I did the math, at least assuming that the tires are stacked at 90 degrees relative to each other. I know they're not, but i guess it gives a reasonable approximation. I also postulate that any other angle than 90 degrees gives suboptimal stacking (for volume filling purposes). In summary: You can utilize anywhere up to about 65% of the "void" in the tires, depending on the ratio of the inner and outer ratio of the tires. I've posted further down, but i'm late to the party and don't want to get burried, so i'm answering you :)

Notation as follows see figure: r = inner radius of tire R = outer radius of tire d = thickness of tire theta = angular span of tire that is inside another tire h = "height" of tire section inside another tire see figure

The angluar span is related to the radii by sin(theta) = r/R, and to the "height" by cos(theta) = (R - h)/R. The volume occupied by the tire section inside the space within any tire is given by

V_1 = 2 [pi R² (theta/2pi) - 0.5(2 r (R - h))] d

where the leading number 2 comes from there being one tire filling from each side. Inserting for h and cancelling som constants we get we get

V_1 = [0.5 theta R² + r R cos(theta)] d

The total volume available inside a tire is clearly

V_2 = pi r² d

The ratio V_1/V_2 tells us how much of the "void" we have filled:

V_1/V_2 = (theta R² + r R cos(theta))/(pi r²⁾

interestingly enough, this does not depend on d! Simplifying the ratio gives:

V_1/V_2 = (theta / pi) (R / r)² + (R / r) cos(theta)

plotting this function gives this graph. Note, however that this is only valid for * 2h <= d, which implies *2R(1 - cos(theta)) <= d. This figure shows how large d must be for different R as a function of the ratio r/R for the derivation to be valid. Also note that if d is any smaller or larger than R(1 - cos(theta)) stacking is suboptimal.

Though the figure does not show it, clearly d/R >= 1 - cos(theta), so the minimum ratio between the thickness and outer radius is only a function of the ratio between the inner and outer radius.