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## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: Set and drift
From: Gary LaPook
Date: 2011 Nov 9, 01:10 -0800
BTW, you can also do the computation on a standard slide rule. Line up the RWA on the sine scale with the airspeed on the D scale. Then look for the wind speed on the D scale and read out the WCA on the sine scale. If it is a tail wind add the WCA to the RWA and look at that angle on the sine scale and take out the GS on the D scale. The problem with a regular slide rule is that you might run off the end and have to reposition the slide and you don't have this problem with the MB-2A since it is a circular slide rule.

Doing an example with boat speeds. Speed of the boat is 6 knots, the current speed is 2 knots and the angle between the course and the current is 40 degrees. Line up 40° on the sine scale with 6 on the D scale, look at the current speed, 2, on the D scale and take out the correction angle to keep you on course of 12° on the sine scale. If the current is from behind add the 40 and 12 and look at 52° on the sine scale and take out the speed over the bottom of 7.5 K. If the current was from ahead subtract 12 from 40 and look at 28° on the sine scale and take out the speed over the bottom of 4.4 K

gl

--- On Sun, 11/6/11, Gary LaPook <garylapook@pacbell.net> wrote:

From: Gary LaPook <garylapook@pacbell.net>
Subject: [NavList] Re: Set and drift
To: NavList@fer3.com
Date: Sunday, November 6, 2011, 12:01 AM

Actually I meant an MB-2A
I show working this problem in a prior post:

http://fer3.com/arc/m2.aspx?i=105458

gl

--- On Sat, 11/5/11, Gary LaPook <garylapook@pacbell.net> wrote:

From: Gary LaPook <garylapook@pacbell.net>
Subject: [NavList] Re: Set and drift
To: NavList@fer3.com
Date: Saturday, November 5, 2011, 11:51 PM

 The easiest way to do these computations is to use the law of sines which is solved on many flight computers, such as the MB-4 that we have discussed before, since these are standard calculations for pilots.It is easy to solve on a calculator.WD = wind direction (or direction from which the current is coming)C = courseRWA = relative wind angle (or relative current angle)WCA  =wind correction angle (or current correction angle)TAS = true airspeed ( or boat speed)WS = wind speed (or current speed)GS = ground speed (or speed over the bottom)RWA = difference between course and direction that the wind (or current) is coming fromsin WCA = (sin RWA/TAS ) WSif the wind is a head wind thenGS = sin(WCA - WRA) /sin WCA if wind is a tail wind then GS = sin(WCA + WRA) /sin WCA Say the course is 90 and the current is coming from 130 so the RWA = 40TAS = 120WS = 20(sin 40 )/120 = .005 So sin WCA = .005 store in memory.WCA = 6.1sin (40 - 6.1) = .56GS = ..56/.005 = 104if the current was coming from 230 then the RWA would still be 40 but it would be a tailwind so you would add the WCA to the RWA and get 46.1, the sine of which is .72 divided by .005 makes the ground speed 134.Works for boats too.glgl--- On Sat, 11/5/11, P H wrote:From: P H Subject: [NavList] Set and driftTo: "NavList" Date: Saturday, November 5, 2011, 11:03 PMThe Sailings calculations determine the course for a vessel to follow in order to get from the point of Departure to the Destination.  This is course with respect to ground which can differ from the course to steer if the vessel is deflected sideways by currents and/or winds.  This leads to a class of "set and drift" problems described, for example, in the Dead Reckoning chapter in Bowditch.  From what I can tell these problems are typically solved graphically by plotting procedures.  As for the equivalent numerical solution the NavList archive shows discussions a few years back in which some formulae were mentioned but not given.  After that I didn't dig deeper into it; instead, I present the following.The essence of these problems is the relationship between three vectors in which the ground speed is the vessel's speed relative to the water plus the set and drift vector.  The relevant geometry occurs in 2-D, so this relationship translates into two equations for the vectors' components.  Therefore, given four pieces of information on input we can solve for the remaining two.If these two numbers both pertain to the same unknown vector (e.g. its magnitude and direction), then two Cartesian-component equations provide the solution to the problem by direct addition or subtraction of the other two (fully specified) vectors.  This is the case for spreadsheets:http://www.navigation-spreadsheets.com/dead_reckoning.html#set_and_driftground_speed.xls:Calculation of the ground speed from the current’s speed and direction (i.e. set and drift) and the vessel speed relative to the water.course_and_speed.xls:Calculation of the required vessel speed and course from the set and drift and the desired ground speed and track.The situation is a little bit more complicated for:course_to_steer.xls:Given the set and drift, the vessel's speed and the intended direction relative to ground, this spreadsheet calculates the required vessel course and the resulting ground speed.  If the vessel's speed is too small to counteract the current, an error message is displayed in row 4.in which the two unknowns are distributed between two partially known vectors.  This problem can be solved in the following steps encoded into the spreadsheet:1) Decompose the current's vector into components parallel and perpendicular to the prescribed "Sailings" (ground) direction.2) Reverse the sign of the perpendicular component; this becomes the perpendicular component of the vessel's speed with respect to water.  That way the deflecting effect of the set and drift is neutralized.3) Use the Pythagorean theorem to determine the component of the vessel's speed (w.r.t. water) parallel to the ground direction.  The known vessel's speed w.r.t water is the hypotenuse and the result of step 2) is one of the sides.4) Convert the vessel's speed's now known two components (along with the ground direction) into course to steer.5) Add the two parallel speeds' components in order to obtain the ground speed.  If this problem has two mathematically good solutions, this picks the "faster" one.This problem does not always have a solution.  If the vessel's speed w.r.t. water is less than the current's perpendicular component, the vessel is not fast enough to compensate for the deflection off course.  If the current's parallel component is negative and larger in absolute value than the parallel component of the vessel's speed w.r.t. water, then the vessel is not fast enough to progress toward its destination.  In either case, the spreadsheet zeroes out the output and displays an error message.Peter Hakel Browse Files

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