CE Board Exam Randomizer

⬅ Back to Ports and Harbors Topics

Tides

Tides are the periodic rise and fall of the sea caused by the gravitational pull of the moon and sun. When the sun and moon align (new/full moon) their pulls add to give large spring tides; at quadrature they partly cancel, giving small neap tides. Because of friction, the largest tide lags the moon's phase by a constant interval called the age of the tide.

By daily pattern, tides are diurnal (one high + one low per day), semidiurnal (two highs + two lows, period ~12 h 25 min), or mixed. Soundings and elevations are referenced to tidal datums such as MLLW (mean lower low water), MSL (mean sea level), and MHW.

$$\text{Tidal range}=\text{HW}-\text{LW}, \qquad \text{Mean tide level}=\frac{\text{MHW}+\text{MLW}}{2}$$

★ Age of the Tide and Tidal Range

The lag between the moon's phase and the corresponding spring tide is called the age of the tide. At a station, high water is +2.40 m and low water is −0.60 m. Determine the tidal range.

$$\text{Range}=\text{HW}-\text{LW}=2.40-(-0.60)=3.00\ \text{m}$$

Final answer: 3.00 m.

★ Mean Tide Level

Using the same station (MHW = +2.40 m, MLW = −0.60 m), determine the mean tide level.

$$\text{MTL}=\frac{\text{MHW}+\text{MLW}}{2}=\frac{2.40+(-0.60)}{2}$$
$$=\frac{1.80}{2}=+0.90\ \text{m}$$

Final answer: +0.90 m above datum.

★★ Charted Depth from a Sounding

Soundings are referenced to MLLW (0.00 m). A leadline reads a water depth of 5.20 m at an instant when the tide is +1.30 m above MLLW. Determine the charted depth (referenced to MLLW).

The charted depth removes the height of tide above datum at the time of sounding.

$$d_{\text{chart}}=d_{\text{measured}}-h_{\text{tide}}=5.20-1.30=3.90\ \text{m}$$

Final answer: 3.90 m below MLLW.

★★ High and Low Water from MSL

At a port, mean sea level is +1.50 m above MLLW and the mean tidal range is 2.40 m. Estimate the elevations of mean high water (MHW) and mean low water (MLW) above MLLW.

For a symmetric tide, MHW and MLW lie half a range above and below MSL.

$$\text{MHW}=\text{MSL}+\tfrac{1}{2}R=1.50+\tfrac{1}{2}(2.40)=+2.70\ \text{m}$$
$$\text{MLW}=\text{MSL}-\tfrac{1}{2}R=1.50-1.20=+0.30\ \text{m}$$

Final answer: MHW = +2.70 m, MLW = +0.30 m above MLLW.

★★★ Timing of Semidiurnal Tides

At a station the tide is semidiurnal, recurring every 12 h 25 min. If high water occurs at 8:10 AM, when is the next high water, and approximately when is the following low water?

Successive high waters are one tidal period apart; a low water falls about half a period after a high water.

$$T=12\text{ h }25\text{ min}\;\Rightarrow\;\text{next HW}=8\text{:}10\text{ AM}+12\text{:}25=8\text{:}35\text{ PM}$$
$$\tfrac{T}{2}=6\text{ h }12.5\text{ min}\;\Rightarrow\;\text{LW}\approx8\text{:}10\text{ AM}+6\text{:}13=2\text{:}23\text{ PM}$$

Final answer: next high water at about 8:35 PM; intervening low water near 2:23 PM.

Scroll to zoom

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: w100

MSTE - Ports and Harbors / Tides / MSTE November 2019

Owing to the retardation of the tidal wave in the ocean by frictional force, as the earth revolves daily around its axis and as the tides tend to follow the direction of the moon, the highest tide for each location is not coincident with conjunction and opposition but occurs at some constant time after the new and full moon. This interval, which may amount to as much as 2.5 days, is known as:

  1. semi-diurnal tide
  2. age of tides
  3. lunar tide
  4. diurnal tide
The lag interval (up to about 2.5 days) between the new/full moon and the resulting highest tide at a location is called the age of tides.
$\boxed{\text{age of tides}}$