Brain hurts - Working out course to steer

[db9987]

Originally Posted by Pete7

Indeed, there was an even cheaper program that only worked for the South Coast of England, but it seems to have disappeared sadly. The big advantage I think is the ability to change to start time and speed to see if departing earlier or later has an effect and it does. From memory offering 30 - 60 minutes savings for a 10 - 11 hour passage.

For the OP who mentioned multiple tidal changes, this would suggest several days. In the UK (and perhaps the PNW) this gets super complicated because the weather is certainly going to change during that time. Also any forecast is just that, a forecast averaging over quite a big area. Add some headlands with tidal gates plus wind shifts and you almost need a full time nav handcuffed to the chart table. Certainly make a plan and run through a series of options for times and speed etc, then stay flexible. A tweak of the course as you approach your destination so you don't end up down tide during a big Spring would be prudent.

I think that would be this one here - TidePlan2 I used it for a few years but it's gone now and I you try to download the most recent version my Security software says 'dangerous URL'. a shame as it worked really well for the English Channel

[saltfree]

A lot of people pointing out that electronics will allow you to constantly correct for current/leeway as things change, and that this will give you the shortest route (eg. across the English Channel), seem to have overlooked one point:
Sailing the shortest distance is NOT necessarily the most efficient or quickest. If you turn your boat further and further into the oncoming tidal stream in order to stay on the shortest route, your SOG will drop dramatically, which is a huge waste if 6 hours later it will just be pushing you equally hard the other way. Don't fight it, keep your speed up and roll with it! That was the whole point of the Youtube video that someone linked to. The question that the OP put was a good one... how do you calculate that CTS for EFFICIENCY, not for distance?...

[sailingtaiyo]

Course to steer is pretty simple to do with a chart, slide rule and a pencil, maybe a calculator if you really need to.


You first need to figure out the Set and Drift. Then measure from that ( C to D ) and that is your course to steer. Although you may have to do this several times as others have mentioned due to tidal and current changes.


In the image below, the COG is 90 True, with the Set and Drift, you can see that now one would have to point more NE in this example. Check out this video link.

https://www.youtube.com/watch?v=8S1ldErZ2ZQ

[saltfree]

Originally Posted by sailingtaiyo

Course to steer is pretty simple to do with a chart, slide rule and a pencil, maybe a calculator if you really need to.


You first need to figure out the Set and Drift. Then measure from that ( C to D ) and that is your course to steer. Although you may have to do this several times as others have mentioned due to tidal and current changes.


Yes, this method allows me to end up at point "D" when and where I want to... but it means I was steering against the current to do it. If I don't care whether I arrive at point "D", because I'm aiming for somewhere further into the distance/future. With this method, I arrive as planned at point "D", then recalculate and start crabbing the other way, steering against the new current direction when it changes. Yes, I crab along my rhumb line more or less straight, but giving up SOG to do it. In theory, if I were sailing at a perfect right-angle to the current, and sailing exactly through two full tide changes of equal strength, then I would just maintain my compass heading consistently at my destination. I would then sail a big "S" shape across the ground (much more distance), but get there much more quickly because I kept my SOG high, while allowing the power of the current to simply take me out of position, and then push me back into place.

The challenge is figuring out what the heading should be when I am not crossing the current at right angles, and where future tides may be of different strengths than the one I am currently in, between now and when I arrive on the other side.

Yes, this method allows me to end up at point "D" when and where I want to... but it means I was steering against the current to do it. If I don't care whether I arrive at point "D", because I'm aiming for somewhere further into the distance/future. With this method, I arrive as planned at point "D", then recalculate and start crabbing the other way, steering against the new current direction when it changes. Yes, I crab along my rhumb line more or less straight, but giving up SOG to do it. In theory, if I were sailing at a perfect right-angle to the current, and sailing exactly through two full tide changes of equal strength, then I would just maintain my compass heading consistently at my destination. I would then sail a big "S" shape across the ground (much more distance), but get there much more quickly because I kept my SOG high, while allowing the power of the current to simply take me out of position, and then push me back into place.

The challenge is figuring out what the heading should be when I am not crossing the current at right angles, and where future tides may be of different strengths than the one I am currently in, between now and when I arrive on the other side.

[sailingtaiyo]

Ah ok I see. I think you can find the mean after the fact or use tide tables and factor them in using a Distance Speed Time Calculation multiple times.. Ha, this is a bit of a head scratcher. As far as finding the heading, you will need to do a DR from the position you end up in after getting tossed around. That is if I am understanding this correctly. I you are just trying to find a future course to steer for future tides and currents, use the tide table and current vectors to calculate the set and drift and that will give your course to steer. I am sure there is an easier way such as a nice computer set up with racing software but on paper, this is how I would approach it.

[GoaWeigh]

Interesting discussion... The same set of issues apply for "Log towers" crossing the Salish Sea (Strait of Georgia). When departing Nanaimo for the "North Arm". The track observed over time will look more like a series of arcs and not a straight line.
The average speed is anywhere from 1.5 to 2.5 knots, with a "3, 4 or 5 wide" tow. The tides and wind are all important when choosing initial course to steer. They come out of Nanaimo on the flood tide to get pushed out of the harbour, then nearing Entrance Island set up for the last of the push NW from the flood.
As tide changes they are drawn SE towards Thrasher Rock and then are pushed back North again. Dependant on conditions they run a more northerly course to take advantage of the draw-push from Howe Sound or southerly toward Thrasher Rock and then up when nearer to Lower Mainland (Where they can take advantage of the strong push or hug the shore and ride the backeddy if tide opposing) Rarely do you see a log tower pointing straight towards their destination.... Instead they are following a rough CTS to allow for the shortest distance through water.
The decisions that the old timers made for the transit were based on timing and speed of the tidal current as well as wind speed/direction. Some of the younger computer-age towers can get a little overly focused on XTE rather than CTS and this can actually make for longer transit times.
Personally I have only made those tows a few times and make no claims to being an expert open water log tower. The above is based on having sailed with some very talented old school log towers who loved to explain the issues, plus years of observing their transits while barge towing.

Attached 1st photo is just a grab from internet for illustration purposes and shows a typical 4 wide tow. The second is mine, abeam Entrance Island with a 5 wide destined for the North Arm.

[Dockhead]

Originally Posted by saltfree

Yes, this method allows me to end up at point "D" when and where I want to... but it means I was steering against the current to do it. If I don't care whether I arrive at point "D", because I'm aiming for somewhere further into the distance/future. With this method, I arrive as planned at point "D", then recalculate and start crabbing the other way, steering against the new current direction when it changes. Yes, I crab along my rhumb line more or less straight, but giving up SOG to do it. In theory, if I were sailing at a perfect right-angle to the current, and sailing exactly through two full tide changes of equal strength, then I would just maintain my compass heading consistently at my destination. I would then sail a big "S" shape across the ground (much more distance), but get there much more quickly because I kept my SOG high, while allowing the power of the current to simply take me out of position, and then push me back into place.

The challenge is figuring out what the heading should be when I am not crossing the current at right angles, and where future tides may be of different strengths than the one I am currently in, between now and when I arrive on the other side.

Crabbing along the rhumb line is only efficient in case the current is constant. If current is changing -- like a tide over multiple hours -- trying to stay on the rhumb line is NOT indeed sailing a shorter distance, but rather, a longer one. Your boat is attached to water, not to the bottom underneath. That's why your SOG and VMG to the waypoint is less than your boat speed -- because when you're crabbing, you're sailing more miles through water, than the distance you are making good over ground.



To deal with tides which are not perpendicular just do the vector triangle shown in the post you quoted -- one for every hour. Then just add up the vectors for the whole passage and steer that one course for the whole passage. That's the fastest way, because it's the shortest distance THROUGH THE WATER.



A simplified example --


You're sailing approximately three hours at six knots.


Your destination is 18 miles away bearing 180.


Tide runs back and forth at 90, then 270.


Two hours at 4 knots 90, one hour at 4 knots 270 (yes I know that doesn't exist in nature -- it's a simplified example ).


The sum of the vectors gives you 191.8 degrees CTS. If you steer that course and never change it, you will arrive at the destination in 3 hours and 4 minutes. Your course over ground will be a long S curve, but you will sail only 16.381 miles through the water.



If on the other hand you crab along the rhumb line, you will only be making 4.473 knots SOG, and it will take you more than four hours. You will sail more than 24 miles through the water. The fact that you are making only 16 miles over ground doesn't matter -- your boat doesn't have wheels.

[parkstone bay]

I THINK what the OP wanted was to know, say, leaving the South Coast of England in a boat that does 5 knots, going to France (about 60 miles), could a tablet be used to workout the TOTAL offset required caused by, say, 3 hours of west-going tide, (at the end of which you might be, say, 7 miles west of the Rhumb line to Cherbourg), followed by 6 hours of east-going which might push you 12 miles east, followed by a further 3 hours of West-going on the French side, which is stronger, let's say 9 miles. In this example we'd have a total of 16 miles westward set, minus 12 miles east = nett 4 miles west. But, as we all know, the Man upstairs doesn't always follow the rules. You pick up a lump of discarded net around your keel; or the stream on the UK side is a little stronger than you had foreseen; or the breeze is just a tad kinder than you'd dared to hope ....... that's why you do a fix every hour in a sailboat, and in a fast MB, I would do one every 30 mins. That way you can see what's happening early. But - and maybe this is what he meant to ask, the "cross-track error" component of a tidal current is proportional to the sine of the angle between the current and the boat's track (allowing for leeway). Thus a head-on current is at an angle of 0 degrees to your heading (sin 0 = 0), so it is not setting you off course, it's 100% slowing you down, and a beam on current is 90 degrees to your heading, it doesn't affect your speed, but sets you off course 1 mile for every knot of flow, per hour. The sin of 90 degrees is 1 It'd be a piece of piff to work up a spreadsheet on excel . (To work out the retardation/acceleration effects use the cosine of the angle)

[Dockhead]

Originally Posted by parkstone bay

I THINK what the OP wanted was to know, say, leaving the South Coast of England in a boat that does 5 knots, going to France (about 60 miles), could a tablet be used to workout the TOTAL offset required caused by, say, 3 hours of west-going tide, (at the end of which you might be, say, 7 miles west of the Rhumb line to Cherbourg), followed by 6 hours of east-going which might push you 12 miles east, followed by a further 3 hours of West-going on the French side, which is stronger, let's say 9 miles. In this example we'd have a total of 16 miles westward set, minus 12 miles east = nett 4 miles west. But, as we all know, the Man upstairs doesn't always follow the rules. You pick up a lump of discarded net around your keel; or the stream on the UK side is a little stronger than you had foreseen; or the breeze is just a tad kinder than you'd dared to hope ....... that's why you do a fix every hour in a sailboat, and in a fast MB, I would do one every 30 mins. That way you can see what's happening early. But - and maybe this is what he meant to ask, the "cross-track error" component of a tidal current is proportional to the sine of the angle between the current and the boat's track (allowing for leeway). Thus a head-on current is at an angle of 0 degrees to your heading (sin 0 = 0), so it is not setting you off course, it's 100% slowing you down, and a beam on current is 90 degrees to your heading, it doesn't affect your speed, but sets you off course 1 mile for every knot of flow, per hour. The sin of 90 degrees is 1 It'd be a piece of piff to work up a spreadsheet on excel . (To work out the retardation/acceleration effects use the cosine of the angle)

Sure, but a simple vector triangle will give you the same result with simple trig.


If you're doing it by hand, mapping it out on the chart to take account of your speed of advance towards the destination, will deal with the issue you're talking about.



In practice, however, Channel sailors (I've crossed the Channel to or from Cherbourg at least 100 times, myself) don't bother to do it so finely. It's quite enough to eyeball the tidal atlas and adjust by eye for any deviation from perpendicular. Inevitable deviations from planned passage speed, will wash out any error from using this simplified method.

[parkstone bay]

I don't think there's any disagreement on that.

[RayRodriguez]

Really cool discussion. And no politics nor guns!

Here's question number one of a few rattling in my mind...

(Disclaimer: I'm sure that this is very boat dependent and will also depend on the relative direction of the wind and wind speed.)

If I calculate, as in the previous example, that I should steer 191.8 degrees based on current and speed, and the wind is from due south, how much extra crab do I need to apply to account for the velocity of the boat not being exactly parallel to the centerline of the boat?

Since several of you have self identified as pilots, what I'm asking is a "typical" angle of attack AKA the difference between velocity vector and pitch angle that you use for your navigation?

Edit: Originally I meant this to apply to a sailboat, but a powerboat will drift too, eh?

[sailingtaiyo]

It all has to do with the sailing characteristics of your boat as well as sail plan. This is beyond my skill level. I am sure there are some racers who could give you much more info. But to get from point A to point B asap with any given conditions, use the set and drift with the course to steer.



Aloha,



Mike.

Originally Posted by RayRodriguez

Really cool discussion. And no politics nor guns!

Here's question number one of a few rattling in my mind...

(Disclaimer: I'm sure that this is very boat dependent and will also depend on the relative direction of the wind and wind speed.)

If I calculate, as in the previous example, that I should steer 191.8 degrees based on current and speed, and the wind is from due south, how much extra crab do I need to apply to account for the velocity of the boat not being exactly parallel to the centerline of the boat?


Since several of you have self identified as pilots, what I'm asking is a "typical" angle of attack AKA the difference between velocity vector and pitch angle that you use for your navigation?

Edit: Originally I meant this to apply to a sailboat, but a powerboat will drift too, eh?

[sailingtaiyo]

" If I calculate, as in the previous example, that I should steer 191.8 degrees based on current and speed, and the wind is from due south, how much extra crab do I need to apply to account for the velocity of the boat not being exactly parallel to the centerline of the boat? "


None, The course to steer has been determined using Distance / speed*time.




" Since several of you have self identified as pilots, what I'm asking is a "typical" angle of attack AKA the difference between velocity vector and pitch angle that you use for your navigation? "


I am not pilot per-se


The velocity is a known ( current speed in knots ) not sure about the pitch angle, if you mean heeling, reef haha!



This is a pickle, when dealing with tides, currents and leeway, one has to constantly make changes to the boats course. Would be to easy to point and just go straight!

Originally Posted by RayRodriguez

Really cool discussion. And no politics nor guns!

Here's question number one of a few rattling in my mind...

(Disclaimer: I'm sure that this is very boat dependent and will also depend on the relative direction of the wind and wind speed.)

If I calculate, as in the previous example, that I should steer 191.8 degrees based on current and speed, and the wind is from due south, how much extra crab do I need to apply to account for the velocity of the boat not being exactly parallel to the centerline of the boat?

Since several of you have self identified as pilots, what I'm asking is a "typical" angle of attack AKA the difference between velocity vector and pitch angle that you use for your navigation?

Edit: Originally I meant this to apply to a sailboat, but a powerboat will drift too, eh?

[sailingtaiyo]

You can download a copy of - Boditch, The American Practical Navigator - in .pdf format which goes into great detail about all of these questions.

[Stu Jackson]

Originally Posted by RayRodriguez

Really cool discussion. And no politics nor guns!

Here's question number one of a few rattling in my mind...

(Disclaimer: I'm sure that this is very boat dependent and will also depend on the relative direction of the wind and wind speed.)

If I calculate, as in the previous example, that I should steer 191.8 degrees based on current and speed, and the wind is from due south, how much extra crab do I need to apply to account for the velocity of the boat not being exactly parallel to the centerline of the boat?

Since several of you have self identified as pilots, what I'm asking is a "typical" angle of attack AKA the difference between velocity vector and pitch angle that you use for your navigation?

Edit: Originally I meant this to apply to a sailboat, but a powerboat will drift too, eh?

What you really need to so is stop making up terms. CRAB?


There are definitions of all these terms. The previous poster mentioned Bowditch. It's also defined in Chapmans and any other good navigational text.


As I previously mentioned in this thread, we have seen altogether too many instances of skippers mistakenly confusing / using heading and bearing, for example.


Please learn the proper terms and what they mean. Things like heading, bearing, drift, set, leeway, etc. By doing so, your questions will be answered.