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cae? 26 "TAE. MARINE. REVIEW SCIENTIFIC LAKE NAVIGATION BY CLARENCE E. LONG A reference to the laws governing the various. parts of the total deviation teaches us that the constant deviation has the same value on any and all azimuths --that the semicircular deviation is a uni- _form curve varying as the sine of the azimuth, or in proportion to the ratio of its squares as the angle or azimuth is changed. It changes in the same man- ner that the sides of a right triangle change when the angle at the hypotenuse is changed but retains always its one length. The quadrantal deviation is also a uniform curve varying as the sine of twice the azimuth, or what is the same thing as twice the ratio of its squares, or as the sides of a right triangle would change where the angle at the hypotenuse - doubled but retained its one length. - - The deviation curve makes it a simple matter to separate this combination of results into its component parts. The. semicircular deviation consists of two parts; one of the parts being zero in the fore-and-aft line of the ship and its max- imum amount in the, athwartship line of the ship, or at right angles to the fore- andaaft line of the ship. The other por- tion is zero on the athwartship line but has its maximum amount on the fore- and-aft line. The deviation on the north and south points is due to this athwart- ship force and has its maximum amount .when the ship's fore-and-aft line is in one with the magnetic meridian (ship's head north or south correct magnetic), and is zero when the magnetic meridian is at right angles to the fore-and-aft line; in other words, when the ship's head is either east or west correct magnetic, the athwartship force has no action because it is in one with the magnetic meridian. The deviation on north and south has no value on east and west, and the deviation on east and west has no value on north and south. Drawing each to a curve the 'two together will give the total amount of semieircular deviation for any course or point of the compass desired. The quadrantal deviation has its maximum amount on the quadrantal or intercardinal points, and is zero on the cardinal points. "However unsymmetrical the total de- viation curve may be, or its amount on reverse points, it is comprised of three symmetrical curves, because the semicir- cular and quadrantal deviations can have the same name on the same compass point when combined, but when separated they can be opposite in name. The semi- circular deviation gets its name because it has opposite names in reverse semi- circles. If the semicircular deviation is Ely in one-half the compass, it will be Wly in the other half of the compass. Quadrantal deviation has the same name in reverse quadrants, that is, the quad- rantal deviation on NE and SW will have the same name (usually -Ely, and on NW and SE (usually Wly). ; If we take the mean algebraic sum of . . . ¢ the deviation on reverse points, the re- - sult is the constant and quadrantal devi- ation combined. It is a record of the results of superposing the western curve upon the eastern. As the semicircular deviation varies as the sine of the com- pass azimuth, it is evident that the semi- circular curve in the eastern semicircle is equal in value and opposite in sign to 'that of the western semicircle. The su- perposition of* the two curves, then, re- sults in the elimination of the semicircu- lar deviation. 'For example: Supposing that by an observation we found the de- viation on NE (c. m.) to be 15° Wly and on' SW (c. m.) 27° Ely, how much is the semicircular and quadrantal devia- tions on these points? Mark Wly. devia- tions with a minus sign (--), and Ely deviations with a plus (+) sign, and proceed according to the rules of addi- tion in algebra, that is, when the signs are opposite subtract the two quantities and give it the sign of the greater quan- tity, thus: Totat Devon SW (Ely) + 27° Total Dev. on NE (Wly)°-- 15° "lz, the mean of 12° is 6°, which equals the constant and quadrantal deviations on NE and SW, being Ely in name. Sup- posing 'the total deviation on NW _ was 3° Ely and on SE 15° Wly for this same compass, we would then have SE -- 15° NW + 3° -- 12°, the mean, 6° Wly.. We can find the same thing by adding together (algebraically) the deviations determined with ship's head correct. mag- netic NE, SW, SE and NW, reversing the sign of the two latter, and divide. by 4, thus: NE,--15° SW ob 27° NW -- 3° (sign reversed). SE+15° -- 18° +- 42° ale" 4) + 24° + 6° -To determine the semicircular devia- tion alone we take the mean algebraic dif- ference of the total deviation on reverse points. The rule for algebraic subtrac- tion is: change the sign of the quantity to be subtracted, take their difference, and affix the sign of the greater, if they are then of opposite names; but if changing the sign of the subtrahend, make it sim- ilar to that of the minuend, then add arithmetically both quantities together, and prefix the same (or common) sign The minuend is the larger of the two numbers to be subtracted. _The mean al- gebraic difference of the deviations on reverse points shows the result of super- posing the western curve with the signs of the deviation changed upon the east- ern curve. The constant and quadrantal deviations are eliminated, since half the constant curve is now equal in value and opposite in sign to the other half, and the quadrantal curve, varying as twice the sine of the azimuth, has, by changing signs in the. western semicircle, also be- come equal and opposite in value to the curve in the eastern semicircle. « Exam- ple: What is the semicircular deviation on NE and SW? NE 15" SW --27° (sign changed). 2) e420 --21°, the amount of the semicircular deviation on NE and SW. | The total deviation on NE is only 15° while the semicircular de- viation alone is 21°, - Now, to show how the deviation changes from point to point when the total curve is solved into its component parts; thus, if we wish to know. how the deviation which attains its maximum amount on north and south will change _ as the boat's head is turned aside from the magnetic meridian, we may show it graphically as follows: Let the total de- viation on north or south represent the length of the hypotenuse of a right tri- angle; supposing that it was 10 degrees. From this same effect what would the deviation be on N by E, NNE, NE by -N, and NE. First draw a N by E line and lay it off into 10 equal parts, say M%4-inch to the degree. Next erect a perpendicular (a north line) and a base (an east line) to conform to this hy- potenuse; then the length of the perpen- dicular will give the amount of the devi- ation of N by E. The length of this line will be 9.8 as compared with 10. Thus it changes in proportion to the ratio of its squares, since the sum of the