This is part 2 (4mula.pt2 of useful Navigational Formulae. If
you have any questions please leave me a message in the forum or
EMail 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) BarometricPressure 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 0.11.0 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).
******************************************************************
BarometricPressure 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 selfevident 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. 18028128 = 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 (293273), and at the second the light bore 35
deg. (308273). 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. (6020). 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. (050015), while the side "boat" is 3.96
miles. For angle "A" we use 59 deg. (074015), and for angle "LB"
24 deg. (074050). 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. (205140).
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. (180100) 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
northeast 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
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