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A MB-2A in available on ebay now
http://search.ebay.com/search/search.dll?from=R40&_trksid=m37&satitle=mb-2a&category0=

is are E6-Bs


http://search.ebay.com/search/search.dll?from=R40&_trksid=m37&satitle=e6b&category0=

gl



On Jun 18, 11:53 pm, "Gary J. LaPook"  wrote:
> 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 50� 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 90� 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 5.74�, approximately 6� 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 45� angle on the bow of two tenths
> of the boat's speed will require a 8.13� 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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