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Navigation Questions. Can you do them?

15K views 121 replies 16 participants last post by  Boasun 
#1 ·
1. As we ask these questions, lets see if we can stay away from all charts but the universal chart.
2. These are for your edification, so give a try and you may suceed.

A. Convert the following true bearings to relative bearings of a ship who is steering 302°; 009°, 158°, 328°.










Hint: R - S = T
 
#2 ·
1. As we ask these questions, lets see if we can stay away from all charts but the universal chart.
2. These are for your edification, so give a try and you may suceed.

A. Convert the following true bearings to relative bearings of a ship who is steering 302°; 009°, 158°, 328°.

Hint: R - S = T
Your formula works out to:

R = T + S

However, the army.mil site says the equation is:

RB + SH = TB or

RB = TB - SH

RB = Relative Bearing
SH = Ship's Heading
TB = True Bearing

Why the discrepancy? Am I missing something?
 
#5 ·
The discrepancy is from not checking my books first before I wrote the question.
R + S = T or as you said: RB + SH = TB

Must have been a mental lapse from deminta or something...
But I did get people to start thinking...
And when I am teaching I do put errors on the board to see if my students are awake and have studied. But it wasn't suppose to happen this time.:rolleyes:
 
#6 ·
009°T is 67°R (forward the starboard beam)
158°T is 216°R (port quarter)
328°T is 26°R (off the starboard bow)

The army.mil site is correct. What you want to do with the given information is: TB - SH = RB. If the result is negative, you can either consider it an angle towards port (clockwise) or you can add 360° to normalize it to a standard 0°-360° angle.
 
#13 · (Edited)
One other:
the scale of a chart is 1:80,000. What distance at sea does a distance of 3.5 inches on the chart represent? What distance on the chart represents an actual distance of 14.3 NM?






Remember that a NM is 6076.11549 ft thought we use 6000 ft for practical purposes.
 
#24 ·
One other:
the scale of a chart is 1:80,000. What distance at sea does a distance of 3.5 inches on the chart represent? What distance on the chart represents an actual distance of 14.3 NM?
3.5 in at 1:80,000 = 3.84 nm
13.3 nm at 1:80,000 = 13.03 in

Remember that a NM is 6076.11549 ft thought we use 6000 ft for practical purposes.
Okay,
3.5 in at 1:80,000 = 3.89 nm
14.3 nm at 1:80,000 = 12.87 in
 
#15 ·
You do not need the chart.

This an example that I use to teach set and drift, CMG, SMG, CTS, Speed.

On November 18, 2003 we plan on leaving Thieves Bay Marina, Mouat Point Light (Lat 48°46.5'N Long 123°18.75'W) for Isabella Island (Lat 48°43.75'N Long 123°25.75'W). We will use the current information for Swanson Channel. The chart shows the flood direction to be 360. Using our current prediction tables, we determine that that the drift is 1 knot. Our boat speed is 5 knots. Our course to make good is 242T. We need to gauge our speed made good and our course to steer.
 
#17 ·
#22 ·
If the wind is reported as 360T, it's coming from the North. If the current (set) is reported as 360T, it's going to the North. Who decided this, and was it done just to torture me?
I also have the same question and just suppose that it's just convention. Much like a circuit diagram in EE where electrical current is shown going from the positive to the negative post whereas in reality the electrons flow from negative to postive. Not sure how all this "conventional wisdom" got started.
 
#19 ·
The set is 360. You are being pushed north. The wind is not a factor.
 
#20 ·
Sorry -- I wasn't saying the wind was a factor. I was just asking why, when set is reported, it's the direction I'm being pushed. But when wind direction is being reported, it's the other way around. So if the wind is 360T, it's *from* the north, but when current is 360T, it's *to* the north.
 
#21 ·
Last I checked, wouldn't that be 0˚T not 360˚T...
 
#25 ·
You are closer than me.

Vol 5 of Canadian Tides and Currents uses 000 for currents flowing to True North.

I will have to fix that.
 
#23 · (Edited)
The winds are named from where they are coming from because when you face into the wind you are looking in that direction. Thus the North wind is from the North because that is the direction you are facing.
About everything else is the direction they are going; Currents, Your Boat and so forth.
This is not that hard to understand... Not hard at all.

BTW: My post #13 here has not been answered yet... It is about figuring distance with a known chart scale.
 
#26 ·
Boasun and others

Re: Chart scales.

I use Canadian Chart 3462 in my navigation class; it is 1:80,000 chart. However, the 1:80,000 is measured at 49˚ 30'N which is not on the chart. The top of the chart is 48˚ 54'N. As this is a Mercator projection, it seems to me that converting distances measured using the scale would not be exact.

In you view, would this be a correct interpretation?
 
#36 ·
I use Canadian Chart 3462 in my navigation class; it is 1:80,000 chart. However, the 1:80,000 is measured at 49˚ 30'N which is not on the chart. The top of the chart is 48˚ 54'N. As this is a Mercator projection, it seems to me that converting distances measured using the scale would not be exact.

In you view, would this be a correct interpretation?
I also posed this question to the Canadian Hydrographic Service. Here is their response:

Thanks for asking this question which many mariners never notice or have no idea what this is all about.

The challenge when making a map is representing a spherical object (the earth) on a plane surface (piece of paper). There are many potential way of doing this (different projections) and they all have some inherent distortions. In the CHS, for navigating purposes at this latitude, we generally use a Mercator projection since for navigation it has desirable qualities, such as a straight line plotted on a Mercator projection represent a true course or constant bearing, and the shapes of features are preserved.

When we do a series of charts we want some consistency between each chart in that series. When we derive a base projection, we need to determine where the scale is true, and we call that the scaling latitude or mid latitude. That scaling latitude is mentioned in brackets in conjunction with the scale stated on the chart. On chart 3462 you will find this as 1:80,000 (49 30'N). What this means is that the place where you find the scale true is at this latitude. The rest of this chart or any other chart in the series with this latitude will have some distortion in the scale.

The reason we choose one latitude for the series of charts is so that they will all overlap each other with consistency, allowing mariners to transfer there position from one chart to the other with ease. If each chart in a series has a different scaling latitude, then the areas of overlap would not match each other, making it difficult for use.

As I noted earlier, there are always compromises when using different projections. In this case the Mercator projection is used since it has desirable qualities for navigating, such as plotting bearings, and a series scaling latitude is used since it allows easy transfer of positions between charts in that series.

To answer your question directly, nowhere on chart 3462 would the scale be true, though it is close. It is only accurate at the latitude which it is stated, and in this case it would be on the adjoining chart at 49 30'N.

I hope this answers your question. If I can be of more assistance, please just ask.
 
#27 ·
If the chart is 1:80,000 that means that you have 80,000 square inches representing land & water area squeezed into 1 square inch of the chart.

But when using the latitude scale on the chart for measuring distances, I have found that there can be a slight difference from near the bottom of the chart compared to the top of the chart.
 
#30 ·
Jacdale's post was spot on.
Boasun, you probably teach others this:
Chart is a flat paper representing a sphare. It is impossible to make it an accurate presentation.
Your statement is only true around one (reference) latitude. at any other latitude it is either less or more - but wrong.
Nautical charts are pretty accurate in representing angles, but not so good in representing surface areas. That is why we read distances at the closest left of right edge of the chart at about the same latitude as the distance was taken.
If we would only measure that distance and multiply it we could not get a very accurate result.
The error is neglectable for charts representing small areas, but can be significant in a general charts like Ca4700 or US13003.
 
#28 ·
How far will a vessel traveling at 16.5 knots travel in 34 minutes? In 2h 24m?

How long will it take a ship traveling at 14 knots to go 0.8 miles? To go 69 miles?
 
#29 ·
Answers in bold
How far will a vessel traveling at 16.5 knots travel in 34 minutes? 9.35nm In 2h 24m? 39.6nm

How long will it take a ship traveling at 14 knots to go 0.8 miles? 2.98 minutes To go 69 miles? 4.29 hours.
 
#32 ·
You are anchored in Tod Inlet (Lat 48° 34’N, Long 123° 28’W), planning to go to Maple Bay (Lat 48° 48’N Long123° 36’W) via Sansum Narrows. The turn at Sansum Narrows is at 1200.
The distance to Sansum Narrows from Tod Inlet is 14 miles. Assuming a speed of 5 knots, at what time should you leave Tod Inlet to make the turn?
 
#53 ·
You are anchored in Tod Inlet (Lat 48° 34'N, Long 123° 28'W), planning to go to Maple Bay (Lat 48° 48'N Long123° 36'W) via Sansum Narrows. The turn at Sansum Narrows is at 1200.
The distance to Sansum Narrows from Tod Inlet is 14 miles. Assuming a speed of 5 knots, at what time should you leave Tod Inlet to make the turn?
Depart at 0912 assuming zero currents.
 
#33 ·
And I was a Navigator for USN for about as long and it seems as if we both learned it differently. But checking bowditch I find that it is linear in any Direction.
But we both learned our data differently and sometimes it needs to be brought out so that we may refine our knowledge.
 
#34 ·
I'd point out that what Boasun said would have been correct if he had said "SQUARED" instead of "SQUARE". :D
 
#37 ·
I want to be at point A in a narrow channel at slack water, but I have no current tables. I know that high tide is at 0300. At what time should I be at point A and why is slack water not at the high tide mark?
First some terminology: slack water actually refers to a lack of current. Standing water means the depth is not increasing or decreasing.

The reason slack water does not coincide with standing tide is that the horizontal motion of water is due to its being accelerated by differential tides. The point of highest tide moves as the day goes on. Say the tide is flooding and it's almost high tide. But suppose high tide upstream of me happens later in the day, even if only a few minutes later. Obviously the current will have to keep flooding, even when it's high tide where I am, because the water level upstream has to keep rising.

In the given example, the exact amount of time difference between high tide and slack current depends on the upstream volume difference between the time of high tide at A, and the farthest upstream high tide time. If there's a big reservoir upstream of you, it will take a while after high tide before the water at A is slack, but if you're just a mile from the head of an inlet, slack water will coincide almost exactly with stand.
 
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