Re Radius of the Earth
Many thanks to everyone who contributed to the comprehensive response to my query. Having spent ten hours today on that 94 metre hill today - waiting for the sun to come out for most of the time - I can now get to grips with the essentials before sleep overtakes me. The Sundial Mailing List is such a wonderful resource z z z z z z z z z z z z! Thanks again Tony Mo. z z z z z z
Re: Radius of the Earth
At 12:18 3-8-98 +0100, you wrote: -Original Message/Oorspronkelijk bericht-- >Fellow Shadow Watchers, > Information boards are being prepared for a nearly-completed large dial on a hill near the Northumberland (UK) coast which is 94 metres above mean sea level at Latitude 55° 1' 38" North Longitude 1° 30' 16" West. > > The latest question is "How far away is the sea horizon?" School geography taught me that the earth is an 'oblate speroid' so I suppose the true distance varies slightly depending on the direction in which the observer is looking but so little as to be unimportant perhaps? The sea is only visible in a generally easterly direction. > >Can any list member supply the mean sea level radius of the earth at this location on which to base the necessary trig calculation plus any subtleties I may have overlooked as I don't have ready access to specialist reference material of this sort. > >With thanks in anticipation of any helpful response. > >Tony Moss > Tony, I saw already some nice formulas for the calculations. However your starting point is constant changing: your heigth is 94 meters above mean sea level you say. So that is only 4 times a day. All other moments your height above the actual water level will be more or less thanm the 94 meters! Each meter more or less changes the distance to the horizon by ca. 0.1 nautical miles. Is that heigth of 94 meters at ground level or is it the heigth of the eye of the observer? How much is the difference between high tide and low tide at that spot? In Britain the sea-levels are related to the Chart datum, which usually approximates to Mean Low Water Springs (which occurs roughly once a month). Normally a sailor doesn't have to take this into account because he floats on the water, but you have a fixed observationpoint on the land. I am afraid you have to make a few corrections to the 94 meters! The refraction is dependant on the airpressure and the temperature. - Thibaud Taudin-Chabot, home email: [EMAIL PROTECTED] (attachments max 1.2MB, in case of larger attachments contact me)
Re: Radius of the Earth
On Mon, 3 Aug 1998, Tony Moss wrote: > > The latest question is "How far away is the sea horizon?" Like most sundial questions this is also of imporance in navigation. How far away the horizon is at sea depends, of course, on how far above sea leval you are. If you are treading water with your eyes at seal leval your horizon will be very near your face. If you stand in the lifeboat with your eyes 6 feet above sea level the horizon is 3 miles out. Take the distance in feet from the sea surface to your eye level, multiply by 3/2 and then take the square root. This is approximately the distance in miles to the horizon. (I assume they mean statute miles--this is boiled down from Science Digest Feb 1984 p 34)
Re: Radius of the Earth
Navigators use the term "dip" to refer to the effect of height of eye above sea level has on the altitude of a celestial body above the apparent horizon. Similarly , the distance to horizon is greater at higher eye elevations. A sailor on ship deck has an advantage in spotting land over one at sea level; but the sailor in the crow's nest sees land first. Any nautical celestial navigation book will have a table. The Nautical Almanac also. The simple formula is D = 1.169 SQR(h) where D is distance to horizon in nautical miles and h = height of observers eye in feet. [ If using meters the constant is 2.07] Use of this simplified formula is usually all that is needed. It requires of course a clear day to the horizon, perhaps questionable in Great Britain :) For greater accuracy try: D= SQR( [2*r*h] / [6076.1 *B] ) That 'B' will be a beta in the books. Again: D is distance to the horizon in nautical miles; r is the mean radius of the earth ( 3440.1 nautical miles) ; h is the height of eye IN FEET; and B is a contant relating to terrestrial refraction, 0.8279 Seems to me that B could vary depending on atmospheric conditions but I never studied it to that detail. However, Bowditch, "The American Practical Navigator" (a classic tome) states the " error in refraction is generally less than that introduced by nonstandard atmospheric conditions" Good Luck DAVE Tony Moss wrote: > Fellow Shadow Watchers, >Information boards are being prepared for a > nearly-completed large dial on a hill near the Northumberland (UK) coast > which is 94 metres above mean sea level at Latitude 55° 1' 38" North > Longitude 1° 30' 16" West. > > The latest question is "How far away is the sea horizon?" School geography > taught me that the earth is an 'oblate speroid' so I suppose the true > distance varies slightly depending on the direction in which the observer is > looking but so little as to be unimportant perhaps? The sea is only visible > in a generally easterly direction. > > Can any list member supply the mean sea level radius of the earth at this > location on which to base the necessary trig calculation plus any subtleties > I may have overlooked as I don't have ready access to specialist reference > material of this sort. > > With thanks in anticipation of any helpful response. > > Tony Moss
Re: Radius of the Earth
The prime vertical radius of curvature of the ellipsoid (the radius of the osculating circle in the east-west direction) is given by a^2 N = --- (a^2 * (cos phi)^2 + b^2 * (sin phi)^2)^0.5 The ellipsoid of the Geodetic Reference System 1980 has a = 6 378 137 metres b = 6 356 752 metres (to the nearest metre). In computing the distance to the horizon, if you want the visibile horizon (as opposed to the geometric one), you should take refraction into account. -- Richard Langley Professor of Geodesy and Precision Navigation where a and b are the semimajor and semiminor axes of the ellipsoid and phi is the geodetic latitude. On Mon, 3 Aug 1998, Tony Moss wrote: >Fellow Shadow Watchers, > Information boards are being prepared for a > nearly-completed large dial on a hill near the Northumberland (UK) coast > which is 94 metres above mean sea level at Latitude 55 1' 38" North > Longitude 1 30' 16" West. > > The latest question is "How far away is the sea horizon?" School geography > taught me that the earth is an 'oblate speroid' so I suppose the true > distance varies slightly depending on the direction in which the observer is > looking but so little as to be unimportant perhaps? The sea is only visible > in a generally easterly direction. > >Can any list member supply the mean sea level radius of the earth at this >location on which to base the necessary trig calculation plus any subtleties I >may have overlooked as I don't have ready access to specialist reference >material of this sort. > >With thanks in anticipation of any helpful response. > >Tony Moss > > === Richard B. Langley Internet: [EMAIL PROTECTED] or [EMAIL PROTECTED] Geodetic Research Laboratory BITnet: [EMAIL PROTECTED] or [EMAIL PROTECTED] Dept. of Geodesy and Geomatics Engineering Phone:(506) 453-5142 University of New BrunswickFAX: (506) 453-4943 Fredericton, N.B., Canada E3B 5A3 Telex:014-46202 Fredericton? Where's that? See: http://www.city.fredericton.nb.ca/ ===
