I've been faced with an interesting question last game: what load does a Size -1 incapacitated maga generate for the Size +2 magus who's carrying her?

In other words, is there, somewhere where I didn't notice it, a Size to Load relation?

I've been faced with an interesting question last game: what load does a Size -1 incapacitated maga generate for the Size +2 magus who's carrying her?

In other words, is there, somewhere where I didn't notice it, a Size to Load relation?

I would say that a size -2 = load 6 (weight of a full metal chain mail)

For a 0, i would say load 28 (need a 7 strenght to be carried without malus).

For a +2, 66 (so you need a 11 str to carry without malus)

So i would use some formula like:

same math as for a Art increment, but in place of the level Art, the size*2 +7

But it's an interesting question,and my answer here are just my thought now. Maybe in a game i would state otherwise

For most creatures size and strength are related so if it weren't for those pesky exceptions we wouldn't need to make a rule.

ExarKun wrote:

Now, I've never been a storyguide, and I haven't been playing this game very long, but for a size 0 person to carry another size 0 person, are you saying this requires having a strength of 7? That seems to make it impossible for a mortal man to carry another. Maybe i am missing something. I am not familiar with the word malus. Isn't that a kind of apple tree?

Keyword here is "without being encumbered".

It's the antonym of "bonus".

Anyway, we did hand-wave the situation for the sake of keeping the game rolling. I was just being curious about your collective thoughts - I haven't come up with anything yet myself.

Like doing something with "malus aforethought"?

Oh, ok, of course. (Heh.) Now I can play along I guess; I also find this interesting. If full chainmail armor (load 6) weighs, say, 30 pounds (that's a wild guess) and if that's 18% of a size 0 person's weight (another wild guess), then maybe a carried object gives a load of 1 for every 3% of its carrier's weight that it...weighs. Then if you let D = the size of the carried body minus your size, then 2 to the Dth power gives you an approximate weight ratio. So divide 2^D by 0.03 and you get the Load. Too much math?