This is part 2 (4mula.pt2 of useful Navigational Formulae. If you have any questions please leave me a message in the forum or E-Mail me a note. I hope you are able to make as much use of these formulae as I have over the last 25+ years. You can put them in a spread sheet, use them on a hand held calculator, or as I have done way back when, solve your problem on a 10" slide rule. As with part 1 (4mula.txt), this file IS NOT formatted to print out with page breaks it one continuous file. Steven A. Sternberg /CT [72711,603] TABLE of CONTENTS (in order of appearance) 1) Distance by Sextant Angle. 2) Distance Short of the Horizon. 3) Distance Beyond Horizon. 4) Direction and Speed of the True Wind. 5) Barometric-Pressure Conversion Factors. 6) Sailing to Weather. 7) Tacking Down Wind - When the Lee Mark is Dead to Leeward. 8) Time Conversion a) Time to Arc b) Arc to Time 9) DISTANCE by BEARINGS ( 6 different ways) a) Distance off, with 1 bearing & a Run to Beam . b) Distance off at 2nd bearing, from 2 bearings on the bow & a Run. c) Distance off when Abeam by 2 bearings on the Bow and a Run Between. d) Run to a given Bearing and Distance off when on that Bearing. e) Distance Off 2 Fixed Objects. f) Heading to Bring Light to Specified Distance & Bearing and the run thereto. 10) Rate of Change of Altitude 11) Rate of change of Azimuth 12) SIGHT REDUCTION a) Computing Azimuth b) Computing Altitude 13) Table of Conversion Factors ************************************************************** Distance by Sextant Angle The distance between you and object that is not beyond the horizon can be determined with the aid of a sextant. What we measure is the angle between the object and the horizon. Once the is measured, It is corrected for the index error and then corrected for the dip (see 4mula.txt) with the SIGN REVERSED. In other words you ADD THE DIP instead of subtracting. The corrected sextant angle is called the horizon angle "H". Now we can solve for the distance, "D" as follows: D = (HE / sin H) * cos H D = distance, in feet HE = height of eye above the water H = sextant angle EXAMPLE: You measure the angle between the horizon and a buoy, the angle read off your sextant is 1 deg. 4.'3, your sextant has an index error of -1.9 and the height of your eye is 20 feet. How far are you from the buoy? First we must make the corrections to find the corrected sextant angle "H". sextant reading 1 deg. 4.'3 index correction +1.'9 Dip for 20', SIGN REVERSED +4.'3 ------- + 6.'2 + 6.'2 --------------- H = 1 deg. 10.'5 Now we can solve the problem: D = (HE / sin H) * cos H D = (20 / sin 1 deg. 10.'5) * cos 1 deg. 10.'5 D = (20 / 0.020506) * 0.999790 D = 975.324295 * 0.999790 D = 975.12 feet Therefore you are 975 feet away from the buoy. This is a great trick if you are racing and want to know how far ahead or behind you are from the boats around you! **************************************************************** Distance Short of the Horizon You can also use your sextant as a very range precise range finder when you are able to see the base of an object and you know its height. Once you measure the angle between the base of the object and its top, you correct for the index correction, and then use the following formula. D = A / sin H D = Distance, in feet A = the known height of the object, in feet H = corrected sextant angle EXAMPLE: 1) You are approaching shore, when in the distance you see a lighthouse whose height on the chart is listed as 187 feet above the water. You find its angle between the top and the water line to be 0 deg. 45.'7 after making your index correction. How far are you from the lighthouse. D = A / sin H D = 187 / sin 0 deg 45.'7 D = 187 / 0.013293 D = 14,067.34 feet Ok, so you say, but how many miles (nautical) are we from the lighthouse. We have 2 ways to do this. The first is to look at the conversion chart below, and we see 1 nautical mile = 6,076.11549 feet, so we divide the above answer by 6,067.11549 and come up with 2.3 miles. The second method would be used if you know in advance you want your answer in some unit other than feet at the start. Let us say you know you want the answer in nautical miles from the start. Look at the conversion factor chart, at the end of this file, and you see: 1 nautical mile = 6,076.11549 feet. Divide 1 by 6,076.11549 and we get a factor of 0.000164579. Now we solve the problem as follows: D = (A * factor) / sin H D = (187 * 0.000164579) / sin 0 deg. 45.'7 D = 0.030776242 / 0.013293 D = 2.3 nautical miles 2) You are in a race on the downwind leg, you want to know how much of a lead (or are behind) you have on another boat. You know the top of the mast on the other boat is 75 feet above the water. How far are you (in feet) from the other boat if your corrected sextant angle is 4 deg. 1.'8. D = A / sin H D = 75 /sin 4 deg. 01.'8 D = 75 / 0.070279 D = 1067 feet Many years back this method was used by a defender in the America's Cup race. The only other correction used was adjusting the mast height above the water due to the heel of the boat (sine formula). ******************************************************************* Distance Beyond Horizon Are there times you wished you had radar on board in order to determine your distance to the shore when approaching an area of shoals, reefs or other perils. Well want no more (maybe). If the land you are approaching has a high bluff, mountain or any other tall well defined object you can come up with a VERY GOOD GUESSTIMATE of your distance off, even though you can not see the total object from its base to its top, if you know its height. The first step is to get a sextant reading of the object from its top to the water line that you see. Than correct for index error, dip of the horizon and refraction as I am about to explain. For this purpose only the refraction is found by dividing YOUR estimated distance to the object (in nautical miles) by 13.75. The answer will be in minutes and tenths of a minute of arc, which you SUBTRACT from the sextant altitude. The next step is to make a correction for the curvature of the Earth. This is done by squaring your estimated distance and then multiplying your answer by 0.907, your answer will be in feet. Now you subtract your answer from the known height of the object. Then you divide the corrected height by the fully corrected sextant reading, this answer is multiplied by a factor of 0.566. The resulting answer will be your calculated distance in feet from the object. Your first try most likely give you an answer that differs from your initial guess, so recalculate using your answer as your new guess. REMEMBER to change your refraction and curvature of the Earth factors by your new guess. I normally stop calculating when my new answer and my guess are with in 0.1-1.0 mile, this depends on what the danger if any I might be sailing into. EXAMPLE: You are making a landfall to an Island that has a large structure whose top is known to be 1,126 feet above the water. The height of your eye is 9 feet, your sextant has no error and you found the sextant angle to be 1 deg. 7.'2. You estimate you are 15 miles away. How far are you from the object? Alt. by sextant 1 deg. 07.'2 Dip (9') - 2.9 Refraction (15 / 13.75) - 1.1 ------- correction - 4.0 - 04.'0 --------------- 1 deg. 03.'2 = (1 * 60 + 03.2) = 63.'2 Object height 1,126' corr. for the curvature, est. dist. 15 mi. squared * 0.907 - 204' --------- 922' = corrected height Computed Distance = (Corrected Height / corrected sextant alt., in minutes) * 0.566 Computed Distance = (922/63.2) * 0.566 = 14.6 * 0.566 = 8.2 MILES Since we are not very close we resolve the problem using as our estimated distance 8.2 miles. Alt. by sextant 1 deg. 07.'2 Dip (9') - 2.9 Refraction (8.2 / 13.75) - 0.6 ------- correction - 3.5 - 03.'5 --------------- 1 deg. 03.'7 = (1 * 60 + 03.5) = 63.'5 Object height 1,126' corr. for the curvature, est. dist. 8.2 squared * 0.907 - 61' --------- 1,065' = corrected height Computed Distance = (Corrected Height / corrected sextant alt., in minutes) * 0.566 Computed Distance = (1,065/63.5) * 0.566 = 16.8 * 0.566 = 9.5 MILES Since we are still not very close we resolve the problem again using as our estimated distance this time 9.5 miles. If you notice each time we redo the problem we are coming closer to our estimate. Alt. by sextant 1 deg. 07.'2 Dip (9') - 2.9 Refraction (9.5 / 13.75) - 0.7 ------- correction - 3.6 - 03.'6 --------------- 1 deg. 03.'6 = (1 * 60 + 03.6) = 63.'6 Object height 1,126' corr. for the curvature, est. dist. 9.5 squared * 0.907 - 82' --------- 1,044' = corrected height Computed Distance = (Corrected Height / corrected sextant alt., in minutes) * 0.566 Computed Distance = (1,044/63.6) * 0.566 = 16.4 * 0.566 = 9.3 MILES This is a VERY GOOD answer for this problem, only 0.2 miles different. ************************************************************** Direction and Speed of the True Wind Knowing our course & speed of our boat and knowing the direction and speed of the RELATIVE wind, we are able to calculate the direction and speed of the TRUE wind very quickly. C Side "a" ---> . . True Wind blowing . . <-- Side "b" from "C" to "B . . . . Apparent Wind B . . Blowing from "C" | . to "A" side "c"--> | . | . Boat going | . from "A" | . to "B" |. . A --------------------------------- C Side "b" ------> . . Apparent Wind blowing . . <---Side "a" from "A" to "C" . . . . True wind blowing A . . from "B" to "C" | . Side "c" ---> | . | . Boat going | . from "B" to "A" | . |. . B If we look at the above 2 triangles (fill in the dots and dashes to form a "real" triangle), we can see what is happening better (I hope). In both drawings the boat is moving from the bottom to the top. In both drawings side "a" is the speed and direction of the TRUE WIND relative to the boats heading. Side "b" is the speed and direction of the APPARENT WIND relative to the boats heading and side "c" is the course and speed of the boat. The speed of the TRUE WIND is found with the following formula: a = sq. root of ( [b*b] + [c*c] [+ or -] (2 * [b*c] * cos A ) the sign [+ or -] is "-" if the APPARENT WIND is forward of the beam, and "+" if it is abaft the beam. After we get the speed of the TRUE WIND we can solve for its direction, relative to the boats heading. To do this we solve for angle "B" if the wind is forward of the bean or the supplement if the wind is abaft the beam. Then we apply angle "B" to the boats heading, and presto we now have the speed and direction of the TRUE WIND. The formula for the TRUE WIND speed is: sin B = ( b / a ) * sin A EXAMPLE: 1) We are on a course of 090 degrees at a speed of 11.0 knots. The apparent wind is blowing from 120 degrees True at 29 knots. Relative to our heading, 120 degrees True is 030 degrees. What is the speed and direction of the TRUE WIND. Since the apparent wind is forward of our beam we use the sign "-". The first formula is: a = sq. root of ( [b*b] + [c*c] [+ or -] (2 * [b*c] * cos A ) = sq. root of ( [29*29] + [11*11] - (2*29*11 * cos 30 deg.) ) = sq. root of 841 + 121 - (638 * 0.866025) = sq. root of 962 - 552.523950 = sq. root of 409.48 a = 20.24 knots of TRUE WIND Now we solve for the direction of the TRUE WIND. sin B = ( b / a ) * sin A = ( 29 / 20.24) * sin 30 deg. = 1.4328 * 0.500000 = 0.7164 B = 45 deg. 45.'5 call it 46 deg. The 46 degs. is the supplement of angle "B" and since the apparent wind is on our starboard bow we add the 46 deg. to our heading of 090 & get a TRUE WIND direction of 136 degrees. So we now know the TRUE WIND is 20 knots and coming from a direction of 136 degrees. EXAMPLE: 2) We are on a course of 305 degrees at a speed of 15 knots, the APPARENT WIND is from 230 deg. relative to our course at a speed of 8 knots. What is the speed and direction of the TRUE WIND? cos A = 50 deg. (230 deg. - 180 deg.) a = sq. root of ( [b*b] + [c*c] [+ or -] (2 * [b*c] * cos A ) a = sq. root of ( [8*8] + [15*15] + (2 * 8 * 15 * cos 50 deg.) ) a = sq. root of 64 + 225 + (240 * 0.642788) a = sq. root of 289 + 154.269 a = sq. root of 443.269 a = 21.05 Formula 2 is: sin B = ( b / a ) * sin A = (8 / 21.05) * sin 50 deg. = 0.380048 * 0.766044 = 0.291133 B = 16 deg. 55' call it 17 deg. The TRUE WIND is blowing at 21 knots and is coming from 197 deg. (180+17) relative to our heading or from 142 deg. TRUE (305- 180+17). ****************************************************************** Barometric-Pressure Conversion Factors To convert from millibars to inches of mercury: IM = 0.02953 * Mbs To convert from inches of mercury to millibars: Mbs = IM / 0.02953 To convert from millimeters of mercury to millibars: Mbs = Mm / 0.75 IM = pressure in inches of mercury. Mbs = number of millibars. Mm = millimeters of mercury. **************************************************************** Sailing to Weather It is self-evident that most sailing races are won or lost by a boats ability and performance while sailing to weather. When the mark is dead to weather, the closer to the wind you can sail the shorter the distance you have to travel. Conversely, if you have to bear off because of a bad head sea the further you will have to sail. The distance you must sail, to reach the mark when it is dead to weather, I have been taught, is best expressed as a percentage of the straight line distance to the mark. This percentage is found as follows: D = (200 / sine of the angle between tacks) * sine attack angle D = percentage of the distance to the mark. attack angle = the angle off the wind when sailing. EXAMPLE: What is the increase in the distance we must sail to reach the mark that is 4.5 miles dead to weather if we tack 1) 84 deg. 2) 88 deg. 3) 94 deg. D = (200 / sine of the angle between tacks) * sine attack angle 1) D = (200 / sin 84 deg.) * sin 42 deg. = (200 / 0.994522) * 0.669131 = 201.1 * 0.669131 = 134.6% Distance = 4.5 * 134.6% = 6.1 miles 2) D = (200 / sine of the angle between tacks) * sine attack angle = (200 / 0.999391) * 0.694658 = 200.1 * 0.694658 = 139.0% Distance = 4.5 * 139.0% = 6.3 miles 3) D = (200 / sine of the angle between tacks) * sine attack angle = (200 / 0.997564) * 0.731354 = 200.5 * 0.731354 = 146.6% Distance = 4.5 * 146.6% = 6.6 miles From the above you can see that a considerable speed increase is needed if we bear off from our "normal" attack angle. If we normally tack through say 90 degrees, we would have to increase our speed by at a minimum 6%. ***************************************************************** TACKING DOWN WIND When the Lee Mark is Dead to Leeward In light going, a sailboat running can normally increase her speed if she hardens her wind. The problem is to determine whether the increase in speed will offset the additional distance you will travel. When looking at this problem, erratic actions of the weather must be ruled out, and it must be assumed that the wind will remain constant in both speed and direction. When there is a change is the winds speed or direction we must redo our solution to account for the changes. The first step is to determine the increased speed for a given angle of divergence from the base course. Normally we use 10 deg. increments. For now we will assume we will only have 2 legs in running to the mark. Most likely you will have more than 2 legs in order not to wander to far from the rhumb line. For any given angle of divergence from the rhumb line the distance sailed we remain the same, no matter how many legs. The total distance sailed for a given angle of divergence from the base line can be determined as follows: Total distance sailed = 2 * base distance * sine divergence angle / (sine (divergence angle * 2) Knowing the speed for each divergence angle and also the total distance to be sailed if that divergence is used, the divergence angle that will get you to the lee mark in the least amount of time can be determined. EXAMPLE: The lee mark is exactly 10.0 miles away, and when we sail directly for it, our speed is 5.0 knts.. When we harden up 10 deg. our speed increases to 5.25 knts., at 20 deg. our speed is 5.65 knts., and at 30 deg. our speed is 6.0 knts. What is our optimum divergence angle? Without doing all the math (use above formula), below is a chart of the solutions: Divergence | Speed in | Distance to | Time required angle in deg. | Knts. or MPH | sail in miles | in Hours ----------------|---------------|---------------|---------------- 0 | 5.0 | 10.0 | 2.0 10 | 5.25 | 10.15 | 1.935 20 | 5.65 | 10.64 | 1.88 30 | 6.00 | 10.54 | 1.923 From the above chart we can see that to get to the mark in the least amount of time we should harden our wind 20 deg. ******************************************************************** Time Conversion a) Time to Arc Time can very easily be converted into its arc equivalent in a series of steps. To aid you just remember to following: 1 hour of time = 15 degrees 4 minutes of time = 1 degree or 60' (60 minutes of arc) 1 minute of time = 0.'25 or 15' (15 minutes of arc) 4 seconds of time = 1' (1 second of arc) 1 second of time = 0.'25 (1/4 second of arc) 1) Multiply the number of hours by 15, and note the resulting number as degrees. 2) Divide by 4 the number of minutes, and note the resulting whole number as degrees. 3) multiply by 15 the number of minutes remaining, and note the resulting number as minutes of arc. 4) Multiply by 0.25 the number of seconds, and note the resulting number as minutes of arc. 5) Add together the number of degrees and minutes of arc in the above steps. EXAMPLE: Convert 17 hours 33 minutes 51 seconds into its arc equivalent. degrees minutes -------- ---------- 1) 17 hours * 15 = 255 2) 33 minutes / 4 = 8 with 1 left 8 3) 1 (leftover from above) * 15 = 15 4) 51 seconds * 0.25 = 12.75 ------------------- 263 27.75 Therefore: 17 hours, 33 minutes, 51 seconds of time = 263 deg. 27.'75 minutes of arc. ************ ************ ********** ********** ****** b) Arc to Time Look at the chart above to see the relationship of arc and time. The step are as follows: 1) Divide the number of degrees by 15, and note the resulting whole number as hours. 2) Multiply the remaining number by 4 and note the result as minutes. 3) If the remaining number of minutes is greater than 15, divide it by 15, and note the whole number in the dividend as minutes of time. If it is less than 15, goto step 4. 4) Multiply by 4 the remaining number of minutes of arc and the decimal of minute, and note the answer to the nearest second. 5) Add together all hours, minutes and seconds from the above step. EXAMPLE: Convert 329 deg. 59.'6 to time. hours minutes seconds 1) 329 deg. / 15 = 21 hours + 14 deg. 21 2) 14 deg. * 4 = 56 3) 59.'6 / 15 = 3 minutes + 14.'6 3 4) 14.'6 * 4 = 58.4 seconds 58 --------------------------- 21 59 58 Therefore: 329 deg. 59.'6 = 21 hours, 59 minutes, 58 seconds of time. ********************************************************************** DISTANCE by BEARINGS Distance off, with 1 bearing & a Run to Beam You are able to calculate how far away a fixed object when it is abeam by taking a bearing on the bow, running "R" miles until the object is abeam. D = (R * sin A) / cos A D = Distance off when abeam R = miles traveled A = initial angle on the bow EXAMPLE: We sight on the shore a tower bearing 319 degrees RELATIVE, after running 6.5 miles the tower is abeam. How far are we from the tower when it is abeam? First, 319 deg. Relative is 41 deg. on the bow, so our problem looks like this. D = (R * sin A) / cos A = ( 6.5 * sin 41 deg.) / cos 41 deg. = ( 6.5 * 0.656059 ) / 0.754710 = 4.264384 / 0.754710 D = 5.7 miles away when abeam ****************************************************************** Distance off at 2nd bearing, from 2 bearings on the Bow & a Run The distance from a fixed object can be computed by taking two bearings on the bow and a run between the two bearings of a known distance. D = (R * sin A) / sin (A [abs. diff.] B) D = distance off at second angle R = distance of the run A = 1st angle on the bow B = 2nd angle on the bow EXAMPLE: A prominent landmark bears 27 deg. on the bow, after running 3.6 miles it bears 66 deg. on the bow. How far are we from the landmark at the time of the second bearing. D = (R * sin A) / sin (A [abs. diff.] B) = (3.6 * sin 27) / sin [27 abs. diff. 66] = (3.6 * 0.453990) / sin 39 = 1.634364 / 0.629320 D = 2.6 miles ****************************************************************** Distance off when Abeam by 2 bearings on the Bow and a Run Between The distance off a fixed object when abeam can be computed when we have 2 bearings on the bow and a run of a known distance between them. To solve this problem, the first step is to find the distance we are of at the bearing. This can be computed when we get our 2nd bearing and than we determine the angle the object formed by the two bearing lines as the apex of a triangle. (ie. 1st bearing = 30 deg. & the 2nd bearing = 50 deg. - the object angle = 20 deg. ( 180 - (130+30) ). The distance off the object at the time of the first bearing is then found as follows: D1 = (R * sin B2) / sin C D1 = Distance off at the first bearing. R = The run between the 1st & 2nd bearing. B2 = The second bearing. C = The angle formed at the object by the 2 bearing lines. After solving the above, we can solve the problem of, how far off we will be when abeam with: D2 = D1 * sin A D2 = Distance off when abeam D1 = Distance off at the time of the first bearing. A = The first angle on the bow. EXAMPLE: We sight an object bearing 28 deg. on the bow. After sailing 6.5 miles the second bearing is 52 deg. on the bow. How far off will the object be when it is abeam? 1) Determine the angle formed at the object by the 1st & 2nd bearing lines. Take the 2nd angle of 52 deg. and subtract it from 180 deg.. This gives you its compliment.. this is 128 deg.. Now subtract from 180 deg. the 1st bearing and 128 deg. in order to find the angle formed at the object. 180-28-128 = 24 deg. = angle "C". So to solve the first part we say: D1 = (R * sin B2) / sin C = (6.5 * sin 52 deg.) / sin 24 deg. = (6.5 * 0.788011) / 0.406737 = 5.1 / 0.406737 D1 = 12.5 miles 2) Solving for when abeam: D2 = D1 * sin A = 12.5 * sin 28 deg. = 12.5 * 0.469472 D2 = 5.9 miles Therefore when abeam the object will be 5.9 away. NOTE: This does _NOT_ take into account any outside factors such as set, drift, tides, currents etc.. SO BE FOREWARNED ****************************************************************** Run to a given Bearing and Distance off when on that Bearing At times it is necessary for us to determine the distance we need to sail to bring a fixed object to a given bearing and to compute its distance off when on that bearing. This problem has three steps. The first step is to obtain 2 bearing on the bow and to note the distance sailed between the 2 bearings. The distance off at the time of the first bearing can be computed as follows: D1 = (R1 * sin B) / sin C D1 = Distance off at the first bearing. R1 = The run between the 1st & 2nd bearing. B = The second bearing on the bow. C = The angle formed at the object by the 2 bearing lines. The second step is to compute the distance we have to sail to bring the object to the second or given bearing. This formula is: R2 = (D1 * sin E) / sin F R2 = The run between the first and second bearing. D1 = Distance off at the first bearing. E = The angle formed at the object by the 2 bearing lines. F = The bearing on the bow of the object when on the given bearing. The third step is to compute the distance off at the time we at the given bearing. D2 = (D1 * sin A) / sin F D2 = Distance off given bearing. D1 = Distance off at the first bearing. A = The first bearing on the bow F = The bearing on the bow of the object when on the given bearing. EXAMPLE: We are sailing on a course of 273 deg., speed 12.0 knots when we sight a navigational light on shore. We want to change our course to 305 deg. when the light bears 333 deg.. At 2207 hrs. the light bears 293 deg. and at 2257 hrs. it bears 308 deg.. At what time will we change course and how far off will the light be at that time. Since we are on a course of 273 deg., at 2207 the light was 20 deg. on the bow (293-273), and at the second the light bore 35 deg. (308-273). The angle formed by the 2 bearing lines "C" is 15 deg. and at the speed of 12.0 knots in 50 minutes we sailed 10.0 miles. The first step: D1 = (R1 * sin B) / sin C = (10.0 * sin 35 deg.) / sin 15 deg. = (10 * 0.573576) / 0.258819 = 5.74 / 0.258819 D1 = 22.18 miles off at the time of the 1st bearing. Step 2: We are to change our course when the light bears 333 deg. The bearing on the bow will then be 60 deg. and the angle at the light will therefore be 40 deg. (60-20). So let us solve the second step. R2 = (D1 * sin E) / sin F = (22.18 * sin 40 deg.) / sin 60 deg. = (22.18 * 0.642788) / 0.866025 = 14.26 / 0.866025 R2 = 16.5 miles The run from the first bearing to the turning bearing is 16.5 miles. At our speed of 12 knts. it will take us 82.5 minutes to sail this distance ( [16.5 / 12] * 60), lets call it 83 minutes. Since our first bearing was at 2207 hrs. at 2330 hrs. we should change our course. The last step is to compute our distance off at the time of the course change. D2 = (D1 * sin A) / sin F = (22.18 * sin 20 deg.) / sin 60 deg. = (22.18 * 0.342020) / 0.866025 = 7.586 / 0.866025 = 8.76 miles off when we change course. ******************************************************************* Distance Off 2 Fixed Objects I hope you will be able to understand the "drawing" I have provided. I find it a little hard to explain this formula and procedure in just words. When the distance between two FIXED objects and the bearing from one to the other is known, we are able to compute our distance from each object with out plotting by our true bearing from each object. The solution is by the law of sines: side "boat"/sin "LS" : "a"/sin "A" : "b"/sin "B" In this ratio, "LA" is the angular difference between the true bearing of "A" from our boat and the bearing of "B" from "A". "LB" is the angular difference between its bearing from "A" and its true bearing from our boat. Our boat is located at "Boat". "LS" is the angular difference between the true bearings of "A" and "B". The known distance between "A" and "B" is "side 'boat'". The distance our boat from "A" is "side 'boat'" and the distance our boat is from "B" is called "side 'a'" NORTH | |<------> Bearing of "B" from "A" | | | | | . | | . | . LA side "boat" "A"| . _____________ . "B" . "LB". side "B" . . . . NORTH . . | . . side "a" | . . | . "LS". . . "Boat" EXAMPLE: Cuttyhunk Light bears 074 deg. TRUE from Buzzards Light and it is 3.96 miles apart. We get a bearing of 015 deg. TRUE on Buzzards, which we will call "A" and bearing of 050 deg. on Cuttyhunk, which we call "B". What is our distance from both lights. Angle "LS" is 35 deg. (050-015), while the side "boat" is 3.96 miles. For angle "A" we use 59 deg. (074-015), and for angle "LB" 24 deg. (074-050). Our distance off buzzards light is side "b" and our distance off Cuttyhunk light is side "a". The ratio is: side "boat"/sin "LS" : "a"/sin "A" : "b"/sin "B" 3.96/sin 35 deg. = "a"/sin 59 deg. = "b"/sin 24 deg. 3.96/0.573576 = "a"/0.857167 = "b"/0.406737 6.904055 = "a"/0.857167 = "b"/0.406737 so 6.904055 * 0.857167 = "a" 5.9 miles = "a" and 6.904055 * 0.406737 = "b" 2.8 miles = "b" From where our boat lies Buzzards Light is 2.8 miles away and Cuttyhunk Light is 5.9 miles away. ****************************************************************** Heading to Bring Light to Specified Distance & Bearing and the run Thereto At night when we have good visibility and normal refraction, it is possible to estimated the correction needed to bring a light to a specified distance and bearing, and the estimate the time at which we will arrive at that point. The first step is to calculate the correction to be made to our present heading so as to be the required distance off when we are on the specified bearing. The course correction is found with the following formula: sin C = (D2 * sin B) / D1 C = The bearing on the bow relative to our present heading to which the light must be brought so that it will be at the specified distance. D2 = The specified distance from the light when the desired bearing is reached. B = The bearing on the bow relative to our PRESENT heading when the light is on the desired bearing. D1 = The range of visibility of the light for the observers eye height. The correction to our present heading is then made by bringing the light to the bearing on the bow,"C", from above The run to the point where the light is on the required bearing and at the specified distance, "D3" is found as follows: D3 = (sin E * D1) / sin F D3 = Specified distance. E = The angle at the light between the first bearing and the bearing when at the specified distance. D1 = The range of visibility of the light for the observers eye height. F = The lights bearing on the bow relative to the corrected heading when at the specified distance. EXAMPLE: Our present course is 140 deg., speed 13.0 knts.. Our course is to be changed when Light "X" is 9 miles away, and bears 205 deg. For our eye height, Light "X" will become visible at a range of 18.6 miles. At 2217 we sight Light "X", bearing 160 deg. True, or 20 deg. on the bow. What is the change required to our present heading to bring us to the required distance off the light when it bears 205 deg. Also, at what time will we be at that point. The first step is to find out what the relative bearing of the light should be when the boat is headed for the point where the course is to be changed. Angle "B" = 065 deg. (205-140). Therefore: sin C = (D2 * sin B) / D1 = (9.0 * sin 65 deg.) / 18.6 = (9.0 * 0.906308) / 18.6 = 8.156770 / 18.6 sin C = 0.438536 C = 26 deg. To bring the boat to the point where the course is to be changed, the light should bear 26 deg. on the bow. So we alter our course to the left by 6 deg.. (we sighted the light 20 deg. on the bow -- see above) to a new heading of 134 deg. Next we compute the run to the turning point: E = 45 deg. (205 - 140) F = 71 deg. (205 - 134) D3 = (sin E * D1) / sin F = (sin 45 deg. * 18.6) / sin 71 deg. = (0.707107 * 18.6) / 0.945519 = 13.152186 / 0.945519 D3 = 13.9 miles The distance from where the light first came into sight to the point where the course is to be changed is 13.9 miles. At our present speed of 13.0 knts. it will take us 64.2 minutes ( [ 13.9/13 ] * 60) to arrive at the turning point. Since we sighted the light at 2217 hrs. we will be at the point for our course correction at 2321 hrs. ****************************************************************** Rate of Change of Altitude There are times when the navigator may need to determine a body's rate of change of altitude. If a sequence of sights of the same body has been taken, the rate of change provides a check on the accuracy of the observations. Also, if a star finder has been used to predict the altitudes and azimuths and the navigator has not been able to shoot the stars because of poor visibility, correction of the sextant settings will compensate for the delay. The formula for computing the rate of change of altitude, delta H, in minutes of time is: delta H = 15 * cos L * sin Z OR To compute delta H in seconds of time we use: delta H = (cos L * sin Z) / 4 L = Latitude of the observer. Z = The angle between the body and the meridian. EXAMPLE: We are in Lat. 30 deg. and the predicted azimuth of the body is 100 deg.. What is the body's rate of change of altitude in minutes of time? Since the body's predicted azimuth is 100 deg.: Z = 80 deg. (180-100) and the formula is: delta H = 15 * cos L * sin Z = 15 * cos 30 deg. * sin 80 deg. = 15 * 0.866025 * 0.984808 delta H = 12.'8 per minute of time. ******************************************************************* Rate of change of Azimuth A stationary observer may find the rate of change of azimuth of a body by the use of two formulae. The first one determines the parallactic angle, M, and the second provides the actual rate of change of azimuth. In the celestial triangle, PZM, the parallactic angle, M, is the angle formed at the body. Even tho this formula is for a stationary observer it works very well on board a slow vessel. To find the angle "M", the formula is: sin M = (cos L * sin Z) / cos d L = observers latitude. Z = Azimuth angle. d = Declination. After finding angle "M", we go on to find the rate of change of azimuth per minute of time as follows: delta Z' = (15 * cos d * cos M) / cos H delta Z' = Rate of change of azimuth in minutes of arc per minute of time. d = Declination. M = (from above) H = Computed altitude or the corrected sextant altitude. EXAMPLE: We are in Latitude 40 deg. N. and the body's dec. is N 27 deg. 30'. The azimuth, Zn, is 163.9 deg. (163 deg. 54'), which we convert to an azimuth angle of 16 deg. 06' (180- 163 deg. 54'), the corrected altitude, Ho, is 77 deg. 04.'2. What is the rate of change of azimuth in minutes of arc in one minute of time? First we compute sin M. sin M = (cos L * sin Z) / cos d = (cos 40 deg. * sin 16 deg. 06') / cos 27 deg. 30' = (0.766044 * 0.277315) / 0.887011 = 0.212435 / 0.887011 = 0.239496 M = 13 deg 51' Next we solve for delta Z: delta Z' = (15 * cos d * cos M) / cos H = (15 * cos 27 deg. 30' * cos 13 deg. 51') / cos 77 deg. 04.'2 = (15 * 0.887011 * 0.970926) / 0.223760 = 12.918331 / 0.223760 Z' = 57.'7 Therefore the azimuth is changing at a rate of about 57.'7 per minute of time. ****************************************************************** SIGHT REDUCTION In order to plot the line of position, "LOP", resulting from an observation of a celestial body, two computations are required, both the altitude of the body, "H", and its azimuth, "Z". In order to solve "H" we must first compute "Z". One VERY important point I must make before going on. The quadrant in which the object lies CAN NOT be determined by these calculations, you must obtain an approximate azimuth at the time of your sight. For computing azimuth or altitude the following are constant: L = Latitude d = Declination t = Meridian angle Ho = Fully corrected sextant angle ********************************************************************* Computing Azimuth There are three formulae for computing the azimuth of a body, each one requires different in put, so based on which data you have you can opt for the formula of your choice. The first formula is the easiest. 1) sin Z = (cos d * sin t) / cos H 2) cos Z = (sin d [+ or abs. diff.] sin H * sin L) / (cos H * cos L) 3) tan Z = sin t / (cos L * tan d [+ or abs. diff.] sin L * cos t NOTES: For #1 above: In the divisor you can use "H", "Hc" or "Ho" For #2 above: The sign is "+" when "L" & "d" have opposite names and when "L" & "d" have the same name the sign is "abs. diff." and the smaller amount is subtracted from the larger. For #3 above: a) When "L" & "d" have the same name AND "t" is less than 90 deg. the sign is "-". If "t" is greater than 90 deg. the sign is "+". b) If "L" & "d" have opposite names the sign is "+", "t" being less than 90 deg. EXAMPLE: 1) You are in Lat. 34 deg. N and observe to the Southwest a body whose dec., "d" is S 22 deg. and meridian angle, "t" is 35 deg. W. "Ho" is 24 deg. 51.'0. Formula 1) sin Z = (cos d * sin t) / cos H = (cos 22 deg * sin 35 deg) / cos 24 deg. 51' = 0.927184 * 0.573576 / 0.907411 = 0.531810 / 0.907411 = 0.586075 Z = 35 deg. 53' The azimuth angle is S 35 deg. 53' W, and the azimuth "Zn" is 215 deg. 53 minutes (180+35 deg. 53'). ******************* 2) You are in Lat. 34 deg. N and observe to the Northeast a body whose "Ho" is 20 deg. 08.'0 and dec. is N 19 deg. 43.'7. Formula 2) cos Z = (sin d [+ or abs. diff.] sin H * sin L) / (cos H * cos L) = ( [sin 19 deg. 43.'7 [abs. diff.] sin 20 deg. 08.'0] * sin 34 deg.) / cos 20 deg. 08.'0 * cos 34 deg. = 0.144778 / 0.778379 = 0.185999 Z = N 79 deg. 17' E = Zn of 79.3 deg *********************** 3) You are in Lat. 31 deg. N and observe to the East- north-east a body whose "d" is N 29 deg. and "t" is N 30 deg. E. 3) tan Z = sin t / (cos L * tan d [+ or abs. diff.] sin L * cos t = sin 30 deg. / (cos 31 deg. * tan 29 deg. [abs. diff.] sin 31 deg. * cos 30 deg. = 0.500000 / (0.857167 * 0.554309) [abs. diff.] 0.515038 * 0.866025 = 0.500000 / 0.475135 [abs. diff.] 0.446448 = 0.500000 / 0.028687 = 17.429452 Z = 86 deg. 43' = N 86.7 deg. E Zn 86.7 deg. ****************************************************************** Computing Altitude As in computing azimuth there are three formulae we can use, and as above each has it own time and place, and in part by the data you have to in put. The first one is a classic and the one used to produce H.O.229. 1) sin H = sin L * sin d [+ or abs. diff.] cos L * cos d * cos t NOTE: a) If "t" is less than 90 deg. AND "L" & "d" have the same name the sign is "+". If "L" & "d" have opposite names the sign is "abs. diff.". b) If "t" is greater than 90 deg. AND "L" & "d" have the same name the sign is "abs. diff.". If "L" & "d" have opposite names the sign is "+". c) When "t" is greater than 90 deg. use (180 deg. - t) which we call "t1" EXAMPLE: Given - L = 14 deg N, d = N 29 deg. and t = 94 deg. E. Since "t" is greater T1 = (180 - 94) = 86 deg. sin H = sin L * sin d [+ or abs. diff.] cos L * cos d * cos t = sin 14 deg. * sin 29 deg. [abs. diff.] cos 14 deg. * cos 29 deg. * cos 86 deg. = 0.241922 * 0.484810 [abs. diff.] 0.970296 * 0.874620 * 0.069756 = 0.117286 [abs. diff.] 0.059198 = 0.058088 Hc = 3 deg. 19.'8 2) cos H = (cos L * sin d ["+" or abs. diff.] sin L * cos d * cos t / cos Z NOTE: a) Before you can use this formula you must calculate "Z" by means of the tan Z formula (#3 above). b) The sign is "+" if "L" & "d" are of opposite names. If "L" & "d" are of the same name AND "t" is less than 90 deg. the sign is "-", but if "t" is greater than 90 deg. the sign is "+". EXAMPLE: Our D.R. Lat. is 23 deg. 57.'5 N and we observed the morning Sun to have a corrected altitude of 56 deg. 30.'5 and it was bearing about 160 deg. TRUE by our compass. At the time of the sight the Sun's dec. was S 8 deg. 15.'1 & "t" = 9 deg. 29.'3 E. We want to find the computed "Z" & "H" as well as "Zn". First, we compute the value for "Z" using the tan Z formula from above and find "Z" to be S 17 deg. 10' E (17.2 deg.), which we will use in computing "H", Zn = 162.8 deg. (180 - 17.2). cos H = (cos L * sin d + sin L * cos d * cos t / cos Z = (cos 23 deg. 57.'5 * sin 8 deg. 15.'1 + sin 23 deg. 57.'5 * cos 8 deg. 15.'1 * cos 9 deg. 29.'3 / cos 17 deg. 10' = (0.913841 * 0.143521) + 0.406072 * 0.989647 * 0.986319 / 0.955450 = 0.131155 + 0.396370 / 0.955450 = 0.527525 / 0.955450 = 0.552122 Ha = 56 deg. 29.'2 Ho = 56 deg. 30.'5 intercept = 1.'7 Towards The last formula we can employ is tan H, to use this we first must compute "M". "M" is the angle at the geographic position of the body in the navigational triangle, PZM. The other name is the parallactic angle. Before you can compute "M", you must find "Z" using the tan Z formula from the above section. Because of the complexity of this solution, I will only give the formulae. I am providing this solution only as a last resource if you should ever need it! 1) Solve for tan Z from above. 2) tan M = sin Z / (cos Ho * tan L) [+ or abs. diff.] (sin Ho * cos Z) 3) tan H = ( (sin M / tan t) [+ or abs. diff.] sin d * cos M) / cos d NOTES: a) In #2 above (tan M), the sign is "+" when "L" & "d" are the same name AND "t" is less than 90 deg.; when "t" is greater than 90 deg. the sign is [abs. diff.]. The sign is also "abs. diff." if "L" & "d" have opposite names. b) In #3 above (tan H), the sign is "+" when "L" & "d" have the same name AND "t" is less than 90 deg.. If "t" is greater than 90 deg. the sign is [abs. diff.]. The sign is also [abs. diff.] if "L" & "d" have opposite names. ****************************************************************** CONVERSION FACTORS ---------------------- SPEED 1 statute mile per hour = 88.0 feet per minute * = 29.333333 yards per minute = 1.60934 kilometers per hour = 1.466667 feet per second = 0.86898 knots = 0.44704 meters per second * 1 knot per hour = 101.26859 feet per minute = 33.75620 yards per minute = 1.852 kilometers per hour * = 1.68781 feet per second = 1.15078 statute miles per hour = 0.51444 meters per second 1 kilometer per hour = 0.62137 statute miles per hour = 0.53996 knots 1 meter per second = 196.85039 feet per minute = 65.61680 yards per minute = 3.6 kilometers per hour * = 3.28084 feet per second = 2.23694 statute miles per hour = 1.94384 knots 1 yard per minute = 0.03409 statute miles per hour = 0.02962 knots = 0.01524 meters per second * DISTANCE 1 statute mile = 5,280.0 feet * = 1,609.344 meters * = 0.86898 nautical miles 1 nautical mile = 6,076.11549 feet = 2,025.37183 yards = 1,852.0 meters * = 1.15078 statute miles 1 kilometer = 3,280.83990 feet = 1,093.61330 yards = 0.62137 statute miles = 0.53996 nautical miles 1 meter = 3.28083990 feet 1 meter = 1.09361330 yards 1 yard = 0.9144 meters * 1 foot = 0.3048 meters * NOTE: "*" = absolute measurement