and of the above the relative strength of the series are—
| a – a′ = 0.952 | a – a′ = 1.000 |
| b – b′ = 0.932 | b – b′ = 0.937 |
| c – c′ = 1.00 | c – c′ = 0.937 |
We find, also, the relative transverse strength of the three lengths is as follows, viz.:—
| Top-length | =0.870 | |
| Mid-length | = | 0.947 |
| Butt-length | = | 1.000 |
and the specific gravity—
| Top-length | =0.980 | |
| Mid-length | = | 0.946 |
| Butt-length | = | 1.000 |
The tables show that the maximum transverse strength lay in the outer series marked c′ – c. It is not, however, certain whether the tree from which they were taken, although reduced to 22 inches square, would not have yielded a much larger square log, say 28 or 30 inches; and thus it seems probable that the point c, although nearer to the outside of this log than in the other, may, after all, be in about the same position in the tree. The experiments for the tensile strength show that the series a′ – a were the strongest.
Table CLXII. shows that the vertical strength of Kauri timber is about 2.8665 tons per square inch of base.
