(TF heading this way)
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Small anti-missile corvettes (2,000 tons and harder to spot than the rest)
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Anti-missile frigates
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Beam area defense (by being ahead of the main body, they can fire their lasers at missiles flying past them, thereby doubling their firing opportunities)
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Main Body (missile destroyers and colliers)
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Sensor ship(s) - easy to spot, should be protected by every defense in the TF
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Support ships (tankers, jump ships etc.) - waaay behind the main body
Theokrat there are a couple of very important things you do not appear to addressing correctly.
- beam fire control hit probability at intercept range
- tracking bonus is not a constant, it's cumulative
I am not sure what you mean. I do explicitly cover the effect of range on the hit probabilities. That is after all the entire point that makes a difference between final-fire and area-defence. The whole derivation of the hitchances quoted above shoves everything else into the “A” factor; the rest of the formulas is only concerned with the hit probability due to range.
Take final fire for example: Intercept occurs at 10,000 km, so the hit probability due to the range of the firecontroll is P = 1 – 10,000 km / R, where R is the range of the firecontroll. So for the example above its P = 1 – 10,000km / 48,000km = 79%. Of course this is not the final hitchance, which would include other effects (most importantly tracking speed), but it’s the separable factor due to the range of the BFC.
For area defence the derivation is a bit more complex, as essentially you have to take an integral over the hitchance at range x (P = 1 – x/R) times the probability density function ( uniform distribution, i.e. rho(x) = 1/(5v)). With a bit of algebra you get to the formula above.
Could you explain that in more detail? I was under the impression that the hit chance is computed as follows P = p_range * p_trackingspeed * p_crewgrade * p_trackingbonus * p_missileECM. So I basically wrote that as P = p_range * EverythingElse in order to compare final fire versus area defence. If the tracking bonus is an additive bonus rather than a multiplicative one, then that would change the calculations (in favour of area defence). Is that what you are saying?
I understand that the tracking bonus must accumulate over time until it reaches the maximum possible amount at the current tech level. So I implicitly assumed that this highest possible level had been reached at the maximum range of the area-defence variant (and that thereafter the tracking bonus was constant). That is prudent I would say, as its quite easy to have missile search sensors with a range of millions of km, while the maximum range of the 10cm laser from above was a mere 90k km. Moreover it is a conservative assumption in the sense that the assumption favours the area-defence variant (which still loses out).
Sweet Jesus. The man has a degree in auroralogy. Anyway, that's great stuff. I'll post it to the wiki when I understand it.Hehe, I am merely an analyst, so breaking things down to numbers is kind of what I do. Although it greatly helps that in Aurora systems come with a number and formula tag attached. In real life the painful thing is trying to get people to give you a reasonable answer how much an increase of this or that variable by x% would help them. Here (http://www.rand.org/pubs/occasional_papers/2008/RAND_OP223.pdf)is a rather remarkable paper on system analysis. Ok I am getting quite off-topic.
But although I know that area defense is less effective, there are two reasons I still build laser turrets:Both of these would also apply to other final fire weapons, like railguns and gaussguns though.
1) for the unlikely event that I came across an unarmed target (civilian, crippled ship, etc.) and
2) for the time when my missile magazines have run dry, or exploded.
Whether beam point defense is set to area or final the formula for hit probability is the same. The only things that are different are intercept range and tracking time.
I am not sure what you mean. I do explicitly cover the effect of range on the hit probabilities. That is after all the entire point that makes a difference between final-fire and area-defence. The whole derivation of the hitchances quoted above shoves everything else into the “A” factor ; the rest of the formulas is only concerned with the hit probability due to range.
Take final fire for example: Intercept occurs at 10,000 km, so the hit probability due to the range of the firecontroll is P = 1 – 10,000 km / R, where R is the range of the firecontroll. So for the example above its P = 1 – 10,000km / 48,000km = 79%. Of course this is not the final hitchance, which would include other effects (most importantly tracking speed), but it’s the separable factor due to the range of the BFC.
For area defence the derivation is a bit more complex, as essentially you have to take an integral over the hitchance at range x (P = 1 – x/R) times the probability density function ( uniform distribution, i.e. rho(x) = 1/(5v)). With a bit of algebra you get to the formula above.
Could you explain that in more detail? I was under the impression that the hit chance is computed as follows P = p_range * p_trackingspeed * p_crewgrade * p_trackingbonus * p_missileECM. So I basically wrote that as P = p_range * EverythingElse in order to compare final fire versus area defence. If the tracking bonus is an additive bonus rather than a multiplicative one, then that would change the calculations (in favour of area defence). Is that what you are saying?
I understand that the tracking bonus must accumulate over time until it reaches the maximum possible amount at the current tech level. So I implicitly assumed that this highest possible level had been reached at the maximum range of the area-defence variant (and that thereafter the tracking bonus was constant). That is prudent I would say, as its quite easy to have missile search sensors with a range of millions of km, while the maximum range of the 10cm laser from above was a mere 90k km. Moreover it is a conservative assumption in the sense that the assumption favours the area-defence variant (which still loses out).
your failing to account for is the modifier for fire control and/or turret tracking speed being below the target speed. Against missiles, unless you have a significant tech advantage, there will always be a negative modifier. That is what the tracking bonus, if the tech has been developed, offsets. The modifier is tracking speed/target speed.
Area defense mode is the least likely to be effective. But it is a function of beam range vs the distance a missile travels in a single game cycle.Yeah. And if you have a close look at the formula above you will see that these factors are in there: Beam range (r), distance a missile travels in a single game cycle (5*v), and even range of the beam fire-controll (R).
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
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.
Now we plug-in weapon and fire control values for the area defense example:Well you went through great length through about the tracking speed and was it really worthwhile? We could have saved ourselves the trouble and just abbreviated the factors as “A” and thereby have arrived at the same relative result. Oh wait ;-)
4*P=(90/(1(5*24)))(((0.01)+(1-10/192))/2)(12/24+0.2) = 1.01 missiles intercepted not 4.6.
The final defense example becomes:
8*P=(1-10/48)(12/24+0.2) = 4.43 missiles intercepted not 6.3.
Especially with you own examples have fire controls with significantly different tracking speeds against the same target speed(3000 vs 12000).Actually I am a bit surprised by this statement. Not only is it not true, because my example entails the same exact tracking speed in both scenarios, but also because you also then use the same tracking speed of 12,000 for both and you must be aware of the firecontroll’s speed rating because it enters your calculations on the weight as well. I do not understand how you could interpret my post as to say the firecontrolls have different tracking speeds:
So for the area-defence version we would have to spend 800t on the firecontroll (size 16, 4x range, 4x tracking ). […]
Conversely, what happens for the final-fire version? The firecontroll is only 200t (size 4, 1x range, 4x tracking ),
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.
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)
Quote from: Charlie BeelerYou 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.
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.
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
So for the area-defence version we would have to spend 800t on the firecontroll (size 16, 4x range, 4x tracking)
Maybe the last bit is the most controversial here, because you write:Quote from: Charlie BeelerBut 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.
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.
Quote from: Charlie BeelerThe 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.Quote from: Charlie BeelerSince 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.
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
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
Quote from: Charlie BeelerSomething 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)
You guys are arguing way too strenuously over methodology when your results both indicate that final fire is more effective.
Looks like I didn’t forget anything. It is explicitly excluded with a stipulation as to why.
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.
*claps* I humbly turn my geek badge over to you, you clearly deserve it more.*bow
...I'll be looking forward to this!
Now time to earn it back. =D
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) = 2.2969The 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).
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).
...
(http://latex.codecogs.com/gif.latex?P=2\cdot%20\varrho%20\cdot%20\widetilde{p}_{other}\cdot%20\frac{-r_{FC}}{2}\cdot\left%20[%20(1-\frac{R}{r_{FC}})^2%20-%201\right])
Writing out the quadratic term in the the brackets and canceling the factors of 2 and 1/2:
(http://latex.codecogs.com/gif.latex?P=%20-\varrho%20\cdot%20\widetilde{p}_{other}\cdot%20r_{FC}\cdot\left%20[%201-2\cdot%20\frac{R}{r_{FC}}+\left%20(\frac{R}{r_{FC}}%20\right%20)^2%20-%201\right])
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/rFC) 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=pother*0.5*(2-r/rFC)=pother*(1-0.5r/rFC) - 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.
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.rFC 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:
If r_FC is a constant, then:
(http://latex.codecogs.com/gif.latex?\int_{0}^{R}(1-\frac{r}{r_{FC}})dr=\int_{0}^{R}1%20dr-\frac{1}{r_{FC}}\int_{0}^{R}rdr)
(http://latex.codecogs.com/gif.latex?\int_{0}^{R}1\,%20dr-\frac{1}{r_{FC}}\int_{0}^{R}r\,%20dr=\left%20[%20r%20\right%20]_{0}^{R}-\frac{1}{r_{FC}}\left%20[%20\frac{r^2}{2}%20\right%20]_{0}^{R}=R-\frac{R^2}{2r_{FC}}=R(1-\frac{R}{2r_{FC}}))
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?"he, the semi-AAR makes an amusing read, and its nice to see some differential equations applied to analysis in Aurora. It would also have been fun to device a build-strategy that just barely makes use of the existing mineral stocks (i.e. so that the integral of the usage until break even plus existing stocks is just 0). Also curious to see the Federation dispersing its fighting assets so broadly, while seemingly little protection is retained on earth. Anyway, getting a bit off-topic here.
My job doesn't use my math degree, so I spend a lot of time doing cost/benefit analysis (http://forums.somethingawful.com/showthread.php?threadid=3474164&userid=85279&perpage=40&pagenumber=6#post405719579) 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.
Theokrat thank you for that last explination, I was misreading what your were doing with Pother.Glad we could clear this up then :-)
Also curious to see the Federation dispersing its fighting assets so broadly, while seemingly little protection is retained on earth. Anyway, getting a bit off-topic here.