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Yachtmaster School / Tides

Work through depth and clearance

Practise a simple depth calculation, then identify what the number leaves out.

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Our goal today

Calculate a simplified under-keel clearance with explicit assumptions.

What you’ll practise

  • Combining compatible soundings and tidal heights
  • Calculating static under-keel clearance
  • Testing falling-water assumptions and practical margins

Each question includes a custom diagram, the reasoning behind the answer, an explanation of the other choices and a follow-up exercise. Try the quiz first, or use the study view to read every answer.

01 / Learn

Keep a common reference

For these fictional exercises, charted depth and tide height share the same vertical datum and all values are metres. Estimated water depth = charted depth + height of tide. Static under-keel clearance = estimated water depth − vessel draught. Check actual chart and tide-table references before using real data.

02 / Learn

A worked example

A fictional sounding is 2.4 m, the predicted tide height is 1.1 m and draught is 1.8 m. Estimated water depth is 3.5 m. Static clearance is 3.5 − 1.8 = 1.7 m. That is a calculation, not a declaration that a route is safe.

03 / Learn

Allow for the real world

Predicted tide levels can differ from observations, including because of weather. A practical clearance assessment must also consider waves, vessel movement, squat where relevant, survey uncertainty and the skipper’s required margin. Do not apply this addition rule to bridge clearance: overhead structures can use a different reference.

Your two-minute exercise

With the same fictional 2.4 m sounding and 1.8 m draught, calculate static clearance at tide heights of 0.3 m and 1.3 m. List what you still need to know.

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Source check: . Scenarios and quiz questions are original YouSail practice material. Editorial policy.

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Have a go

Tidal Height & Under-Keel Clearance Quiz

10 illustrated questions. Worked explanations and practical follow-ups for every answer.

A 2.4 m sounding plus a 1.1 m tide gives an estimated depth of…
A fictional seabed cross-section labelled charted depth 2.4 metres and tide height 1.1 metres, with total depth unknown.
Estimate water depthIllustrative · not to scale · tap to enlarge ↗
Read the answer and explanation

Answer: 3.5 m

Why this answer works

Using the shared datum in the exercise, 2.4 + 1.1 = 3.5 m.

Why the other choices fall short

Subtracting the tide gives 1.3 m and treats a positive rise as a fall; 4.6 m counts the 1.1 m tide twice. A sounding and a tide height can be combined only on compatible references.

Put it into practice

Draw chart datum, the seabed and the water surface. Show the 2.4 m sounding below datum and the 1.1 m tide above it before adding them.

Water depth is 3.5 m and draught is 1.8 m. Static clearance is…
A fictional yacht with draught 1.8 metres in 3.5 metres of water; under-keel clearance is unlabelled.
Find the static clearanceIllustrative · not to scale · tap to enlarge ↗
Read the answer and explanation

Answer: 1.7 m

Why this answer works

3.5 − 1.8 = 1.7 m before considering additional allowances.

Why the other choices fall short

Adding draught to depth gives 5.3 m, the opposite of the clearance calculation. Repeating 1.8 m reports the boat's draught rather than the water left below its keel.

Put it into practice

Sketch a 1.8 m keel in 3.5 m of water. Label the remaining 1.7 m as static clearance and keep it separate from any practical safety allowance.

In the same static exercise, the tide falls by 0.5 m. Clearance…
The same fictional yacht and seabed at two water levels, the second 0.5 metres lower.
A falling tideIllustrative · not to scale · tap to enlarge ↗
Read the answer and explanation

Answer: Decreases by 0.5 m

Why this answer works

With other values unchanged, a lower water level reduces clearance.

Why the other choices fall short

With the boat and seabed unchanged, lower water means less room below the keel. Reversing the sign or leaving clearance unchanged ignores the falling surface.

Put it into practice

Starting from 1.7 m, subtract 0.5 m to obtain 1.2 m static clearance. Explain why the same change applies to both water depth and clearance in this simplified model.

Does positive calculated clearance alone prove a passage is safe?
A static cross-section showing calculated water depth and yacht draught, asking what else needs consideration.
The number and its limitsIllustrative · not to scale · tap to enlarge ↗
Read the answer and explanation

Answer: No; practical allowances and uncertainties remain

Why this answer works

The simplified number does not include all real-world effects or required margins.

Why the other choices fall short

A small positive result can be consumed by vessel motion, squat or uncertainty. A phone changes the calculation tool, not the reliability of its inputs or the required margins.

Put it into practice

List the assumptions excluded from the arithmetic before making a real passage decision. Use suitable current information and instruction to establish the necessary allowances.

Before combining a sounding and a tide height, check…
A chart sounding card and a tide-prediction card with blank units and reference lines.
Use compatible referencesIllustrative · not to scale · tap to enlarge ↗
Read the answer and explanation

Answer: Their units and vertical reference

Why this answer works

Values must use compatible units and references for the calculation to make sense.

Why the other choices fall short

Adding values with different units or zero references can produce a plausible-looking but meaningless total. Route length and boat colour do not reconcile those references.

Put it into practice

Find the chart's sounding units and vertical datum, then check the tide source. Do not combine the numbers until you know they measure from compatible levels.

Why might observed water level differ from a prediction?
A tide graph with two different level traces and a cloud above the water.
Prediction and observationIllustrative · not to scale · tap to enlarge ↗
Read the answer and explanation

Answer: Weather can affect actual sea level

Why this answer works

BOM explains that weather conditions can cause actual levels to differ from predicted tides.

Why the other choices fall short

Draught describes the boat and speed describes motion; neither explains the difference between predicted and actual sea level. Weather can alter the water level relative to an astronomical prediction.

Put it into practice

Compare a relevant observation with the prediction using matching references and times. Reassess clearance if observed conditions differ from the assumption in the plan.

In a fictional static calculation, charted depth is 2.4 m, tidal height above the same datum is 1.1 m and draught is 1.8 m. What is the calculated under-keel clearance?
Water-depth cross-section with charted depth 2.4 metres, tide height 1.1 metres and draught 1.8 metres; clearance is unknown.
Calculate static clearanceIllustrative · not to scale · tap to enlarge ↗
Read the answer and explanation

Answer: 1.7 m

Why this answer works

Water depth = 2.4 + 1.1 = 3.5 m. Subtract 1.8 m draught to get 1.7 m static clearance. This simplified value does not include motion, squat, uncertainty or the required safety allowance.

Why the other choices fall short

3.5 m is total water depth, not clearance. 0.6 m omits the tide, while 0.7 m does not follow from the stated addition and subtraction.

Put it into practice

Write two lines: depth = 2.4 + 1.1 = 3.5 m; clearance = 3.5 − 1.8 = 1.7 m. Then list the real-world allowances deliberately excluded from this static exercise.

For a paper exercise, you require 0.5 m static clearance with 2.0 m draught over a charted depth of 1.2 m. What minimum tidal height above the same datum satisfies that arithmetic?
A static depth exercise labels charted depth 1.2 metres, draught 2 metres and a specified half-metre allowance, with tidal height unknown.
Find the required tideIllustrative · not to scale · tap to enlarge ↗
Read the answer and explanation

Answer: 1.3 m

Why this answer works

Required total depth = 2.0 + 0.5 = 2.5 m. Required tide = 2.5 − 1.2 = 1.3 m. The 0.5 m allowance is only a stated exercise assumption, not a recommended real-world minimum.

Why the other choices fall short

2.5 m is the required total depth, not the tide contribution. 3.7 m adds the sounding instead of subtracting it; 0.7 m does not meet the stated depth requirement.

Put it into practice

Check the result forward: 1.2 + 1.3 − 2.0 = 0.5 m. The half-metre margin belongs only to this paper problem and should not be adopted as a universal allowance.

Your fictional clearance calculation assumes 1.4 m of tide. A reliable local observation referenced to the same datum shows 1.0 m. With other inputs unchanged, what happens to calculated clearance?
Two water-level cards compare a predicted 1.4 metres with an observed 1 metre, for a clearance reassessment.
Prediction versus observationIllustrative · not to scale · tap to enlarge ↗
Read the answer and explanation

Answer: It decreases by 0.4 m

Why this answer works

The observed water level is 0.4 m lower than assumed, so calculated clearance is 0.4 m less. Reassess the plan and its margins; a prediction is not a guarantee of actual water level.

Why the other choices fall short

The relevant change is 1.4 − 1.0 = 0.4 m, not the entire predicted height. A published prediction cannot keep clearance unchanged when the observed level is lower.

Put it into practice

Replace the assumed water level with the compatible observation and recalculate. Explain which other inputs and allowances still need checking before making a real decision.

A fictional plan gives 1.2 m static clearance before a predicted 0.7 m fall in tide. An additional 0.3 m lower-than-predicted level is then allowed for. What clearance remains in this simplified calculation?
A clearance exercise lists an initial 1.2 metres, a 0.7 metre fall and another 0.3 metre reduction, leaving the remainder unknown.
Two reductions in clearanceIllustrative · not to scale · tap to enlarge ↗
Read the answer and explanation

Answer: 0.2 m

Why this answer works

Apply both reductions: 1.2 − 0.7 − 0.3 = 0.2 m. The result is an arithmetic remainder, not a conclusion that the passage is safe; other uncertainties, vessel motion and appropriate margins still matter.

Why the other choices fall short

0.5 m ignores the extra reduction. 0.8 m uses the wrong sign for the lower-than-predicted level, while 1.6 m adds an effect that should reduce the clearance.

Put it into practice

Write each adjustment with its sign and reference. Explain why an apparently positive remainder can still be inadequate for a real passage decision.

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Practice only: this quiz is not an official RYA examination or an assessment of practical competence. Leaving mid-quiz starts that attempt again.

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