NavList:

Gary LaPook writes:

This is just another example of the standard computation done by all pilots to correct for the wind, see my post 5459 of 6/15/08 on "Canned survival problem" thread which contains an example of a 20 knot cross wind at 50º from the nose (bow) and a 120 knot TAS. This example could also represent a 12 knot boat speed and a 2 knot current.
  

I had written:

"In case you were wondering how you can solve the wind triangle with trig on the MB-2A without a vector diagram  the answer is simple, the law of sines. TAS/sin RWA = WS/sin WCA = GS/sin (RWA +/- WCA). To do this on a digital calculator first figure the Relative Wind Angle (the angle the wind is coming from compared the true course, WD - TC. Next divide the True AirSpeed  by the sine of the RWA  and save that value in a memory as you will use this constant twice.  Next divide the Wind Speed by this constant, take the inverse sine and you have the Wind Correction Angle. Finally add or subtract the WCA from the RWA, subtract if the wind is a head wind and add if a tail wind, take the sine of this angle and multiply by the constant to give you Ground Speed."

Simply substitute the current "angle on the bow" for relative wind angle ( RWA) and current speed for wind speed (WS) and ship's speed for true airspeed (TAS.) Also substitute "speed over the bottom" for ground speed (GS) and current heading correction angle for wind corruption angle (WCA.)

Using a calculator or a MB-2A shows that the rule of thumb works and produces answers within the level of possible accuracy, the accuracy of the input  data. A 90º cross wind or a current on the beam of one tenth of the TAS or boat's speed will require a correction of the heading into the current of 5.74º, approximately 6º and the speed over the bottom will be reduced by .6% (a 10 knot boat speed will be reduced to 9.94 knots over the bottom.) Current at a 45º angle on the bow of two tenths of the boat's speed will require a 8.13º heading correction and the speed over the bottom will be reduced by 15.2% (a 10 knot boat speed will result in a 8.48 knot speed over the bottom.)

You do not need a MB-2A to do this calculation, you can use a regular slide rule or a digital calculator of draw a vector diagram.

gl

Lu Abel wrote:

Tony:

I haven't heard of this rule-of-thumb, but geometrically it makes a lot
of sense -- a 6 degree course offset will take a vessel 0.1 nm sideways Ground speed, GS.
for every 1.0 nm of forward progress, which is what you'd need for a
beam current of 10% of vessel speed.   And if I work out the plot for a
bow or quarter current, 4 degrees is about right, too.

Lu Abel

Tony wrote:
  

For every 10% of tide to speed

Adjust your course by ± 6° if the tide is on your beam
                       Use ± 4° if the tide is on your bow or quarter

E.G.    If you have 2 knots of tide and your speed is 10 knots; the tide
is 20% of your speed.
    Adjust your course by ± 12° if the tide is on your beam
             ±  8° if the tide is on your bow or quarter

    

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