Navigation formulae

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.

out with page breaks it one continuous file.

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.

Between.

Bearing.

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

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".

index correction +1.'9

Dip for 20', SIGN REVERSED +4.'3

-------

---------------

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

ahead or behind you are from the boats around you!

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 = 187 / sin 0 deg 45.'7

D = 187 / 0.013293

D = 14,067.34 feet

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

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).

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 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.

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

(15 / 13.75) - 1.1

-------

correction - 4.0 - 04.'0

---------------

1 deg. 03.'2 = (1 * 60 + 03.2) = 63.'2

curvature,

est. dist. 15 mi.

squared * 0.907 - 204'

---------

922' = corrected height

in minutes) * 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.

(8.2 / 13.75) - 0.6

-------

correction - 3.5 - 03.'5

---------------

1 deg. 03.'7 = (1 * 60 + 03.5) = 63.'5

curvature,

est. dist. 8.2

squared * 0.907 - 61'

---------

1,065' = corrected height

in minutes) * 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.

(9.5 / 13.75) - 0.7

-------

correction - 3.6 - 03.'6

---------------

1 deg. 03.'6 = (1 * 60 + 03.6) = 63.'6

curvature,

est. dist. 9.5

squared * 0.907 - 82'

---------

1,044' = corrected height

in minutes) * 0.566

= 16.4 * 0.566

= 9.3 MILES
This is a VERY GOOD answer for this problem, only 0.2 miles

different.

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"--> | .

| .

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"

| .

| .

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 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.

"-". The first formula is:

= 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

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 ( [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

= (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).

******************************************************************

To convert from millibars to inches of mercury:

IM = 0.02953 * Mbs
To convert from inches of mercury to millibars:

Mbs = IM / 0.02953

Mbs = Mm / 0.75

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.

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

angle

= 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

= 200.5 * 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%.

speed if she hardens her wind. The problem is to determine

whether the increase in speed will offset the additional distance

you will travel.

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:

angle / (sine (divergence angle * 2)

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.

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:

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.

a) Time to Arc

series of steps. To aid you just remember to following:

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.

-------- ----------

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

minutes of arc.

whole number as hours.

2) Multiply the remaining number by 4 and note the result as

minutes.

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.

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

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.

= ( 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.

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

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.

we will be when abeam with:

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:

= (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

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.

bring the object to the second or given bearing. This formula is:

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.

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.

= (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.

= (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"

| |

| .

"A"| . _____________ . "B"

. "LB".

side "B" . .

. .

NORTH . .

| . .

.

.

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.

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:

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

5.9 miles = "a"

2.8 miles = "b"

Cuttyhunk Light is 5.9 miles away.

and the run Thereto

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.

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:

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.

bearing is reached.

the light is on the desired bearing.

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:

bearing when at the specified distance.

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.

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:

= (9.0 * sin 65 deg.) / 18.6

= (9.0 * 0.906308) / 18.6

= 8.156770 / 18.6

sin C = 0.438536

C = 26 deg.

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.

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. 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:

OR
To compute delta H in seconds of time we use:

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?

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.

azimuth per minute of time as follows:

of time.

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.

minute of time?

= (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'

= (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

minute of time.

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

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.

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

"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'

deg. 53 minutes (180+35 deg. 53').

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

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.

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).

"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 *

= 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 *