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1 Introduction	9 Altitude Intercept	17 Latitude by Meridian Altitude
2 Corrections to Sextant	10 Using Position Plotting Sheets	18 Latitude and Azimuth by Polaris
3 Altitude Corrections	11 Plotting Lines of Position	19 Running Fixes
4 Time	12 Summary of 1 thru 11	20 Time of Sunrise Sunset
5 Finding GHA and Declination	13 Finding Deviation or Gyro Error	21 Star Identification/Selecting for Fix
6 Assumed Position and Local Hour Angle	14 Finding Azimuth by 229	22 Time Tick Problems
7 Computed Altitude and Azimuth	15 Amplitudes	23 Deviation Table Construction
8 Interpolation	16 Time of Local Apparent Noon	24 Sample Final Exam

Interpolation

  Interpolation is just a word for the procedure used to read between the lines in a set of tables. It will be necessary for you to interpolate in four of the tables you'll use in this course: 229, tables 27 and 28 from Bowditch, and the Polaris and sunrise and sunset tables in the almanac. One simple routine can be used for all the tables, and you can punch out the answers on the simplest hand held calculator. You can do this arithmetic by hand, of course, but a calculator will save much time and many mistakes. 

  If you have a complete volume of 229 and a complete almanac, you may find that they have interpolation tables for certain things in them. These tables were helpful in the days when very few could afford a calculator, but the method we'll describe here is easier and faster than using the interpolation tables. When you get to the tables you'll use from Bowditch you'll have to do your own interpolating anyway, so it makes sense to learn one method which will work for all tables.

Look at this section of the 229 tables. View image

  If your LHA were 16˚, assumed latitude were 22˚ and declination were 4˚,exactly, then Hc would be 59˚ 39.1'and Z would be 147.0˚ exactly. No interpolation would be necessary.

  Suppose the declination was 4˚30'. The values of Hc and Z for that declination aren't printed in the tables. Hc must therefore be something less than Hc 59˚39.1' (at 4˚) and more than the Hc just beneath it at 5˚ (58˚47.1') because the value of Hc is decreasing as declination increases in this part of the table. The value of Z is increasing as declination increases. Therefore, Z for declination 4˚30' must be greater than 147.0˚ but less than 148.0˚.

  You can estimate what Hc and Z should be and come pretty close. The actual declination is halfway between the tabulated declinations of 4 and 5 degrees, so the actual Hc and Z must be halfway between those tabulated values. That would put Hc at around 59'20' and Z about 147.5˚.

  An experienced navigator in real life would probably do it just that way and no one would ever know the difference. But, you aren't an experienced navigator and the precision required to pass the coast guard test isn't so much like real life. To be sure your answer is within 0.1' of the book answer, you should do a formal interpolation.

  To do the arithmetic of formal interpolation on a plain four function calculator, you need two simple formulas:

1. Difference x increment / interval = correction

2. Base (+) or (-) correction = answer

  Before that will make any sense, we must define some terms. As we do, let's use the formulas to find the Hc for a Dec. Of 4˚30'. Then we will use them to find Z for the Declination.

Take another look at the first formula:

Difference x increment / interval = correction

 

  The difference in the formula is the difference between two numbers next to each other within the table. Look at the 229 excerpt.

 View image

  you can see that Hc for 4˚30' is somewhere between 59˚39.1' and 58˚47.1', which are the two Hc values next to each other within the table. (From now on, we'll call any values next to each other within the table by their proper name respondents.)

  The difference between these respondents is 52', as indicated by the number in the d column next to 59˚39.1'. Just as important, it is a minus 52', as shown by the minus sign above it in the d column. The minus sign tells you that Hc decrease by 52' between a Dec. Of 4˚ and a Dec. of 5˚. In other parts of 229, you will find that the difference between two tabulated values (respondents) increases, so be sure to check the first + or - sign you find above the d number to determine how to handle this difference.

  The format of 229 makes it easy to find the difference between two values of Hc because the math is already done and the answer is listed. You will have to do your own arithmetic to find the difference between respondents in other interpolations, however. With the difference of -52' that we obtained in the d column, our formula now reads:

-52' x increment / interval = correction

  Note: a negative number multiplied or divided by a positive number gives a negative answer. Plain four function calculators will not keep track of the minus signs for you, however. You will have to do the math as if all numbers are positive, then put the minus sign in front of the correction.

  We now need to understand what an increment is in order to insert the next number in our formula. The increment is the difference between the exact number you are working with and the nearest value in the table which is less than the exact number. In our example, we are looking for the Hc of an exact Dec. Of 4˚ 30'. The increment in this case is 30', because that is the difference between 4˚ 30' and the nearest tabulated value for declination less than the exact value (4˚). If you were looking up a Dec. Of 6˚ 22', the increment would be 22'. For a Dec of 17˚55', the increment would be 55', because that is the difference between the "exact" number and 17˚ , the nearest lesser-tabulated value for Dec. With that increment of 30' established, our formula now reads:

-52' x 30' / interval = correction

  The interval in the formula is simply the difference between any two values that you use to enter the table. (Values that are used to enter the table, such as Dec., LHA, or Latitude are called entering arguments.)

  In our example, we have entered the table through the declination column. The interval between numbers in the declination column (from 4˚ to 5˚ of declination, for example) is one degree. Throughout 229, the intervals between the entering arguments of LHA, Latitude, and Declination are one degree. To get the correct answer the increment and interval must be in the same units. Since the increment is already expressed in minutes, it will be easier to convert the one degree to 60 minutes and put that number in the formula, as follows:

-52' x 30' / 60' = correction

  When you punch those numbers into a calculator, you should get a correction of -26'. That is not the final answer, however. With this correction, you can now begin the second formula:

Base (+) or (-) correction = answer

  Because the correction is a minus number, you put it in as base (-) 26' = answer. The base is your starting point, the Hc value you need to correct. This whole process began with an Hc of 59˚39.11 for a Dec. Of 4˚. Entering that base Hc in the formula gives us: 59˚39.11 (-) 26' = answer. Some simple arithmetic (followed by a careful double check) results in an answer of 59˚13.1', which is the Hc for Dec. 4˚ 30'.

  Are you ready to follow the interpolation steps to find Z for a Dec of 4˚ 30' now? Remember the formulas:

1. Difference x increment / interval = correction

2. Base (+) or (-) correction = answer

  The difference, you will remember, is the difference between two respondents (values next to each other within the table). Because we are looking for Z, this time we need the difference between numbers in the Z column. View image

  Z for Dec. 4˚ is 147˚ and Z for Dec. 5˚ is 148˚, so the difference will be one degree. But don't forget to check on whether the difference is (+) or (-). Checking the table we see that z is increasing from Dec. 4˚ to Dec. 5˚, so the difference is (+), making the formula:

1˚ x increment / interval = correction

  Do you know what increment to use? Again, we are entering the table with a Dec. Of 4˚30', so the increment will be 30', the difference between 4˚ 30' and the nearest lesser tabulated value.

Our formula now reads:

1˚ x 30' / interval = correction

  The interval is the difference between the two values you use to enter the table (entering arguments), in this case the difference between 4˚ and 5˚ of Dec. is 60'. All intervals in 229 are 60', but that is not the case in other tables. Be sure to check every time. Now our formula can be completed:

1˚ x 30' / 60 = 0.5˚

  Using the correction of 0.5˚ in the second formula, we get:

Base + 0.5 = answer

  The base z that we started with at a dec of 4˚ was 147˚, so the final answer of z is:

147˚ + o.5˚ = 147.5˚

 

  Let's do another interpolation problem for practice. Suppose you had a LHA of 343˚, assumed latitude of 23˚s and a declination of 08˚ 43'. The LHA and assumed latitude will get you to this part of 229, which is the same name (Dec And Latitude are both south) page for LHA 343˚.   

View image

  But now you have to do the interpolation to find Hc and Z for Dec. 08˚43'. Do you remember the two formulas? Write them on a scratch pad.

View image 

Let's use the formulas to find Hc first.

  What is the difference? Using the declination of 8˚ as the entering argument, you see that the number you need is supplied for you in the d column. Is it a plus or minus difference? You should have found a difference of +41.4, which can now be entered in the formula on your scratch pad.

  What is the increment if the declination is 08˚43'? The next lesser entering argument is 08˚, so the increment is 43' enter this in the formula.

  Now what about the interval? We said that all intervals in 229 are 60', so use that number in the formula and find the correction.

  Now that you have the correction (it should be +29.7'), use it in the second formula to find the answer. Do your work on scratch pad. What are you going to use as a base?

  At our starting point of Dec. 08˚ , the base Hc is 67'50.0', so your final solution should be 68˚19.7'.

Now use the same process to find Z at a Dec. Of 8˚43'.

View image

   This was kind of a toughie. Did you catch the fact that Z decreased by two degrees, thereby giving you a difference of minus 2˚?

  Are you starting to feel comfortable with this process? Here is (one to do on your own, using the 229 excerpts provided with this course. The answer and explanation follow in case you run into trouble.

  Problem:

With an LHA of 274˚, a DR Latitude of 28˚ 06.2' N, and  Dec. Of N 19˚ 39.7', what is Hc and Z? Begin by writing down the formulas, then use LHA and assumed latitude as entering arguments to get to the right page of 229. 

View image for 229,

 View image for solution

Here is another problem to reinforce the interpolation process.

 Problem:

 LHA is 344˚, Assumed Latitude is 16˚ N, and Declination is N 5˚17.2'. Required: Hc and Z.

View image  for 229,    

 View image for solution.

  We should cover one more point on arithmetic before you start on the practice problems--rounding off numbers. If you worked the last example on your calculator you found the correction for Z to be (-) 0.74533333. You may have wanted to round that to 0.75 and then round the 0.75 to 0.8. Many people do that because they learned to round the way it's done in stores and banks. If the bill comes out in fractions of cents they drop the fraction of a cent if it's less than half a cent, and round up to the next penny if the fraction is half a cent or more. To round off a number properly, follow the convention "round to an even number" when the last digit is 5. For example, in rounding the following numbers to the nearest tenth:

0.76 rounds to 0.8 (last number is greater than 5)

0.74 rounds to 0.7 (last number is less than 5)

0.75 rounds to 0.8 (last number is 5, so you round to an even number)

0.85 also rounds to 0.8 (last number is 5, and you round to an even number)

  Also, when you round off a number with a long string of decimal places (as often happens when you're using a calculator you pay attention only to number immediately to the right of the place you are rounding to and ignore the rest. 0.74599999999, rounded to the nearest tenth, still rounds to 0.7.

  Why are we being so fussy about something that can only make a difference of 0.1' in an answer? The answers to these practice problems have been rounded as described. If you agree with the book answer, you know you've done the whole problem right. If your answer differs by one tenth, it could be merely that you've rounded off incorrectly, but then you may have made a mistake elsewhere. Two or three of those little mistakes can throw your answer far enough off to cause you to fail a test.

  Here are some problems to practice your interpolation skills with the 229 . As we have said, a good work form helps a lot.  This form saves a little writing and keeps everything orderly.

  View image for form,

View image for practice problems

One more thing left to do before we are finished with the 229. Hc  is ready to go, but azimuth angle,  Z,  must be converted to azimuth,  Zn. Rules for doing that appear at the top right of every left-hand page and bottom left of every right-hand page.

 An English translation of the one on the left-hand page would read:

 If you are in north latitude and the LHA is greater than 180˚, then Zn and Z are the same thing. Zn = z

      But, If you are in north latitude and the LHA is less than 180˚, then   Zn = 360˚-Z.

      The one on the other page can be translated much the same way:

       If you are in south latitude and LHA is greater than 180˚, then   Zn = 180˚ -Z. 

        But,    If you are in south latitude and LHA is less than 180˚, then Zn = 180˚+ Z.

 There are three out of four situations in which z has to be converted to zn. Too often people leave out that last easy step and blow the whole thing. It is an easy one. Convince yourself by converting the Z's you interpolated for in the last set of problems into zn's .

  View image

 Here are the answers:   

1.  137.7˚  2.  066.8˚  3.  224.6˚  4.  325.1˚  5.  073.4˚  6.  037.9˚  7.  133.0˚

ALTITUDE INTERCEPT