Taking currents into account

Taking currents into account

In the previous post, I showed an example of vector subtraction related to current.

After all that work with vectors, we might want to get something out of it.  The concepts in the last three postings were intended to get you up-to-speed on a few important vector operations that one might need to do on the water.   One of the important aspects in sea kayaking is dealing with currents.   These are almost invariably due to tidal flow.   Typically one can get information on local current flow from websites, and I’ll get into this in a bit.   For now, let us assume that you’ve found the current, which is specified as both a set (direction) and drift (speed, typically in knots).   How do work with this information?

Say, you find that the current is 2 knots in a direction of 330^o.   You paddle at a speed of 3 knots at the heading of 30^o, but the current carries you along.   What is the resulting course bearing?  

You may recall that I introduced the term course bearing a few posts back – this is the net result of your heading and where the current takes you.   You can use vectors to solve the above problem.   In this case the vectors represent velocity and can be represented as a direction and magnitude just like the displacement vector.   The magnitude for velocity is speed.   The units for speed that you choose can be anything, although most nautical applications use knots (nautical miles per hour).  

In representing a velocity vector, we can choose a scale.   So, let’s say that one-inch equals on knot.   This seems strange at first – you’re equating a distance to a speed.   This is OK as long as you are consistent that whatever vectors you’re representing on paper have the same scale and remember that the result can be interpreted in terms of knots.

In the figure below, I represent the problem and solution through vector addition.   The kayaker is paddling at 3 knots at a heading of 30^o and the current has a set of 330^o and a drift of 2 knots.   I set up a scale where 1 inch equals one knot.   Then I first drew the paddling velocity vector, put the drift vector at its tip (arrowhead) and then drew the resulting course bearing vector, and then measured it.   From measurement, it’s 4.4 knots at a heading of 7^o.    The speed is greater than the paddling speed because both vectors have components of flow going northward, so they add together.

You can now use this course bearing velocity vector to figure out where you are using dead reckoning by multiplying the velocity times travel time.   Note that the units work out:

                                                                               (nautical miles)

                        Distance (nautical miles) = Speed  ------------------  * time (hours)

                                                                                 (hour)

Here I used something called dimensional analysis.   If you look at the crossed out hours, you can see that (nautical miles/hour)/(hours)=nautical miles.   By explicitly writing out the units when you mix units, you can see how the units on both sides work out.

Figure 1 Summation of paddling velocity vector and current velocity vector to get the course bearing velocity vector.   Note that the scale is one inch equals one knot.

Course in current

One issue that arises when kayaking in the presence of tidal currents is what heading to take in order to compensate for the current.   A word is in order about crossings.   When going from one point of land to another, there are many different things you can try to accomplish.   In the presence of a lot of boat traffic, you might want to minimize the amount of time in a channel.   In this case you just might paddle on a heading directly across the channel and let the current carry you.   This will minimize the amount of time in the channel, although it might mean paddling against the current later on.   A GPS with a way-point on the other side of the crossing can give you a path that automatically compensates for the current. 

One of the more regular cases, however, is making a crossing in the fog, where you want to take a direct course bearing in the presence of current and need to compensate your heading for the current.    In this case, you want to hold a direct (a line) course bearing from one point to another. There are multiple ways to solve this, including a fairly fancy trigonometric calculation, which is difficult to remember on the fly.   There is a graphical solution using vectors that I’ll demonstrate below. 

Here’s the set-up.   Let’s say you want to get a course heading with an azimuth of 20^o.   You find that the current runs at 2.5 knots in a direction of 80^o, and you can paddle at a speed of 3 knots.   At what heading should you paddle to achieve a course bearing of 20^o?

Figure 2 Step 1 – draw the desired course heading as an extended line.   In this example, the desired course bearing is 20^o.

Step 1

The first step is to again get a set of velocity-based coordinates.   I chose one inch equals one knot, because it’s easy to remember.   Then, draw an extended line in the direction of the desired course bearing.   The reason for making the line as long as you can is that you don’t know what the resultant course velocity is until you solve it, so you can only draw a long line.   In figure 2, I show the coordinated and the extended course-bearing line.

Step 2

Draw in the current vector.   This is shown in Figure 3.  Remember that in this example, the current is 2.5 knots at a bearing of 80^o.  

Figure 3 Step 2 – draw the current vector.   In this example, it’s 2.5 knots at a bearing of 80 degrees.

Step 3

This one is a little tricky, but doable.   To find the heading to achieve the course bearing, take the ruler and put the “zero” on the tip of the current vector.   Note that length of the required to represent the paddling speed.   In this case 3 inches = 3 knots.   Rotate the ruler until the 3 inch point just touches the course bearing line.   This is your solution, and is illustrated in Figure 4.

Figure 4 Step 3 – Place the ruler with the “zero” on the tip of the current vector and then find the length associated with the paddling speed – in this case 3 inches (=3 knots). 

Step 4

Draw the vector from the tip of the current vector to the course bearing line along the ruler.   You can then look at the azimuth of this to find what your heading should be in order to make course bearing you desire.   In this example the course bearing is 325^o.  

I deliberately chose a somewhat difficult situation, where the current is nearly as fast as the paddling speed.   As a result you get a heading that’s at quite an angle to the current and even quite different from the resulting course bearing.   Note that there are times when there is no solution.   You rotate the ruler and the length of the paddling vector cannot touch the desired course bearing line.   This is because your paddling speed is too slow or the current too strong to find a heading that will give you your desired course bearing.   You have to find another way through. 

Figure 5 Step 4 draw the heading vector along the ruler and measure its azimuth.   In this case it’s 325 degrees. 

Step 5

Draw the course bearing vector by making a line from the origin to the point where the heading vector touches the course bearing line.   You can then measure its length to find out the speed made good.   The course bearing, by construction, is at the bearing you chose, which is 80

Figure 6 Step 5 Measure length of course bearing vector.   In this example, it’s 2.7 inches=2.7 knots. 

Figure 6 - Step 5, measure the length of the course bearing vector. 

Special cases

Many times, a crossing will be at right angles with respect to the current.   In this case, there is a rule of thumb that works.   Most people paddle at about 3 knots, and this is only for that case.   The table below shows what heading you need to make in order to cross a given current at an angle of 90^o.   

Typically, if you add 10^o of offset of every 0.5 knots of current, you can get a pretty good solution up to around 2 to 2.5 knots where that breaks down.   In other words, let’s say you want to paddle at a course bearing of 80^o and the current is 1.0 knots running to your right.   In that case, you want an offset of 20^o to your left, or a heading of 60^o.  

In addition to that, the typical speed that you “make good” (that is, with respect to land) is diminished by about 0.2 knots per every 0.5 knots of current.   So, in that case of the 1.0 knots of current, you’d be making good 2.6 knots, approximately.   This rule of thumb breaks down with the speed of the current begins to approach the paddling speed.

Figure 7  Table of angular offsets for a paddler who goes at 3.0 knots with the current perpendicular to the crossing.   E.g. for 1.5 knots, you paddle with a 30 degree offset.  

Exercise

You want to paddle from Swan Island to Placentia as shown in the chart below, but there's a thick fog.   You hand-rail along the north coast of Placentia and then reach the point marked on the chart with "??".    The water from Bluehill Bay is draining through the passage between Placentia and Swan at a speed you reckon to be 0.5 knots and you paddle at 3.0 knots.   On what heading should you paddle to first intercept the buoy marked G C "1" on the chart and continue on to the shore of Placentia on the same course heading?   Give your answers both in true heading and magnetic, given that the magnetic variation is 17 degrees.

Figure 8 - Exercise: you handrail along the shore of Swan Island.

In the next set of posts, I’ll give some information on where to look to get information on currents, and also how to figure currents at different locations and times of the lunar cycle.