Reworking Level Adjustment: Fractional LA

[StevenC21]

So, I was thinking about the many issues LA has. Since as you level up, higher LA becomes a larger burden, especially when many effects on the template/race aren't scaling to keep up, making the template/race less useful. So, I came up with an idea, and I'd like to hear what you all think!

Fractional LA! Here's what I mean. Instead of 'LA +X', you would have something like 'LA +X/level', where X is the character level where the template can be assumed to be not very useful, and level is character level (Not ECL!)

This system allows templates to have a lesser cost as you level up, in a scaling way that is more flexible than LA Buyoff, and is automatic.

Thoughts?

[JNAProductions]

Not a bad idea.

Part of why LA is bad is that it eats your HIGHEST level, not your lowest. If you're LA 2/Wizard X, you're always a spell level behind a straight Wizard X+2. While less of an issue for a Fighter or similar, it royally sucks for those with nice class features. This kinda addresses that? Could we get some examples of how exactly it'd work, to clarify it?

[StevenC21]

I can't really give a great example since the system still has some quirks; at some levels & LAs, your ecl may actually drop with this system. More work is needed, which is part of why I am turning to the community!

[NNescio]

So, I was thinking about the many issues LA has. Since as you level up, higher LA becomes a larger burden, especially when many effects on the template/race aren't scaling to keep up, making the template/race less useful. So, I came up with an idea, and I'd like to hear what you all think!

Fractional LA! Here's what I mean. Instead of 'LA +X', you would have something like 'LA +X/level', where X is the character level where the template can be assumed to be not very useful, and level is character level (Not ECL!)

This system allows templates to have a lesser cost as you level up, in a scaling way that is more flexible than LA Buyoff, and is automatic.

Thoughts?


I can't really give a great example since the system still has some quirks; at some levels & LAs, your ecl may actually drop with this system. More work is needed, which is part of why I am turning to the community!

The minimum is given exactly by character level = sqrt(X). (Solve by setting the first derivative to zero)

Though this assumes levels follow a continuous variable, which isn't the case. But practically speaking, ECL will decrease when the the new character level is ≤ sqrt(X), while increasing otherwise (when new CL > sqrt(X) ). The decrease is sharp (the inverse function dominates), while the increase is predominantly 1:1 linear (and gets even more linear the further one gets above sqrt(X) ). Also, practically speaking, this turns into LA + 1 at X (and then continues to be LA + 1 above X if we round upwards, or LA + 0 otherwise).

So, effectively? This has LA buyoff built in. Good job. That said, the LA buyoff can be too drastic for higher values of X. Take X = 8, where, if we assume ECL is rounded up, we get 9, 6, 6, 6, 7, 8 (...), becoming LA + 2 at CL 5 and then LA + 1 at CL 8. But note the first four levels: you start at ECL 9, immediately drop down sharply to ECL 6 after getting a single level, and then stay there for three straight levels. The first buyoff is too drastic.

Basically, mathematically, the buyoff is horribly distorted before the inflection point (the point where curvature changes. Which, interestingly for this function, is also where ECL will start to stick around a certain value), given by EL = (2*X)^(1/3). [Solve by setting the second derivative to 0.] But for small values of X the distorted buyoff might not be visible. The 'magic' number is X = 4 because that sets the concavity point at 2 (which interestingly is also the minimum point), such that we no longer see any distortion after CL 2.

So... overall? I'd say your method works, but for X ≤ 4. Past that the inverse function no longer models buyoff satisfactorily. Though X = 5 and X = 6 still sort of works.

Edit:

Wait, hang on, you can flatten the curve by setting ECL = CL + (k+1)*X/(CL + k), where k is some constant. This lets it work for greater ranges of X. ECL = CL + 2*X/(CL + 1), rounded down, looks okay but I'm just 'eyeballing' the mental graph.