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  1. #1
    Little Five is offline Registered User
    Location : Co.Dublin, Ireland
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    Default gps speed question

    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?

  2. #2
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    Quote Originally Posted by Little Five View Post
    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?
    Whatever is shown on your log.
    You never know, I might be right!

  3. #3
    BlowingOldBoots's Avatar
    BlowingOldBoots is offline Registered User
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    5.19 kts
    Having time is unavoidable.

  4. #4
    Little Five is offline Registered User
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    Quote Originally Posted by BlowingOldBoots View Post
    5.19 kts
    Thanks B.O.B. How do you do the maths?

  5. #5
    john_morris_uk is offline Registered User
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    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...
    Wishing things away is not effective.

  6. #6
    Little Five is offline Registered User
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    I am interested in how to do the calculation if, say your log isn't working or you suspect it is giving false readings.

  7. #7
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    Quote Originally Posted by Little Five View Post
    I am interested in how to do the calculation if, say your log isn't working or you suspect it is giving false readings.
    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

    Last edited by JayBee; 23-02-12 at 11:17. Reason: typo
    Better roughly right than completely wrong

  8. #8
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    Quote Originally Posted by Little Five View Post
    I am interested in how to do the calculation if, say your log isn't working or you suspect it is giving false readings.
    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
    Last edited by chockswahay; 23-02-12 at 11:26.

  9. #9
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    5.196 knots, approximately.

    Better roughly right than completely wrong

  10. #10
    NickRobinson is offline Registered User
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    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....

    Nick (Rivendell)

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