Minimum depth to anchor?

Worked example

At 2310 NP DST (zone 0) on Wednesday 16th October, A yacht is anchoring for the night near Port Fraser in the Northern Peninsula. The skipper would like a minimum clearance under the keel of 1 metre. The echo sounder reads from the surface of the water and the yacht has a draft of 1.8m from the waterline

What depth of water at 2210 NP DST (zone 0) will the echosounder show when the vessel drops its anchor.

Step 1

We know that we must work out the height of tide at 2310 NP DST so must consult the tide table for the 16th October. We can see from the tide table that we need to work in NP Standard Time. So we must convert 2310 NP DST into 2210 NP to work with the tide table and curves. We should also note that in this example we are anchoring in a ebbing, lowering tide which is carrying over into the next day, Thursday 17th October.

Fig 5.0 show the time and ranges of tide we are interested in 2039 UT 3.6m- 0241 UT 1.4m = 2.2m, and this is closer to mean neaps and indicates to us that we will be using the dashed blue line on the tidal curve.

Step 2

We can now mark the tidal range heights for HW (3.6m) and LW (1.4m) on the tidal curve and draw a diagonal line between them (Fig 5.43). Remember we are using the first LW on the 17th October (1.4m) as our LW mark. Always double check the axis scale bar to make sure you have marked it correctly, in this case we have 10 devision per metre, sometimes we have 5 and in other case it may be a composite scale bar.

Step 3

As always the timeline scale bar at the base of the tidal curve runs form left to right increasing an hour at a time with each box as we move to the right. The tidal curve above is describing how the tide flood and ebbs during the tidal cycle. The centre HW box is our reference box and must enter the High Water tide time that is closest to our target time of 2210 NP (zone 0). As we mentioned before the tide is falling and we can enter the LW from the tide table in the HW+6 box.
We now need to find the height of tide at 2210 NP and need to find where it located on the time line axis. The difference between 2110 and 2139 is 30 mins, so we can count 3 subdivisions form +1hr box to find our mark for the target time, disregarding the 1 minute.

Step 4

We then draw a line form our target time on the timeline up to the blue dashed tidal curve and where it intersect draw another line along to the tidal range line drawn earlier. Finally draw a line up or down to the tidal heights axes.

We can read off the height of water above chart datum at 2210 NP), remember the subdivision on this scale are 1/10th of a metre each.

Step 5

At 2210 NP standard time the tide level will be 3.2 metres above chart datum. So the fall of tide is 3.2 metres height of tide at 2210 NP minus the low water level of 1.4 metres:

Fall of tide is 3.2m – 1.4 m = 1.8 metres

So after our calculations in the steps above we know that from 2210 NP standard time to the next low water the tide will fall 1.8 metres (Fig 5.46a).

In the last step we relate the fall of tide to the position of the vessel and the depth see on the echo sounder. We now can relate our problem to the real situation and seabed rather than to chart datum.

Depth required when anchoring is the Fall of tide + clearance + draft (Fig 5.56b).

To put it into context, 4.6 metres is the minimum depth the vessel could anchor in to leave a clearance of 1 m under the keel when the tide falls to the next low tide level.

Remember, so far we have converted time to NP standard time and must add 1 hour so that we are now back in Northern Peninsula Summer Time (DST).

So we must anchor in 4.6m of water at 2310 NP DST to ensure we leave 1 metre under the keel at the next low water.

Until now we have used HW as our reference point when working up tidal curves. We will now use a tidal curve that centres on LW and demonstrate how it is used. This type of tidal curve illustrates what is called a double high water and is found in the Solent and Southampton areas.