CR, EL, and Lanchester's Law

diaglo said:
lanchester's law is based on all things being equal.
Actually, no, Lanchester's Law includes a coefficient for effectiveness (actually, the ratio of effectiveness between the two forces) -- but that would have complicated the message: "The strength of a military unit — planes, artillery, tanks, or just soldiers with rifles — is proportional not to the size of the unit, but to the square of its size."

If Ogres kill Goblins four times as effectively as Goblins kill Ogres, then we can put that coefficient into our calculations -- and we'd need twice as many Goblins to match a force of Ogres, not four times as many.
 

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Actually, that handles the Tarrasque v Kobolds question exceedingly well, then.

A Tarrasque kills a Kobold infinitely better than a Kobold kills a Tarrasque. Therefore, you would need the square root of infinity in Kobolds to kill a Tarrasque.

...presumably by suffocating the poor bastard in them.
 

Yes, that does work quite well - thanks for the addition!

One interesting thing I was considering is that, for a wizard, four kobolds at one time use up fewer resources than four kobolds encountered singly. In the first case, a single sleep spell will suffice... in the latter, four spells must be used.

I find that interesting. :)

Cheers!
 

Well, as mmadsen said,


mmadsen said:
In barrage combat, the rate of attrition isn't proportional just to the number of attackers but to the number of defenders too. If you cram enough defenders into an area, one fireball can kill all of them.
Unfortunately, it seems that Lanchester never got round to naming this principle. Perhaps we could call it Lanchester's Fireball Law? :)
 

If you treat the wizard as having a 400% kill rate of kobolds, then against one kobold, a lot of the wizard's capacity is wasted. But that's an issue of ammunition - if just <i>getting into a fight</i> costs you a certain minimum amount, then certain foes are hurtful to fight, because the cost isn't worth the kill.

Of course, when you start getting into 100% and above, Lanchester's Law is overrulled by the Law of Initiative: he who has it wins. :)
 


The offsetting factor here is area of effect spells, which are the great equalizer.

1+1 does not equal 2, because of area of effect spells.

The more bunched up opponents you fight, the more powerful your area of effect spells become.

If you fight one monster, your fireball only does a total of 50 points of damage.

But if you fight 2 monsters, your fireball is twice as powerful, dealing 50 points to each opponent, or a total of 100 points.

Fighting 4 monsters means you do 200 points of total damage, and so on.
 

mmadsen said:
If Ogres kill Goblins four times as effectively as Goblins kill Ogres, then we can put that coefficient into our calculations -- and we'd need twice as many Goblins to match a force of Ogres, not four times as many.
Ogre
Hit Points: 26
AC: 16
Atk: +8
Dmg: 14

Goblin
Hit Points: 4
AC: 15
Atk: +1
Dmg: 3.5

An Ogre hits a Goblin 70% of the time, and "kills" on a hit with certainty. Ogres eliminate Goblins at a rate of .7 Goblins per Ogre per round. A Goblin hits an Ogre 30% of the time, and "kills" after seven or eight hits. Goblins eliminate Ogres at a rate of .04 Ogres per Goblin per round.

It appears that Ogres really have 17.5 times the relative combat efficiency of Goblins -- and this means that we need roughly four times as many Goblins to match a force of Ogres, not 17.5 times as many.

If we assume that every Goblin can find a target -- and that Lanchester's Square Law thus holds -- a battle of 10 Ogres against 175 Goblins ends quickly: 10 vs 175, then 3 vs 168, then 0 vs 166. The Goblins overwhelm the Ogres. A battle of 10 Ogres against 41 Goblins ends after 15 rounds, with two Ogres left.
 
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A goblin kills an ogre after 7-8 hits. An ogre is 17.33x what a goblin can manage, so 4x goblins are needed.

However, the ogre can fight defensively. That increases it to 18.5x a goblin.

If the goblins fight defensively, they're just screwing themselves, though.
 


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