Question for any instructor/examiner?

[Spirit (of Glenans)]

First mate is about to do her DS theory. Her instructor has told her that any innaccuracy in plotting a fix will lead to fail. in revising ahead of tomorrow's exam, she has just obtained a 100 yard "cocked hat" and is now convinced she will fail.

I am trying to say that a 100 yard margin of error (from an object 3.5m away) is acceptable. It's a fraction of a degree, the thickness of the lines on the plotter. A big cocked hat is a prompt that you made a significant plotting error, then you check again.

More importantly in real world, if visibility is such that you can see 3.5 miles, 100 yards in open water (which you've covered several times over in the time taken between fixes) is simply neither here nor there. If you are within 100 yards of an unmarked hazard and you don't know it's there or which way lies safe water, a far-off fix isn't much good to you.

Am I right? Or is the instructor right to indicate that all 3 lines must actually cross? Trying to help 1st mate through something she is finding quite tough.

To drag you all back, however unwillingly, to the OP's actual question:
There is a difference between what happens in the real world, (or a Practical Assessment for DS), and what happens in answering a question on a DS Theory paper. In the former case, a properly executed Three-Point Fix will give an approximation of the position of the boat to within perhaps 100 yds, or so, which was perfectly acceptable and workable in the days before GPS became widely available.
In the latter case the bearings of the Charted Objects are GIVEN in degrees M, along with figures for the Variation and Deviation. If the correct mathematical workings are applied, a very precise and predictable result in degrees True is produced, which results in a "Cocked Hat" of a particular size and shape. This is what the Assessor will be looking for. Whether or not the boat is inside or outside the triangle is absolutely irrelevant in this discussion.
To further bolster this point, in the real world, the Rule of Twelfths is enough for Tidal Calculations whereas a Theory assessor will be requiring an answer correct to two decimal places.

 

[Iliade]

Take the bearings several times, then filter the data to produce the location you desire.

What could possibly go wrong?

P.S. Small triangles with a certain amount of error in the initial measurements have smaller areas of probability of position than bigger triangles. Larger triangles with the same amount of error have larger areas of probability of position outside of them than do the smaller triangles. Thus if you subtract the area where the probability of position is less than, say, 0.5 from the total area outside of the triangles you end up with an expression proving that God must exist because there is evidence of her sense of humour...

 

[Iliade]

in the real world, the Rule of Twelfths is enough for Tidal Calculations whereas a Theory assessor will be requiring a answer correct to two decimal places.

But then one should add, or subtract, a healthy quarter fathom (~0.0005 km) or more to account for the storm surge you didn't include in the original calculation ;0)

 

[Sharky34]

Read my post again, I was talking about an area outside the perimeter of the plotted triangle.

"As your cocked hat gets bigger, that area gets bigger"

 

[[163233]]

"As your cocked hat gets bigger, that area gets bigger"

The point is that each bearing line could be out to either side.
The inside-the-cocked-hat idea presumes the error to be towards the centre of the hat, but this isn't true.

 

[penberth3]

"As your cocked hat gets bigger, that area gets bigger"

You've selectively quoted one sentence. My previous sentence refers to the area outside the triangle , "that area" means the area outside the triangle.

 

[Sharky34]

You've selectively quoted one sentence. My previous sentence refers to the area outside the triangle , "that area" means the area outside the triangle.

Not selective quoting at all, simply quoting what you posted.
PS if the triangle is getting bigger, then the area outside it must be getting smaller, unless the globe has changed in area.

 

[penberth3]

Not selective quoting at all, simply quoting what you posted.
PS if the triangle is getting bigger, then the area outside it must be getting smaller, unless the globe has changed in area.

Time has come to ignore, I think.

 

[RJJ]

The point is that each bearing line could be out to either side.
The inside-the-cocked-hat idea presumes the error to be towards the centre of the hat, but this isn't true.

No it doesn't. Starting with the assumption (1) that you are on or near each line, with equivalent probability of error each side. This assumes the fix is technically correctly plotted i.e. the potential error is down to the thickness of the pencil. If that assumption isn't true, the whole fix process is a waste of time; it is simply the formal statement of the bearing line being our best effort to represent our bearing over the ground. If the error wasn't 50/50 either side, we'd be doing it differently.

If the triangle is very small, it's unlikely you are in it. The bit outside the triangle is still quite close to all three lines. At the limit, the triangle is a pinprick and the probability tends toward zero. result (A).

As the triangle grows, then the area (numerator) inside it grows. The area (denominator) in which you might possibly be (whether with 95 or 99 or 99.99999 pc confidence before starting your fix) does not grow, nor is it infinite (assumption 2). Per comment about Bembridge and Portsmouth. Before you even start, you know you are not in Cherbourg or the Southern Ocean. If your triangle is the size of the Eastern Solent and approaches, it's certain you are in it, albeit that fact is of little use. At the limit, if your triangle is 100 pc of the possible area, the probability tends toward 100pc. Result (B).

Mathematically, the Open University problem is irrelevant as it is based on an infinite area (counter to assumption 2), and entirely unknown position i.e. entirely random across the chart and with the three lines independent of each other (counter to assumption 1).

If you are outside a very large triangle then you might be on or very close to both line A and line B. But if you are at the apex of line A and line B, you are a long way from line C. Under assumption (1) the distance to line C represents diminishing probability; as the triangle grows, that probability is pulling your reasonable estimate of your position within its boundaries.

It's not a case of, as your post said "presuming" that the error lies in the direction of the centre. It stems directly from assumption (1) that the error is increasingly likely to be in the direction of line C as the triangle becomes larger. If that's not right, the whole process of taking a fix is wrong.

This is the mathematically formal representation of the problem. If you are still debating it, do so formally - show either that my assumptions (1) and (2) are wrong or that the logic linking those assumptions to results (A) and (B) is invalid.

 

[[163233]]

No it doesn't. Starting with the assumption (1) that you are on or near each line, with equivalent probability of error each side. This assumes the fix is technically correctly plotted i.e. the potential error is down to the thickness of the pencil. If that assumption isn't true, the whole fix process is a waste of time; it is simply the formal statement of the bearing line being our best effort to represent our bearing over the ground. If the error wasn't 50/50 either side, we'd be doing it differently.

If the triangle is very small, it's unlikely you are in it. The bit outside the triangle is still quite close to all three lines. At the limit, the triangle is a pinprick and the probability tends toward zero. result (A).

As the triangle grows, then the area (numerator) inside it grows. The area (denominator) in which you might possibly be (whether with 95 or 99 or 99.99999 pc confidence before starting your fix) does not grow, nor is it infinite (assumption 2). Per comment about Bembridge and Portsmouth. Before you even start, you know you are not in Cherbourg or the Southern Ocean. If your triangle is the size of the Eastern Solent and approaches, it's certain you are in it, albeit that fact is of little use. At the limit, if your triangle is 100 pc of the possible area, the probability tends toward 100pc. Result (B).

Mathematically, the Open University problem is irrelevant as it is based on an infinite area (counter to assumption 2), and entirely unknown position i.e. entirely random across the chart and with the three lines independent of each other (counter to assumption 1).

If you are outside a very large triangle then you might be on or very close to both line A and line B. But if you are at the apex of line A and line B, you are a long way from line C. Under assumption (1) the distance to line C represents diminishing probability; as the triangle grows, that probability is pulling your reasonable estimate of your position within its boundaries.

It's not a case of, as your post said "presuming" that the error lies in the direction of the centre. It stems directly from assumption (1) that the error is increasingly likely to be in the direction of line C as the triangle becomes larger. If that's not right, the whole process of taking a fix is wrong.

This is the mathematically formal representation of the problem. If you are still debating it, do so formally - show either that my assumptions (1) and (2) are wrong or that the logic linking those assumptions to results (A) and (B) is invalid.

Yep.

Though assuming you are inside the cocked hat is presuming the error lies towards the centre.

 

[RJJ]

Yep.

Though assuming you are inside the cocked hat is presuming the error lies towards the centre.

That's what we're doing every time. If the cocked hat is small, we should recognise there's strong probability we are outside. The larger it is, the more probably we are inside, but the less useful that is.

 

[[163233]]

That's what we're doing every time. If the cocked hat is small, we should recognise there's strong probability we are outside. The larger it is, the more probably we are inside, but the less useful that is.

I think we basically agree then....

 

[Elessar]

Time has come to ignore, I think.

I can’t decide.
Deliberate troll or just not very bright.
Either way your solution is wise.

 

[Elessar]

That's what we're doing every time. If the cocked hat is small, we should recognise there's strong probability we are outside. The larger it is, the more probably we are inside, but the less useful that is.

NO.
Wrong.
The probability does not change.

 

[penberth3]

That's what we're doing every time. If the cocked hat is small, we should recognise there's strong probability we are outside. The larger it is, the more probably we are inside, but the less useful that is.

Is this helpful? From Skysail Training. Because of errors in the sights, you could be anywhere inside the larger triangle formed by the dashed red lines. Larger cocked hat means larger red triangle.

 

[lw395]

That's what we're doing every time. If the cocked hat is small, we should recognise there's strong probability we are outside. The larger it is, the more probably we are inside, but the less useful that is.

Not sure that's true.
I suggest plotting a few fixes adding a systematic error for starters. A bit of deviation will give you a hat, and your position will often not be in it,

 

[Elessar]

Is this helpful? From Skysail Training. Because of errors in the sights, you could be anywhere inside the larger triangle formed by the dashed red lines. Larger cocked hat means larger red triangle.

Yep that’s an excellent diagram to illustrate the +/- error possibilities.

 

[RJJ]

NO.
Wrong.
The probability does not change.

If you guys want to continue disputing this, can I please ask you to go through the formal exposition and explain which assumption is wrong, or where the logic is wrong? Otherwise it's a pantomime, not a discussion.

The red dotted line example is fine. But that's not the point. If the black triangle grew, while the plotting error remained constant, the black triangle would occupy a growing proportion of the red triangle, and also a growing proportion of the whole sea area. Ultimately, as I stated in the formal exposition, the black triangle would approach 100pc of the possible area of sea you could conceivably be in. At that point, you are approaching 100pc certain to be inside it, but that piece of information is of diminishing value in itself.

I think you are confusing statistical concepts, not the geometry. For any size triangle (regardless of size), the single most probable point is the centre of the triangle. But there are lots of possible points. If the triangle is small, few of those possible points lie within it (small numerator), even though they include the most probable ones. If the triangle grows, the numerator grows. The denominator remains constant - the sea, or the conceivable area of sea, does not change.

 

[Elessar]

If you guys want to continue disputing this, can I please ask you to go through the formal exposition and explain which assumption is wrong, or where the logic is wrong? Otherwise it's a pantomime, not a discussion.

The red dotted line example is fine. But that's not the point. If the black triangle grew, while the plotting error remained constant, the black triangle would occupy a growing proportion of the red triangle, and also a growing proportion of the whole sea area. Ultimately, as I stated in the formal exposition, the black triangle would approach 100pc of the possible area of sea you could conceivably be in. At that point, you are approaching 100pc certain to be inside it, but that piece of information is of diminishing value in itself.

I think you are confusing statistical concepts, not the geometry. For any size triangle (regardless of size), the single most probable point is the centre of the triangle. But there are lots of possible points. If the triangle is small, few of those possible points lie within it (small numerator), even though they include the most probable ones. If the triangle grows, the numerator grows. The denominator remains constant - the sea, or the conceivable area of sea, does not change.

See the diagram posted by pen berth.

If you still don’t get it - “it’s behind you”.

“Oh no it isn’t.”

 

[RJJ]

See the diagram posted by pen berth.

If you still don’t get it - “it’s behind you”.

“Oh no it isn’t.”

Diagram seen, understood, and correctly makes the point it is supposed to make, which (per my last post) is not the point under discussion. Either you can address the correct problem, and do so formally, or not.