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squares of perpendicular and base equal the square of the hypotenuse. Next draw a NNE line of the same length and then draw a perpendicular and base to conform to it. The length of the per- 'pendicular will be the amount of the de- Dev. oONNXE. 9.8° DevV.ONNNE 92° : NN.E. Neb Fee FIG. | viation on NNE if it were 10 degrees on north: The deviation will be 9.2 de- grees; on NE by N it will be 8.3 degrees, and on NE it will be 7.1 degrees. For the first point (N to N by E) the change in the deviation for a change in the ship's head. of 1 point was but 2 degree, the next.point .8 degree, for the next 9 degree, and for the fourth point 1.2 degrees. See Figs. 1, 2, 3 and 4. . If it were still desired tio get the effects NEE KE. Dev oONNEXN.83° » ONE XA, -tiG 3 of this deviation for NE by E, ENE, E by N and East, all we have tq do is to follow up the same rule. As will be seen by the diagrams the relations of the two elements are simply reversed. The per- pendicular of a right triangle for a 3- ' ematics. . TAeE Marine Keview - point course becomes the perpendicular in the same triangle for a 5-point course; or it may be stated thus: In a 3-point course the diff. lat. is thee same as the dep. in a 5-point course; in other words, there is the same relation between the sides of the right triangle for a course of NE by N as there is for a course of NE by E. NNE and ENE conform to each other, since NNE is a 2-point course its complement is a 6-point course (8-- 2), and ENE isa 6-point course. As has been stated a number of times before the Traverse Tables for solving the various sailings are based on the right triangle, so that two of its sides conform to the difference of latitude (meaning a true north or south line) on the earth and departure, (meaning a true east or west line) on the earth. The hypotenuse con- forms to course and distance in the sail- ings. The right triangle is also em- ployed in solving problems in the com- position and resolution of forces. Two forces acting at right angles to each other will produce a result equal to the Dev onN.E.ZI° NE hypotenuse of that triangle because the square on the hypotenuse is equal to the . , sum of the squares on the other two sides. The two side forces, or the two sides of the right triangle are called the components, and these two forces produce the resultant force, and this is equivalent to the hypotenuse. of the compass is the result of a combi- ration of causes, which, when resolved into its component parts become simple laws based upon a simple rule of math- It is by this process that the total deviation of a compass is resolved into its ecmponent parts and the value of the coefficients which enter into each. Any agent that unites its action with 'that of some other agent for the pro- curing of a certain result is called a co- efficient. One of several forces which acting in fixed directions are the equiva- lent of a given force is called a compo- nent. The semicircular deviation of the compass has its maximum effects on the cardinal points. This semicircular devia- tion 'is. divided into two parts The total deviation 27 Dev. on NEXE. NEC, FIG.> or forces, one being an athwart- ship force' and the other a fore- and-aft force. The deviation with the ship's. head N or. S correct magnetic is due to this athwartship force, the direc- tion of the compass needle being the re- sultant (hypotenuse) force, and the fore- and-aft and athwartship lines of the ship represent the' component forces; or imagine the two sides of a right triangle that would conform to the direction of the compass needle as hypotenuse. The resultant force is the -direction of the compass needle with the ship's head N or S, and the deviation for all other points of the compass is based upon this resultant force measured by the fore-and- aft component, or that line of the right triangle that is coincident with. the fore- and-aft line of the ship. As will be seen by Figs. 1, 2, 3, 4, that the deviation on a 5-point (NE by E.) course is 5.6 degrees if it were 10 degrees on N, since if the ship's head: was NE by E the line marked 5.6° (Fig. 3) would be coincident with the fore-and- aft line of the ship. The deviation on ENE would*be 3.8° and on E by N 2° dwindling to nothing on East. Bear in mind that the sum, of the three angles of every triangle are equal to 16. com- pass points (180 degrees). In the right 'angle there is always 8 points (90 de- | grees) so that the other two together should always make 8 more. If one of the angles at the hypotenuse is 3 the angle at the other end of the hypotenuse . will be what it lacks of*8 points. If you~ were to draw a NE by E line at a length of 10 and erect a perpendicular 5.6 the base line of same would measure 8.3, see Fig 5. Figs. 3 and 5 are practically the same thing, the conditions being merely reversed. Now, the second part of the semicir- cular deviation has its maximum effect with the ship's head E or W correct mag- netic, and dwindles to zero on N and §S, just the reverse of the maximum devia- tion for N and S. In this case the devi- ation for each point is taken from the © athwartship line (the base or departure line of the right triangle), since the re- sultant force is due to a force running