Why does shallow water surge and deep water does not? There is a differential equation that describes the balance of wind and hydraulic forces when strong winds blow across water. I present it here for the edification of any readers of my book *Between Migdol and the Sea* who are interested in the math behind Exodus 14.

**Figure 1. Pressure diagram of storm surge.**

Consider a tank of water diagrammed in **Figure 1** (right). There is a porous divider down the middle of the tank which allows water to pass freely between left and right. The ten vertical levels are simply for calculating; there are no physical horizontal dividers. When no wind is blowing, we have the water level marked in blue. The pressure from the right and left sides balances at the middle, and the surface of the water is level. Simple enough.

When a strong wind blows from the left, it pushes water into the right side of the tank, thereby piling up water in the red cell. But the water doesn't pile up very high, and for this reason we can draw rectangles instead of diagonal lines. The red cell represents an **incremental** change in water level. If we let the system achieve a steady state, the pressure on the right and left sides of the tank will again balance. The water surface will acquire a low-angle tilt shown by the red cell.

What is that balance in mathematical terms? Since Figure 1 is a cross section, a real physical tank of water would have a **Y**-dimension extending into the page. I'll include that dimension so that the units work out correctly (but the **Y**-dimension will cancel out). We already know that the pressure of just the blue cells across the porous divider is equal, so we only have to calculate the pressure exerted by the red cell pressing down on the blue column.

Let **Fwind** be the force of the wind exerted on the water surface, measured in Newtons per square meter. The top of the entire tank measures **2 X** by **Y**, and the top of the left column measures **X** by **Y**. So the wind force acting on the left side is:

Force of wind =Fwind * X * Y; N/m^{2}* m * m = Newtons

The red cell is held in place by the wind. But the red cell also weighs on every blue cell below it in the water column. Remember that water pressure is equal to the weight of the water column alone; **X** is not involved here. For each level in **H**, the pressure is **un**balanced by the weight of the red cell at the top. What is the pressure exerted at the bottom of the red cell? Let **g** be the force of gravity and **rho** be the density of water. The height of the water column (**dH**) determines the **pressure** per square meter:

Pressure of red cell =dH * g * rho; m * m/sec^{2}* kg/m^{3}= N/m^{2}

The **force** acting from right to left on the central divider at each level is:

Force of water at each cell = pressure * surface area at divider Force per cell =dH * g * rho * dH * Y; N/m^{2}* m * m = Newtons

The number of blue cells is equal to **H / dH**, where **dH** is the (small) height of each calculating level (blue and red). The total pressure balance in the entire tank is:

Force of wind = Force per cell * number of blue cells (H / dH)Fwind * X * Y = dH * g * rho * dH * Y * H / dH; Newtons

or simplified:

Fwind * X = g * rho * H * dH; Newtons(1)

If **Fwind** and **X** are held constant, and **H** (the depth of water) increases, then **dH** must **decrease** in order to maintain the equality and the balance of forces in the tank. Therefore deeper water will acquire a smaller angle of tilt when the wind blows. **Shallow water surges more**. *Quod Erat Demonstrandum*.

Now let's fiddle with the left side of the equation. **X** represents the fetch distance, the length of water along which the wind blows. Suppose **X** increases and **H** remains constant; our tank of water then represents a long inlet such as Long Island Sound. Equation (1) also states that the surge height **dh** will be large for Long Island Sound, and this result is exactly what was observed at Kings Point during Hurricane Sandy 2012.

And **that** is why the Gulf of Aqaba does not surge (and setdown) nearly as much as the Gulf of Suez. The Gulf of Suez is shallow. Equation (1) describes in mathematical terms why Moses and the Israelites did **not** cross the Red Sea at the Gulf of Aqaba.

Students of Calculus will recognize the notation **dH** as a **differential**, meaning that the quantity **dH** refers to a **small change** in the depth of water. Equation (1) uses the quantity **X** to refer to the length of the tank of water, from left to right. But **X** could also be a differential in Figure 1. Equation (2) below specifies a very narrow tank of water (narrow with respect to the depth of water **H**), but the balance of forces and units does not change.

Fwind * dX = g * rho * H * dH; Newtons(2)

Equation (2) relates small changes in the position **dX** to small changes in the water depth **dH**, also accounting for the total depth **H** and the constants **Fwind**, **g**, and **rho**. The marvelous thing about differential equations is that we can use them to calculate a smooth curve representing the water surface. None of this "piecewise linear" stuff here!

To solve differential equation (2), we **integrate**. Integration takes the differential equation and transforms it into a new equation that represents the **entire** water surface in terms of the variables **X** and **H**. The first step is to add integral signs before the variables, and place the constants out front:

Fwind * ∫ dX = g * rho * ∫ H * dH; Newtons(3)

Here comes the magical part. Remember your high school Calculus class? That integral sign means that you take the exponent, increase it by 1, and also multiply by the new exponent as a fraction, like this:

Fwind * X = g * rho * H; Newtons^{2}/ 2(4)

I have set the constant of integration to zero for simplicity. The final step is to solve equation (4) for the depth (height) of water **H** in terms of the distance **X** from the left side of the tank (the origin).

H = √(2 * Fwind * X) / (g * rho); meters(5)

**Figure 2. Water depth under strong wind stress.**

Blue = Water depth calculated from differential equation (5).

Red = Water depth calculated from ROMS ocean model.

Click for larger image.

Aside from the constants, equation (5) says that the depth of water **H** increases according to the **square root** of the distance **X** from the left side of the tank. Let's plot that with some realistic values (**Figure 2**, right, in blue):

- Wind stress
**Fwind**= 2.0 Newtons / meter^{2}for wind blowing at 28 meters/second, measured at 10 meters above surface - acceleration of gravity
**g**= 9.8 meters / second^{2} - density of water
**rho**= 1000 kg / meter^{3}

Note that the vertical scale of Figure 2 is greatly exaggerated. A human observer on the left would perceive the edge of a shallow lake, not a wall of water rising up in front of her.

As a check on the differential equation, I have also plotted in red the water surface calculated with the ROMS ocean model at a grid resolution of 100 meters (Figure 2). Yes, that red curve also looks like a square root. But something is wrong! The two surface profiles of Flat Lake do not match. What happened? The ocean model presumably has all the equations of fluid mechanics built into its Fortran code. Should not the analytical calculation match the ocean model? The numerical difference is small, but it's annoying.

**With differential equations, it's all about the slope**. The original differential equation (2) does not say anything about the

Remember that "constant of integration"? The value that I set to zero in equation (4) for simplicity? Yeah, that one. The **constant of integration** is used in Calculus theory and practice to adjust the overall height of the solved differential equation (5). By inspection of the numerical values in Figure 2 we can determine that the blue analytical curve needs a decrease of 0.1 meters (10 centimeters) in order to match the red curve calculated by the ROMS ocean model. Equation (5) becomes:

H = √(2 * Fwind * X) / (g * rho) - 0.1; meters(6)

**Figure 3. Comparison of calculated surge heights.**

Blue = Water depth calculated from differential equation (6),

including the constant of integration.

Red = Water depth calculated from ROMS ocean model.

Click for larger image.

**Figure 3** (right) shows the results of the adjustment. The two lines are on top of each other, which is what we want. I really did plot both curves! You can see a tiny bit of red peeking out from under the blue curve at the left end of the plot (the origin).

ROMS is the **Regional** Ocean Modeling System. The water surface in Figures 2 and 3 approaches vertical as the depth nears zero meters, and the ROMS numerical model was not formulated to model near-vertical surfaces. Consequently, the discrepancy occurs at the near-vertical portion of the curve at the plot origin (left side). The remainder of the curves in Figure 3 are nearly identical.

__Question:__ Which way should we have made the adjustment - to the analytical equation (5) or to the ocean model? If we constructed a 40-kilometer long Flat Lake uniformly 2 meters deep and somehow blew a strong wind across it for 12 hours, what would happen? Which configuration is closer to **the Truth**?

__Answer:__ Probably the ROMS ocean model. Equation (5) depends on uniform wind stress across the entire water surface, and that's just not going to happen at the plot origin. A real wind is not going to blow as strongly down into that last centimeter where the water surface meets the ground, because of boundary effects. A real lake will not stack up quite vertically there, and the actual result will look more like the adjustment in Equation (6). I have personally spent many hours verifying that the ROMS ocean model does produce accurate heights for storm surge.

By the way: My book *Between Migdol and the Sea* contains **NO** derivations as complicated as this one. Chapter 8 calculates an estimate of the number of Israelites who took part in the Exodus using basic algebra.

**Posted:** August 12, 2015

**Updated:** August 12, 2015

**Author:** Copyright 2015 by Carl Drews.

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