Each of our course pages has a nano-forum (called SeaTalks) attached so that students can comment on the content of that page. Recently a student asked a question on the nano-forum about the distance off/double the angle method and pointed out something which when we did the geometry analysis it turns up something quite revealing.
The distance off method uses basic geometry to determine the distance from an object without doing a fix on a chart. It is quite brilliant. First I’ll explain the concept then show you the failings of it.
The concept is basic triangle geometry.
Distance-off or double the angle method
It says that when you have traveled a distance so that the angle to an object has doubled that the distance you traveled is the same as the distance from the object. All you have to know is your speed and the time you traveled. This comes from the fact that when the angle is doubled then the inside angle next to it is 180 – 2 x the angle. Thus since the sum of all angles in a triangle must = 180 then the unknown angle becomes the same. Thus it is an equilateral triangle and b must equal a.
So for example, say you are traveling at 5 knots and you spot a lighthouse ahead and to port @ time 1030. You measure the angle from your bow to the light house which you find to be 33 degrees. Over time you monitor the angle, when it reaches 66 degrees you note the time of 1105.
The time you traveled was: 35/60 = 0.583 hours
The distance then was speed x time = 5 knots x .583 hours = 2.92 nm
So simple! Right?
Here is the catch. What is your course? Do you know it? Not really, there is leeway (sideways pushing of your boat on the wind) and quite possibly current. The geometry equation fails when the heading (direction your boat is pointing) and the course over ground do not match. It is because you are measuring the angle from the bow to the object, which is not necessarily your course over ground. Often times these can be 10-20 degrees or more in a sailboat. Less so in a power boat where the speed is higher so current is less of a % and leeway is reduced.
All well if you know your course, but where did you get it from? GPS? Well then you definitely know where you are so you don’t need this method. 3 point fixes? Well then you definitely again have a chart and know where you are. So hmmmm what is one to do with this seemingly useless geometric wonderland that in practice does not work?
You could estimate your leeway and apply the correction to your heading and you could consult the tide current table to estimate the current effect of your heading. For example, say your leeway was 10 degrees to starboard in the example above (no current). Then the real angle from your course to the object would be 43 degrees. You should wait until the object then is at 86 degrees off your course which is 86-10 =76 degrees off your heading off the port bow.
Distance off with correction
We have this same discussion in our Coastal Navigation Course.
Take the NauticEd Coastal Navigation Course online NOW. Get real practical training and examples. Do it in your own time and take the tests as many times as you like – forever.
At 2245 your GPS fixes your position at LAT 41 deg 01.75′ N and LONG 72 deg 48.40′ W. You are steering course 086 deg psc at a speed of 6.0 knots. At 2400 you fix your position at LAT 41 deg 04.2′ N and LONG 72 deg 38.85′ W. What were your set and drift?
Use the following
(1) Here is a pdf of the chart for you to work on
(2) On the chart, the variation is 14 deg W
(3) Since the problem says psc (per ships compass) we need to account of the ships compass deviation. In the NauticEd Coastal Navigation course exercises we used the following table.
Ships Compass Deviation Table
Set up the TVMDC table
D 2E C 086
Thus, your True heading on the chart is 074 deg T. Your water speed along this line is as given is 6 knots.
The time difference is 1 hour and 15 minutes = 1.25 hours. Thus in 1 hour and 15 minutes, you would travel 7.5 nautical miles.
Scribe a line 7.5 nm from the origin along 074 deg T line. This is your water position. The ground position is described by the GPS coords. Draw a line from your water position to the ground position. This is your 1.25 hour long current vector. It is headed due north and is 0.5 nm long. Since this happened in 1.25 hours the current speed is .5/1.25 = 0.4 nm/hr (knots).
Set (or Direction) is 0 deg T
Drift (or Rate) is 0.4 knots
Note: Current is always expressed in deg True and always expressed in the direction it is heading whereas wind is expressed as where it comes from. Note and remember the difference – important.
ATTN: The NauticEd Coastal Navigation Course has been upgraded and updated. See below.
If you like that we update things for free, LIKE us over there —->
(1) At Eastport, Maine. What was the max spring high tide height after the eclipsed super full moon on September 28th 2015. What was the min spring low tide and was it below the datum? What will be the height of the tide at noon today – Oct 5th?
(2) You live in SanFrancisco. You’ve got friends in town and you want to take them sailing today. What are the best times to take them out of the Bay under the Golden Gate bridge and back?
(3) How often does a spring tide occur and how could you predict it?
(4) Can the water level ever get below the chart datum? Why so or why not?
ANS: posted below – see where we got these plots (in less than 10 seconds)
(4) The USA sets the datum at MLLW which is the mean of the spring low tides over the 19 year cycle lunar solar. UK and the rest of the world set the datum at LAT which is the lowest astronomical tide meaning it should be the lowest it could ever get over the 19 year cycle. Thus often using MLLW the water level can drop below the datum. Using LAT it is less likely but can still happen.
These questions are a breeze when you know what you are doing and the data answers are at your finger tips on your phone or on the Internet within seconds, if you know what you are doing.
One of the really cool things about eLearning software is that you can upgrade a course on demand – you can do a big update or a little one and the update goes instantly to your students. You don’t have to wait until the inventory is sold out and you don’t have to leave schools holding old inventory to be thrown out.
Last week we did a huge upgrade to the Coastal Navigation course. Mainly because we added in lots of new technology about tides and currents but we also added better explanations of plotting courses using animations.
Understanding of tides and currents have come a long way and websites have been automated to include instant data and tide predictions. Older courses and textbooks make you rely on looking up charts (on paper) – but why would you do that on a daily basis when the exact data is at your finger-tips. Off course, you must understand the fundamentals and we teach that but now we also give you access and knowledge on how to use apps and websites for instant data. It’s what a modern sailing course should do!
Students who have taken our older course now have the benefit of the new course at no cost. Just sign in to NauticEd now and go. You can retake the tests and get up to date on latest coastal navigation techniques and understanding.
Learn the theory of course plotting, how to do it and make it second nature, how to measure distances, predict ETAs, account for current flow in course plotting, calculate current flow rate and direction, determine water depth relating to tide, best times for harbor entry, understand GPS, using parallel rulers, bretton plotters, buoys-markers-ATONS (aids to navigation), lights etc etc. Lots and lots of real examples and plotting challenges. You use a real chart. At the end of this course you will have completed the World’s most up to date Coastal Navigation Course and will fly through any other required course like the USCG Commerical Captains License navigation course.
Get free updates for life. Access the course for life. Take the test as many times as you want.
Oh and the other cool thing we did was to add in the requirement to have passed either the FREE Navigation Rules course or the Navigation Rules Module in the Skipper (or RYA Day Skipper) course. This ensures everyone taking this course is up to date on Navigation Rules. It was the responsible thing to do. We did this by adding this piece of code to our software.
IF FREE Navigation Rules Course = Passed
OR IF Skipper Course OR RYA DAY Skipper Course = Passed
AND IF Coastal Navigation Course = Passed
THEN Add Coastal Navigation to the Certificate and the Resume
If you like our problems and solutions, LIKE us on facebook and g+1 us. It really helps us grow – thanks for that. AND you’ll get instant notification when we make a new problem or post a solution. How cool is that?
Course to Steer Calculation
Given the last Rate and Direction problem, now calculate the CTS (Course to Steer) to go to the safe water mark RW “NH” and the TTD (Time to Destionation)?
Answer is posted below. BUT please give it a go first to test your current knowledge.and post your answer to our
Here is the Answer plot (no cheating – give it a go first) (no really) (oh come on really – give it a go first)
CTS and TTD plot
In this answer plot we used 1/2 hour for the time frame. Thus with current flowing at 1.8 knots in 1/2 hour the distance the current will push you off course is 0.9 nm at 17 deg T. Your speed is 5 knots through the water so in 1/2 hour you would travel 2.5 nm.
In the answer plot I have done it two ways. Both work equally as well. First I drew the desired track from the eFix through the buoy RW “NH”. The first vector I plotted was from the eFix to Point A using the current vector (0.9 nm at 17 deg T). Then from Point A, I scribed a 2.5 nm arc to cross the desired track. This gives me position C. I drew a line from Point A to Point C. This gives me what is called a water track. Designated by 1 arrow (mnemonic “water has one”). I measured this direction which is 118 deg T. This is the CTS (Course to Steer). It means if your boat heading is 118 T then your ground track will be from the eFix position towards point C AND towards the Buoy RW “NH” since it lies on the same line.
The distance from the eFix Position to Point C is 2.5 nm. Since you will travel at 5 knots this will take you 30 minutes. NOTE: it is just by pure coincidence that this is the same speed as the boat through the water in this case. i.e. the way you find the boat speed over ground is to measure the distance from the eFix to point C and divide by the time. In this case it just so happened that the distance from eFix to Point C is the same distance as Point A to Point C. You can imagine it would be totally different if the current direction was 30 deg T instead of 17 deg T.
NOTE: also I have solved the problem using the vectors in a different manner. First I scribed an arc 2.5 nm out from the eFix Position. Then I brought in the current vector so that the end was touching the desired track line and moved it until the start touched the scribed arc. This result satisfies the condition that the boat must move 2.5 nm whilst the current brings the boat back to the desired track. The start of the current vector creates Point B. Then I drew a line from eFix to B to create the water track. The water track is the heading of the boat. The heading is (and must be) exactly the same as before at 118 deg T.
This second method, just to me, seems to make more sense of a vector triangle because I can see the boat starting from the eFix and heading out at 118 and getting dragged back to the desired track line. Maybe it’s just me. In any case the triangles are exactly the same. I believe however, that using the first triangle method may be more accurate in real live plotting because you draw the lines from exact points. In the second method, you’re moving that current line to satisfy the two conditions. My brain however, thinks the first method looks weird. i.e. the current drags you all the way out and you have to crawl back. In either case this is not reality. In reality, your boat just follows the desired track. If your mind can handle the first method do it that way.
Next part of the problem is to calculate TTD – the time it takes to get to the buoy.
That’s easy – the distance is 3.2 nm. At 5 knots this will take 0.64 hours = 38.4 minutes.
Now could you solve this problem if the was ALSO 10 degrees of leeway with the wind out of the North?
Permission for a rant? (if you know me, I break out in rants every now and then. It’s a collection of thoughts and I tell you its a rant so as not to offend – ie don’t read this if you’re sensitive)
This problem should be second nature to you. In reality, you’re probably not solving these problems everyday whilst sailing and it’s why some people think they can get away without knowing this stuff. However, this is fundamental to sailing and I think it is irresponsible (strong word – I said it was a rant) to not be able to lay out the method to solve this problem. Laying out the method means you grasp the concept which is the most important to understanding and keeping you out of trouble.
Example: Last year whilst sailing from St. Lucia south to St. Vincent we saw a sailboat way to the west almost on the horizon about half way across. He had left St Lucia and held a compass heading towards St. Vincent. In the meantime, we calculated a CTS and sailed in a straight line over ground to St. Vincent. His path was a complete arc which took him miles off course. Our path was the shortest distance between two points ( a straight line). I’d call this guy a crappy sailor – I know this because of another rant that I wrote about the same guy when it comes to a crossing and give way situation later on as we approached our bay on St. Vincent [see that blog article and story].
Don’t be a crappy sailor – sail with knowledge. It might seem like a selfish rant to get you to buy more courses – maybe or maybe it is and an attempt to reduce the number of crappy sailors out there. NauticEd courses are ridiculously amazing value and after taking at least our Bareboat Charter Master Bundle of courses, I guarantee you will not be on Neptune’s naughty list.
Personally, I’m impressed with the European community and requiring the ICC for all sailors. The ICC requires the above type knowledge. When you’re sailing in Croatia, Greece, France etc, on a Bareboat Charter, you’re nervous enough. You don’t want all the other charter skippers having limited sailing knowledge. If you know they all have the ICC, you’re going to be a lot more comfortable in a crossing situation.
NauticEd has taken it a step further by not only providing the ICC but offering the Bareboat Charter Master bundle of courses. These courses really ready you for a proper safe Bareboat Chartering Experience.
Do you have you United Nations Sailing License (the ICC) yet?
If you’re looking for the ICC license, visit NauticEd and take the RYA Day Skipper Course.
In the USA, the terms “set and drift” are often used when it comes to specifying current flow. It is found that this is confusing to many students and in the rest of the world the terms rate and direction are used. At NauticEd, we adopt the term “rate and direction” in favor of the student.
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RATE AND DIRECTION
Rate is the flow rate or speed of the current in knots aka “drift” in the USA
Direction is the direction the current is flowing towards expressed in True degrees aka “set” in the USA
Here is a simple problem to calculate Rate and Direction based on how a vessel went off course over a period of time.
Print out the PDF provided below or if you are a NauticEd Coastal Navigation student you can use your Chart 12354.
Your motor vessel has cleared out of the Channel and an electronic GPS fix at 1020 places your position east of the mark G “1” Fl G 2.5s at 41 deg 9.4 min N and 73 deg 5min E. You are making way at 5 knots towards the safe water mark RW “NH” approx SSE of the outer channel marks from at New Haven. At 1120 you take another electronic fix and you find your position to be 41 deg 12.7 min N and 72 deg 58 min W. What is the rate and direction of the current?
You’ll need a set of dividers and a protractor and a calculator.
Click to download
Before you look ahead to the answer, give it a go – it’s actually pretty easy.
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On Wednesday October 16th you are going to sail past this port in the morning. There is a shallow area you’d like to pass over. The tidal information you have obtained is as such.
The chart says the depth of the water of the shallow area is 1 meter. You draw one and a half meters and you would like 1 meter below the keel for safety. Summer daylight savings is in effect.
The Tidal Curve for this port is as follows: (click image for downloadable PDF)
Between what times in the morning can you safely pass over the area?
If you’re just reading this for the first time – think about the question and try to answer it before cheating and dropping straight to the answer.
Sorry about this but it was sort of a trick question to get you thinking. Many who emailed us before the deadline of Jan 31st 2015 to win $10 credit towards a NauticEd class got it right – congrats.
Next – apologies for a little ambiguity – the problem did not list if the tide height took into account daylight savings in the tide height listings. Normally they don’t but sometimes they do so Kudos to those who accounted for/discussed this in their answer. We developed the problem using tide heights of NOT using daylight savings – and thus 0131 is 0231, 0752 is 0852, 1427 is 1527 and 2039 is 2139. Actually it doesn’t matter in the answer really because we were only discussing the morning.
To solve: (don’t cheat if you have not answered yet)
First off you need the water to be 2.5 meters deep. 1.5 meters in draft and 1 meter for safety.
Next you need to realize that the reported tide height numbers in a tide table are always listed as the height above the MLLW (mean low low water) datum. So at 0231 the height of the low tide water was 1.6 m above the MLLW datum.
Next realize that chart depths are always listed as the depth of the water at the MLLW datum.
So tide heights and chart depth numbers use the same datum.
Note: USA uses MLLW while most other places use LAT (lowest Attainable Tide). Regardless both are using the same datum in this problem.
This means that at 0231 (low tide) the height of the water in the shallow area was 2.6 m (1 meter depth plus 1.6 meters tide). This is already deeper than the required depth of 2.5 m.
So since 0231 is low tide and the problem asked in the morning then for any time midnight to 0231 the water height is higher than this (0231 is low tide).
Next take a look at the next low tide of 1527 (1427) which is 1 meter. This would not fit the 2.5 meter depth requirement as the depth of the water in the shallow area would be only 1m tide plus 1 meter depth = 2.0 meters. But again the question was “in the morning”. So the question is would there be a time prior to noon where the tide height is lower than 1.5 m?
See below for the plot or the ebbing tide. The tide drops from 3.3 meters to 1.0 meters. And the curve to use is neaps since the difference between high and low is 2.3 meters (close to the mean range for neaps at 2.5m) (neaps are when the tide range is less due to the sun and moon not combining their effects – spring is when the sun and moon combine their effects to make the tides higher).
Draw a sloped line on the chart from 3.3 meters to 1 meter. Now drop a line down from 1.5 meters to the sloped line. Bring this intersection point across to the neaps tide curve for the descending tide. Add in the times starting with 0852 being the high tide and adding 1 hour per section. When you hit the neaps curve drop down to the time. The time shows 1301 (each tick is 10 minutes). Thus at 1301 the tide height will be 1.5 meters which is the threshold. Thus you’d have to be clear of the shallow area before 1301.
Since the question asked for the morning – the answer is anytime in the morning. If you made the assumption that the tide height time did include daylight savings then your answer would have been 1201 which still meets the question answer for anytime in the morning.
ANSWER – ANYTIME IN THE MORNING!
Answer to the tide curve problem is anytime in the morning.
Understanding tides is essential.
Common mistakes made in the sent in answers to this question were:
Using the rule of twelves (no the tide curve was provided)
Not knowing how to use a tide curve
Assuming linear dropping of the tide
Not realizing that the same datum for tides and chart depth are the same.
Not reading the problem properly
And pure not understanding the concept of tides e.g. neaps, springs, sinusoidal type rise and fall
Here is a great Comment regarding extra practical thoughts around this problem given to us by Michael Sisley Instructor and free lance yacht skipper. Thanks Michael!
Our sport is fun when we plan in order to make it safe! 1)Read the question. – in a harbour right? Flat water. 1m clearance to allow for uncalculated variables such as atmospheric pressure, on shore wind, shifting sands – good seamanship built into the plan. Yes 2m waves on the sandbar leading to the harbour entrance, an ebbing tide and a strong on shore wind would lead to a very different contingency. It’s all a matter of preparation and planning. 1) Before setting out, check the your depth echo sounder with a lead line. 2) Harbours silt up and sand bars shift with the tide. So contact the harbour master and ask “Where is the shallow patch now?” “How deep?” 3) The hydrographical service publishes valuable information – use it! You can then use your calculation to help decide when it is safe to go. – And enjoy!
This week we uploaded our updated Coastal Navigation course to Apple for publishing on iTunes as an interactive eLearning App. We’re very excited about this update because of the HTML 5 animations we’ve used to explain some of the concepts.
Specifically in regards to this post, no where on the web have we seen a decent explanation of how to do a very simple and elegant position fix using only one land position. The concept is called a running fix. In fact, all we found was very poor, long, complicated and sometimes wrong explanations of how it works. Certainly we found no animated interactive explanations. So as usual, at NauticEd we have broken down the seemingly complicated to the very simple.
Play this animation below then read the explanation and solution below – then watch the animation again.
(NOTE: If you like this animation, please LIKE it on facebook and g+1 it. Thanks it really helps us grow and pay for all the free stuff we give away)
The Example Problem is:
You are sailing along on heading 57° psc (61° Mag) (47° T) your knot meter reads 5 knots. You are passing Horton Point Light to your starboard. At 1548, a hand held bearing shows that the bearing to Horton Point Light is 119°Mag (105°T). You decide to do a running fix. At 1615, the bearing to Horton Point Light is 160°M(146°T). Determine your running fix position.
The time elapsed is
1575 -1548 = 27 minutes (/60) = .45 hours
@ 5 knots you will travel 5 x 0.45 = 2.25 NM
You draw the true bearing line of 105° T to Horton Point Light. You then draw a vector 2.25 NM long in a direction of your heading 47° T starting anywhere on the 105° line. You then draw another line parallel to the first 105° T bearing that intersects the end point of the 2.25 NM vector. Finally, you draw in your second bearing line of 146° T. Where the 146° T line intersects the parallel 105° T line – you mark as your running fix position.
The theory behind this is simple but not usually explained. Initially you must lie on the 105 degree line somewhere but you don’t know where. You know that over the time elapsed you will travel the 2.25 NM from some where off the initial 105 deg line but you don’t know from where – yet. The parallel 105 deg line projected forward means that you will also lie somewhere on that projected line, again – somewhere. By doing the second bearing, off any object, the intersection of that bearing with the projected line means you must be at that point (given that your speed and heading were accurate).
LAT 41° 07.55’ N and LONG 72° 28.8’W
You can then draw your running vector from the fix position back to the original sighting line if you like to find your original position. This will also be your track.
Pretty slick ah? You’ll never be confused about a running fix again – and I bet you’re now wanting to get out on the water and practice next time out. Imagine trying to explain this using text and a paper book – yuk. Ahhhh no wonder people don’t like slogging through books anymore. In what?, less than a minute you grasped this concept fully.
eLearning is where it is all at.
Show off to your friends next time you’re out about your new found knowledge.
Or – if you’re really ready to get going properly why not invest in the BareboatCharter Master bundle of courses. The Coastal Navigation Course is included and you get a bonus of the Electronic Navigation course for free AND you save a ton of $ over the A La Carte Prices. That’s the best deal ever!!!
Posted by Director of Education on February 27, 2012 under Coastal Navigation | Comments are off for this article
Currently we’re working on a cool project with SimRad to create a training simulator for a gps device. But to do that and to write the code and the statement of work for the programmer we had to be very clear about the math behind it all. At least to me, it’s all pretty interesting and so I thought I’d share it here. Kinda like “the making of…” in movie talk.
With traditional coastal navigation you are plotting positions on a chart then measuring the angles and distances. That’s all pretty easy and well described in the NauticEd Coastal navigation course. But once you start to do it electronically, you’ve got to have the mathematics of it all well understood in order to write the code. Turns out it’s again pretty easy but you’ve got to reach deep into some brain cells to drag out the trigonometry you’ve so desperately tried to forget from high school days and thought/hoped you’d never use again.
But this is a worthwhile read because you’ll exit out understanding the principles, which is really the point of the sailing blog entry.
So here’s one snippet of what we did:
We simulated a vessel moving at 20 knots that can be controlled using the autopilot. We input a static current to the east at 2 knots. For this scenario and for any direction the vessel is moving, we want to present the following variables in real time as they changed:
Bearing to Destination – BTD
Distance to Destination – DTD
Course to Steer to reach destination – CTS
Speed Over Ground – SOG
Time to Destination – TTD
Position of the vessel
The difficulty comes in with the current. You can’t just head towards your destination because the current will drag you off. So you’ve got to solve for a triangle where you have limited information and the triangle is not a right angle triangle.
We decided early on in the simulation to work in meters and meters per second because it makes the math a lot simpler. Also just by universal convenience meters per sec is ½ speed in knots. Thus 20 knots = 10m/s.
In a first example, the destination is at a point 500m to the north and 1000 m to the east of the vessel. Since we know the instantaneous position of the vessel and the position of the waypoint, we know the instantaneous distance to the waypoint. I say instantaneous because the vessel is moving.
Solving for Distance and Bearing to Destination
That’s a good start however the vessel is not moving at 20 knots towards the target. Due to the current the resultant vessel speed over ground is slightly different. To visualize what’s happening we drew this triangle. We added the current vector to the bearing vector to create a Course To Steer vector.
Vessel heading versus resultant track
There is a further complication however, since the whole process of getting to the destination doesn’t necessarily take 1 hour. We can’t say that the current vector is 2 nautical miles long. Thus in the distance triangle above we only know two variables, the angle (a) and the distance to destination – that’s not enough information. Instead we have to draw a velocity triangle whereby the current is 2 and the vessel speed is 20. Fortunately this triangle is exactly the same shape as the distance triangle. In that triangle we know three variables, vessel velocity, current velocity and the angle (a). Thus we can proceed to solve for the entire situation.
Solving for velocity triangles and distance triangles
Here’s the painful part from the ol days
A/sin(a) = B/sin(b) = C/sin(c) it’s called the sin rule.
The lower case letters are the angles and the upper case letters are the triangle side lengths opposite the same lower case letter. IE length A is opposite the angle a.
Just bare with us if that scares you. Actually it’s pretty simple from here you just have to plug and play. What it means is that if you know 3 pieces of information about any triangle you can find the others.
The first thing well solve for is the angle b. To do this, use the velocity triangle. The three pieces of information that we know are the current speed vector length, the vessel speed vector length and the angle to destination (a).
In this quadrant ie angle to destination is less than 90 deg, the Bearing to Destination is 90 minus the angle to the target ie BTD = 90-(a)
Solving the sin rule for angle (b) then
(b)= arcsin ( current x sin (a)/(vessel speed)) = arcsin (B x sin (a)/A)
= 2.6 deg
Since all angles in a triangle add to 180 the angle c must = 150.9 deg. This is the course that you would steer to arrive at the target due to the current.
The luck of it is that this is the same angle to plug back into the distance triangle
IE (c)= (c*).
Now we can return to the distance triangle because we know enough variables. If we plug the two known angles (a) and (c) in to distance triangle we can solve for the distance A*. We need to know A* because the vessel attempts to travel along this path at 10m/s but ends up at the destination. Solving for the time it would take the vessel to attempt A* is the same time it takes for the vessel to actually travel to the destination. IE the vessel is helped by the current.
From the sin rule:
A* = C* x sin(a)/sin(c) = 1027.2 meters
and at 10 m/s the time elapsed is 102.72 sec = ~ 103 seconds
The vessel actually travelled 1128 meters in 103 seconds thus the
SOG = 1128/103 = 10.9 m/s = 21.8 knots
The distance that the vessel was pushed to the east by the current = Set =B*
Set= current speed(drift) x time = 1 *103 = 103 meters.
So for this tiny moment point in time:
BTD= 63.4 degT
CTS=60.9 deg T
DTD = 1128 m
TTD = 103 sec
SOG = 10.9 m/s
In coding all this up and since the vessel is moving we have to solve all these at each instant in time using location information from the last moment in time. Again not too difficult it just takes a bit of thought in laying it all out.
If you know the velocity of the vessel over ground (solved from above) and the increment in time, you can calculate the change in position. This is done in Cartesian coordinates Ie x and y directions
Change in X = velocity in x direction times the time increment = Vx x dt
Change in Y= velocity in y direction times the time increment = Vy x dt
Where Vx and Vy are solved from the angles calculated from above and SOG
So the position at this instant is now the position at the last instant plus the distance travelled in the increment in time.
We have to solve all this for each instant in time because the vessel is also changing direction from input from the autopilot. In a real situation the position information is being gathered from the actual gps not the last instant. Altho some smarter systems will use this for predictive situations.
Another level of complexity is added in when you consider other quadrants. For example in the below drawing, the current works against the vessel. Note the SOG has slowed, the angle b reduces the CTS, and the triangle is quite different. Thus in the code we have to add a lot of if/then statements to determine what quadrant were working in to determine which formula to use.
Solving the velocity and distance triangles
OK ADMITTEDLY THAT MIGHT HAVE BEEN A BIT HEAVY GOING
Hopefully this gave you some insight into:
Some principles behind coastal navigation
Help in understanding electronic coastal navigation
Some insight into what we do at our sailing school to provide great sailing education
We can’t wait to deliver the SimRad Electronic Navigation simulator and training tool to you.
If you’re interested in properly understanding how to navigate a vessel using charts then take the NauticEd Coastal Navigation Course. Once we finish the electronic navigation simulator we’ll embed it into the Navigation sailing course.
The posting here is not a course in celestial navigation by any means. However it’s meant to simplify a few principles for you so that you’ll at least have some sort of celestial orientation. And… perhaps it’ll inspire you to learn the aging art.
This was written by Grant Headifen, Educational Director of NauticEd. NauticEd provides online sailing courses and Sailing Certifications accepted by charter companies worldwide.
Latitude: In the northern hemisphere, finding latitude is simple using one of the greatest gifts to human kind – The North Star. What ever angle the northern star is at from the horizon, that’s your latitude.
Imagine you’re an ant sitting on the top of an apple looking at a spot directly above you on the ceiling then the spot is 90 degrees from the surface you’re standing on. If you’re standing half way around the apple then you’d barely see the spot but it would be horizontal to the surface you’re standing on and so the spot would be at zero degrees. And if you were ¼ of the way down the apple then the spot would be at 45 degrees etc. ie the northern star is the spot on the ceiling to us.
You can also find latitude using other celestial sightings but they involve table lookups and are slightly more complicated. Not meant for this post and also note that there are a few more complicated variables not taken into account during this simplistic explanation like the height of your eyeballs above the earths surface etc etc. But at least you’ve now got the principle.
Longitude: Now this is a fun one and in an incredibly easy principle. But years ago (early 1700’s) while the principle was easy then the execution was difficult. Read on to see why.
The earth rotates through 360 degrees in 24 hours. That’s 15 degrees per hour. By convention, when the sun is at it’s highest point in Greenwich, it is noon in Greenwich. That means that at a place that is 15 degrees to the West of Greenwich the sun will be at it’s highest point one hour later. Six hours after Greenwich the sun will be at it’s highest point somewhere in over the USA and 12 hours later the sun will be at it’s highest point in New Zealand.
Animation of time zones
So if we know the time in Greenwich and sun just reached its highest point where we are then we can calculate our longitude.
Lets do a few examples. If it is 6 pm in Greenwich and the sun just peaked overhead here, then I am 6 x15 degrees to the west of Greenwich which is 90 degrees West which is right near St Louis Mo.
If the sun peaked overhead in Los Angeles what time would it be in London.?Well LA is 118.15 degrees West (from Google earth). Divide that by 15 degrees per hour and we get 7 hrs 53 minutes. Now since the times zones are created in bands this would round up to 8 hours. Thus it would be 8pm in London.
You’re sailing in the Greek islands in the Mediterranean and a little bird just told you your latitude is 34 deg 54 minutes north but failed to tell you the longitude. Fortunately you have your handy sextant and just as you take a shot, the sun just reached its apex overhead. You look at your watch and the local time is 12:10:48 pm. Where are you?
Since you’re in time zone B you are 2 hours ahead of Greenwich. Thus the time in Greenwich is 10:10:48 am. And since the sun peaked just now (=noon) then you are 12:00:00 minus 10:10:48 = 1 hour 49 minutes and 12 seconds from Greenwich. Putting this into decimal time this is 1.82 hours. Multiply this by 15 degrees per hour and we have 27.3 degrees East or 27 degrees, 18 minutes East.
You’re in the harbor north of the town of Kos on the Island of Kos.
That was incredibly easy, so why all the hoopla back in the 1700’s? The King of England even offered up a ₤10,000 reward to anyone who could solve the issue of Longitude. The above math was well known but the issue was telling the time. No one could accurately keep time at sea. After 27 years of work on the project, John Harrison, finally invented the Chronometer more commonly known as the watch. The watch was not susceptible to the sudden crashes of waves at sea and thus kept proper time.
James Cook on his second trip around the world in 1772 sailing on Rendezvous, took Harrison’s watch with initially much skepticism. Stating that he’d give it a try. After six months at sea, Cook stated that the Chronometer would almost certainly become the way of the future for Navigators. Cook then went on to reposition many of the Islands in the Pacific including Tahiti, his favorite island. His map of New Zealand astounds people even today with its accuracy.
Again there were a few simplistic assumptions taken in that explanation. But now, at least you understand the principle of longitude determination from a noon shot of the sun. You can also determine your latitude from a noon shot of the sun as well using tables and a bit of math. Again beyond this posting.
If you’d like to delve deeper into these topics, NauticEd provides online sailing lessons and an Introductory Celestial Navigation Sailing Course, or maybe you’re just happy with your handy boring ol GPS.
We had a blast putting this Ocean Navigation Discussion video together.
The big time fun part was coming up with the dialog. By the constraints of the program to create it we had to use an astronaut and Napolean. Take a view of the video and let us know if it was at least mildly funny.
We’re going to embed it into the start of the NauticEd Coastal Navigation Sailing Course to get the student in the right mood for learning the basics of ocean navigation. Our thoughts are if some one is in a good mood they can absorb more info. So you’ll find that, through out all NauticEd courses, we try to inject a little humor. It’s something that I think can be done more effectively in online learning as opposed to a paper book.
We used Google Earth to find the distance from the Taj Mahal to Auckland New Zealand. The line it drew was accurately a “great circle” which on a sphere is the shortest distance between two points.
Great Circle Distance from Taj Mahal to Auckland New Zealand
Please enjoy the video but also the NauticEd Coastal Navigation Course.