A clockmaker wants to design a clock such that the area swept by each hand (second, minute, and hour) in one minute is the same (all hands move continuously). What is the length of the hour hand divided by the length of the second hand?
1. Let's denote the lengths of the hour hand, minute hand, and second hand as \( L_h \), \( L_m \), and \( L_s \) respectively.
2. The area swept by a clock hand in a given time is proportional to the square of its length and the angle it sweeps through. The area \( A \) swept by a hand of length \( L \) sweeping an angle \( \theta \) is given by:
   \[
   A = \frac{1}{2} L^2 \theta
   \]
3. In one minute:
   - The hour hand sweeps through \( \frac{1}{60} \) of an hour, which is \( \frac{1}{60} \times 30^\circ = 0.5^\circ \). Converting to radians, \( \theta_h = 0.5 \times \frac{\pi}{180} = \frac{\pi}{360} \).
   - The minute hand sweeps through \( 6^\circ \). Converting to radians, \( \theta_m = 6 \times \frac{\pi}{180} = \frac{\pi}{30} \).
   - The second hand sweeps through \( 360^\circ \). Converting to radians, \( \theta_s = 360 \times \frac{\pi}{180} = 2\pi \).

4. The areas swept by each hand in one minute are:
   \[
   A_h = \frac{1}{2} L_h^2 \cdot \frac{\pi}{360}
   \]
   \[
   A_m = \frac{1}{2} L_m^2 \cdot \frac{\pi}{30}
   \]
   \[
   A_s = \frac{1}{2} L_s^2 \cdot 2\pi
   \]

5. According to the problem, these areas are equal:
   \[
   \frac{1}{2} L_h^2 \cdot \frac{\pi}{360} = \frac{1}{2} L_m^2 \cdot \frac{\pi}{30} = \frac{1}{2} L_s^2 \cdot 2\pi
   \]

6. Simplifying the equations:
   \[
   L_h^2 \cdot \frac{\pi}{360} = L_m^2 \cdot \frac{\pi}{30}
   \]
   \[
   L_h^2 \cdot \frac{1}{360} = L_m^2 \cdot \frac{1}{30}
   \]
   \[
   L_h^2 = 12 L_m^2
   \]

   And:
   \[
   L_m^2 \cdot \frac{\pi}{30} = L_s^2 \cdot 2\pi
   \]
   \[
   L_m^2 \cdot \frac{1}{30} = L_s^2 \cdot 2
   \]
   \[
   L_m^2 = 60 L_s^2
   \]

7. Substituting \( L_m^2 = 60 L_s^2 \) into \( L_h^2 = 12 L_m^2 \):
   \[
   L_h^2 = 12 \cdot 60 L_s^2
   \]
   \[
   L_h^2 = 720 L_s^2
   \]
   \[
   L_h = \sqrt{720} L_s
   \]
   \[
   L_h = 12\sqrt{5} L_s
   \]

The final answer is \( \boxed{12\sqrt{5}} \).