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Re: Course to steer. Has anybody come across this little rule of thumb before?
From: Gary LaPook
Date: 2008 Jun 19, 22:55 -0700
From: Gary LaPook
Date: 2008 Jun 19, 22:55 -0700
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 --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To , email NavList-@fer3.com -~----------~----~----~----~------~----~------~--~---