Page 1 of 3 123 LastLast
Results 1 to 10 of 32

Hybrid View

Previous Post Previous Post   Next Post Next Post
  1. #1
    Join Date
    Aug 2010
    Location
    glasgow
    Posts
    1,012

    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
    Join Date
    Mar 2004
    Location
    Southampton
    Posts
    5,798

    Default

    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
    Join Date
    Aug 2009
    Location
    Scotland.
    Posts
    9,187

    Default

    5.19 kts
    All pictures in my posts are hosted on Flickr
    ⓒ BlowingOldBoots 2014

  4. #4
    Join Date
    Aug 2010
    Location
    glasgow
    Posts
    1,012

    Default

    Quote Originally Posted by BlowingOldBoots View Post
    5.19 kts
    Thanks B.O.B. How do you do the maths?

  5. #5
    Join Date
    Jul 2002
    Location
    Home near Exeter, work in Hampshire, boat in Plymouth
    Posts
    13,931

    Default

    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...
    [B][I]Wishing things away is not effective.[/I][/B]

  6. #6
    Join Date
    Aug 2010
    Location
    glasgow
    Posts
    1,012

    Default

    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
    Join Date
    Sep 2004
    Posts
    843

    Default

    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
    Join Date
    Jul 2003
    Location
    Glasgow
    Posts
    199

    Default

    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
    Join Date
    Sep 2004
    Posts
    843

    Default

    5.196 knots, approximately.

    Better roughly right than completely wrong

  10. #10
    Join Date
    Dec 2007
    Location
    Near Burton-on-Trent
    Posts
    1,391

    Default

    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)

Page 1 of 3 123 LastLast

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •