7. The dial of a watch lies in a plane. The 12 numbers 1, 2, $\cdots$, 12 are evenly distributed on the circumference of a circle with radius $\frac{\sqrt{2}}{2}$. The vector from number $i$ to number $(i+1)$ is denoted as $\boldsymbol{t}_{i} \boldsymbol{t}_{i+1}$. Then $\boldsymbol{t}_{1} \boldsymbol{t}_{2} \cdot \boldsymbol{t}_{2} \boldsymbol{t}_{3}+\boldsymbol{t}_{2} \boldsymbol{t}_{3} \cdot \boldsymbol{t}_{3} \boldsymbol{t}_{4}+\cdots+$ $t_{12} t_{1} \cdot t_{1} t_{2}=$ $\qquad$ .
ii. $7.6 \sqrt{3}-9$.
The line segments connecting adjacent scales form a regular 12-sided polygon inscribed in a circle with a radius of $\frac{\sqrt{2}}{2}$. The angle between two adjacent sides, which is the exterior angle of the 12-sided polygon, is $30^{\circ}$. The length of each side is
$$
2 \times \frac{\sqrt{2}}{2} \sin \frac{\pi}{12}=\sqrt{2} \times \sqrt{\frac{2-\sqrt{3}}{4}} .
$$

Thus, $t_{1} t_{2} \cdot t_{2} t_{3}=\left(\sqrt{2} \times \sqrt{\frac{2-\sqrt{3}}{4}}\right)^{2} \cos \frac{\pi}{6}=2 \times \frac{2-\sqrt{3}}{4} \times \frac{\sqrt{3}}{2}$.
Therefore, there are 12 equal terms.
Hence, the sum of the dot products is $6 \sqrt{3}-9$.