The science of Navigation is based on mathematical principles, mainly from Geometry and Trigonometry. It has been shown throughout this study, however, that it is possible to practice Navigation with no more mathematical knowledge than the ability to add and subtract, and to look up figures in books of Nautical Tables, using simple formulae. The Ocean Navigator, however, owing to the small scale of his navigational charts, will find it infinitely more convenient to calculate his courses and distances along rhumb-line or Great Circle tracks, and the advanced section of this study is devoted to this subject. This chapter covers the basic mathematical principles around which the whole subject of Navigation revolves, most of which students will have previously learned at School, but which from years of non-use may possibly have forgotten. They are, however, essential towards understanding of the succeeding chapters.
The “angle” between two lines is the inclination of one line to the other and this inclination is measured in degrees and sub-divisions of a degree. In fig. 58-1 (a), imagine two lines, AB and AC exactly coincident, that is, one on top of the other so that they appear as one single line. Suppose we move the line AB away from the line AC, pivoting at point A, as shown in fig. (b). The line AB is now inclined away from AC by the amount of the angle at A between the two lines, and if we take a protractor and measure this angle we find it to be 25º.
If we now again move the line AB further away from AC until AB is perpendicular to AC, the angle at A will measure 90º, and this is what we call a right-angle fig. (c).
We can further move line AB away from AC, still pivoting at point A, until the angle between them measures 180º, and it will be seen that this forms a straight line BAC fig. (d).
Continuing to move line AB anti-clockwise about point A, when we reach 270º, line AB is again perpendicular to line AC fig. (e), the right-angle of 90º being formed on the opposite side of the lines to that of fig. (c).
Finally, if we move line AB further anti-clockwise until it again coincides with line AC; in completing one revolution around point A we have measured 360º, the total number of degrees in a circle fig. 58-1 (f).
It will be noticed from fig. (d) that the angle between any two parts of a straight line is 180º, so that in fig. 58-2; if angle CXB is 30º, angle CXA must be 180º – 30º = 150º.
Similarly, as the total number of degrees in one revolution is 360º, in fig. 58-3 if ∠CXA is 150º, then the angle on the opposite side of the lines is about the point X must be 360º – 150º = 210º.
The word “trigonometry” – often frightening to a beginner – means nothing more than “triangle measurement”. A triangle consists of six parts – three sides and three angles. In each of the triangles in fig. 58-4 the three sides are AB, BC, and CA, and the three angles are ∠ABC, ∠BCA, and ∠CAB (∠ is the symbol used to indicate that an angle is being referred to, the actual angle being the one at the middle letter).
There are two types of triangle, and fig. 58-4 shows one of each of these types. fig. 58-4 (a) is called a “right-angled triangle” because one of its angles (∠ACB) is a right-angle (i.e., measures exactly 90º), or two of its sides are perpendicular to each other (side AC is perpendicular to side BC).
Fig. 58-4 (b) contains no right-angles and none of its sides is perpendicular to one of the other sides, and this type of triangle is called an “oblique-angled triangle”.
Now one of the properties of all triangles, right-angled or oblique, is that the sum of their angles equals 180º, so that if two of the angles are known, the third angle can always be found. For instance, in the triangle in fig. 58-4 (a), if ∠ABC = 25º, we can find the value of ∠BAC because, as it is a right-angled triangle ∠ACB = 90º, and taking the sum of the angles ABC and ACB, i.e., 90º+25º = 115º, and subtracting this from 180, i.e., 180 – 115º, we get 65º, the value of ∠BAC.
Similarly, in the triangle in fig. 58-4 (b), if angle ∠CAB is 20º and ∠BCA is 135º, then ∠ABC = 180º – (135º+20º) = 25º.
Of the six parts of a right-angles triangle, if three are known, providing at least one of the parts is a side, the unknown parts may be found by calculation. This is done by using what are called “trigonometrical functions”. The “function” of an angle in a right-angled triangle is the ratio between two of the sides of that triangle.
Consider the right-angled triangle in fig. 58-5, in which ∠CBA is the right-angle. In any right-angled triangle, the side opposite to the right-angle is always called the hypotenuse. When referring to one of the other two angles, the side nearest to the angle (other than the hypotenuse) is called the adjacent side, and the third side is called the opposite side. Thus, in fig. 58-5, for ∠CAB, side AB is the adjacent side, and side CB is the opposite side. When referring to ∠ACB, however, side CB would be the adjacent side.
To find the value of ∠CAB, which for convenience we will abbreviate to Â, there are six functions of Â, and each of these is given a name. The first three are of fundamental importance. The last three are inversions of the first three.
A useful aid for memorising the three fundamental functions, Sine, Cosine and Tangent is “SOH CAH TOA” – Sine equals Opposite over Hypotenuse, Cosine equals Adjacent over Hypotenuse, Tangent equals Opposite over Adjacent. The three remaining functions can be remembered by inverting each of these if it is recalled that the Cosecant is the opposite of the Sine, Secant the opposite of Cosine, and Cotangent the opposite of Tangent.
With these functions of the angle Â, providing we know the length of any two of the sides of the triangle we can find the value of angle Â. For instance, suppose the length of the hypotenuse AC in fig. 58-5 is 48 parts and the length of side CB is 20 parts, then the
The value of Â for a Sine of 0.4167 can be found by looking up the Sine 0.4167 in any mathematical or nautical tables. The table to look for is the “Table of Natural Functions of Angles”. In this, running an eye down the column for Sines, we find that the angle corresponding to Sine 0.4167 is 24º 38’ which is therefore the value of Â in this case. We could have solved this problem equally well by using the Cosecant function of Â.
Looking this up in the Table of Natural Functions, under the Cosecant column we find exactly the same value for Â opposite the Cosecant of 2.400 that is 24º 38’.
The value of one of the sides of a triangle may also be found by using the functions of an angle. In fig. 58-5, suppose Â is 30º and the hypotenuse AC is again 48 parts, and we wish to find the value of the adjacent side AB.
From the functions of Â seen earlier, we see that
Looking up the Cosine of 30º in the Table of Natural Functions., we find that Cos. 30º = 0.8660, so side AB = 0.8660 x 48 = 41.57. This could also have been calculated by using Secant Â = AC to give the same result.
The value of ∠ACB (Ĉ) can also be found by using the trigonometrical functions. Suppose side CB = 20 and side AB = 42; for angle Ĉ, side AB is the opposite side to the angle and side CB is the adjacent side.
Looking in the Tangent column of the Table of Natural Functions, we find that the angle corresponding to a Tangent of 2.100 is 64° 34’, which is therefore the value of Ĉ.
Answers to self-test exercise
It is frequently necessary in calculations to multiply or divide trigonometrical ratios. Multiplying and dividing are most tedious mental processes and mathematicians many years ago evolved a method of simplifying them, this method being the use of Common Logarithms (usually abbreviated simple to “logs.”). The log. of any number can be found by looking it up in tables of Common Logarithms. To multiply any two numbers it is only necessary to add their respective logs. Together, extracting the natural number corresponding to the sum of the logs. From the same log. Tables. Similarly, to divide any two numbers, it is only necessary to subtract their respective logs. This will become clearer in the following examples: - Multiply 13964 by 278.5
Turn to the Logarithms Table in any Book of Mathematical or Nautical Tables.
We must find the logs. of 13964 and 278.5 : Now every log. consists of two parts, separated by a decimal point. The part to the right of the decimal point is extracted from the tables, the part to the left of the decimal point, called the characteristic", must be decided by the operator according to simple rules. For numbers greater than 1, the characteristic is the number one less then the number of figures before the decimal point in the original (natural) number.
For instance, in this example, we wish to find the logs. of 13964 and 278.5. The first number, 13964, could have been written 13964.0, in which case there are five figures before the decimal point, so the characteristic of its log. is one less than this, i.e., 4. the second number, 218.5, has three figures before the decimal point, so the characteristic of its Log. is 2. We can now look up the second part of the logs. of these numbers in the Logarithmic Table. The first three figures of each number, in this case 139 and 278, will be found in the left-hand column of the table, the fourth figure (here 6 and 5 respectively will be found along the top of the table, and if there is a fifth figure (4 in the case of 13964) the correct log. is found by interpolating for this between 6 and 7 of the fourth figure. To make this interpolation easier, the difference between successive logs. is given in the right-hand column of the table.
To find the log. of 13964, therefore, run down the left—hand column to 139, then move across the table to the column under "6" (the fourth figure), where the log is found to be 14489. The difference between the logs. in columns "6" and "7" (given in the extreme right-hand column) is 31, and the final figure of the number we require is 4, so multiplying 31 by 4 we get 12.4 or nearly 12, which is added to log. 14489 (from column "6") to give 14501. Combining the characteristic with the extracted figure from the tables, we have the log. of 13964 = 4.14501. Similarly, the log, of 278.5 will be found to be 2.44483.
1. To multiply. the natural numbers 13964 and 273.5 together, we merely add their respective logs. together:
This resultant log. is called the anti-log., and to find the natural number corresponding to it, it is just necessary to glance through the Logarithm Tables until the second part of the log. (i.e, 58984) is located, extracting the natural figures from the top and sides of the table in the reverse process to finding a logarithm. In this case, 58984 gives a natural number of 3889. To place the decimal point in this natural number, as the log. characteristic represents one less than the number of figure’s before the decimal point in the natural number, and the characteristic is 6 in this case, there must be seven figures before the decimal point in the product of this multiplication, so 13964 x 218.5 = 3,889,000.
Although this process may seem slightly involved to a beginner, with a little practice logs. can be extracted very rapidly from the tables, .and the process of multiplication, particularly of large numbers, becomes a quick and simple procedure involving only a simple addition.
2. Divide 354.92 by 3.6781
Look up the anti—log. 98451, which gives natural number 96496. The characteristic 1 indicates there are two figures before the decimal point, so — 354.92 ÷ 3.6781 = 96.496.
3. Multiply 26.47 by 0.453
(This example introduces a number less than 1, in other word there are no figures before the decimal, as in 0.453. The
characteristic of the log. In this case becomes negative and the Rule for determining it is: The characteristic for a number less than 1 is one more than the number of noughts immediately to the right of the decimal point. In the number 0.453 there are no noughts after the decimal point, so the characteristic must be one more than 0 = 1, and as it’s a negative it is written ¯1, and called "bar one". The minus is written over the characteristic in this way because the remaining logarithm (after the decimal point), extracted from the tables, is always positive. The log. of 0.453 is therefore ¯1.65610. Had the log. of 0.0453 been required this would have been ¯2.65610, the log. of 0.00453 would have been ¯3.65610, and so on, in. each case the characteristic being one more than the number of noughts after the decimal point.
Having written down the logs. of the two numbers 26.47 and 0.453, to multiply we add the logs. together as before. Do not be afraid of negative figures in the characteristic when adding logs. The part to the right of the decimal point is simply added in the usual way. If there is a “bar” in one or both the characteristics, treat it as a simple negative; in other words, add up all the parts which are positive first, then subtract the negative part(s). In this example, adding the part to the right of the decimal point leaves us with 1 to carry over into the characteristic; this is always positive, so this 1 is added to the positive 1 of the log. of 26.47 to give + 2, from which we subtract the negative “barb one” of the log. 0.453, leaving us with a positive 1 : + 2 - 1 = + 1. The anti-log. 1.07885 gives us a natural number of 11.99, so 26.47 x 0.453 = 11.99.
4. Divide 107 by 0.00284.
from 8 leaves 6, 3 from 3 leaves 0, 3 from 9 leaves 6, 5 from 2 leaves 7 with 1 to carry over, 5 from 0 leaves 5 with 1 to carry over to the characteristics. As always, this 1 is positive, thus making the ¯3 become ¯2 (- 3 + 1 = - 2). The rule for subtracting a negative quantity is to change the sign of the bottom line and add, so the ¯2 becomes + 2, which, added to the + 2 on the top line, gives the correct characteristic for the anti-log., + 4. Looking up the anti-log in the tables gives us 37675 and characteristic 4 indicates there are five figures in the natural number so 107 ÷ 0.00284 = 37675.
Those students who have not used logs before may like a little practice in their use before continuing further and the following self-test exercise is included for their benefit. Check your answers with those given below but if you have any difficulty with the problems, please try some more exercises.
Self-Test Exercise: Decide the characteristic and final log of the following natural numbers:
Answers to Self-Test Exercise:
In navigation, it is frequently necessary to multiply and divide trigonometrical ratios and, of course, logs. can be used for this purpose. It would be possible to look up the numerical value of the ratio and then enter the log. tables and find the logarithm, but it would be more convenient if we have a table of log sines, log. tangents, log. secants, etc., so that the log. of the ratio can be selected with one entry into the tables. Such a table will be found in all volumes of Nautical Tables, and it is called "Logs. of Trigonometrical Functions".
Many trig. ratios are less than one, so the characteristic of their log. is negative. To avoid negative characteristics, ten is added to all the log. ratios tabulated in Nautical Tables. In normal calculations these tens will cause no difficulty, and tens may be added or discarded at any time to keep the characteristic between 0 and 10 inclusive. For instance, the log. sine of 38°20’ is really ¯1.79256, but in Nautical Tables it is given as 9.79256 to keep the characteristic positive (-1 +10 =+9). Suppose it is required to multiply sine 38°20’ by 6143. Then:
from which the added 10 must now be discarded to give: 0.6007; which is extracted from the log. tables to give the natural number 3.988. Therefore sine 38°20‘ x 6.43 = 3.988.
Referring back to section on “right-angled triangles”, in which we solved the parts of right-angled triangles by using the table of "Natural Functions of Angles" let us now show how this can be simplified by using instead the table "Logs. of Trig. Functions".
Suppose in triangles ABC in fig. 58-7 Ê = 90°, Ĉ = 76° 13.0, and side AC = 382.8, and we require to find the lengths of the other two sides AE and CB, and the value of the angle A.
We are given that AC = 382.8 and that ĉ = 76° 13’. Therefore we can write:
Enter the table of Logs. of. Trig. Functions and extract the sine of 76° 13’
Looking up this anti.log. in the logarithm table we can extract the value of side AE = 371.78.
To find the third side CE, suppose we choose to use the Tangent function, then:
In this subtraction, 1 is carried over into the lower characteristic, making it 11 (10+1). If the surplus 10 is now discarded, this makes the lower line characteristic which, subtracted from the upper line characteristic 2, leaves the characteristic in the anti-log.)
Looking up the anti.log. in the Logarithmic table gives us the side CE = 91.2.
There is no need to calculate the value Â by formulae since we already know Ê = 90° and Ĉ = 76° 13', therefore A = 180° — (90°+76.13') = 13° 47’.
Students should practice using the tables “Logs of Trig. Functions” in the following self-test exercise, the answers to which are given below.
With reference to the accompanying figure, solve the following right-angled triangles using log. tables and the logs. of trig functions:
The plane trigonometry outlined in the above section forms the basis of navigation by the Rhumb Line Sailings to be described in the following two chapters of this Course. The next section of this Study shows how Spherical Trigonometry can be applied to the reduction of sights, while the last section summarises the basic navigational concepts which will be used in the final sections.
Students should practice solving right-angled triangles using these trigonometrical functions in conjunction with Nautical Tables before continuing with this study, in order to become familiar with their concept and use, as they occur frequently in the practice of Navigation. The following exercise can be used as a Self-Test; the answers are given below.