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OK, tell me if I am off the wall...
EDITED:The heading was 319 degrees M with a variation of 15 degrees W which would be 334 True. (The variation is taken from the True not the compass.) The boat traveled for 48 minutes at 9.9 knots. That would be about 7.9 nautical miles.( using the formula 60 D St. 60xDistance= speed x time (in minutes)through algebra, distance= speed x time (in minutes) divided by 60.) Plotting the actual coordinates on paper, and the DR course, the current would be 3 knots at 236 degrees True. r.w.landau
EDITED:The heading was 319 degrees M with a variation of 15 degrees W which would be 334 True. (The variation is taken from the True not the compass.) The boat traveled for 48 minutes at 9.9 knots. That would be about 7.9 nautical miles.( using the formula 60 D St. 60xDistance= speed x time (in minutes)through algebra, distance= speed x time (in minutes) divided by 60.) Plotting the actual coordinates on paper, and the DR course, the current would be 3 knots at 236 degrees True. r.w.landau
Corrections!
Variation 9W Deviation 0!!! My math didn't work because I grabbed the variation from the wrong chart. I'm doing the problem again.
Variation 9W Deviation 0!!! My math didn't work because I grabbed the variation from the wrong chart. I'm doing the problem again.
I found a page in my ADC chesapeake bay
chart book that matches this problem. I plotted the two GPS fixes They are 7.1 nm apart. If you travel for 48 minutes (.8hrs) at 9.9 knots you will cover 7.9 miles. Therefore the set of the current must be south east and multiplying the reciprocal of .8hrs times .8 nm equals 1.25 times .8 = 1kt drift the set must be to the south east. If the set were 275 then the distance between fix 1 and fix 2 would be greater than 7.9 nm.
chart book that matches this problem. I plotted the two GPS fixes They are 7.1 nm apart. If you travel for 48 minutes (.8hrs) at 9.9 knots you will cover 7.9 miles. Therefore the set of the current must be south east and multiplying the reciprocal of .8hrs times .8 nm equals 1.25 times .8 = 1kt drift the set must be to the south east. If the set were 275 then the distance between fix 1 and fix 2 would be greater than 7.9 nm.
Elapse time
The first fix was at 1712 and the second was at 1800, the elapsed time was 88 minutes. The time duration should be 88 minutes. The DR speed was 9.9 knots; therefore the DR distance should be 14.52 Nm on a compass heading of 319 degrees with 9 degrees of magnetic variation. Based on the Lat & Lon of first fix and the second fix, the vessel actually traveled 7.07 Nm on a heading of 317 degrees. There is no way there is only 2 knots at 275 degrees. Something doesn't add up. Fair Winds, Clyde
The first fix was at 1712 and the second was at 1800, the elapsed time was 88 minutes. The time duration should be 88 minutes. The DR speed was 9.9 knots; therefore the DR distance should be 14.52 Nm on a compass heading of 319 degrees with 9 degrees of magnetic variation. Based on the Lat & Lon of first fix and the second fix, the vessel actually traveled 7.07 Nm on a heading of 317 degrees. There is no way there is only 2 knots at 275 degrees. Something doesn't add up. Fair Winds, Clyde
I'm stupid!!! 
I know what I did wrong!!! I should not play with my computer or calculator after a couple of beers. I kept pushing buttons and entering numbers, instead of looking at what I was entering. Fair Winds, Clyde
I know what I did wrong!!! I should not play with my computer or calculator after a couple of beers. I kept pushing buttons and entering numbers, instead of looking at what I was entering. Fair Winds, Clyde
According to my chart,
the first plot is about 100 yards south of the warwick river hard aground!Edited: I am wrong. My chart stops at 37 degrees 10 minutes. Ross I reworked my plot in the area of the positions so that long and lat differences were close. I come up with 213 true and 2.3 knots. If you use dividers and adjust from the DR to the Fix,and take it directly to the latitude scale on the side of the chart you should come up with 2.3. I plotted my first set on piece of graph paper (equal lat and long). that did not allow for the mercater (spelling) corrections for long and lat. I agree with te 7.1. The position as south west ofthe DR, Dead Rec. Is the set not like a heading where they are recorded in the direction they are headed? r.w.landau
the first plot is about 100 yards south of the warwick river hard aground!Edited: I am wrong. My chart stops at 37 degrees 10 minutes. Ross I reworked my plot in the area of the positions so that long and lat differences were close. I come up with 213 true and 2.3 knots. If you use dividers and adjust from the DR to the Fix,and take it directly to the latitude scale on the side of the chart you should come up with 2.3. I plotted my first set on piece of graph paper (equal lat and long). that did not allow for the mercater (spelling) corrections for long and lat. I agree with te 7.1. The position as south west ofthe DR, Dead Rec. Is the set not like a heading where they are recorded in the direction they are headed? r.w.landau
Ok I put down the Beer.
The numbers I inputted into my calculator is DR speed 9.9 knots for 48 minutes gives me 7.92 Nm at 328 degrees from the first fix. The actual is 7.07 Nm at 317 degrees from the first fix to the second fix. I get a current of 2.1 knots at 22 degrees . It makes sense; you are heading NNW and getting pushed south. The current should be from the NNE. Fair Winds, Clyde
The numbers I inputted into my calculator is DR speed 9.9 knots for 48 minutes gives me 7.92 Nm at 328 degrees from the first fix. The actual is 7.07 Nm at 317 degrees from the first fix to the second fix. I get a current of 2.1 knots at 22 degrees . It makes sense; you are heading NNW and getting pushed south. The current should be from the NNE. Fair Winds, Clyde
It is my understanding that if the set is 180 True
you will move south even if you have no other motion. So from the first fix to the second fix we moved 6 minutes west and 5.2 minutes north. The distance from f1 to f2 is 7.1 miles northwest. But the speed of our boat is 9.9 kts for .8 hrs which equals 7.9 mile. Therefore we have been set southeast about .8 miles. The only way to account for the missing .8 mile is for a southeasterly set with a velocity of one knot.
you will move south even if you have no other motion. So from the first fix to the second fix we moved 6 minutes west and 5.2 minutes north. The distance from f1 to f2 is 7.1 miles northwest. But the speed of our boat is 9.9 kts for .8 hrs which equals 7.9 mile. Therefore we have been set southeast about .8 miles. The only way to account for the missing .8 mile is for a southeasterly set with a velocity of one knot.
Ross you are not taking into account the
the original DR (Dead Rec.) course. You have to plot the DR coarse from Fix 1 to the DR. Then plot the Fix 2. The distance and direction between the two plotted points will be the set and drift. OK asking again, is set recorded in the direction it is moving like the boat course or is it recorded like wind, the direction it is coming from? Clyde, how do you get 328 degrees from 319 M with 15 degree variation? r.w.landau
the original DR (Dead Rec.) course. You have to plot the DR coarse from Fix 1 to the DR. Then plot the Fix 2. The distance and direction between the two plotted points will be the set and drift. OK asking again, is set recorded in the direction it is moving like the boat course or is it recorded like wind, the direction it is coming from? Clyde, how do you get 328 degrees from 319 M with 15 degree variation? r.w.landau
Sometimes we can try to be too precise
Traditionally there were but thirty-two points on the compass Then they got a little bit better and could steer east by northeast one quarter east or one half east. Finally they put numbers on the card. I'll bet if our protractors just had thirty-two points some of these problems would cause a lot less worry over one or two degrees. Just try using the charts for around Martha's Vineyard lots of currents, big tides. The lat-lon for the problem you posted is just off the Virginia coast south of Cape Henry. Probably no easterly current.
Traditionally there were but thirty-two points on the compass Then they got a little bit better and could steer east by northeast one quarter east or one half east. Finally they put numbers on the card. I'll bet if our protractors just had thirty-two points some of these problems would cause a lot less worry over one or two degrees. Just try using the charts for around Martha's Vineyard lots of currents, big tides. The lat-lon for the problem you posted is just off the Virginia coast south of Cape Henry. Probably no easterly current.
CALL Seatow We're still on the rocks....
Guys, this is fun. It would be nice if this were a weekly event posted by anyone that wants to set one up, instead of putting the burden on one person. What do you think? r.w.landau
We're still on the rocks....Guys, this is fun. It would be nice if this were a weekly event posted by anyone that wants to set one up, instead of putting the burden on one person. What do you think? r.w.landau
Thanks Malcolm
Absolutely, a min of longitude does not equal 1 nm except at the equator. Lets see the correction factor at the equator (angle = 0) would be 1 the correction factor at either pole (angel = 90) would be 0 My trigonometric instincts tells me that this is a COSIN function or via math speak nm of longitude = minutes of longitude * COS("center" latitude) for small changes in latitude. The center latitude = (start lat + end lat)/2 I suspect that if you have a long route there are going to be "great circle" issues. But we should be able to ignore these for this problem. So restating the problem Time = 48 min = 0.8 hours DR course 310T DR distance 7.9 nm For the actual COG Difference in latitude between the start and end points is 36 54.8'N - 37 00.0'N = 5.2' in a northerly direction or 5.2 nm north Difference in longitude is 75 39.8W - 75 45.8'W = 6' in a westerly direction. We have to correct for the latitude so 6' longitude = 6* COS((36 54.8' + 37 00.0)/2) = 6*COS(36.957) = 4.8 nm (Malcolm, a pretty big error!!!) I still don't care how far it is as that does not need to be known for this problem For the DR course The northerly component of movement would be: 7.9*SIN(310-270) = 5.1' or 5.1 nm north The westerly component of movement would be: 7.9*COS(310-270) = 6.1' or 6.1 nm west. Since the units are in nm and not degrees of longitude I do not need to correct here The difference between the two differences (Still weird here) will be the vector we are looking for and that works out to For north and south movement we should be (DR) 5.1 nm north but we are actually 5.2 nm north so there is a northerly component speeding us up of 0.1 nm in 48 minutes. No change here. For the east west movement we should be (DR) 6.1 nm west but we are actually 4.8 nm west so there is an easterly component to the current slowing us down of 1.3 nm in 48 minutes. So the direction of the vector is north easterly and the actual value of the angle from East is; Set = INVerse TANgent(rise/run) = tan^-1(0.1/1.3) = 4.4 degrees north from east or 90-4.4= 85.6T. The set is the other direction or 85.6+180= 265.6. But since I can't even measure a single degree on my compass I'll settle for 266T Drift is still by Prognathous ((0.1^2+1.3^2))^(1/2)= 1.3 nm in 48 minutes. normalizing to 1 hour units by 1.3*48/60=1.0 knots. Set 266T Drift 1.0 knots. Arrrrgggga?!?!?!? I think that knowing the current position would be enough for me as how I got there is kinda meaningless given I'm there now and have to deal with the fact. Clearly, we (I am certainly ncluding myself) are having difficulty rapidly getting an consistent answer. Shame on me. Bad Sailor, Bad Sailor. If you don't shape up I'm going to ground your boat. This is yet another example of how safe it is going out on the wide blue ocean.
Absolutely, a min of longitude does not equal 1 nm except at the equator. Lets see the correction factor at the equator (angle = 0) would be 1 the correction factor at either pole (angel = 90) would be 0 My trigonometric instincts tells me that this is a COSIN function or via math speak nm of longitude = minutes of longitude * COS("center" latitude) for small changes in latitude. The center latitude = (start lat + end lat)/2 I suspect that if you have a long route there are going to be "great circle" issues. But we should be able to ignore these for this problem. So restating the problem Time = 48 min = 0.8 hours DR course 310T DR distance 7.9 nm For the actual COG Difference in latitude between the start and end points is 36 54.8'N - 37 00.0'N = 5.2' in a northerly direction or 5.2 nm north Difference in longitude is 75 39.8W - 75 45.8'W = 6' in a westerly direction. We have to correct for the latitude so 6' longitude = 6* COS((36 54.8' + 37 00.0)/2) = 6*COS(36.957) = 4.8 nm (Malcolm, a pretty big error!!!) I still don't care how far it is as that does not need to be known for this problem For the DR course The northerly component of movement would be: 7.9*SIN(310-270) = 5.1' or 5.1 nm north The westerly component of movement would be: 7.9*COS(310-270) = 6.1' or 6.1 nm west. Since the units are in nm and not degrees of longitude I do not need to correct here The difference between the two differences (Still weird here) will be the vector we are looking for and that works out to For north and south movement we should be (DR) 5.1 nm north but we are actually 5.2 nm north so there is a northerly component speeding us up of 0.1 nm in 48 minutes. No change here. For the east west movement we should be (DR) 6.1 nm west but we are actually 4.8 nm west so there is an easterly component to the current slowing us down of 1.3 nm in 48 minutes. So the direction of the vector is north easterly and the actual value of the angle from East is; Set = INVerse TANgent(rise/run) = tan^-1(0.1/1.3) = 4.4 degrees north from east or 90-4.4= 85.6T. The set is the other direction or 85.6+180= 265.6. But since I can't even measure a single degree on my compass I'll settle for 266T Drift is still by Prognathous ((0.1^2+1.3^2))^(1/2)= 1.3 nm in 48 minutes. normalizing to 1 hour units by 1.3*48/60=1.0 knots. Set 266T Drift 1.0 knots. Arrrrgggga?!?!?!? I think that knowing the current position would be enough for me as how I got there is kinda meaningless given I'm there now and have to deal with the fact. Clearly, we (I am certainly ncluding myself) are having difficulty rapidly getting an consistent answer. Shame on me. Bad Sailor, Bad Sailor. If you don't shape up I'm going to ground your boat. This is yet another example of how safe it is going out on the wide blue ocean.
Because I earn my living as a carpenter
I tend to draw diagrams. For one reason you can see the error of your ways and the other is they make more sense for me. Bill, 1 min of latitude at 37 north equals.7986355 miles if you want to be precise.( it is the cosine i=of the latitude angle) so 6 min west equals about 4.8 miles combined with the 5.2 miles north gives a course of about 317 and a distance of 7.07 miles. If you plot it out with engineers scale and protractor you get the same answers. Then without any stress or strain you draw a line from the DR point to the GPS fix and you have the slope of the set and the distance between the DR position and the GPS fix is the distance of the drift in .8 hours. I can see a good reason for being able to work this out. There are several major ocean currents in the world and knowing the set in one hour will allow a correction in the course steered before we have traveled too far.
I tend to draw diagrams. For one reason you can see the error of your ways and the other is they make more sense for me. Bill, 1 min of latitude at 37 north equals.7986355 miles if you want to be precise.( it is the cosine i=of the latitude angle) so 6 min west equals about 4.8 miles combined with the 5.2 miles north gives a course of about 317 and a distance of 7.07 miles. If you plot it out with engineers scale and protractor you get the same answers. Then without any stress or strain you draw a line from the DR point to the GPS fix and you have the slope of the set and the distance between the DR position and the GPS fix is the distance of the drift in .8 hours. I can see a good reason for being able to work this out. There are several major ocean currents in the world and knowing the set in one hour will allow a correction in the course steered before we have traveled too far.
Ross, 1 minute of Latitude is 1 NM anywhere.
It is longitude that is manipulated to make charts lay flat. r.w.landau
It is longitude that is manipulated to make charts lay flat. r.w.landau
R.W. Ok . The east -west distance between the
meridans changes. But it also depends on the projection used for making the chart. It is one mile from 37 degrees north to 37d01m north but it is .798 miles from 75w to 75d01m west at 37 north on the surface of the earth. Edit to add: I just measured a 37 north chart, the distance between the lines of latitude are elongated by the reciprocal of the cosine of the latitude. The diagram drawn on the chart or drawn on paper with the allowance for the difference will yield the same results.
meridans changes. But it also depends on the projection used for making the chart. It is one mile from 37 degrees north to 37d01m north but it is .798 miles from 75w to 75d01m west at 37 north on the surface of the earth. Edit to add: I just measured a 37 north chart, the distance between the lines of latitude are elongated by the reciprocal of the cosine of the latitude. The diagram drawn on the chart or drawn on paper with the allowance for the difference will yield the same results.
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