NAVIGATION FORMULAE by: Steven A. Sternberg /CT (72711,603) Up until just a short time ago I would rely on either a small hand held calculator with trig functions or my trusty 10" slide rule. Now with the speed and compactness of the PC and a spreadsheet program like LOTUS 123 or EXCEL most, if not all, of my calculations can be done in the time it takes to input the data and press the enter key while afloat. All the formulae were originally meant to be worked out on a 10 inch slide rule by navigators on board U.S. Navy ships and Naval aviators. I hope, what I am about to present to you will help you as much as it has helped me. None of the formulae are original, most were taught to me while I was in the U.S. Navy, and solved with only a 10" slide rule, some were shown to me by friends, and some were "liberated" while watching real "Navigators" at work, while on a couple of TRANS-PAC races in the early '70s (I was the cook). You can incorporate the formulae into a spreadsheet program on your computer, use them on a hand held calculator that has the ability to do trig and has at least 1 register for memory, or if you can still find one... you could use a 10" slide rule. NOTE: This file is NOT formatted for any printer (no page breaks etc.). Just call it up in your favorite word processor, it is pure ASCII. Format it as you wish and print it. Or you can print it out as one continuous sheet with the "type" or "copy" command with a redirect to your printer. To the best of my knowledge all of the following formulae are within the public domain so you may do with then as you wish. PLEASE NOTE: I have noticed in a number of messages in this forum questioning the use of (Pi/180). This is a conversion factor to convert degrees into radians. Most spreadsheet programs (Lotus 123 to name one) and some programming languages (BASIC to name one) require the user to manually convert degrees into radians. This is done by converting degrees (DDD), minutes (MM) & seconds (SS) into degrees.degrees (DDD.dddd) and then multiplying this decimal number by ( (Pi/180) which by the way is 0.017453293 ). PLEASE CHECK THE MANUAL FOR YOUR SPREADSHEET TO SEE IF YOU MUST ADD THIS EXTRA STEP. *** THIS EXTRA STEP IT IS NOT SHOWN IN ANY OF THE FORMULAE involving trig functions *** *************************************************************************** TABLE of CONTENTS (in order of appearance) 1) DISTANCE to the HORIZON 2) DIP of the HORIZON 3) DIP SHORT of the HORIZON 4) MEAN REFRACTION 5) REFRACTION for Non-STANDARD TEMPERATURE 6) REFRACTION for Non-STANDARD BAROMETRIC PRESSURE 7) REFRACTION for COMBINED Non-STANDARD TEMPERATURE and Non-STANDARD BAROMETRIC PRESSURE 8) SEA-AIR TEMPERATURE CORRECTION 9) OFFSHORE NAVIGATION 10) MID-LATITUDE SAILING 11) PLANE SAILING 12) GREAT-CIRCLE SAILING *************************************************************************** DISTANCE to the HORIZON On a clear day, with standard temperature and barometric pressure, the distance to the horizon would be as follows: In Nautical Miles: D = 1.144 * (sq. root of [HE] ) In Statute Miles: D = 1.317 * (sq. root of [HE] ) D = Distance & HE = Height of the eye above the surface of the water, in feet. EXAMPLES: 1) You are floating on the surface of the water on your back, your eyes are 3 inches above the water. What is the distance to the horizon. 3 inches = 3/12 = .25 feet D = 1.144 * ( sq. root of [.25] ) D = 1.144 * .5000 D = .5720 Nautical Miles or D = 1.317 * .5000 D = .3293 Statute Miles 2) You get out of the water and climb to the top of your mast which is 71 feet above the water. What is the distance to the horizon now? D = 1.144 * (sq. root of [71] ) D = 1.144 * 8.4261 D = 9.6395 Nautical Miles or D = 1.317 * 8.4261 D = 11.0972 Statute Miles Taking the above one step further. You are approaching the coast at night. Your chart shows a lighthouse on the coast that is 167 feet above the surface of the water. This light has a range of 20 miles. At what distance will you be able to _first_ see the light if you are on deck and your eye is 9 feet above the waters surface. (in Nautical Miles) D1 = 1.144 * (sq. root of [9] ) + D2 = 1.144 * (sq. root of [167] ) D1 = 1.144 * 3 = 3.432 N.M. + D2 = 1.144 * 12.923 = 14.784 N.M. --------- 18.216 N.M. D1 = The distance you can see to the horizon. D2 = The horizon distance of the object. Even tho the light has a range of 20 miles, you can only see it when you are 18.2 miles away. NOW if the light had a visual range of 15 miles, you would not see the light until you were 15 miles from it. The reason for this is even tho you can see 18.2 N.M. to the horizon, the light can under ideal conditions be seen for 15 miles because of its intensity. *************************************************************************** DIP of the HORIZON As we know the Earth is a sphere and because of this the angle by which the visible horizon differs from the true horizon is called the Dip of the Horizon. This angle increases as the observers eye is raised above the surface of the water. In all sight reduction, this value must be accounted for when adjusting to Ho (Height [altitude] of Observed body fully corrected). Remember the dip is always subtracted from the sextant altitude. The below formula is for "normal" atmospheric conditions. Later on we will show you how to correct for the anomalies. D = 0.97 * ( sq. root of [h] ) D = Dip, in minutes of arc & h = Height of observers eye above the surface of the water, in feet. EXAMPLE: You are standing on deck taking a sight of a celestial object. Your eye height is 13 feet above the water. What is the dip correction? D = 0.97 * (sq. root of [13] ) D = 0.97 * 3.61 D = (-) 3.5 ************************************************************************** DIP SHORT of the HORIZON Sometimes it may be necessary to view a celestial object even tho we do not have a visible horizon due to an obstruction in our way, be it land or another vessel. Under these conditions if we know how far (in Nautical Miles) we are from the obstruction we can calculate the dip short of the horizon using the waterline of the obstruction as our horizontal reference point. Ds = 0.416d + ( 0.566 * h/d ) Ds = Dip short of the sea horizon, in minutes of arc. d = distance to the water line of the obstruction, in Nautical Miles. h = height of the observers eye above sea level, in feet. This formula is an adaption of the one given in Bowditch and works quite well as long as the height of the eye is not great or the obstruction is not very close. Using the above formula if the height of the observers eye was 100 feet and the obstruction was 0.1 N.M. away the dip would work out to be 566.'0, verses 565.'8 using the (Bowditch) long method. EXAMPLE: Your eye height is 24 feet and the distance to the obstruction is 0.75 N.M. What is the DIP SHORT of the HORIZON? Ds = (0.416 * d) + ( 0.566 * h/d ) Ds = (0.416 * 0.75) + ( 0.566 * (24 / 0.75) ) Ds = 0.312 + (0.566 * 32) Ds = 0.312 + 18.112 Ds = (-) 18.424 Thus the DIP SHORT of the HORIZON is (-) 18.'4 when correct to the nearest tenth of a minute. Remember the dip is always subtracted from the sextant altitude. ************************************************************************** MEAN REFRACTION As I mentioned above the dip & dip short were calculated for "normal" conditions. "Normal" is based on a standard of 50 deg. F (+10 deg. C) for the temperature and 29.83 inches of mercury (1010 millibars). When the conditions are not normal we have to calculate correction factors for the "non-standard" conditions. It is of the utmost importance that we correct the mean refraction when we observe a low altitude body, because this is where our biggest errors are produced. As a matter of practice, it is a good idea to apply the correction for dip and index error before applying the correction for refraction. Once all the corrections are made we have what is known as ha, the apparent altitude. I have found the following 2 formulae to be very good in computing the mean refraction, Rm. Rm is always a subtractive factor and is calculated to the nearest 10th of a minute of arc. 1) Rm = ( 0.97 / tan ha ) When the body has an altitude of 15-40 degrees. 2) Rm = (0.93 / tan ha ) When the body has an altitude of 8-15 degrees. Rm = mean refraction ha = apparent altitude of body EXAMPLE: 1) After applying the index & dip corrections the altitude of the observed body is 47 deg. 56.'2. What is the Rm? Rm = ( 0.97 / tan ha ) Rm = 0.97 / tan 47 deg. 56.'2 = (0.97 / 1.108147) Rm = (-) 0'.875 REMEMBER the lower the altitude of the body sighted the HIGHER the correction factor. Do not use a body that has an observed altitude of less then 8 degrees with this formula. Your answer will be off by a considerable amount. ************************************************************************** REFRACTION for Non-STANDARD TEMPERATURE Once the temperature deviates from the "standard" temperature of 50 deg. F, the first step is to determine the mean refraction, Rm. Then we find the correction factor "C". The formulae are: 1) C = (500 or 520 / 460 (plus or minus) T). 2) Rt = Rm * C C = correction factor T = temperature, in Fahrenheit Rt = Refraction, corrected for Non-Standard Temperature Rm = mean refraction (see above) NOTE: A) Use 500 if the temperature is ABOVE 60 deg F and use 520 if the temperature is BELOW 40 deg. F. B) The temperature is ADDED to 460 if the temperature is ABOVE 0 deg. F and SUBTRACTED if it is LESS then 0 deg. F. EXAMPLES: 1) The mean refraction is (-) 7.'4, the temperature is 81 deg. F. What is the mean refraction corrected for temperature? Formula 1) C = 500 / (460 + 81) = 500 / 541 C = 0.924 Formula 2) Rt = (-) 7.'4 * 0.924 Rt = (-) 6.839 The refraction, corrected for temperature is (-) 6.'8. 2) The mean refraction is (-) 7.'4, the temperature is (-) 2 deg. F. What is the mean refraction corrected for temperature? Formula 1) C = 520 / (460 - 2) = 520 / 458 C = 1.135 Formula 2) Rt = (-) 7.'4 * 1.135 Rt = (-) 8.402 The refraction, corrected for temperature is (-) 8.'4 . ************************************************************************** REFRACTION for Non-STANDARD BAROMETRIC PRESSURE The Non-STANDARD BAROMETRIC PRESSURE affects the mean refraction less then the Non-Standard Temperature but it should be adjusted for. The formula for mean refraction corrected for barometric pressure, Rp, at the standard temperature is as follows: Rp = Rm * ( EP / 29.83 ) Rp = refraction for non-standard barometric pressure Rm = mean refraction (see above) EP = existing barometric pressure EXAMPLE: The mean refraction is (-) 7.'4, the temperature is 50 deg. F and the barometric pressure is 30.33 inches of mercury. What is the refraction for the barometric pressure? Rp = (-) 7.'4 * ( 30.33/29.83 ) Rp = (-) 7.'4 * 1.017 Rp = (-) 7.'5 The refraction, corrected for barometric pressure (-) 7.'5 . ************************************************************************** REFRACTION for COMBINED Non-STANDARD TEMPERATURE and Non-STANDARD BAROMETRIC PRESSURE When the mean refraction is to be corrected for both the non-standard temperature AND non-standard barometric pressure, the difference between Rm and the correction for non-standard barometric pressure, Rp, is applied to the correction for non-standard temperature, Rt. EXAMPLE: Mean refraction is (-) 7.'4, the barometric pressure is 30.60 inches of mercury and the temperature is (-) 20 deg. F. The pressure being 30.60 inches of mercury, Rp is (-)7.'6, and the difference between Rm, (-)7.'4, and Rp is (-)0.'2. The Rm (-)7.'4, corrected for temperature is (-) 8.'7. To this we apply the Rp correction (-) 0.'2 giving us a total correction of (-) 8.'9 for both Non-STANDARD TEMPERATURE and Non-STANDARD BAROMETRIC PRESSURE. ************************************************************************** SEA-AIR TEMPERATURE CORRECTION When there is a difference in the temperature of the water and the air in contact with it the dip correction is affected. The Japanese Hydrographers Office found after much testing that the dip would be affected by 0.'11 minutes of arc for each degree Fahrenheit of difference between the temperature of the air and water. The formula to correct for this effect is: Sea-Air temp. correction = 0.'11 * temperature difference in degrees Fahrenheit between the air and water. NOTE: If the air is cooler than the water the correction is subtracted and if the air is warmer than the water the correction is added. EXAMPLES: 1) The air temperature is 91 deg F and the water temperature is 84 deg. F. What is the correction factor? Air = 91 (warmer = +) Water = 84 ------ + 7 Correction factor = 0.'11 * 7 = 0.'77 The correction to the sextant altitude for sea-air temperature is (+) 0.'77 2) The air temperature is 35 deg. F and the water temperature is 49 deg. F. What is the correction factor? Air = 35 (cooler = -) Water = 49 ------ 14 Correction factor = 0.'11 * 14 = 1.'54 The correction to the sextant altitude for sea-air temperature is (-) 1.'54. ************************************************************************** OFFSHORE NAVIGATION MID-LATITUDE SAILING Mid-latitude sailing is based on approximations which greatly simplify solutions to more complex problems. When the distance is greater than about 1,200 miles the more complex method (Great-Circle Sailing -- see below) should be employed. When we know the distance and the course, mid-latitude sailing allows us to determine the difference in latitude (l) and the departure (p), stated as a difference of longitude (DLo), in minutes of arc. Departure in very simple terms is the distance East or West we must travel when going from here to there, (point A to point B). With mid-latitude sailing we are also given the option when we know the coordinates of our starting point and of our destination to determine the rhumb line course and the distance (D) between them. In mid-latitude sailing, departure & the difference of longitude may be inter converted, using the mean, or mid, latitude "Lm". The three formulae are: 1) p = DLo, in minutes * cos Lm 2) DLo, in minutes = p / cos Lm and 3) tan C = p /l Lm = mid-latitude l = difference of latitude, in minutes of arc C = course angle (see note below), from N or S towards east or west to 90 degrees EXAMPLE: What is the course & distance from Brenton Reef Light (off Newport, R.I.) -- Lat. 41 deg 26 min NORTH, Long. 71 deg 23 min WEST to St. David's light, Bermuda -- Lat. 32 deg 22 min NORTH, Long. 64 deg 39 min WEST. L1 41 deg 26 min N Long 71 deg 23 min W ~ (absolute difference) ------------> same (abs. diff.) L2 32 deg 22 min N 64 deg 39 min W ------------------ ----------------- l = 9 deg 04 min S = 544 min S DLo = 6 deg 44 min E = 404 min. E 1/2 l 4 deg 32 min S ~ (absolute difference) L1 41 deg 26 min N ------------------ Lm = 36 deg 54 min N 1) p = DLo (in minutes) * cos Lm p = 404 * cos 36 deg 54 min p = 323 miles 3) tan C = p / l tan C = (323 / 544) = 0.593883 = S 30 deg 42 min E = 30.7 deg S Thus the course to reach St. David's Light is: 149.3 degrees (180 degs - 30.7 degs) 2) DLo (in minutes) = p / cos Lm DLo = 544 / cos 30 deg 42 min = 632.7 miles THUS to go from Brenton reef light to St. David's light you would set a TRUE course of 149.3 degrees and travel 632.7 miles. Solving the same problem using the great-circle method you come up with an initial course of 147 deg 47.4 min and a distance of 632.2 miles. NOTE: In plane sailing, the course is reckoned as a course angle from North or South to 90 degrees East or West. Thus a course (Cn) of 162 deg. would be written as S 18 deg. E and Cn 341 deg. as N 19 deg. W. The quadrant is derived when you calculate "l" & "DLo". In the above example you can see "l" is South & "DLo" is East -- of your starting point, thus this places us in the South East quadrant. Once you do the #3 formula (tan C), you must apply this "correction factor". Keeping the above in mind, S 30.7 deg. E MUST be subtracted from 180 degrees to move us into the proper quadrant. Since the quadrant is S & E we must subtract to move us towards the East by 30.7 degrees. If by chance you work the problem backwards you would then be in the NW quadrant & you would subtract you answer (tan C) from 360 degrees. ************************************************************************** PLANE SAILING As we know the Earth is round, and because of this, when solving problems involving distances on the Earth spherical trigonometry should be employed. BUT, when we are looking at small distances we can consider that part of the Earth as a flat object. If we consider the Earth flat for these short distances we are then able to utilize the less complex plane trigonometry formulae to solve our navigational problems. In plane sailing, the triangle formed by the meridian passing through the point of departure (P1), the parallel of latitude passing through the destination (P2) and the course line can be considered for our purposes a plane right angled triangle. (see below) p ---------.P2 | . | . l | . | . D | . --> |C . <-- | . |. . P1 In the above triangle, "P1" is the point of departure and "P2" is the destination. "p" is the "departure" -- the distance in nautical miles made good East or West while heading to our destination. Side "l" is the meridian from our starting point (point of departure) to the latitude of our destination. Side, "l", is measured in nautical miles and is the difference in latitude which is equal to minutes of arc along the meridian. Side "D" is the distance sailed in nautical miles, and "C" is the course angle. Remember course angles are measured from North or South to East or West. There are 7 formulae to solve all the problems you will encounter, they are: 1) Given C and D, find l l = cos C * D 2) Given c and D, find p p = sin C * D 3) Given l and p, find C tan C = ( p / l ) REMEMBER -- If "l" is greater than "p", "C" will be less than 45 degrees. If "l" is less than "p", "C" will be greater than 45 degrees. 4) Given C and l, find D D = ( l / cos C ) 5) Given C and p, find D D = ( p / sin C ) Knowing "p", it is often necessary to convert it to a difference in longitude "DLo". "L" represents latitude. 6) DLo = ( p / cos L ) 7) p = DLo * cos L EXAMPLES: 1) We have sailed 123 miles on a course (Cn) of 321 deg. and want to find the difference of latitude, "l". A) Convert the course to a course angle. (360 - 321 = N 39 deg W) B) Use formula #1: "l" = cos C * D l = cos 39 deg. * 123 l = 0.777145961 * 123 l = 95.59 = ( 95.59 - 60 = 1 deg 35.59 min) The difference in latitude is: 1 deg 35.59 minutes North 2) Using the above data, find "p". p = sin C * D p = sin 39 deg. * 123 p = 0.62932 * 123 p = 77.41 The departure is 77.41 nautical miles West. 3) Given an "l" of 69.'0 South and a "p" of 57.'9 miles West, find "C". tan C = ( p / l) tan C = 57.9 / 69.0 = 0.83913 C = 40.00 The course angle is South 40.00 degrees West which is a Cn of 220 degrees (180 + 40 = 220 degrees) ************************************************************************** GREAT-CIRCLE SAILING As we know, the shortest distance between to points is a straight line. On a sphere (the Earth) this line is called a great-circle. It is impossible to sail a great circle on the same course unless you are traveling due North, South, East or West. It is the custom to select a number of points along the great-circle, usually 5 degrees apart, and sail the rhumb-line course between them. The great-circle route should not be considered if the tract taken crosses over land, into dangerous waters or into to high a latitude if you are unable to provide the proper watch precautions. Another factor to take into consideration is the location of the vertex, or the point of greatest latitude through which the circle passes. The vertex might lie beyond you destination, behind the point of departure or between the two. When solving great-circle sailing problems you must first establish the distance along the great-circle tract to your destination, the initial heading and the latitude and longitude of the vertex. This must be done even if the vertex is beyond your destination because it is used to obtain the coordinates of the intermediate points at which rhumb-line course changes will be made. This set of formulae is tricky, I hope I don't loose you. If by chance you do get lost or bogged down, E-Mail me or post an open message to me in the forum and I will _try_ to help you work it out. There are 2 formulae for finding the distance (D). The first one is the cosine D formula (#1 below). This is the best formula when the distance is over 4,050 miles (45 degrees). If the distance is less than 4,050 miles the use of the sine D formula (#11 below) is suggested. #1 cos D = sin L1 (+ or abs. diff.) sin L2 (abs. diff) (cos L1 * cos L2 * cos DLo) L1 = Latitude of the point of departure L2 = Latitude of your destination DLo = Difference in Longitude between the to places If DLo is less than 90 degrees, and A) L1 & L2 have the same name, the sign is (+) B) L1 & L2 have opposite names, the sign is (the absolute difference). If DLo is more than 90 degrees, and A) L1 & L2 have the same name, the sign is (the absolute difference) B) L1 & L2 have opposite names, the sign is (+). If "DLo" is more than 90 degrees, use ( 180 deg. - DLo ) instead of "DLo". The distance may be greater than 90 degrees (5,400 miles) in which case the angle found by the cos D formula is subtracted from 180 degrees to obtain the distance angle. Distance, in nautical miles, is found by multiplying the number of degrees by 60 and then adding the number on minutes. If you ever have a question as to whether or not the distance, "D", is greater than 90 degrees do the following: A) Find the value of the auxiliary angel "R" with the following: #2 sin R = sin DLo * cos L2 B) After you have "R", find a second auxiliary angle, "K", as follows: #3 sin K = sin L2 / cos R "K" will be more than 90 degrees if "DLo" is greater than 90 degrees, in which case use "K1", where K1 = (180 degrees - K). "D" is greater than 90 degrees if: A) "L1" & "L2" or of the same name AND "K" or "K1" is greater than (L1 + 90 degrees) or B) "L1" & "L2" are of opposite names AND "K" or "K1" is greater than (90 degrees - L1). The initial great-circle heading,"C", is found as follows: #4 L2 sin heading C = ( cos L2 * sin DLo) / sin D "D" is the distance expressed in angle. The heading found will be similar in nature to azimuth angle and will have to be applied to 000/360 deg. or 180 degrees in an easterly or westerly direction. The initial heading can also be found with the tangent formula. This is a better formula to use if the heading is greater than 45 degrees. #5 tan heading C = sin DLo / (cos L1 * tan L2 [ + or abs. diff ] sin L1 * cos DLo) In the tan heading C formula, when DLo is less than 90 deg. AND "L1" and "L2" have the same name the sign in the divisor is subtractive. If "DLo" is greater than 90 deg. the sign is additive. When "L1" & "L2" have opposite names and "DLo" is less than 90 deg. the sign is additive. Our next step is to determine the latitude of the vertex, "Lv". #6 cos Lv = cos L1 * sin C after the "Lv" is determined we will calculate the longitude of the vertex. This is done by first determining the DLo between the longitude of the point of departure and that of the vertex, "DLo-v", as follows: #7 sin DLov = cos C / sin Lv then using "DLo-v" the angular distance from the point of departure to the vertex, "Dv" is computed as follows: #8 sin Dv = cos L1 * sin DLov The next to the last step is to find the latitude, "Lx", of each point along the great-circle track where we will change course. The formula is as follows: #9 sin Lx = sin Lv * cos Dv-x Dv-x = The angular distance to the vertex less the angular distance along the great-circle at which course is to be changed. The last step is to find the longitude of these change points. #10 sin DLov-x = sin Dv-x / cos Lx "DLov-x" is the difference between the meridian of "DLov" (as found above in #5) and that of each course change point. IF I HAVE NOT LOST YOU, let us pull it all together. EXAMPLE #1: We are going to sail from: San Francisco, Lat. 37 deg. 47.'5 N, Long.1 122 deg 27.'8 W to Sydney, Australia , Lat. 33 deg. 51.'7 S, Long.2 151 deg 12.'7 E. ----------------- ---------------- 70 deg 99.'2 S 273 deg 40.'5 W <-- 71 deg 39.'2 S | | | What is the great-circle distance, our initial heading, the latitude & | longitude of the vertex, and the latitude & longitude of the first | turning point where we will alter our course. We want to space our | turning points 6 degrees (360 miles) apart on the great-circle. | | | DLo = ( 360 deg - [long. 1 + long. 2] ) <-------------------| DLo = 360 deg. - 273 deg 40.'5 <---------------------------| DLo = 86 deg 19.'5 cos D = sin L1 (+ or abs. diff.) sin L2 (abs. diff) (cos L1 * cos L2 * cos DLo) cos D = sin 37 deg 47.'5 * sin 33 deg 51.'7 S (abs. dif) cos 37 deg 47.'5 * cos 33 deg 51.'7 * cos 86 deg 19.'5 cos D = (0.612792 * 0.557190) [abs. diff] (0.790244 * 0.830385 * 0.064097) cos D = .341442 -.042061 = .299381 = 72 deg 35' Since the distance to Sydney is greater than 5,400 miles (90 deg), we subtract 72 deg. 35' from 180 degrees and obtain 107 deg. 25' as the angular distance. This converts to 6,445 nautical miles (107*60+25). The next step is to find the initial great-circle heading. We can use formulae #4 or #5 above. sin C = cos L2 * sin DLo / sin D sin C = (cos 33 deg 51.'7 * sin 86 deg 19.'5) / sin 72 deg 35' sin C = 0.830385 * 0.997944 / 0.954153 sin C = .868496 = C 60 deg 17' the course angle of S 60 deg 17 ' W gives us an initial course of 240 deg 17' (180 deg + 60 deg 17') Since the course is greater than 45 degrees we will recalculate using the "tan C" formula (#5) in order to derive a more precise initial course. #5 tan heading C = sin DLo / (cos L1 * tan L2 [ + or abs. diff ] sin L1 * cos DLo) tan C = sin 86 deg. 19.'5 / ( [ cos 37 deg. 47.'5 N * tan 33 deg. 51.'7 S ] + [ sin 37 deg. 47.'5 N * cos 86 deg. 19.'5 ] ) tan C = 0.997944 / 0.790244 * 0.671001 + (0.612792 * 0.064097) tan C = 0.997944 / 0.530255 + 0.039278 tan C = 0.997944 / 0.569533 tan C = 1.752215 = S 60 deg. 17' W or a Cn of 240 deg. 17'. Now we calculate the latitude of the vertex with formula #6. cos Lv = cos L1 * sin C cos Lv = cos 37 deg. 47.'5 * sin 60 deg. 17' cos Lv = 0.790244 * 0.868487 cos Lv = 0.686317 = 46 deg. 40' S The latitude of the vertex is named South because we are headed in a southerly direction. NOTE the vertex also lies beyond of destination. Now we calculate the longitude of the vertex, but first we must find the difference in longitude between our point of departure (San Francisco and the vertex, "DLov", and convert this to longitude of the vertex by using formula #7. sin DLov = cos C / sin Lv sin DLov = cos 60 deg 17' / sin 46 deg. 40' sin DLov = 0.495711 / 0.727374 = 0.681508 sin DLov = 42 deg 58' As noted above the vertex lays beyond our destination and "DLov" is greater than the DLo of our destination. Therefore "DLov" is equal to 137 deg 02'W (180 deg - 42 deg 58'). "DLov" takes the name West because our destination is west of us. To arrive at the longitude of the vertex we add the longitude of our departure point to the "DLov" and get 259 deg 29.'8 (122 deg 27.'8 + 137 deg 02'). The longitude of the vertex therefore will be 100 deg 30.'2 E (360 deg - 259 deg 29.8). Having found "DLov", we can find the distance to the vertex, "Dv" with formula #8. sin Dv = cos L1 * sin DLov sin Dv = cos 37 deg. 47.'5 * sin 42 deg. 58' sin Dv = 0.790244 * 0.681573 = 0.538609 sin Dv = 32 deg. 35' therefore Dv = 147 deg 25' (180 deg - 32 deg 35'). Next we need to find the latitude of the first turning point where we will change our course, "X1". This is done with formula #9. sin Lx = sin Lv * cos Dv-x sin Lx = sin 46 deg 40' * cos 38 deg 35' sin Lx = 0.727374 * 0.781702 = 0.568590 sin Lx = 34 deg 39' In the above formula "Dv-x" is obtained from Dv, which we computed to be 147 deg 25'. The turning points are to be 6 deg. apart, "Dv-x" therefore is 141 deg 25'(147 deg 25' - 6 deg), which we write as 38 deg 35' (180 deg - 141 deg 25'). The latitude of "X1" is therefore 34 deg. 39'. To find the longitude of point "X1", we need to find the difference in longitude between the vertex and point "X1". This is done with formula #10. sin DLov-x = sin Dv-x / cos Lx sin DLov-x = sin 38 deg. 35' / cos 34 deg 39' sin DLov-x = 0.623652 / 0.822641 = 0.758110 sin DLov-x = 49 deg. 18' which is converted to 130 deg 42' (180 deg - 49 deg 18'). The longitude of point X1 equals 259 deg 32.'8 W ( the longitude of the vertex which we found above [see #7] expressed westward) minus 130 deg. 37' equals 128 deg. 55.'8. Now we would proceed to figure out the other "X" points in a similar manner and then plot them on our Mercator chart(s). The heading for each leg can also be determined by mid-latitude sailing or if the legs are short enough by plane sailing. However, the best method of finding the initial rhumb-line course to the first point "X1", is to apply the conversion angle to the initial great-circle heading. In this example the mid-latitude between San Francisco and point "X1" is 36 deg. 11.'8 N, and one half the difference of longitude is 3 deg. 14.'0, so the conversion angle becomes: tan conversion angle = sin 36 deg. 11.'8 * tan 3 deg. 14.'0 tan conversion angle = 0.590559 * 0.056492 = 0.033362 conversion angle = 1 deg. 54.'6 We now subtract the conversion angle from our initial great-circle course of 240 deg 15' (see #5 above) to derive our initial course of 238.3 deg. to be plotted on a Mercator chart. When the distance is less than 1,800 miles (30 degrees) or more than 9,000 miles (180 deg. - 30 deg) the accuracy may often be improved by first solving for the initial heading by using formula #5. If this used the distance may than be calculated with the following formula: #11 sin D = (cos L1 * sin L2 [+ or abs. diff.] sin L1 * cos L2 * cos DLo) / cos C where, with DLo less than 90 deg., the sing in the dividend is (+) when L1 and L2 are of opposite names, and (abs. diff.) when they are of the same name. When DLo is greater than 90 degrees the reverse is true, and the sign is (abs. diff.) if L1 and L2 are of opposite name and (+) if they are the same name. EXAMPLE #2: What is the initial heading and distance from: Cape Race, Newfoundland Lat. 46 deg 39' N Long. 53 deg 05' W to Fastnet Rock, off the south coast of Ireland Lat. 51 deg 23' N Long. 9 deg 36' W ------------- ------------- l 4 deg 44' N DLo 43 deg 29' E Our 1st step is to find DLo (see above) and the we can proceed to find the initial heading, "C" and apply formula #5. tan heading C = sin DLo / (cos L1 * tan L2 [ + or abs. diff ] sin L1 * cos DLo) tan C = sin 43 deg. 29' / ( cos 46 deg. 39' * tan 51 deg. 23' [abs. diff.] sin 46 deg. 39' * cos 43 deg. 29') NOTE that the sign is [abs. diff.] because DLo is less than 90 deg. and L1 & L2 are of the same name. tan C = 0.688144 / ( 0.686453 * 1.251931 [abs. diff.] 0.727174 * 0.725575 tan C = 0.688144 / 0.859392 [abs. diff.] 0.527619 tan C = 0.688144 / 0.331773 = 2.074141 C = 64 deg. 15.'6 Since Fastnet Rock is to the North and East of our point of departure point the course "Cn" is 64 deg 15.'6 (000 + 64 deg 15.'6). Now we now solve for the distance using formula #11. sin D = (cos L1 * sin L2 [+ or abs. diff.] sin L1 * cos L2 * cos DLo) / cos C sin D = (cos 46 deg 39' * sin 51 deg 23' [abs. diff] sin 46 deg 39'* cos 51 deg 23' * cos 43 deg 29') / cos 64 deg 15.'6 sin D = 0.686453 * 0.781339 [abs. diff.] 0.727174 * 0.624107 * 0.725575 / 0.434288 sin D = 0.536353 [abs. diff.] 0.329291 / 0.434288 sin D = 0.207062 / 0.434288 = 0.476785 sin D = 28.475634 degrees D = 28.475636 * 60 = 1,708.53 miles ************************************************************************** - The End -