CELESTIAL NAVIGATION Section 17

The SailingsThe Sailings

Prepared by: John C. HudsonSaanich Peninsula Squadron, VISD Jan 2006 / Rev. Jan 2007

Sailings used for such things as:

- determining distance and course between 2 known points

- finding point of arrival after travelling given distance and

course

- for short distances, information can be obtained by plotting on

Mercator Chart

- for longer distances, easier to use mathematical solution with

calculator and trigonometrical functions

These methods are called The Sailings

- types of sailings are:

- Plane Sailing (this section)

- Mid-Latitude Sailing (this section)

- Traverse Sailing (section 18)

- Great Circle Sailing (section 19)

The Sailing Triangle

- boat sails distance “D” from point “X” to “Y” on course angle “C”

- boat moves north a distance of “XZ”

- if Latitude of“X” and distance“XZ” known,Latitude of“Y” can becalculated

The Sailing Triangle

- not as easy to calculate Longitude of “Y”

- distance covered by a degree of Longitudevaries withdistance fromequator(section 15.1)

Latitude and Departure

- distance in miles “XZ” known as difference in Latitude “ ”

- distance inmiles “ZY” iscalleddeparture “p”

Latitude and Departure

- units of measure are as noted in triangle

- if distance “D” and course angle “C” known, “ “ and“p” can becalculated usingtrigonometricalfunctions

Course in a Quadrant: azimuth angle or direction between 2 points as measured from either a northerly or southerly meridian in an

E or W direction

Course in a Quadrant- angle can never exceed 90º

- 4 quadrants are NE, SE, SW and NW- actual course sailed is “Cn” to indicate

measured clockwise from N

Plane Sailing (use if sailing < 100 miles)

triangle XP1P2 formed by:

- meridian throughpoint of departure P1

- destination atparallel of Latitude P2

- course P1 to P2

Plane Sailing

Sin C = p / DCos C = / DTan C = p / where

D = distance travelled (miles)

= difference in Latitude (miles or arc)

p = departure (miles) or difference in Longitude (arc) C = vessel’s course (in quadrant)

Plane Sailing Example

- find difference Latitude “ “ and departure “p”, if D = 15.6 miles = D Cos C = 15.6 Cos 27º = 13.9 miles

p = D sin C = 15.6 Cos 27º = 7.1 miles

Mid-Latitude Sailing (use if sailing < 1000 miles)

- meridians of Longitude converge towards North and South Poles

- distance covered between meridians of Longitude is dependent on Latitude

- Mid-Latitude sailings more accurate than plane sailing but less accurate than

other methods

Mid-Latitude Sailing

- to convert departure “p” (miles) to difference in Longitude “DLo” (degrees

and minutes of arc), first necessary to calculate Mid-Latitude “Lm” (degrees)

- “Lm” is arithmetical mean of initial Latitude L1 and final Latitude L2

Mid-Latitude Sailing

Mid-Latitude (Lm) = (L1 + L2)/2 or

= L1 +/- (1/2) *Example:

if L1 = 33º N and L2 = 35º N

Lm = (33º + 35º)/2 = 68º/2 = 34º

Mid-Latitude Sailing

- having derived Mid-Latitude “Lm” (degrees), difference in Longitude “DLo” calculated by: (answer in minutes or miles and has to be converted to

arc degrees and minutes)

DLo = p / Cos Lm

where: - departure Latitude = L1

- destination Latitude = L2

(L1 & L2 in degrees and minutes)

- departure = p (miles)

- diff. of Latitude = (miles)L1

L2

Lm

DLo (degrees & minutes)

(miles)

1) Mid-Latitude Sailing - find Cn and D, given point of departure L1 and Lo1,destination L2 and Lo2,

steps: Lm = (L1 + L2)/2

calc DLo & convert to minutes

p = DLo Cos Lm

= L1 ~ L2

C = Tan-1 p / then get quadrant Cn

D = / Cos C

Lm

DLo

L1,Lo1

L2,Lo2

1) Mid-Latitude Sailing – Example:- point of departure L 43º 13.5’ N, Lo 57º 18.3’ W

- point of arrival L 15º 27.1’ N (error in text), Lo 19º 22.1’ W - find Cn & D:

Lm = (43.225 + 15.452)/2 = 29.339º N

DLo = (57º 18.3’ - 19º 22.1’) = 37º 56.2’ = 2276.2’ to E

p = DLo Cos Lm = 2276.2 Cos 29.339º = 1984.2 miles to E

= 43.225 – 15.452 = 27.773 = 1666.4 miles to S

C = Tan-1 p / = Tan-1 1984.2/1666.4 = S 50.0º E

Cn = 180º - 50.0º = 130.0º

D = / Cos C = 1666.4 / Cos 50.0º = 2592.5 miles

2) Mid-Latitude Sailing - find destination

L2 and Lo2, given point of departure L1 and Lo1, Cn and D

steps:find C quadrant from Cn = D Cos C (miles to degrees)

L2 = L1 ± (degrees & Min.)

P = D Sin C (miles)

DLo = p / Cos Lm

Lo2 = Lo1 ± DLo

Lm

DLo

L1,Lo1

L2,Lo2

2) Mid-Latitude Sailing – Example: point of departure L 48º 20.2’ N, Lo 7º 02.1’ W

course 253º, distance 2382 miles, find destination:

C = 253º - 180º = S 73º W = D Cos C = 2382 Cos 73º = 696.4’ or miles

= 11º 36.4’ to S

L2 = 48º 20.2’ N - 11º 36.4’ N = 36º 43.8’ N

p = D Sin C = 2382 Sin 73º = 2277.9 miles to W

Lm = (48.337º + 36.730º) / 2 = 42.533º

DLo = p / Cos Lm = 2277.9 / Cos 42.533º = 3091’ = 51º 31.3’ to W

Lo2 = (7º 02.1’ W + 51º 31.3’) = 58º 33.4’ W