7. As shown in Figure 1, the edge length of the regular tetrahedron $ABCD$ is 2, and $A_{1}, B_{1}, C_{1}$ are the midpoints of edges $DA, DB, DC$ respectively. With $D$ as the center and a radius of 1, arcs $\overparen{A_{1} B_{1}}$ and $\overparen{B_{1} C_{1}}$ are drawn in the planes $DAB$ and $DBC$ respectively, and each arc is divided into five equal parts, with the points being $A_{1}, P_{1}, P_{2}, P_{3}, P_{4}, B_{1}$ and $B_{1}, Q_{1}, Q_{2}, Q_{3}, Q_{4}, C_{1}$. A beetle wants to travel from point $P_{1}$ to point $Q_{4}$ along the surface of the tetrahedron. The shortest distance it can travel is $\qquad$
7. $2 \sin 42^{\circ}$.

Make two expansions, then compare.
Note that, arc $\overparen{A}_{1} B_{1}$ is divided into five equal arcs by points $P_{1}, P_{2}, P_{3}, P_{4}$, each corresponding to a central angle of $12^{\circ}$; arc $\overparen{B_{1} C_{1}}$ is divided into five equal arcs by $Q_{1}, Q_{2}, Q_{3}, Q_{4}$, each corresponding to a central angle of $12^{\circ}$.

If $\triangle D B C$ is rotated around line segment $D B$ to coincide with $\triangle D A B$, these two arcs will coincide with the circumference of a circle centered at $D$ with a radius of 1, then the central angle corresponding to arc $\overparen{P}_{1} Q_{4}$ is $8 \times 12^{\circ}=96^{\circ}$, at this time, the straight-line distance between points $P_{1}$ and $Q_{4}$ is $2 \sin 48^{\circ}$.

If $\triangle D A B$ is rotated around line segment $D A$ and $\triangle D B C$ is rotated around line segment $D C$ to coincide with $\triangle D A C$, in the resulting figure, the central angle corresponding to arc $\overparen{P_{1} Q_{4}}$ is $7 \times 12^{\circ}=84^{\circ}$, at this time, the straight-line distance between points $P_{1}$ and $Q_{4}$ is $2 \sin 42^{\circ}$. In summary, the shortest distance sought is $2 \sin 42^{\circ}$.