Distinct Activities: Shackled by a Common Name?

[Seaworthy Lass]

Originally Posted by goboatingnow

forget the little triangles , the large one is what I reworked, That is a 4 hour CTS plot
Dave

Sorry, I don't understand your CTS graph.
The vertical component is 10, which is the total displacement due to current (that looks good).
But:
What does the horizontal axis of 12.5 units represent? How did you arrive at this figure?
What are you saying is the required course to steer?

[goboatingnow]

Seaworthy heres the actual graphical solution

The track method time is 6 hours 24

The CTS method is 5 hours 7 ( approaching the destination at a rate of 12.5/16)

[goboatingnow]

Originally Posted by Seaworthy Lass

Sorry, I don't understand your CTS graph.
The vertical component is 10, which is the total displacement due to current (that looks good).
But:
What does the horizontal axis of 12.5 units represent? How did you arrive at this figure?
What are you saying is the required course to steer?

The hover craft traveled 12.5 ( 12.48) miles( towards the destination) in 4 hours, it is approaching the destination at a rate of 12.5/16 *240 minutes.

The vector is a 4 hour plot.

[Seaworthy Lass]

Originally Posted by goboatingnow

The hover craft traveled 12.5 ( 12.48) miles in 4 hours, it is approaching the destination at a rate of 12.5/16 *240 minutes.

The vector is a 4 hour plot.

Huh????

You have the hovercraft reaching the rhumb line at the four hour mark. At this point there is no more cross wind for the rest of the journey (there was only cross wind for 4 hours). So how can the hovercraft continue on its constant course to steer until it reaches the destination (which you said was 16 miles from the departure point) if you have it arriving on the rhumb line at 12.48 miles?

[Seaworthy Lass]

Originally Posted by goboatingnow

Seaworthy heres the actual graphical solution

The track method time is 6 hours 24

The CTS method is 5 hours 7 ( approaching the destination at a rate of 12.5/16)

I have no idea how you arrived at these times.

For method a) I agree with your graph, but not the time you calculated.
I see you have the hovercraft 9.98 miles away from the destination point when the current stops after 4 hours as I did. There is 6.02 miles to go. The hovercraft is travelling at 4 miles per hour so it takes a further hour and a half.
That makes total time for method a) 5 hours 30 minutes. Where did you get 6 hours 24 minutes?

For method b) I disagree both with your graph and your time and can't see how you have the hovercraft steering a constant course the whole way.

[Andrew Troup]

Actually, Dockhead, CaptForce did (with prompting from me) supply precise current information.

It varies linearly as a function of distance from midchannel, where it is four knots, to each 'bank', where it is zero. After initially feeling that it was a bit contrived, I now think it's actually a very interesting scenario ... more akin to a Gulf Stream situation where the phenomenon is not tidal, but essentially geographical.

I've already indicated I've done the analysis based on that exact scenario, including calculating the position of the boat. I had to assume a distance from "bank to bank" of 20 nm, though, given that CaptForce has resolutely kept that information to himself.

If anyone's interested I'll work up a post, but at present it looks as though it would sink in a welter of scenarios; nobody seems interested in anyone else's scenario, so they generate a fresh one of their own. It's like a conversation where no-one's listening very much, because talking is so much more satisfying.

I'm not sure if CaptForce is amenable to persuasion, but I am pretty sure the only way with any chance of success is to address him exactly on his own terms, because he has demonstrated that he is impervious to an unlimited number of excellent examples by others, showing his central contentions to be flawed.

This might seem a lot of trouble to go to, but AFAIK he's the only holdout. And it was an interesting and instructive scenario to dissect, right down to calculating the track across the ground.

And I for one owe CaptForce a debt of gratitude: I know now that if I'm crossing such a current, and I'm well upstream of the rhumbline midway across, it's not business as usual, in fact I've screwed up my CTS.

Conversely if I'm planning such a crossing and the chart shows a shoal (say) just downstream of the midpoint, I would no longer blithely assume that my constant-heading optimum track will loop upstream,carrying me well clear.

I think the "koolaid" of current sailing is the contention that ground track is just a distraction. The trap we all fell into with this scenario seems to me to demonstrate that we've all drunk plentifully of that koolaid.

At the time of first encountering the topic, ignoring the ground track is a useful mindset, but at some point we have to trust ourselves to have officially understood the importance of the concept of sailing through the water, and reintroduce the equally important realities of sailing across the ground: it's what the chart reflects, and it's what most of the dangers are attached to.

[goboatingnow]

Sorry yes the math in (a) is incorrect, since we have no further info

As to (b) you dont join the course to the destination as its a 4 hour vector, ie the craft is only subject to a deviation for part of its course. and yes it has to cover 3.5 miles at 4 knots as there is no futher info, joining the vector to the destination would indicate you increased speed

[Seaworthy Lass]

Originally Posted by Andrew Troup

.......
If anyone's interested I'll work up a post, but at present it looks as though it would sink in a welter of scenarios; nobody seems interested in anyone else's scenario, so they generate a fresh one of their own. It's like a conversation where no-one's listening very much, because talking is so much more satisfying.

I'm interested and listening LOL. I just haven't had a chance to go through Dockhead's last long post yet. I am working on going through the example GoBoating created with him.
Only so much multitasking a person can do .

Originally Posted by Andrew Troup

I'm not sure if CaptForce is amenable to persuasion, but I am pretty sure the only way with any chance of success is to address him exactly on his own terms, because he has demonstrated that he is impervious to an unlimited number of excellent examples by others, showing his central contentions to be flawed.

I think CaptForce is amenable to persuasion if he recognises the flaw in his assumptions about addition of vectors. I really really hope my demonstration of that may have been understood and accepted. I don't think he is amenable to any more repetition about the principle that the boat needs to travel a straight line through the water for the quickest journey. That is just falling on deaf ears.

Originally Posted by Andrew Troup

And I for one owe CaptForce a debt of gratitude: I know now that if I'm crossing such a current, and I'm well upstream of the rhumbline midway across, it's not business as usual, in fact I've screwed up my CTS.
Conversely if I'm planning such a crossing and the chart shows a shoal (say) just downstream of the midpoint, I would no longer blithely assume that my constant-heading optimum track will loop upstream,carrying me well clear.

+1 for me too .
The chartplotter would have shown me otherwise, but then I would have been concerned when the track was not being produced as expected. The awareness of dangers along the track is critical too. All these calculations and vectors drawings have really improved my visualisation of what is happening relative to the ground when a constant heading is chosen .

[Seaworthy Lass]

Originally Posted by goboatingnow

Sorry yes the math in (a) is incorrect, since we have no further info

As to (b) you dont join the course to the destination as its a 4 hour vector, ie the craft is only subject to a deviation for part of its course. and yes it has to cover 3.5 miles at 4 knots as there is no futher info, joining the vector to the destination would indicate you increased speed

Constant heading to steer means just that, a constant heading from departure to arrival, not until the current/crosswind runs out. It is slower to have more than one compass heading over a journey, I thought we had agreed on that.
My calculations were for a constant heading over the entire course you set. The time taken for a constant compass heading all the way was 4 hours and 43 minutes.

[Dockhead]

Originally Posted by Andrew Troup

Actually, Dockhead, CaptForce did (with prompting from me) supply precise current information.

It varies linearly as a function of distance from midchannel, where it is four knots, to each 'bank', where it is zero. After initially feeling that it was a bit contrived, I now think it's actually a very interesting scenario ... more akin to a Gulf Stream situation where the phenomenon is not tidal, but essentially geographical.

I've already indicated I've done the analysis based on that exact scenario, including calculating the position of the boat. I had to assume a distance from "bank to bank" of 20 nm, though, given that CaptForce has resolutely kept that information to himself.
.

I somehow missed that. I agree with Dave that we should try to all model the same scenarios, so I'll model Capt Force's.

It will be worse for the GPS track method. With more "granularity", like in my analysis, the GPS track method looks better - because every hour is optimized as a micro constant heading passage. More frequent change of heading will reduce VMG towards waypoint.

You understand, Andrew, that 20 miles "bank to bank", is incompatible with the 4 hour assumption. With average 2 knot current the longest distance "bank to bank" possible in 4 hours is 18.33 miles.

[Andrew Troup]

Originally Posted by Dockhead

I somehow missed that. I agree with Dave that we should try to all model the same scenarios, so I'll model Capt Force's.

It will be worse for the GPS track method. With more "granularity", like in my analysis, the GPS track method looks better - because every hour is optimized as a micro constant heading passage. More frequent change of heading will reduce VMG towards waypoint.

You understand, Andrew, that 20 miles "bank to bank", is incompatible with the 4 hour assumption. With average 2 knot current the longest distance "bank to bank" possible in 4 hours is 18.33 miles.

Thanks, Dockhead, I await your modelling with great interest.

My granularity was 0.1nm strips: because I used a spreadsheet, it was very little extra work. I tried coarser and finer increments before settling on that: Finer didn't achieve much except make the spreadsheet harder to 'patrol' ...
Coarser granularity (eg 0.5nm) misreported the times for both tactics, overstating them by four or minutes in somewhat less than five hours. If you choose 0.5nm, I can revert to the coarser analysis for comparison purposes.

- - - - -

I'm sorry I forgot to point out that in asking for a distance across in nautical miles, I was implicitly rejecting the scenario which stated the distance in terms of time, which makes no sense for several reasons, in addition to the obvious one:

Firstly (as an earlier post of yours suggests) it presupposes the truth of the proposition it sets out to prove, namely that both boats will arrive in the same time.
Secondly the currents are specific to geographic location, not time.

So my analysis works out the time of transit for each boat for the current speeds and boatspeeds given, and I was forced to assume a distance across.

This distance does not matter to the question, but I hope you'll choose the same distance I did so our results are comparable.

[Seaworthy Lass]

Originally Posted by Dockhead

......
I’m going to make up the essential missing data. I will calculate every half hour, and I’m going to use the following streams:

0 hour - 0:30 0 knots
0:30 – 1:00 2 knots
1:00 – 1:30 2 knots
1:30 – 2:00 4 knots
2:00 – 2:30 4 knots
2:30 - 3:00 2 knots
3:00 – 3:30 2 knots
3:30 – 4:00 0 knots

OK, average 2 knots, as per the assumptions. The current is perpendicular to the rhumb line. Let’s say BTW is 180 and current runs E to W.

First, let’s calculate a valid CTS. Without doing this correctly the whole exercise is useless. In this case, it’s a simple calculation – we simply add up the set of the current, which amounts to 8 miles to the W. So we need a CTS which will put us 8 miles further E.

Next, we will need to know the distance, which Capt Force did not give us. The distance over ground is not 20 miles (sorry Andrew), because we cannot make 5 knots made good towards the waypoint with a unidirectional current across our path. Average VMG towards the waypoint must be less than boat speed. I can calculate how far I can get in 4 hours by solving a right triangle with hypotenuse of 20 miles, and C side of 8 miles – simples. So the rhumb line has a length of 18.33 miles, and we have a distance to waypoint which can be sailed in the presumed 4 hours at the presumed 5 knots, provided we sail the correct path.

We also have my course correction 23.58 degrees, and angle of the apex of the triangle. Subtract that number from 180, the bearing to waypoint, and we have our CTS.

So now we have a complete picture of the constant heading passage:
* CTS constant heading of 156.42
* Distance made good towards waypoint 18.33 miles
* Distance covered through water 20 miles
* Time on passage 4 hours.

Now let’s calculate the GPS track passage. Since this vessel is moving in a straight line over the ground, this calculation is much easier if we do it in a ground-referenced manner.

0 hour – 0:30 HDG 180 SOG 5.0 Distance covered over ground 2.5
0:30 – 1:00 HDG 156.42 SOG 4.58 Distance covered over ground 2.29
1:00 - 1:30 HDG 156.42 SOG 4.58 Distance covered over ground 2.29
1:30 – 2:00 HDG 126.87 SOG 3.0 Distance covered over ground 1.5
2:00 – 2:30 HDG 126.87 SOG 3.0 Distance covered over ground 1.5
0:30 – 1:00 HDG 156.42 SOG 4.58 Distance covered over ground 2.29
1:00 - 1:30 HDG 156.42 SOG 4.58 Distance covered over ground 2.29
0 hour – 0:30 HDG 180 SOG 5.0 Distance covered over ground 2.5

So in four hours the GPS track boat has sailed the same 20 miles through the water as the constant heading boat (obviously -- 4 hours * 5 knots), but has only covered 17.16 miles over ground along the rhumb line, which is its ground track. So after four hours, it has not arrived yet – it has another 1.17 miles to go......

Dockhead, I made nearly the same assumptions as you (the current every half hour was slighly different) and presented the data in post #371:

Originally Posted by Seaworthy Lass

A constantly changing current is an infinite number of vectors, so I have made the current constant for eight half hour blocks (as per your example the direction of the current is constant and from 90 degrees to the course). It varies as follows:
1st half hour: 1 knot
2nd half hour: 1 knot
3rd half hour: 2 knots
4th half hour: 4 knots
5th half hour: 4 knots
6th half hour: 2 knots
7th half hour: 1 knot
8th half hour: 1 knot
It averages 2 knots over the four hours as you specified.
For the rest of the journey however long it is, there is zero current

Dockhead wanting to follow a constant heading calculates there is an average of 2 knots current. If he aims to end up on the "rhumb" line (the initial destination point) after the four hours of current, his distance reached on the line is the square root of (5x4 squared minus 2x4 squared) which equals the square root of 336 which is 18.33 nm.

Meanwhile you change course ever half hour for the four hours and your progress along the straight line to the destination is as follows:
1st half hour: square root of (5 squared minus 1 squared)/2 = 2.45 nm
2nd half hour: square root of (5 squared minus 1 squared)/2 = 2.45 nm
3rd half hour: square root of (5 squared minus 2 squared)/2 = 2.29 nm
4th half hour: square root of (5 squared minus 4 squared)/2 = 1.5 nm
5th half hour: square root of (5 squared minus 4 squared)/2 = 1.5 nm
6th half hour: square root of (5 squared minus 2 squared)/2 = 2.29 nm
7th half hour: square root of (5 squared minus 1 squared)/2 = 2.45 nm
8th half hour: square root of (5 squared minus 1 squared)/2 = 2.45 nm

total distance travelled along the straight chartplotter line over four hours = 17.38 nm

So CaptForce, you are 0.95 nm behind after 4 hours if you change course every half hour to follow the straight line on the chartplotter.

Dockhead, our figures were slightly different for the boat following a straight chartplotter course as I selected a slighly different gradient to the current although it still averaged 2 knots with a peak in the centre .

I made my calculations before CaptForce posted that the current was linear, so these figures do not actually follow the parameters he set

[conachair]

Can I join in?

Hardly ever play around with numbers these days, but here goes.

Aiming for a vector analysis with 360 steps to calc the actual distance through the water if you keep a constant course over the ground.

Distance over the ground=G, distance through the water=B, tidal set=T.

T
_____
\ |
\ |
B \ | G
\ |
\ |


So if we assume the boat juggles the throttle and heading so that course over the ground is always due north and speed over

ground is always 5Kn, then in one hour G=5

Next assume a 12 hour passage due north through a perfect sinisoidal tide which takes exactly 12 hour to cycle setting east

then west.

And assume that the maximum flow is 1Kn.

And X is the angle between B and g.
As T is 1 at maximum flow, sin(x)=T (distance of tidal set over one hour)
So b^2=t^2+G^2
B=sqrt(sin(x)^2+G^2) (G is always 5)

Sticking all that into a spreadsheet into 1 degree increments.. for x = 0 to 180 back down to 0

Total distance over the ground is 5 x 12=60
Distance G for each degree G=60/360

Summing 360 increments, it comes out at 60.76224Nm distance through the water instead of 60Nm.
Or if the tide gets up to 2Kn it's 62.499Nm

Not much in it really

Does any of that make sense or do the numbers sound right?

Calculus next, that really could take a while

Edit. Look what to formatting did to my nice triangle , hope you get the drift

[Bash]

Originally Posted by Dockhead

Am I the only person here who, rather than inventing some method based on rusty college trig, actually learned how to do it? It's very easy to do it with a water-based reference, which is the way they teach you to do it.

I used to do it that way, but these days I program my plotter to project both a course and a heading arrow, which shows my vector angle. Better yet, if I keep the course arrow on the mark, I am usually able to fetch it.

As far as all the trig is concerned, I still use a nautical slide rule and a course protractor. Anyone still using tools like that?

[Astrid]

As far as all the trig is concerned, I still use a nautical slide rule and a course protractor. Anyone still using tools like that?

of course