SeaTalks about Coastal Navigation

* figured it out.

I have a question from Module 11, Leeway and Current. Why, when plotting the Example of sailing out of Port Fitzroy, do you only apply a 6 deg Variation to the Magnetic bearing (67 deg) to obtain the true bearing (61 deg) when the Variation shown on the Compass Rose on Chart 3 is 7’25”W? Am I missing something really simple here?

If one leaves the same plot on the chart, distances are not changed by changing times. So if current vector is 1.5 nm long. And that 1.5 nm is covered in 1/2 hour then the current vectors speed should be 3nm/hr not .75. What am I interpreting incorectly? Thanks.

I had trouble understanding the first example and had to re-read and ponder a lot. If I understand it correctly now, here is another way of looking at it. Start with this: T = ? V = 5W (we know this from charts that tell us the variance for the area we are in) Magnetic = ? D = 2E (we know this from predetermined deviation information about our boat) Compass = 270 (we know this from reading the ship's compass direction toward the buoy) Now we add the deviation of 2 to the compass bearing to get the magnetic bearing because it is an East deviation we are going from up the TVMDC, from Compass to Magnetic to True. Then we subtract the variance of 5 from magnetic bearing to get the true bearing because it is a West variance and we are still going up the TVMDC. This helped me a lot, so I thought I would share it in case it might help anyone else.

Why not apply the 10 degree leeway in addition to heading so you track down your "track" line rather than just allowing it to play out? Be proactive and make the boat go where you want with all things considered?

No idea where they got this info from? Any help?

In the LOP section it states "Breton Plotter is used to draw those on the chart - from the objects toward the ship's position". This is confusing, when and why would you reverse the bearing toward the boat? Thank you

After going back and forth and reading a bunch, I figured out where my trouble is coming from. In Module 3, 4 and 5 the triangle is labelled (and the desired COG is in even different) differently each time! My head was exploding. But I'm good now. According to "Labelling Convention" The vectors are Current + Leeway = COG, with Heading being the angle the boat is/was pointing to get there, But all of the following are mostly acceptable Vectors should be: 1: "Current" or "Rate and Set" (-->>>--), 2: "COG", "CMG", or "Track"(-->>--), 3: "Leeway" (-->--); and then an adjustment angle to give you your vessels "Heading" (called final Heading in Module 5), also labelled with -->-- (See module 3, though I much prefer --=>-- (looks like a little boat, shows me its the heading and not part of the vector solution...so shoot me for breaking with a convention that uses the same symbol for two different angles...) In comparison: Module 3, the vector triangle is made of Current, COG and Leeway, with the free angle labelled "Heading". Yes, OK Module 4, the vector triangle is made of Current, Heading, and Track, with "Leeway" being both the free angle and desired COG!! HUH?? Not OK. Module 5, the vector triangle is made of Set&Drift, Desired Track, and Heading, with the "Leeway" added in after for "Final Heading"(free angle). Um...ya, not perfect but it works. SO, according to that I deduce the examples in Module 3 are correct (in principle at least), with only measurement error, Module 5 is mostly correct, except that the vector labelled heading is actually the leeway, and after the leeway angle is removed you have the heading, so no need to call it "final heading", and again, some measurement error (I suspect) in the given answer. BUT Module 4 (specifically leg 3 of the practical example) is just plane WRONG. On the same page is stated "Leeway line is applied after current and is downwind x degrees from the track line. Thus, your course over ground is on the leeway line", however in the example, the COG is not in fact on the leeway line, but on the heading angle line (because it's mislabelled it appears to be right, but it isn't...) Here are the correct solutions for the examples in Module 3, 4 and 5, figured to 2 decimals, solved with a calculator. Module 3a: Given: Heading 082°, Speed through water 6kts, current 1.5kts 140°T, then COG is 93.46 at 6.93kts Module 3b: Given: Track/COG 072°, Speed through water 6kts, current 1.5kts 140°T, then desired Leeway is 60.03, giving a Desired Heading of 50.03°T (not 48°T heading as given in the solution...sloppy measuring perhaps since the diagram is labeled correctly?) Module 4: Given: Track/COG 056°, Speed through water 5kts, current of 1.5kts 085°T, then Desired Leeway is 47.64, giving a Desired Heading of 57.64°T, 6.26kts resultant SOG Module 5: Given: Track/COG 045°, Speed through water 5kts, current of 2kts 090°T, then Desired Leeway is 28.57, giving a Desired Heading of 38.57°T (not 40°T heading as given in the solution...), 6.21kts resultant SOG. Enjoy.

If I have the lat and long how do I find it with the RYA chart plotter?

In working through the final question 3 a) I had the degrees at 141 degrees taking this off the tidal diamond A, 3 hours after the High Water. The answer was 137 degrees? When i took a bearing off the almanac i came up with 137? I thought you went to the tidal diamond for the degrees?