A question about CRs

Lord Pendragon

First Post
Is there a section of the SRD that has the formula for adding up multiple small CRs into a final EL?

Specifically, I'm wondering how many CR 1/2 critters it would take to get an EL 5 encounter, but knowing where to find the formula would be great...
 

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basic rule is to group fractional CR critters into whole CR and progress from there:

So 2 of the critters are CR1, double that (4) is +2CR, double it again (8) is another +2 CR for total CR5



Cheers
 

Is the Grim Tales Gamemastering Document still for free at RPG Now? Nope - it's 1.95 again. Let me run some numbers through the spreadsheet for you...
8 or 9 CR 1/2 creatures is EL 5, as Plane Sailing says, but the spreadsheet predicts that to be an "very easy" encounter for a party of 4 level 5 characters. Once you get too far beyond 10 creatures or so things get pretty hard to manage. I might recommend making a swarm of these CR 1/2 guys as an alternative.
 

Seeing as I had a long chat with Wulf Ratbane about this recently, I'd suggest you take a look at Grim Tales: Gamemastering! ;)

Other than that, there are a few encounter level generators on the web that'll allow you to do this. One of them can be found here:

Encounter Level Calculator

The answer to your question from this calculator is 12, which is slightly different to what the DMG would suggest in 8.

Pinotage
 


I figured how to add CRs according to the WotC method (which is not the same used in Grimtales):

CR uses a logarithmic scale. So, first bussiness is to convert CR to a lineal scale we'll call "power":

You calculate power by doing:

power(CR) = 2^(CR/2)

That is TWO raised to half the CR... and not the CR squared. When CR is an odd number, do the calculation for CR-1 and CR+1 (both of which are even numbers) and then average both results.
An exception would be CR 1, which has power equal to 1.
If CR is a fractional number, then the power is that same fractional number.

To convert power to an EL, you invert the operation:

EL(power) = log2(power) * 2

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So... you want an EL5 encounter... the power of an EL5 would be:

Since 5 is odd... we calculate for 4 and 6 and then average:
Power(4) = 2^(4/2) = 2^2 = 4
Power(6) = 2^(6/2) = 2^3 = 8
So, an EL5 has a power of (4+8)/2, which is 6.

Since CR1/2 creatures have a power of 1/2, I'd say 12 CR1/2 creatures make an EL5 encounter... of course, YMMV... they might make a very easy encounter nonetheless, depending on the specifics.

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Disclamer: I've found this system to work just fine when calculating stuff like the EL of CR7+CR7+CR8 when you compare it to the numbers suggested by WotC in published adventures. The bit about "the power of a fractional CR is the same fraction" I'm not too sure about, because that's just my best interpretation of what fractional CRs are supposed to mean... but I wasn't able to compare it with WotC's numbers.
 
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