4. (8 points) A cruise ship travels from upstream location $A$ to downstream location $B$ in 1 hour. On the return trip, the ship doubles its speed and still takes 1 hour. Therefore, if the cruise ship also doubles its speed when departing from $A$, it will take $\qquad$ minutes to reach $B$.
【Analysis】According to the problem, the boat's speed is doubled on the return trip, and the time required is the same as the original 1 hour from upstream to downstream. This means the increased upstream speed = original downstream speed, i.e., original boat speed + water speed = original boat speed × 2 - water speed, from which we can deduce that the original boat speed = water speed × 2. When the cruise ship departs from point A, it also uses double the boat speed, so its speed is: original boat speed × 2 + water speed = water speed × 5. The original speed from point A to downstream point B is: original boat speed + water speed = water speed × 3. The speed ratio when the cruise ship departs from point A using double the boat speed to the original speed from point A to downstream point B is: water speed × 5 : water speed × 3 = 5 : 3. Given that the distance is constant, the speed and time are inversely proportional, so the time ratio is 3 : 5. Based on this, we can calculate the required time.

【Solution】According to the problem,
original boat speed + water speed = original boat speed × 2 - water speed
original boat speed = water speed × 2
When the cruise ship departs from point A, it also uses double the boat speed, so its speed is:
original boat speed × 2 + water speed = water speed × 5
The original speed from point A to downstream point B is:
original boat speed + water speed = water speed × 3
The speed ratio when the cruise ship departs from point A using double the boat speed to the original speed from point A to downstream point B is:
water speed × 5 : water speed × 3 = 5 : 3
1 hour = 60 minutes
60 × 3/5 = 36 (minutes)
Answer: It takes 36 minutes to reach point B.
Therefore, the answer is: 36.