I don’t understand why everyone still wants to debate this mess. First of all you are all way off course. His formula was based on Classic not on G40 so the numbers are all off to begin with. The 2 games are vastly different in almost all respects so even if the costs and values were the same the numbers would still be bogus. Throw in the others here who want to change the values of units instead of debating the formula which was based on a game which none of you are talking about in the first place and based on OOB values to begin with. So…WHAT THE SAM HELL ARE ALL OF YOU WASTING YOUR TIME HERE FOR?!?
Put your stupid calculators away and learn how to play the game using your intelligence and creativity. This game was created to be played between people, not machines. If you want to talk about changing the values then go to the House Rules section where there are already several threads to discuss that.
This game is actually a lot of fun to play. Try not to forget that while you are staring at your calculators.
I’ve stressed this in another thread.
@Baron:
I tried to apply your formula to a special case.
For instance, the defensive picket made usually by Russia to prevent Germany’s blitzkrieg while retreating main force behind this front line.Usually, you need to commit 1 ground unit, as much as possible the cheapest: 1 infantry.
But your formula seems to show that it can be more cost effective to put 3 Infs, even if it costs 9 IPCs to inflict more casualty to the attacker.
At 1 or 2 units, it keeps the same ratio for the investment (1.25) but beginning with 3 units the stack get stronger per IPC invest (2.00).
Often, the picket Inf brings no result except blocking but if you block and get 1 casualty or more, it is far better.So, does this formula is showing a better use of unit in this tactictal retreat?
Simple answer is no, it does not. For two reasons, 1, your goal with the scirmishers is not to inflict dammage, but to prevent blitzing. So the purpose is different. And the other reason is covered below.
Longer answer is: The formula does only give the strength of the stack, you need a different formula to figure out what damage that will result in. If the stack will die anyways (like your example) it might be that adding the 2 extra inf not will inflict the needed 6 ipc of damage to the germans attacking it.
On an extra note:
@Baron:
But your formula seems to show that it can be more cost effective to put 3 Infs, even if it costs 9 IPCs to inflict more casualty to the attacker.
At 1 or 2 units, it keeps the same ratio for the investment (1.25) but beginning with 3 units the stack get stronger per IPC invest (2.00).
Often, the picket Inf brings no result except blocking but if you block and get 1 casualty or more, it is far better.It actually shows that a bigger stack is more costeffective than a smaller. So if you have 50 units, it it better to have them in 1 big stack, rather than 2 smaller. it is a more costeffective way of fighting :) But I guess everybody already knew that ;)
I might widen my line of questioning then.
Simply put, your formula specific changes when reaching 3 Infantry (for 9 IPCs), like the start of an exponential curve (there is a square operation in the formula x^2), makes me wonder if there is some circumstances in which a stronger picket of 3 Infs is more potent.
Probably if your totally outwhelmed, in any case your picket will only roll once, but if limited number of units are involved maybe there is a threshold of enemy units in which throwing 3 Infs get you more than a single roll on defense and you get the opportunity to make more casualty on enemy before the decisive battle.
Is it possible?
Why are you guys still discussing this? Didnt we already decide that we aren’t interested? Or am I nuts?
After showing many shortcomings on Vann formula, Kreutzfeld compared with his own formula (which can be more easily use on game without using AACalc, to measure two stacks strength).
Then I tried to showed or validate some tactical combat (for a different optimal tactical retreat) implications from his formula which cannot be implied by Vann formula either.
If someone said he does not want to talk about pratical impact of measuring combat values, it does not mean everyone is clueless. And I cannot see why it was not a correct line to investigate an idea.
Aside some different combats values thrown in (which may condamned this thread to HR even if this was not OP intent) because Vann formula is still able (or unable) to point a few obsolete or weak units in OOB roster.
I would never have think about a possible different way of optimizing a tactical retreat if this specific comparison about mathematical formulas had not been push forward.
And don’t tell me TripleA players never check odds on BattleCalc before sending their final moves in PBF.
:-D
@Baron:
I might widen my line of questioning then.
Simply put, your formula specific changes when reaching 3 Infantry (for 9 IPCs), like the start of an exponential curve (there is a square operation in the formula x^2), makes me wonder if there is some circumstances in which a stronger picket of 3 Infs is more potent.Probably if your totally outwhelmed, in any case your picket will only roll once, but if limited number of units are involved maybe there is a threshold of enemy units in which throwing 3 Infs get you more than a single roll on defense and you get the opportunity to make more casualty on enemy before the decisive battle.
Is it possible?
I guess it would be possible to construct some examples where stronger pickets might be better. But it would require some limitations on your opponent.
If we assume that germany and ussr are picketing each other on the east front. And we assume that they are both strong enough to deadzone the area between their main stackes. And we also assume that they have 3 or 4 areas where then can scirmish. And we also assume that most of the short range luftwaffe are tied up west and cant participate, only the two german bombers can, and 1-2 fighters. If you then put out 3 inf in each area, germany would have to send in at least 3 inf + bomber to reliably take the terr. So, if you are able to picket back with 3 inf and ussr plane, you might be able to force an increase in the losses to scirmishing on the east front. germany would very quickly run out of inf, and would have to use mechs instead. This could be very favourable for the soviet. Therefore the german might go for only attacking 2 of the terrs with 2 planes and 2 infs instead of picketing all 4. But then, you are preventing the german from getting income from 2 of the terrs and you only have to pay with 1 extra inf in each of the two areas that does get hit. So you can picket right back with the red air force and some infs.
So in this very spesific situation, you might want to picket with 3 infs. However, if germany have 2 planes for each terr (so 8 in total), instead of 1 for each, then this picketing strat is getting worse.
Others might be able to think of other scenarios where having more troops scirmishing is beneficial.
This thread is still here also !!!
I also see VANN DANN still can go on site here but is banned from speaking.
So, as a rule of thumb, if the enemy can bring 2 aircrafts in addition to the same number of Infantry picket, then you are toasted. Right?
2 Fgs & 3 Infs^2 * A9/5 = 25*1.8 = 45
3 Infs^2 * D2 = 18
It is more than double value of picket stack.
45 vs 18 = 5 : 2 ratio
But, if there is only 1 additional plane in attack, it is workable for defender?
(1A3+3A1)^2 *A6/4 = 16 * 1.5 = 24
24 vs 18 = 4 : 3 ratio
So, whenever the attacker get only 4:3 ratio or less, the defender should try it.
Is it a right way to apply your formula?
Thats why you use the Larrymarx formula to solve these issues.
@Baron:
So, as a rule of thumb, if the enemy can bring 2 aircrafts in addition to the same number of Infantry picket, then you are toasted. Right?
2 Fgs & 3 Infs^2 * A9/5 = 25*1.8 = 45
3 Infs^2 * D2 = 18
It is more than double value of picket stack.
45 vs 18 = 5 : 2 ratioBut, if there is only 1 additional plane in attack, it is workable for defender?
(1A3+3A1)^2 *A6/4 = 16 * 1.5 = 2424 vs 18 = 4 : 3 ratio
So, whenever the attacker get only 4:3 ratio or less, the defender should try it.
Is it a right way to apply your formula?
@Imperious:
Thats why you use the Larrymarx formula to solve these issues.
For now, this is Kreuzfeld formula that I’m applying here.
Someday, there maybe a way to use both but for now it is two different beasts.
And the Baron-Larrymarx formula does not help, not accurate or too many calculations to get to something.
Kreuzfeld formula is simpler for a F-2-F game, and few numbers.
Kreuzfeld formula.
As long as its not named after a member… better.
@Imperious:
Kreuzfeld formula.
As long as its not named after a member… better.
For now, it is the case all Vann, Kreuzfeld, Baron and Larrymarx are members.
If you have an idea for a more suitable name for Baron-Larrymarx, I’m opened to it, as long as others agree.
I cannot deny Vann, Larrymarx then me filiation (or line of succession) to this Baron-Larrymarx formula.
But Kreuzfeld bring his own much earlier. And it is a different beast, for a different purpose.
You guys are forgetting something important: SIZE MATTERS
The back and forth nature of A&A battles means that very slight advantages will snowball into huge advantages when you scale up the number of units. To demonstrate, I pulled up a battle calculator and tried throwing infantry against half their numbers in tanks (stacks with equivalent values).
Infantry Tanks Attacker wins Avg swing Total value % swing
2 1 55% 0.1 12 0.8
4 2 60% 1.2 24 5.0
8 4 66% 4.2 48 8.8
16 8 74% 11.6 96 12.1
32 16 83% 29.4 192 15.3
64 32 92% 68.3 384 17.8
128 64 98% 147.5 768 19.2
You start with what is nearly a fair fight at the beginning and wind up with an overwhelming advantage as you draw out the battle and increase the number of turns over which the advantage is played out. In contrast, if the battles are only for one round, the % swing rises only to 4.166% (1/24) and then stays there no matter how large the stacks are.
This, combined with my earlier assertion that units are almost always better in mixtures means we really should be looking at how to determine the relative worth of stacks and mixtures of units as opposed to individual units. Units mean nothing by themselves - they always must be analyzed in context.
I propose analyzing different mixtures and quantities of units to determine their worth in attacking vs. stacks of units we already know to be strong on defense - infantry, destroyers, fighters and carriers (because they carry fighters). A compete analysis will increase the size of the defender’s stack if the attacking units are slow because the defender will “see it coming” and have more time to turtle up.
The back and forth nature of A&A battles means that very slight advantages will snowball into huge advantages when you scale up the number of units.
And speaking of formulas, the point mentioned by Larrymarx ties in nicely with Lanchester’s Square Law, a differential equation which models the attrition effects of modern combat involving long-range weapons. As I understand it, the equation basically states that the rate of attrition is proportional to the square of the number of weapons which are shooting, and therefore that any numerical advantage by one side will increase exponentially as the battle progresses.
Yep Marc,
The original thread started with this Lanchester’s Square Law, then found a specific formula for stacks of A&A units.
I made it on AACalc then I revised numbers by applying this formula derived from above Stack formula:
√(P2 / P1) = N1 / N2
To get break even (meaning 50% vs 50% odds) number of unit ratio for different Power relative to each other:
Power 1
vs 2 = 14 units :10 units (140:99)
vs 3 = 19:11 (189:109)
vs 4 = 2:1
Power 2
vs 1 = 10:14 (99:140)
vs 3 = 11:9 (109:89)
vs 4 = 14:10 (140:99)
Power 3
vs 1 = 11:19 (109:189)
vs 2 = 9:11 (89:109)
vs 4 = 15:13 (149:129)
Power 4
vs 1 = 1:2
vs 2 = 10:14 (99:140)
vs 3 = 13:15 (129:149)
So, if you want to see if your on the right side, find the average Power and compare your number of units with the ratio above.
For example, you have 3 Tank and 1 Inf vs 2 Inf 1 Art.
Avg less than 3 compared to 2.
4 units vs 3 units 4:3
The table for 3 vs 2 say 9:11, so, if you drop to 2 tanks vs 3 units during combat resolution, you know you are below 50% odds of survival. You may then retreat, or try your luck.
This is very interesting, and I think it validates the cost formulas: a quadrupling in strength is required to justify a doubling in cost.
I also like that you’ve removed the factor of cost for the above ratios. It’s always better to bring more than you need to win the fight, and if you know what break even is for a fight, you can focus on whether or not you can bring more than that to the battle. For this purpose, unit cost might be a factor or it might not.
With large battles, as I demonstrated earlier, it just takes a slight advantage to be certain of victory. If it is a large battle, then winning it is probably more important than preserving your units with high values. If it’s a small battle, cost matters more because you want your good units left for future battles, but in the case of small battles these formulas and ratios don’t really matter anyways because anyone can look at the units and tell which side will win.
This is very interesting, and I think it validates the cost formulas: a quadrupling in strength is required to justify a doubling in cost.
I also like that you’ve removed the factor of cost for the above ratios. It’s always better to bring more than you need to win the fight, and if you know what break even is for a fight, you can focus on whether or not you can bring more than that to the battle. For this purpose, unit cost might be a factor or it might not.
With large battles, as I demonstrated earlier, it just takes a slight advantage to be certain of victory. If it is a large battle, then winning it is probably more important than preserving your units with high values.
If it’s a small battle, cost matters more because you want your good units left for future battles, but in the case of small battles these formulas and ratios don’t really matter anyways because anyone can look at the units and tell which side will win.
I agree.
Would you agree Larrymarx, about this formula being renamed ENIGMA formula ? To consider all contributors without having a too cumbersome name.
Here is the table based on Baron-Larrymarx formula completed on effective cost vs combat points ratio:
For all 1 hit units, you use : 36 Power/(cost^2) = offense or defense factor* based on cost
For 2 hits and 3 hits unit : 36 Power/(cost^2) {1+[(nb hit -1)/11.62] }= offense or defense factor* based on cost
To get the cost of a 1 hit unit for a given factor of reference: √(36*Power of unit / Offence or defence Factor)= Cost.
For a 2 hits unit for a given factor of reference:
√(36*Power of unit {1+[(nb hit -1)/11.62] } / Offence or defence Factor)= Cost.
For combined arms and multiple units you have to average both combat points per unit and cost per unit.
Then you can add it into the formula.
@Baron:
I made it on AACalc then I revised numbers by applying this formula derived from above Stack formula:
√(P2 / P1) = N1 / N2To get break even (meaning 50% vs 50% odds) number of unit ratio for different Power relative to each other:
Power 1
vs 2 = 17 units :12 units (170:120) 1.42
vs 3 = 19:11 (189:109) 1.73
vs 4 = 2:1 2.00Power 2
vs 1 = 12:17 (120:170) 0.70
vs 3 = 11:9 (109:89) 1.22
vs 4 = 17:12 (170:120) 1.42Power 3
vs 1 = 11:19 (109:189) 0.58
vs 2 = 9:11 (89:109) 0.82
vs 4 = 15:13 (149:129) 1.15Power 4
vs 1 = 1:2 0.50
vs 2 = 12:17 (120:170) 0.70
vs 3 = 13:15 (129:149) 0.87So, if you want to see if your on the right side, find the average Power and compare your number of units with the ratio above.
For example, you have 3 Tank and 1 Inf vs 2 Inf 1 Art.
Avg less than 3 compared to 2.
4 units vs 3 units 4:3The table for 3 vs 2 say 9:11, so, if you drop to 2 tanks vs 3 units during combat resolution, you know you are below 50% odds of survival. You may then retreat, or try your luck.
It is my first shot working with Table feature of the Forum.
If someone can do better, I will appreciate.
With this small 5 x 5 table, you get in a glimpse what is the 50%-50% break even according to the ratio of units and the average power of a given stack.
This table may also be written in decimal instead of a ratio, below. But ratio are better to understand how each ratio is paired to another one which is simply reversed.
A more complete table would include average power: 1.5, 2.5 and 3.5
But, in F-2-F, you can always round up the average enemy’s power, so you do a safer battle to be sure you are above break even ratio.
X means the ratio is 1:1. I did not want to overcharge this table with obvious infos.
| Power
1
2
3
4 | 1
X
12:17
11:19
1:2
| 2
17:12
X
9:11
12:17
| 3
19:11
11:9
X
13:15
| 4
2:1
17:12
15:13
X
| Power
1
2
3
4 | 1
X
0.70
0.58
0.50
| 2
1.42
X
0.82
0.70
| 3
1.73
1.22
X
0.87
| 4
2.00
1.42
1.15
X
For instance, reading from left row to the right, if you have an average power of 1.3 for 17 units and the defender has 2.2 average power for 10 units.
You may cross-referenced the 1 row with the 2 column, saying you need 17: 12 ratio, or 1.42 more units than defender.
So, your 17:12 ratio or 1.72 more units give an above 50-50% odds of winning.
Of course, during battle, ratio of units and average power may changes, especially when fodders are done.
So, you may decide at critical moment to recheck your odds of success.
For example, the defender may have only 3 Tanks left, while attacker only 4 Infs and 1 Tank.
This give a 5: 3 units ratio. And the 1 (A7 / 5= 1.4) row compared to 3 column, says: 19 to 11 or 1.73.
If looking the 2 row, it says 11:9 or 1.22.
So, assuming this rounding up or down, it reveals you are above or below the break even point.
Thus, it is still near 50%-50%.
In fact, AACalc says:
Overall %*: A. survives: 70.6% D. survives: 24% No one survives: 5.5%
But, Punch formula might be better to anticipate results?
A7 for 5 hits vs A9 for 3 hits.
Stack formula is more revealing IMO:
251.4= 35 vs 93= 27 |
|
X means the ratio is 1:1. I did not want to overcharge this table with obvious infos.
| Power
1
2
3
4 | 1
X
12:17
11:19
1:2
| 2
17:12
X
9:11
12:17
| 3
19:11
11:9
X
13:15
| 4
2:1
17:12
15:13
X
| Power
1
2
3
4 | 1
X
0.70
0.58
0.50
| 2
1.42
X
0.82
0.70
| 3
1.73
1.22
X
0.87
| 4
2.00
1.42
1.15
X
This table can also be memorized with 5 basic ratios and you can reverse at will, or 6 if we include 1:1 ratio:
1:2 (4 vs 1), 11:19 (3 vs 1), 12:17 (4 vs 2 or 2 vs 1), 9:11 (3 vs 2), 13:15 (4 vs 3)
OR same order but from reverse ratio:
2:1 (1 vs 4), 19:11 (1 vs 3), 17:12 (1 vs 2 or 2 vs 4), 11:9 (2 vs 3), 15:13 (3 vs 4) |
|
Sure, enigma works fine as a name.
For the tables you are developing, why don’t we call them Lanchester tables? They are very clearly a derivative of his work.
Can we steer this discussion into an actual application towards developing strategy?
Take a German push into Russian territory. Germany needs to be concerned with a stack of units in Moscow whose average power is slightly higher than 2. Russia is going to be building a lot of infantry and artillery so there is very little skew - the exception would be if UK / US have landed a bunch of fighters there.
Germany is going to bring a bunch of tanks and its air force to the battle, but it will also have infantry attacking at one. The average power will likely be higher than the Russian stack’s, and there will be a significant skew as well.
It makes intuitive sense that you should just try to match the number of units that they have over there, and your greater power as well as the skew in your favor will win you the battle. Can we use the Lanchester tables to be a little bit more precise in our assessment of this strategic objective?
