marine navigation :: basic geographical notions |
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Marine Navigation Basic geographical notions |
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A diameter around which the Earth is turning is an axis of the Earth, ends of that axis are poles: north and south. The zero parallel i.e. the equator are coming into existence by cutting in two the Earth with the perpendicular to the axis of the Earth and going through its middle. The equator is dividing also the globe into two hemispheres: north and south.
Geographical longitudes and latitudesLongitude λ (upper and bottom scale on the map) The 180° meridian is international date line.
Latitude φ (right and left scale on the map)
This way so putting the yacht at sea is determining two coordinates: the length and the latitude. In practice he often enrols coordinates of the position into this way: The longitude and the latitude are being measured up in steps, minutes and seconds, where:
The horizonHorizon of the observer - it is the plain distant from the surface of the Earth at the so-called optic height (that is distance equal of raising eyes of the observer above the surface of the Earth) and perpendicular to the clearing of the sector running across the place of the observer and the centre of Land. True horizon - it is a parallel to the horizon of the observer and going through the middle of Land plain. The sight of the observer is limited by the ruler of the horizon and he is depending on the optic height. ![]() Knowing the height of raising eyes of the observer (h) it is possible easily to calculate the distance to the horizon what in certain situations can be essential (e.g. assessment of distance dividing the yacht from the edge). We are calculating the distance from the model.
We can use the line and to calculate the distance to the object about the known height (e.g. lights of the lighthouse) using the moment in which the light is hiding behind the horizon. Best to capture such a moment when the light is visible for the observer standing aboard and invisible for the sitting observer. ![]() Remarks:
Calculating the Distance to the Horizon
Using the Pythagoras theorem for a right-angled triangle OBC it is possible to deduce the formula to the distance to the horizon. AB = h (The height of the observers eye above see level in metres [m])
OA = OC = R (Earth radius)BC = d (Distance of sea horizon in Nautical Miles [Mm]) BO = h + R CO = R so BO2 = BC2 + CO2 (R + h)2 = d2 + R2 d2 = (R + h)2 – R2 = R2 + 2Rh + h2 – R2 So the total distance to the horizon is given by: On account of the minimum value to the Earth radius we are omitting the "h" in the expression (h+2R), then we will receive Calculating in nautical miles (Mm) Mean of the Earth radius R = 6370 x 103 m and we are sharing the result through 1852 m in order to get nautical miles (Mm). ![]() So As a result of the refraction, the average distance to the horizon is enlarged about 1/13. The height of the observer's eye above see level in metres [m] and feet
The text is in the preliminary translation. I am apologizing for mistakes in the text and I am asking for understanding.
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