Hourglasses for Nine Minutes

Start both glasses together at time $0$ . Write each glass’s state as the amount of sand, in minutes, left in its upper bulb.

time	4-glass	7-glass	action
$0$	$4$	$7$	start both
$4$	$0 \to 4$	$3$	4-glass empties: flip it
$7$	$1$	$0 \to 7$	7-glass empties: flip it
$8$	$0$	$6 \to 1$	4-glass empties: flip the 7-glass
$9$		$0$	7-glass empties

From time $0$ to time $4$ the 4-glass runs out. Flip it at once. The 7-glass now has $3$ minutes left.
At time $7$ the 7-glass runs out. The 4-glass was restarted at time $4$ , so it has run for $3$ minutes and has $1$ minute left. Flip the 7-glass, which restarts it full.
At time $8$ the 4-glass runs out. The 7-glass has run for exactly $1$ minute since time $7$ . Its lower bulb therefore holds $1$ minute of sand.
Flip the 7-glass at time $8$ . That $1$ minute of sand is now on top, so the glass runs out exactly $1$ minute later, at time $8+1=9$ .

Every flip is triggered by a glass running out, so all the timings are exact.

\boxed{\begin{array}{c} \text{Start both; flip the 4 at minute }4\text{, the 7 at minute }7\text{,}\\ \text{the 7 again at minute }8\text{; it empties at minute }9. \end{array}}

Key idea. The 4-glass marks minute $8$ while the 7-glass holds a known $1$ -minute residue. Flipping the 7-glass then reuses that residue, giving $9 = 8 + 1$ .