what is the formula to work out distance of the horizon from sea level and 300 feet above sea level?
544990
Well as far as a formula i can't help you there. i just happen to own a program that allows me to draw perfect shapes of any size and diminsion them accordingly (auto CADD). and at the top of your 300 ft house, or where ever it is that your measuring from the horizon at sea level is a tad over 21 miles away (21.20389 to be analy exact). first you must know the the earths circumfrance (i know i can't spell) which is 20,890,637 ft, next you have to know the distance you are away from the earth (which you do, 300 ft). now you have to know the formula for finding a line that intersects your hieght and crosses tangentally across the earths surface. know you have your triangle and apply the Pythagorean theorem. you have an answer.
670636
Google has a very nice built-in measurement converter.  Simply type the following into Google's search engine:\n\nconvert 21000 feet to miles\n\nThe answer:\n\n21 000 feet = 3.97727273 miles
643214
Elevation: 138 feet      \nLatitude: 78 11S   \nLongitude: 162 12W
915176
First of all, latitude first, then longitude.\n\nThere are TWO common ways:\n\nUse the numer of degrees followed by N(orth) or S(outh),\nthen the numer of degrees followed by E(ast) or W(est).\n\nEx) 30*N, 90*W\n\nUse only the number of degrees (positive for North-negative for South), followed by the number of degrees (positive for East-negative for West)\n\nEx) 30*, -90*
823839
Step 1: find a place where the horizon is at the same altitude above sea level as the place you are standing. Standing on the coast or a boat, looking out to sea is best.\nStep 2: Look South from the northern hemisphere and north from the southern hemisphere\nStep 3: Take the reading at 12 noon (no Day Light Saving allowed.)
188418
32.81 feet.\nSo rounded up it's about 33 feet
323470
not really a question. All points of land are based on sea level. If you standing on the beach with your feet in the water your elevation would be 0. walk up it might be 1ft or more. standing on a mountain it might be 2500ft above sea level.
764566
a) 5000*,4 = 2000 feet, so a = 1000 feet\n\nb) x^2/a2 -y2/b^2 =1, where;\n   where the two foci are located at (-c,0) and (c, 0) so c= 4000 ft\n   2a = difference in distances between foci and whale\n   b^2 = c^2-a^2\n\nc) Closest when y = 0 which give x= abs(a)\n\nd) y=+ or - b/a[sqrt ( x^2-a^2)], so if x < a, then y is imaginary. Therefore the region between x=a and x=-a must be excluded, that is, the whale can never be in that region. The whale will always be on the hyperbola branch that is associated with the mike that first receives the signal. The whale can be on either branch of the hyperbola, but never (obviously) on both at the same time. The distance from one foci to the whale will always be < or > than the distance from the other foci to the whale. The distances can never be equal. If they were, this technology would not work.\n\nI believe I included all the info you need to answer the question completely. It would be nice if a diagram could be included.
490274
What several answerers are failing to remember is that St. Louis is not at sea level.  The St. Louis Gateway Arch is 192 meters (630ft) high.  But it is sitting on land that is 132 meters (433ft) above sea level.  This comes to a total height above sea level of 324 meters (1066ft).\n\nI'll assume you are trying to figure out the straight line distance to the horizon (the point where a viewing line is tangent with the curve of the earth).  The curved distance along the surface of the earth is almost the same when close to the earth, so it really isn't worth noting, but if you were in a satellite, there would be notable difference.\n\nAnyway, using triangles, you are trying to find the leg of a right-triangle with sides r, r+h and d.\n\nTo derive a formula figure out that d^2 + r^2 = (r+h)^2.\n\nr = radius of the earth (6378.14km or 6,378,000m)\nh = height above sea level (324m)\n\nAnyway, plug this all in and you get that the distance to the horizon is 64.29km or about 40 miles.\n\n40 miles is the correct answer
10131
Location: 37.0936N 113.2021W\nElevation: 3,250' (991m)
1076556
Altitude\n(feet)   Adjustment\n(inches)      Altitude\n(feet)   Adjustment\n(inches)  \n\n50  \nAdd 0.06      \n550  \nAdd 0.59   \n\n100  \nAdd 0.12     \n600  \nAdd 0.64   \n\n150  \nAdd 0.17     \n650  \nAdd 0.69   \n\n200  \nAdd 0.22     \n700  \nAdd 0.74   \n\n250  \nAdd 0.27     \n750  \nAdd 0.79   \n\n300  \nAdd 0.33     \n800  \nAdd 0.84   \n\n350  \nAdd 0.38     \n850  \nAdd 0.89   \n\n400  \nAdd 0.43     \n900  \nAdd 0.94   \n\n450  \nAdd 0.48     \n950  \nAdd 0.99   \n\n500  \nAdd 0.54     \n1000  \nAdd 1.04
285557
3,835 feet\n\n34º12'33 N\n117º17'37 W
1183908
There is no such equation. \n\nThe post office keeps a database with the latitude and longitude of each post office -- each zip code. You can look up the latitudes and longitudes of the zip codes and use them to compute the distance. For a spherical earth:\n\nLet Lat1 and Lon1 be the latitude and longitude of the first post office. Let Lat2 and Lon2 be the latitude and longitude of the second post office. Then compute the angles:\n\ndLon = | Lon1 - Lon2 |   (make sure to account for any difference in the east- or west- of Greenwich longitudes)\n\nD = 2 arcsin[ sqroot{ sin^2((Lat2 - Lat1)/ 2)+ cos(Lat1) cos(Lat2) sin^2(dLon/2) } ]\n\nIf you computed D in radians, convert D from an angle to a distance (in miles) using the equation: \n\nMiles = D * 12450 / pi\n\nIf you computed D in degrees, use the equation:\n\nMiles = D * 69.167
415302
You need to use Pythagorean theorem.\nMeasure the lenght of the lotus plant which is above the water.(X)\nBend the plant to any one side until it is just submerged under the surface of water.\nMeasure the Dist. between the initial and final positions of the plant(Y)(i.e. the initial position and final position where it is submerged)\n\nLet the actual lenght be L.\nThen (L x L) = (Y x Y) + (L - X)(L - X)
501968
1060 ft and 571 ft, respectively.\n\nLearn to do your own searches, it's much quicker.
559805
9 miles\n\nThat's if you were standing at the edge of a calm stretch of water (and you're approximately 5 foot 8 inches tall) - then the horizon is approx. nine miles away.\n\nObviously, if you gain more height you'd see further, and you could actually see things beyond this horizon - if the things are tall enough!
1170794
you let the navy do the work for you!!!  check out this awesome altitude/azimuth calculator!\n\nhttp://aa.usno.navy.mil/data/docs/AltAz.html