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How to Compute the Distance and Bearing from Two Positions

on the Earth

How to Compute the Distance How to Compute the Distance and Bearing from Two Positions and Bearing from Two Positions

on the Earthon the EarthPosition Information is needed in Position Information is needed in Latitude and Longitude formatLatitude and Longitude format

Sept 2008John McHaleOct 2008

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What are We Going to Talk About

• How did I find your school?• How does the GPS work – is it magic

or is it math?• Why study math?

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Why Do You Need Math?• Why do you need to add & subtract?

• Why do you need to multiply & divide?

• Why else do you need math?

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This Seminar is an Example of How Math Can Be Used.

• The Law of Squares - Also called the Pythagorean Theorem

• Works for any Right Triangle• A2 = B2 + C2

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CAny Shape of Right Triangle A2 = B2 + C2

AB

C

AB

A

B

C

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How is Math Used to Navigate?

The Earth is approximately a Sphere.Actually, Earth is an Oblate Spheroid.

Earth is squashed at the poles.

A Grid of Latitude and Longitude is used to Divide up the Earth’s Surface.

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DefinitionsPosition is a combination of Latitude, Longitude, and Altitude above Sea Level.

Position is Expressed in Degrees (O), Minutes (‘) and Seconds (“)

This format is better to use for a more accurate determination of Distance and Bearing.

Bearing is sometimes called Course.

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Earth’s ShapeLatitude is the measurement from the

Equator at 0 Degrees to 90 Degrees at the North (N) or South (S) Pole.

A Minute of Latitude Equals a Nautical Mile.

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Latitute -lines that go around

the globe

Latitude -lines that go around

the Earth

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Earth’s ShapeLongitude is the Measurement from Zero

Meridian (Greenwich, England) East or West to 180 Degrees at the International Date Line, or Antimeridian (in the Pacific Ocean).

A Nautical Mile at the Equator Equals a Minute of Longitude, BUT

A Nautical Mile at any other Latitude equals the Cosine of the Latitude Times the Longitude in Minutes.

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International Date Line

Longitude -lines that go up &

down around the Earth

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Notice Convergence of Latitude

The Distance of Longitude between Latitude Lines decreases as Distance is Increased from the Equator and Becomes Zero at the Poles.

Computation of Distance between any two Positions is Required.

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Longitude lines are closer together as you go to the

north pole – called convergence

Latitude calculation must be adjusted for convergence

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So, What about Math & Navigation?

We have to calculate the difference between latitude lines.

We have to calculate the difference between longitude lines.

We have to calculate the adjustment for convergence.

Then, we calculate the distance between two points.

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So What Math Do We Use?• Adding & subtracting• Multiplying• Square root• Cosine function*• Arctangent or inverse tangent function*

* These will be explained.

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What is a COSINE?• The cosine of angle is equal to the

length of B, the side adjacent to the angle, divided by the length of the triangle's hypotenuse A.

A BAngle

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Example: Positions A and BA= Your School in Rockville, MD B = A Ranch in Texas

Latitude Longitude

A: N 39o 5’ 40.20” W 77o 6’ 12.6”

B: N 28o 52’ 19.90” W 97o 15’ 13.11”

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Short Distance Between Two Positions

Square root of the sum of the – Square of the Difference (Delta) between the

Latitudes of the Two Positions and the

- Square of the Difference (Delta) between the Longitude of the Two Positions.

D = (Delta Latitude)2 + (Delta Longitude)2

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Determine Latitude of AA

Distance

is Unknown

B

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Math Used to Convert Latitude 39o 5’ 40.20”to Minutes

Convert 39o latitude degrees to minutes…

There are 60 minutes in a degree.So, we multiply:

39 X 60 = 2,340’.

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Math Used to Convert Latitude 39o 5’ 40.20”to Minutes

convert 40.20” latitude seconds to minutes…

There are 60 seconds in a minute.So, we divide:

40.20/60 = 0.67’

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Math Used to Convert Latitude 39o 5’ 16.68”to Minutes

All three of the latitude parts are converted to minutes

So, we add them together2340.000 from degrees0000.670 from seconds0005.000 minutes ------------2345.67’

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Determine Longitude of A

A

Distance i

s Unknown

B

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Math Used to Convert Longitude 77o 6’ 12.60”to Minutes

Convert 77o to minutesThere are 60 minutes in a degree. So, we multiply :

77 x 60 = 4,620

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Math Used to Convert Longitude77o 6’ 12.60” to Minutes

Convert 12.60” to minutesThere are 60 seconds in a minute.

So, we divide:12.6/60 = 0.21

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Math Used to Convert Longitude 77o 6’ 12.6”to Minutes

All three of the longitude parts are converted to minutes

So, we add them together4620.000 from degrees0000.21 from seconds0006.000 in minutes------------4626.21’

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A

Determ

ine the D

istance

BDetermine Latitude and

Longitude of B

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Do the Same Math to Determine the Latitude for

B in Minutes

Latitude 28o 52’ 19.90”Degrees + Minutes + Seconds(28 X 60) + 52 + (19.90/60) = 1,732.33’

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Do the Same Math to Determine the Longitude

for B in Minutes

Longitude 97o 15’ 13.11”Degrees + Minutes + Seconds(97 X 60) + 15 +(13.11/60) = 5,835.22’

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Positions A and B in Minutes

Latitude Longitude

A: 2,345.67’ 4,626.21’

B: 1,732.33’ 5,835.22’

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Determine the Delta (Differences) in Latitude in Minutes between A and B

Latitude A minus Latitude B:2,345.67

– 1,732.33 ----------------------------------------------

613.34’

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Determine the Delta (Differences) in Longitude in Minutes between A and B

Longitude A minus Longitude B:5,835.22

- 4,626.21---------------------------------------

1,209.01

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So, for Short DistanceD = (Delta Latitude)2 + (Delta Longitude)2

= (613.34)2 + (1209.01)2

= 376,185.95 + 1,461,705

= 183,7891 = 1,355.7 nautical miles

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What…Nautical Miles?

• We use Nautical miles because of the minutes of Latitude and Longitude.

• We need to convert nautical miles to statute miles.

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Converting Nautical Miles to Statute Miles

Distance is 1,355.7 nautical miles

One nautical mile = 1.151 Statute milesTherefore,

1,355.7 X 1.1508 = 1,560.1 Statute miles

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However, We have not considered the Latitude Convergence Problem.

The previous answer is probably wrong because of the large distance between your school and Texas.

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Median Distance Between Two Positions

Square root of the sum of the – Square of the Difference (Delta) between

the Latitudes of the Two Positions and the

- Square of the Cosine of the midpoint of Latitude Times the Difference (Delta) between the Longitude of the Two Positions.

(Delta Latitude)2 + (Cosine (mid Latitude) X Delta Longitude)2D =

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Difference of Longitude

A

B

Difference of Latitude

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Determine the mid Latitude of the Two Positions in Minutes

Latitude of B + Difference of Latitudes1,732.33 + (2,345.67-1,732.33)/2 =

1,732.33 + 306.66 = 2,038.99’

Now, we have the mid latitude in minutes.

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A

Delta Longitude

Del

ta L

atitu

deDistance

Mid Latitude

B

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radius

Circumference = 2 X X radius2 X X radius = 360 degrees

Circumference

How Do We Compute Radians ? One radian is the angle of an arc created by wrapping the radius of a circle around its circumference.

1 radian = 57.29582 deg

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Converting degrees to radians:

• = 3.14159• So, 1 radian = 360/(2 X 3.14159) • 1 radian = 57.29582 degrees of arc• 1 radian = 3,437.75 minutes of arc

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WOW! More Math Steps

Our mid latitude answer of 2,039.0008 minutes must be converted to radians.

Converting to Radians 2,039.0008/ 3,437.75 = 0.59312

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CosineInserting Cosine (0.59312) in Google

Cosine of mid Latitude = 0.8292 RadiansThis is the correction factor for Latitude Convergence.

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Now to Determine North-South Distance

We square the Latitude Difference…613.34 X 613.34 = 376,185.9

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…Determine East - West Distance

We square the Cosine of the mid Latitude X the Longitude Difference.

(0.8292 X 1209.01)2 =

(1002.511)2 = 1,005,028.3

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(Cosine mid Latitude X Delta Longitude)2

(Del

ta L

atitu

de)2

A

Distance

(613

.34)

2

B(1002.51)2

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Square Root of Squares of Latitude and Longitude

DifferencesSquare of Latitude Square of the Cosine

mid Latitude X the Delta Longitude

(376,185.9 + 1,005,028.3)

This gives the answer in nautical miles.

Distance = 1,175.25 nautical miles

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Converting Nautical Miles to Statute Miles

Distance is 1,175.25 nautical miles

One nautical mile = 1.151 Statute milesTherefore,

1,175.25 X 1.1508 = 1,352.5 Statute miles

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But, the Earth’s surface curves. How can we compute the True Distance on the surface?

• Yes, Spherical geometry• Computes the Great Circle Distance

on the Earth’s Surface.

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Spherical GeometryD = ACOS((SIN(Lat A) X SIN(Lat B)+ COS(Lat A) X COS(Lat B) X COS(LongA- Long B))

D = ACOS(0.304490048 + 0.63803824)= ACOS(.340678188) in radians= 1171.17 nm = 1,347.8 Statute miles

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-The THREE Answers are-• Short Distance = 1,560 Statute miles

• Medium Distance = 1,352 Statute miles

• Spherical Distance = 1,348 Statute miles

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Distance is FinishedWhat Else Do We Need?

• Bearing – figure out what angle we travel from our starting point A relative to TRUE North to get to B.

• We need to use the Arctangentfunction.

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BEARING

• Bearing is the Angle from Position A to B.• A New TERM is Needed for Determining

Bearing.• Quadrants

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Quadrants

1st is 0-904th is 270-360

3rd is 180-2702nd is 90-180

North

West 270 East 90

South 180

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Need to Define Our Answer for Quadrants?

Is Position A (your school) Greater or Less than B (the ranch in Texas)?

We have to figure out what Quadrant position B (the ranch) is in so the bearing can be corrected.

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Quadrants

1st is 0-904th is 270-360

3rd is 180-2702nd is 90-180

North

B

B

B

B

AWest 270 East 90

South 180

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Latitude LongitudeA: N 39o 5’ 40.20” W 77o 6’ 12.6”

B: N 28o 52’ 19.90” W 97o 15’ 13.11”

Latitude B is less (<) or south than ALongitude B is greater than (>) or west of A

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Determine Quadrant• LAT B>A (N) Long B<A (E) Quadrant 1• LAT B<A (S) Long B<A (E) Quadrant 2• LAT B<A (S) Long B>A (W) Quadrant 3• LAT B>A (N) Long B>A (W) Quadrant 4

Quadrant 3 is the answer

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B is > or west of A

A

B

B is < or south of A

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Quadrants

1st is 0-904th is 270-360

3rd is 180-2702nd is 90-180

North

West 270 East 90

Bearing

South 180

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Bearing Continued:

Bearing = Arctangent (TAN-1)of Difference of Latitude Divided by the Cosine of the mid Latitude X the

Difference in Longitude

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What is an Arctangent?

The inverse trigonometric functionsare the inverse functions of the trigonometric functions. So, the Arctangent is the inverse of the Tangent function.

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Cosine mid Latitude X Delta Longitude

Del

ta L

atitu

de

A

Distance

B

TAN = Delta Latitude divided by (Cosine mid Latitude X Delta Longitude)

= TAN-1 (Delta Latitude divided by(Cosine mid Latitude X Delta Long))

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Difference in latitude is 613.34

Cosine of the mid Latitude X the Difference in longitude is 1,002.51

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FinallyOur equation from

http://www.analyzemath.com/Calculators_2/arctan_calculator.html is:

Bearing = Arctangent (613.34/1,002.511)= Arctangent(0.611804)

Bearing = 0.549053 radians = 31.46 Degrees

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Correct the BearingBearing = 31.460

Position B is in the 3rd quadrant, so we subtract the bearing from 2700

270 – 31.46 = 238.54 degrees

Thus we have our corrected Bearing from A to B

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238.540

1352.57 statute miles

A

B

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2380

1352.57 statute miles

Difference of Longitude

Difference of Latitude

A

B

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Alternative Problems• For short distances, such as from your

school to your home, you could use seconds of Latitude and Longitude.

• Remember that a minute of distance is about a nautical mile

• A nautical mile equals 1.151 statute miles• OR 1.1508 X 5,280 feet equals 6,076.2

feet

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Converting to seconds • There are 60 arc seconds in a

nautical mile, therefore• An arc second equals about 6,075.2

divided by 60 or• An arc second of distance equals

101.2 feet

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Distance to your home from schoolLatitude Longitude

A: N 39o 5’ 40.20” W 77o 6’ 12.6”

B: N 39o 5’ ????” W 77o 6’ ????”Your problem is to figure out the distance

and bearing from school to your home.

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Use the short distance method from home to school

• For further problems use the medium distance such as to Philadelphia.

• You can get the Latitude and Longitude for Broad and Market Streets in Philadelphia from Google.

• Navigating is FUN.

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Presentation Available At:

www.wetrekmd.com

Presentation Available At:Presentation Available At:

www.wetrekmd.comwww.wetrekmd.com

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Sept 2008

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Thank You!!Thank You!!Thank You!!

Sept 2008

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Summary• Pages 1 to 17 Background• Pages 18 to 36 Distance Method 1• Pages 37 to 49 Distance Method 2• Pages 50 to 54 Distance Method 3• Pages 55 to 70 Bearing• Pages 71 to 75 Homework-Contact