# NavList:

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

**Re: Silicon Sea Leg 72**

**From:**Sam Chan

**Date:**2001 May 29, 20:25 EDT

Bill, The issue I brought up regarding Mercator sailing was in the way I was doing it by hand. Given a departure and a destination, if I only write down the course to one decimal place, the resulting calculation of distance can be in error due to the rounding of the course. My alternate method of calculation was a way to overcome this rounding issue. In mid-latitude sailing, the general approach is to use the mean of the departure and destination latitudes as the mid-latitude. This is not exactly correct. In the book "Geometry of Navigation" by Roy Williams, he derives the equation for the middle latitude. cos (middle latitude) = Diff in Lat/DMP (DMP=difference in merdional parts) Sam Chan ----- Original Message ----- From: "Dan Hogan" <dhhogan{at}XXX.XXX> To: <NAVIGATION-L{at}XXX.XXX> Sent: Tuesday, May 29, 2001 3:35 PM Subject: Re: Silicon Sea Leg 72 > On 29 May 2001, at 15:13, Noyce, Bill wrote: > > > > > > 4) What is the Compass Course(CC)/Course-to-Steer for Aruba Gap > > > > > from the DR position? > > > > > -- ------------------------------------------------------------ > > > > > > > > Using Law of Sines, I find I need to adjust my course North > > > > about 1.5 degrees. > > > > CC = 283.4 + 1.5 + 9 = 293.9d > > > > > > Mmmm..Barely OK. Try a current vector diagram. > > > > I drew a diagram to get the orientation of the pieces right, but I > > have a hard time measuring the resulting course adjustment, because > > it's so small. Thus the attempt at a 'digital' method. > > You can make the vector diagram any convenient size. > > > > TC=283.4, Set=265.0, angle between = 18.4 degrees > > > > sin 18.4 / speed = sin adj / drift > > > > sin 18.4 / 8.5 = sin adj / 0.7 > > > > I did this with log tables before; doing it on the computer now > > says adj = 1.490 degrees. The diagram indicates it needs to be > > added to the course. Is there something inherantly wrong with > > this approach, or have I made a slip somewhere? > > Not being mathematically oriented I can't say, but in Shufeldts's The > Calculator Afloat, he gives the following formulas: > > Finding correction angle and speed of advance > Value of correction angle > sin B = a x sin A /c > Calculate speed of advance > c = a x sin [d -(A + B)] /sin A > > Current Sailing when when track and speed of advance are specified > Value of correction angle > tan B = b x sin A / a - b x cos A > Value of speed to steam > a = b x sin A /sin B > > > > > > > > Mid-latitude using 13d 42' > > > > dLat = 15' N dLon = 183d W dep = 177.8 W > > > > TC = 274.6d dist = 178.3 nm > > > > > > OK. Be aware that above 500 miles at 90d/270d Mid-Latitude can give an > > error. > > > Above 1200 mils it gets unreliable. > > > > I saw a discussion initiated by Sam Chan about Mercator sailing > > near 90 or 270, where the problem arises from rounding the course > > before using dLat/cos(C) to compute the total distance. Is this > > a similar problem? In Mid-Latitude sailing, where we already > > have dLat and departure, the total distance can be computed either > > as dLat/cos(C) or as departure/sin(C), so we don't have to divide > > by a tiny rounded number. > > > > Or are you talking about the error in distance that arises from > > assuming a spherical earth? What method would be > > better? I assumed we weren't planning to sail a great-circle > > course here. > > As I understand it it's caused by the formulas and the how they are derived. > Mid-Latitude gets error prone when you cross the sperical quadrants. Try > calculating using Mid-Latitude from North America to Australia and one N.AS. > to South America of about equal distance. > > GC course comes after the Panama Canal transit ;^) > > > > Dan Hogan WA6PBY > dhhogan{at}XXX.XXX > Nav-L Web Page: http://www.wa6pby.com > Catalina 27 "GACHA" >