be included.

Now I could equally assert that you "forgot" the crew grade factor. That can easily be +30%, so its very relevant right? Yes, of course its relevant, but not for the question posed, as both systems would have the same factor.

You must of missed this part

So with the formula above there are 2 subsections that can be excluded for basic missile intercept analysis:

• (1-(tE-sE))   since missiles vary rarely have ECM.
• cG    since crewgrade is ship specific, but is more common than ECM for missiles.

Looks like I didn’t forget anything.  It is explicitly excluded with a stipulation as to why.

The last point of contention is that your 'density function' falls short, it only accounts for a single 5 second impulse not the multiples that you assert. 

That is not quite true. We can actually compute the number of interceptions by integrating over the probability density function alone: N=2*integral_(0)^(r) dx 1 / (5v) =2r/(5v). Plug in the numbers v=24k, r=90k, so N=1.5. So my formula assumed that you get an expected 1.5 shots versus missiles that stream by. Of course that is an average number; in actual combat you will either get 1 or 2 shots versus any particular missile salvo. Because I integrate over the entire range of the area-defence variant, I do account for multiple interceptions.

Since the area defense analysis needs variable intercept range accounted for substitute ((1-bR/fR)+(1-mR/fR))/2 for (1-r/fR).  This is a crude model, but serves.

Well, you use a crude approximation, while I use an exact formula. When these two do not arrive at the same final value I would not take this as a sign that my formula is broken.

Specifically, the way you account for multiple shots is insufficient. You basically treat it as though you would have multiple shots, all with the hit-probability of the middle of the lasers range. While that may serve as a first approximation it does not account properly for the fact that multiple shots will never occur at an average distance. The example missile flew 120kkm/5s, while the laser had a range of 90kkm (=180kkm in both directions). So you can only get a second shot at the missile if it ended up more than 30kkm from the area-defence variant during the first 5s interval- and the further away it was in the first interval, the closed it will be in the second.

Oh, and somewhat more importantly, you seem to be missing a factor of 2. Remember, the assumption was that the area-defence variant serves as advance picket, i.e. it is at a certain distance of the protected task group and can engage missiles that are streaming towards it, as well as those that have passed the ship and are flying onwards to the task group. So you get to use the full range of the ship twice.

Looks like you were missing the "factor of 2" as well.  Yes, my formula was deliberately build to address a single intercept vector.  Nor do you need to multiply the density function numerator by 2 to represent the shot oportunities within beam range, if you simply multiple the orginal formula by 2 you get the same result.

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).

What neither of us has addressed is that movement of the intercepting ship will have a minor but calculable impact of the actual intercept envelope.  While the impact is small, it does invalidate using a simple factor of 2.  Both of us are assuming a stationary platform precisely on the salvos attack vector.  To keep the formula's simple I can accept that.


Before I respond any further I do have to correct my area defense formula.  In review I’ve found that the maximum range intercept sub-calculation (1-bR/fR) only needs to be replaced with .01 when bR>=fR dropping the test against missile impulse movement.  Whether the bR is less that the missile travel distance has no bearing on maximum range intercept chances.

And by adding in the change for handling multiple shot oportunities it becomes: 
4*P=sO*(bR/(I*(5*tV)))*(((1-bR/fR)+(1-mR/fR))/2)*(sV/tV+tB)*Ab

• sO=number of shot opportunities
• Ab=(1-(tE-sE))*cG = 1 for this discussion as previously established

With those changes my results change too:
4*P=2*(90/(1(5*24)))(((1-90/192)+(1-10/192))/2)(12/24+0.2)=3.11*Ab


Mine is not the only one with changed results.

On the face of it yours changes from:
P = r/(5*v) (2 - r/R) * A = 4.6 * A

To:
P = (2r/(5*v)) (2 - r/R) * A = 9.19 * A

Though I hope you meant it to change to:
P = (2r/(5*v)) (1 - r/R) * A = 3.19 * A

Since we’re not going to agree on including the speed modifier:
A=(sV/tV+tB))*(1-(tE-sE))*cG

That just leaves the question of which method is more accurate: Average the hit probability across it’s linear range or use a single fixed range?

Using: 4*P = (1-mR/fR)*(sV/tV+tB)) results are tabled as:

Code: [Select]

Appoach Egess total
120 0.00 10* 2.65 2.65
110 0.00 10 2.65 2.65
100 0.00 20 2.51 2.51
90 1.49 30 2.36 3.85
80 1.63 40 2.22 3.85
70 1.78 50 2.07 3.85
60 1.93 60 1.93 3.85
50 2.07 70 1.78 3.85
40 2.22 80 1.63 3.85
30 2.36 90 1.49 3.85
20 2.51 100 0.00 2.51
10 2.65 110 0.00 2.65
3.33

Using: 4*P = (1-m/f) esuls ae abled as:

Code: [Select]

Appoach Egess total
120 0.00 10* 3.79 3.79
110 0.00 10 3.79 3.79
100 0.00 20 3.58 3.58
90 2.13 30 3.38 5.50
80 2.33 40 3.17 5.50
70 2.54 50 2.96 5.50
60 2.75 60 2.75 5.50
50 2.96 70 2.54 5.50
40 3.17 80 2.33 5.50
30 3.38 90 2.13 5.50
20 3.58 100 0.00 3.58
10 3.79 110 0.00 3.79
4.75

(10* - at range 0 the minimum range used by the program is 10)

The final figure for each table is the average of each ranges total.

Something else to note:  The above figures assume that the lasers are turret mounted with turret tracking at least 12,000kps.  If not the tracking speed drops to 3,000kps, even with the advanced fire control , and the numbers tank (.47 vs 1.01 and 2.06 vs 4.43||.18 vs .72 and .79 vs 3.17).  Nor are the area defense and final defense examples equal in hs/tonnage.  Area defense FC and Turret total 32.32 hull spaces while the final defense FC and Turrets total 36.64 hull spaces (assuming quad turrets).  This does not take into account the required power plants.  If the PP tech is Pebblebed each laser needs a 1hs reactor as well taking the area defense suite to 36.32hs and the final to 42.64hs.

Yeah true, I ignored the gearing and power plant requirements, both of which should have been considered on a closer look. One rather subtle point though: You do not need such a large reactor for the final fire variant. Implicitly the area-defence calculation assumed that enemy salvos were at least 10s apart (so that you could actually shot at one salvo more than once when the opportunity was there). So it would be sufficient if the final fire variant fired every 10s (or even less frequent) to match that. Hence you could get away with half the powerplants/laser for the final fire variant. ( I recognize you have used 6 PP for the 8 Lasers on the final fire variant, but I am not sure whether this is the reason). This does not invalidate your point at all though, even if it decreases the difference in mass slightly. As a funny coincidence when gearing and powerplants are considered both setups have pretty much the same costs though, so in a way it is still a fair comparison (to some extend)

Minor correction here as well.  The final defense suite should be 44.64hs with 8 1hs powerplants (or 1 8hs powerplant).  The reason being maintaining the ROF of 5 that the 10cm/C3 lasers have when fully powered.  It’s never a good idea to underpower beam weapons if they have the capacitor capacity for a higher rate of fire.

[dgibso29]

You guys are arguing way too strenuously over methodology when your results both indicate that final fire is more effective.

I don't think you said it loud enough.

[Zook (OP)]

I'm not sure they're from this planet. Betelgeuzians, maybe? Just let them continue; after exhausting their mental faculties like that, they fall in a deep coma for months.

[Theokrat]

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:

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:


There is a nice property of integrals which allows us to write the sum of p₁ and p₂ as follows (note the change in the limits of the integral):


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:

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):


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:


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:


Moreover, we can write , 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:


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 , where r_FC is the range of the firecontroll. Thus the equation becomes:


Solving the integral:


Factoring out the fraction and inserting the limits:


Writing out the quadratic term in the the brackets and canceling the factors of 2 and 1/2:


Consolidating the bit in the brackets of 1 and -1 and factoring out the minus:


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.

[Hawkeye]

I´m gonna save this post for when people on other forums start bemoaning the complexity of games other than Aurora 

[jseah]

*claps*  I humbly turn my geek badge over to you, you clearly deserve it more. 

...
Now time to earn it back.  =D

[Garfunkel]

I feel stupid now. 

[Charlie Beeler]

Thoekrat I applaud the amount of work you put into that.  To bad that the results of using that formula imply that you can intercept more missiles than there are weapons to fire. 

P = 2(90/(5*24))(2-90/192) = 2.2969

Your assuming that R>d will always be true when it's almost always the other way around.  That probigates though the rest of your formula and results in (2-R/r_FC) giving a probability over 100%.  This is clearly incorrect.

[Theokrat]

*claps*  I humbly turn my geek badge over to you, you clearly deserve it more. 

*bow

...
Now time to earn it back.  =D

I'll be looking forward to this!

Thoekrat I applaud the amount of work you put into that.

Thanks, appreciated.

To bad that the results of using that formula imply that you can intercept more missiles than there are weapons to fire. 

Let's not be too quick here. It can actually be the case that you can intercept more missiles than you have weapons - if you manage to fire more than once, which is sort of the idea behind area defense. And to be clear, when you can not fire multiple times, the formula is not going to indicate that you can shoot down more missiles than you have weapons. Because that would be incorrect.

P = 2(90/(5*24))(2-90/192) = 2.2969

The formula that I posted already assumes you can shoot at the missile on the inbound and outbound leg of its journey. This resulted in a factor of 2 that later got cancelled by another factor of 1/2. So there is no need to introduce another factor of 2, and actually the result would be 1.1484 instead of 2.2969 (ignoring the "other" factor). So we would expect slightly more than one missile being shot down (per weapon).

let's conduct a sanity check: the missile moves 120k km in 5 seconds, and we have a range of 90k km per direction, so a total engagement coverage of 180k km. So we could expect 1.5 shots per weapon. If the "p_other" factor was 1 (i.e. really fast tracking) then 1.1 shot down missiles would not seem particularly high, especially given the relatively good range of the firecontroll.

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).
...


Writing out the quadratic term in the the brackets and canceling the factors of 2 and 1/2:

Your assuming that R>d will always be true when it's almost always the other way around.  That probigates though the rest of your formula and results in (2-R/r_FC) giving a probability over 100%.  This is clearly incorrect.

I am not actually assuming that R>d. Let's say R=0.5d, then the formula would be P=p_other*0.5*(2-r/r_FC)=p_other*(1-0.5r/r_FC) - which is always smaller than 100%. Which is just what you would expect - R=0.5d means the missile is so fast that it could travel twice the systems range in one 5s increment. This means we can only ever get one shot at it as it streams past the system - and thus could at most expect one hit.

[Bgreman]

Here's a question for you:  Does r_FC (the range of the fire control) depend on r itself?  You seem to be taking your integral over r, but r_FC, naively to me, seems to be a constant.  In which case your integration is incorrect.

If r_FC is a constant, then:

[Theokrat]

Here's a question for you:  Does r_FC (the range of the fire control) depend on r itself?  You seem to be taking your integral over r, but r_FC, naively to me, seems to be a constant.  In which case your integration is incorrect.

If r_FC is a constant, then:

r_FC is a constant, but actually your result is just the same as my result, only in a different (more elegant) form. Let's start with mine and get to yours through a series of simple transformations:


Solving the quadratic bracket:


Canceling +1 and -1 in the bracket:


Getting the minus sign inside of the bracket:


Getting the 1/2 inside the bracket:


...and the r_FC


Now taking out one R factor and we arrive at your solution:

[Bgreman]

Fair enough!  I obviously didn't bother to actually see if they were the same, I just saw the integral and the next step and was like, "Why does that look so complicated?"

My job doesn't use my math degree, so I spend a lot of time doing cost/benefit analysis of converting CI to construction factories vs converting CI to mines in the midst of a corundium crunch to see the limits that the economy can handle.

[Charlie Beeler]

Theokrat thank you for that last explination, I was misreading what your were doing with P_other.