How to Compute a Distance and Bearing from Two Positions

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John McHale Sept 2008 Oct 2008 1 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?

2 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?

3 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

4 C Any Shape of Right Triangle A2 = B2 + C2

A B A B

A B

C 5 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.

6 7 Position is Definitionsa 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.

8 Earth’s Shape Latitude 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.

9 LatitudeLatitute -lines-lines that go around thethe Eagloberth

10 Longitude is the Measurement from Zero MeridianEarth’s (Greenwich, Shape 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.

11 International Date Line

Longitude -lines that go up & down around the Earth

12 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.

13 Longitude lines are closer together as you go to the north pole – called Latitude calculation must be convergence adjusted for convergence

14 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.

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

* These will be explained.

16 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 Angle B

17 Example: Positions A and B A= 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”

18 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

19 Determine Latitude of A A

wn no nk s U i nce sta Di

20 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’.

21 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’

22 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 together 2340.000 from degrees 0000.670 from seconds 0005.000 minutes ------2345.67’

23 Determine Longitude of A A

wn no nk s U e i nc sta Di

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

Convert 77o to minutes There are 60 minutes in a degree. So, we multiply : 77 x 60 = 4,620

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

Convert 12.60” to minutes There are 60 seconds in a minute.

So, we divide: 12.6/60 = 0.21

26 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 together 4620.000 from degrees 0000.21 from seconds 0006.000 in minutes ------4626.21’ 27 A

ce an ist D he e t in rm te De

B Determine Latitude and Longitude of B 28 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’

29 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’

30 Positions A and B in Minutes Latitude Longitude

A: 2,345.67’ 4,626.21’

B: 1,732.33’ 5,835.22’

31 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’

32 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

33 So, for Short Distance

D = (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

34 What…Nautical Miles?

• We use Nautical miles because of the minutes of Latitude and Longitude. • We need to convert nautical miles to statute miles.

35 Converting Nautical Miles to Statute Miles

Distance is 1,355.7 nautical miles

One nautical mile = 1.151 Statute miles Therefore,

1,355.7 X 1.1508 = 1,560.1 Statute miles

36 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.

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

38 A

Difference of Latitude

Difference of Longitude 39 Determine the mid Latitude of the Two Positions in Minutes

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

1,732.33 + 306.66 = 2,038.99’

Now, we have the mid latitude in minutes.

40 A

ce an Mid ist D Latitude Delta Latitude

B Delta Longitude

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

Circumference

radius

1 radian = 57.29582 deg

Circumference = 2 X X radius 2 X X radius = 360 degrees 42 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

43 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

44 Cosine Inserting Cosine (0.59312) in Google

Cosine of mid Latitude = 0.8292 Radians

This is the correction factor for Latitude Convergence.

45 Now to Determine North- South Distance

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

46 …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

47 A

ce 2 an ist D 2 (613.34) (Delta Latitude)

2 B (Cosine mid Latitude X Delta Longitude) (1002.51)2

48 Square Root of Squares of Latitude and Longitude Differences Square 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

49 Converting Nautical Miles to Statute Miles

Distance is 1,175.25 nautical miles

One nautical mile = 1.151 Statute miles Therefore,

1,175.25 X 1.1508 = 1,352.5 Statute miles

50 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.

51 Spherical Geometry D = ACOS((SIN(Lat A) X SIN(Lat B) + COS(Lat A) X COS(Lat B) X COS(Long A- Long B)) D = ACOS(0.304490048 + 0.63803824) = ACOS(.340678188) in radians = 1171.17 nm = 1,347.8 Statute miles

52 -The THREE Answers are-

• Short Distance = 1,560 Statute miles

• Medium Distance = 1,352 Statute miles

• Spherical Distance = 1,348 Statute miles

53 Distance is Finished What 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 Arctangent function.

54 BEARING

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

55 Quadrants North

st 4th is 270-360 1 is 0-90

West 270 East 90

2nd is 90-180 3rd is 180-270