---
title: "Model time scale, prototype day and moon"
subject: "Fluid Mechanics"
chapter: 10
key: "semester_3/fluid_mechanics:model-time-scale-moon-gravity"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/model-time-scale-moon-gravity
site: BCT Engineering
---

# Model time scale, prototype day and moon

Fluid Mechanics · Chapter 10: Similitude and physical modeling

- Asked in: 2079 Baisakh, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/model-time-scale-moon-gravity
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-10

## Question

1:400 model is constructed to study tides. What length of time in the model corresponds to a day in the prototype? Suppose the model could be transported to the moon and tested there. What then would be the time relationship between model and prototype? Given, 'g' of earth = 6 times 'g' of moon.

## Short answer

Froude similarity gives time ratio Tr = √(Lr/gr). On earth gr=1, Tr=1/20, model day = 1.2 h. On the moon gr=1/6, Tr=√(6/400)=0.1225, model day ≈ 2.94 h.

## Answer

Tides are gravity-dominated, so the **Froude law** applies: $\dfrac{V}{\sqrt{gL}}$ is the same in model and prototype. The time scale then follows from the length scale and the gravity ratio.

#### Time scale

Velocity scale $V_r = \sqrt{g_rL_r}$, and time $T = L/V$, so:

$$
T_r = \frac{L_r}{V_r} = \sqrt{\frac{L_r}{g_r}}
$$

#### Case 1: model tested on earth ($g_r = 1$)

$$
T_r = \sqrt{\frac{1}{400}} = \frac{1}{20}
$$

Model time for one prototype day $= 24 \times \dfrac1{20} = 1.2$ h $= 72$ min.

#### Case 2: model tested on the moon

The prototype is on earth, the model on the moon, so $g_r = g_m/g_p = 1/6$.

$$
T_r = \sqrt{\frac{1/400}{1/6}} = \sqrt{\frac{6}{400}} = 0.1225
$$

Model time for one prototype day $= 24 \times 0.1225 = 2.94$ h (about 2 h 56 min).

**Answer: on earth, 1 prototype day corresponds to 1.2 h of model time ($T_r = 1/20$). On the moon, $T_r = 0.1225$ and 1 prototype day corresponds to about 2.94 h of model time; the model time is $0.1225\times$ the prototype time.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which model law is used for a tidal model?
   A. Reynolds law, because tides are viscous
   B. Mach law, because tides are compressible
   C. Froude law, because tides are gravity-dominated
   D. Weber law, because tides are surface-tension driven
   Answer: C. Free-surface gravity flows are modelled with equal Froude numbers. (Key point: Governing law)

2. What is the time scale under Froude law?
   A. T_r = L_r x g_r
   B. T_r = sqrt(L_r x g_r)
   C. T_r = L_r^2
   D. T_r = sqrt(L_r/g_r)
   Answer: D. V_r = sqrt(g_r L_r) and T_r = L_r/V_r, which gives sqrt(L_r/g_r). (Key point: Time scale formula)

3. With L_r = 1/400 and same g, how long in the model corresponds to one prototype day?
   A. 1.2 hours
   B. 24 hours
   C. 0.06 hours
   D. 2.94 hours
   Answer: A. T_r = 1/20, so 24 x 1/20 = 1.2 h = 72 min. (Key point: Model day on earth)

4. If the model is on the moon (g_r = 1/6), what is T_r?
   A. sqrt(1/(400 x 6)) = 0.0204
   B. sqrt(6/400) = 0.1225
   C. sqrt(1/400) = 0.05
   D. 6/400 = 0.015
   Answer: B. T_r = sqrt((1/400)/(1/6)) = 0.1225, so one prototype day is about 2.94 h in the model. (Key point: Model on moon)

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