Triangulos esfericos en la resolución de calculos astronomicos Capitan Yate

To solve astronomical navigation calculations and determine the position of a ship on the high seas by observing the celestial bodies, we will use the formulas of spherical trigonometry. The intersection of three great circles on the celestial sphere determines a spherical triangle, and this is what we have been studying as the position triangle, formed by the intersection of the upper meridian of the location, the hour semicircle of the celestial body, and the vertical semicircle of the celestial body. Thus, the position triangle is formed by three vertices, three angles, and three sides, which are arcs of the great circle and are also measured in degrees. From the spherical position triangle, different formulas or theorems used in astronomical navigation can be derived. Knowing three pieces of information, we can find the missing data, either by using the cosine formula, the cotangent formula, or the sine formula. To know which formula to use, we'll need to draw a triangle in position. We'll write down the known data and the value we need to obtain. If we have three sides and one angle, we'll use the cosine formula. If we have two sides, the angle included between them, and the angle opposite one of them, we'll use the cotangent formula. And if we have two sides and their opposite angles, we'll use the sine formula. So, to solve astronomical calculation problems, to find the estimated height and azimuth, you have the quick reference sheets, where you won't have to think about which spherical trigonometry formula to use. By following the different steps, which are very clear, you won't have any trouble finding the solution. The other way to solve these calculus problems is by drawing the position triangle. Considering the known data and the unknown, you'll need to know which formula to use, whether it's the cosine formula, the cotangent formula, or the sine formula. Now I'm going to solve question 12 of the June 2016 exam for the Madrid exam, first using the quick reference sheets and then using the other method. First, I like to write down all the information for the problem so I don't make mistakes when reading it later. I calculate the true altitude of Alkaid using sheet 3 for correcting the altitudes of celestial bodies. I subtract 0.8 minutes from the instrumental altitude, which is the sextant index error, a figure given to me in the statement. I obtain the observed height, and with the observer's elevation value of 15 meters, I go to Table A, page 387 of the AN, and see that with an observer's elevation between 14.8 and 15.3 meters, the horizon depression value has a value of negative 6.9. I obtain the star's apparent height, and with this value of 27º 30.1', I enter Table C and obtain a correction value of negative 1.8'. The exact value, interpolated, would be negative 1.9' for an altitude of 27º, and negative 1.85' for 27º 30'. However, such accuracy is not necessary to solve the problem. So I get a true height = a 27º 28.3´ Zv is = a Za + the CT Za = a 308º + the ct of positive 5º. From the statement, I get a Zv of 313º. I convert it to a quadrant: 360º - 313 = N47W and so I see that of the 4 answers, only b or d can be correct. To calculate the height difference, I must first obtain the estimated height. I go to worksheet 9. I see that the latitude and declination are known data, but I need to obtain the angle at the pole. http://www.sirocodiez.com/ http://www.sirocodiez.com/ I go to file number 4 and see that the star's local time, which the statement calls the star's time at that location, has a value of 84º 55'W. Since time is counted on the equator from the MSL westward to the star's hour circle, I see that the west value of the datum doesn't raise any doubts. Since the star's local time is less than 180º, it would then be equal to an angle at the western pole of 84º 55 minutes, or PW. I begin by developing part A of the formula and obtain a value of 0.412578. Since latitude is north and declination is positive, then l = d, and A will have a value of +. Now I work on part B and see that the result I get on the calculator is positive 0.0485098. The sine of the estimated height is = a + b = 0.461087. I type shit and sin and get a value of Ae = 27.4573º. I type shit and degrees and get the value in degrees, minutes, and seconds. To convert the value from seconds to fractions of minutes, I type 27 and press the degrees key, and then I type 26.18 and press the degrees key again, and thus I get 27.4 minutes. The difference in heights is always the true height minus the estimated height. I get a value of 0.9 + and see that the correct answer is d.