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Solving Distance By Great Circle Sailing

Your ship departs Yokohama, Japan from position Lat. 35° 27' N; Long. 139° 39' E bound for San Francisco, California, USA. At position Lat. 37° 48.5' N, Long. 122° 24' W. Determine the distance by Great Circle Sailing.   

Given:


   Lat1 = 35° 27' N
   Lat2 = 37° 48.5' N
Long1 = 139° 39' E
Long2 = 122° 24' W

What is asked?

Distance by Great Circle Sailing

Solution:

1. Solve first for the difference of longitude (Dlo). Here's the formula in getting Dlo:
    If same name, subtract
    If different name, plus
    Note: Affix name of direction. If Dlo is greater than 180, subtract it to 360 and then affix name of  Long1. 
                
               Long= 139° 39' E
               Long122° 24' W 
                            261° 63'
                           +   1°-60' 
                            262° 03'
                         - 360°        
                  Dlo =   97° 57' E

2. We can now proceed to solve for Great Circle distance. 
     Note: Plus (+) when not crossing the equator; Subtract (-) when crossing the equator.

          Cos GCD = (Cos Lat1 × Cos Lat2 × Cos Dlo) + (Sin Lat1 × Sin Lat2)    
        Cos GCD  = ( Cos 35° 27' × Cos 37° 48.5' × Cos 97° 57' ) + ( Sin 35° 27' × Sin 37° 48.5' ) 
          Cos GCD= -0.08902 + 0.35548
              GCD    = inv Cos 0.26646
                 GCD =  74.55°
                            ×   60        ( to convert to miles)
                 GCD = 4473 miles
           
Great Circle Sailing
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16 comments:

  1. Where did you get +1°-60' in Dlo?

    ReplyDelete
    Replies
    1. since 60' = 1°
      the answer is 261° 63'..
      63'-60'=3', then add 1° =262°

      Delete
    2. Your vessel current position is in Lat 35°40.0’ N and Long 141°00.0’E going to 37° 48.0’N 122° 40.0’W, the maximum latitude to be is 45° N by composite sailing. Solve for the Longitude where the track meets.

      Delete
    3. I used this formula to solve a problem but the answer is not the same.
      * 41°40'S 175°25'E and 7°N 80°50'W
      Using Norie's Table I go the dis=6294.4mls

      But this formula gave me 5752

      Delete
    4. Subtract not plus because it crossed the equator

      Delete
  2. How about initial course and Final course

    ReplyDelete
  3. Replies
    1. "inv" is the inverse button on the calculator. In some calculator, it is named the SHIFT function button.

      Delete
  4. Which is correct?
    DLo=97°97'E or your answer DLo=97°57'E

    ReplyDelete
    Replies
    1. DLo=97°57'E is the correct answer, not DLo=97°97'E.

      360°
      - 262° 03'

      359 deg 60 mins
      - 262° 03'

      Delete
  5. do you convert the latitude into radians? if not, how do you input it in your calculator?

    ReplyDelete
    Replies
    1. Having followed the formula the latitude is not converted into radians.

      You can follow the solution manually by solving first those inside the parentheses.

      Delete
  6. why it was 77.55 degrees while the answer in the calculator is 74.96363877

    ReplyDelete
    Replies
    1. Where did you get the 77.55?

      The answer is 74.55 degrees

      Delete
  7. What is the initial course and final course?

    ReplyDelete
  8. Pls give me the initial and Final course. Also lay and longitude vertex every 10 degrees of longitude

    ReplyDelete