gps speed question

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Little Five

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If, according to the GPS I am making 6 knots SOG but have a 3 knot cross tide on my beam, how much of the 6 knots is forward speed as against sideways slip due to the tide?
 
The vectors are added together arithmetically

If you drifted your SOG (no wind) would be 3kts but not in the desired direction.

Obviously if you tuned into the current and stood still (SOG=0) you would need to see 3kts on your log (if working!)

So you sail or motor, with the rudder pointing you higher than your next wpt so you describe one of those lovely triangle plots from DS night classes.

I bet this doesn't help.. and it's an a.m. not midnight posting too....:confused:

Nick (Rivendell)
 
If its EXACTLY on the beam, then see above. The trouble is tidal current is very rarely exactly on the beam and a little different either f'wd or aft makes a big difference to the answer as to what you are actually doing 'through the water'. It's one of the reasons we like to have a log as well as a GPS...
 
Little Five said:
If, according to the GPS I am making 6 knots SOG but have a 3 knot cross tide on my beam, how much of the 6 knots is forward speed as against sideways slip due to the tide?
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If everything remained constant you would indeed travel six miles in one hour but of course that won't happen :)

If the cross tide is genuinely at 90 degrees to your track then you will actually have a speed thru the water in excess of 6 knots. If the cross tide is say 30 degrees forward of your track then you will actually being doing in the region of 7.5 knots thru the water, if the cross tide is 30 degrees aft of your track then you will only be doing about 4.5 knots thru the water!

Now where the fun really starts is if you perceive the cross tide to be on the beam but in practice it is forward or aft by 60 degrees...........in this case your boat speed thru the water may be as high as 9 knots or as low as 3 knots..........you see where I'm coming from!!

Chox :)
 
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I just used Pythagorus. They hypotenuse is the SOG at 6 kts, the small sid eof the triangle is the 3 kts, the long side is your water speed. As others have stated this right angle triangle is a simplistic model.

Simplistic yes, but very useful for pilotage situations where a passage length is measured in minutes. It can be a very accurate way of calculating water track if the tide current or river current is known reliably or course to steer. Of course pythagorus is a special case involving right angles but the other trigonometry rules are simple grasp to.

Google them.
 
Little Five said:
I am interested in how to do the calculation if, say your log isn't working or you suspect it is giving false readings.
Click to expand...

In this case you can just use a bit of good old Pythagoras*, as we are looking at a right angled vector diagram (triangle).

The hypoteneuse is the actual SOG, 6 knots
One side is the tidal set OG, 3 knots
The third side is your unknown forward SOG

*In a right angled triangle,The square on the hypoteneuse is equal to the sum of the squares on the other two sides

So: SOG ^2 = 6^2 - 3^2 = 36 -9 = 27

and SOG is the square root of 27, which is 5.1961524....

Hope this helps

;)
 
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Little Five said:
I am interested in how to do the calculation if, say your log isn't working or you suspect it is giving false readings.
Click to expand...

I use a 'rule of thumb' from my day job (flying) however you do need to know your boat speed to apply it.


For practical purposes if the current is up to 30 degrees off the nose then the current is ALL against you

If the current is 60 degrees off the nose then the current is HALF against you and ALL on the beam.

This comes from a 'clock code' that substitutes once around a clock face with degrees and assumes any thing more than 60 degrees is MAX.

SO.......... if current is known to be say 6 knots ......then

On the nose = 6 knots against you (max) and NO cross current

30 degrees off the nose still equals 6 knots and barely any cross current

45 degrees off the nose = 4.5 knots against you and 4.5 knots cross current

60 degrees off the nose = barely any knots against you and 6 knots cross current

These figures are generally accurate until 60 degrees and then maybe about 0.8 kt pessimistic

Does this make any sense :)
 
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chockswahay said:
I use a 'rule of thumb' from my day job (flying) however you do need to know your boat speed to apply it.


For practical purposes if the current is up to 30 degrees off the nose then the current is ALL against you

If the current is 60 degrees off the nose then the current is HALF against you and ALL on the beam.

This comes from a 'clock code' that substitutes once around a clock face with degrees and assumes any thing more than 60 degrees is MAX.

SO.......... if current is known to be say 6 knots ......then

On the nose = 6 knots against you (max) and NO cross current

30 degrees off the nose still equals 6 knots and barely any cross current

45 degrees off the nose = 4.5 knots against you and 4.5 knots cross current

60 degrees off the nose = barely any knots against you and 6 knots cross current

These figures are generally accurate until 60 degrees and then maybe about 0.8 kt pessimistic

Does this make any sense :)
Click to expand...

Yes it does. Brilliant and thanks to you and all who contributed.
The pythagorus stuff makes me feel that I am back at school and the "not listening" default setting in my brain kicks in:o
 
Little Five said:
Yes it does. Brilliant and thanks to you and all who contributed.
The pythagorus stuff makes me feel that I am back at school and the "not listening" default setting in my brain kicks in:o
Click to expand...

You are welcome. :)

In reality a best guess is usually good enough and that is what I suspect most of us do. It might be OK for aviators as well, but I don't suppose it is what the MCA or Yachtmaster examiners want to hear.

Does anyone actually draw tidal vector diagrams, outside of the classroom?
 
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