Percentage air void in bituminous mix

Given: weight in air $W_a = 1231$ g, in water $W_w = 715$ g, SSD weight $W_{ssd} = 1232$ g, and $G_{mm} = 2.56$.

Step 1: Bulk specific gravity of the compacted mix

$$\begin{aligned} G_{mb} &= \frac{W_a}{W_{ssd} - W_w} \\ &= \frac{1231}{1232 - 715} = \frac{1231}{517} = 2.381 \end{aligned}$$

Step 2: Percentage air voids

$$\begin{aligned} V_v &= \frac{G_{mm} - G_{mb}}{G_{mm}} \times 100 \\ &= \frac{2.56 - 2.381}{2.56}\times 100 = 6.99\% \end{aligned}$$

Answer: Air voids $V_v \approx 7.0\%$ ($G_{mb} = 2.381$). This is more than the usual 3-5% limit for a dense mix (NRS 2070/Asphalt Institute), so the specimen is too porous.