Re: Oblate Spheroid correction for computing distances?
Hello toves, quote fromtext below.. "At this level of accuracy, what is meant by the "position" of a station starts to get lost in the noise of the planet's shape." There are (of course) programs that will try to tell you to the nearest inch what the distance is between any two points on the globe, and throw in the azimuth for good measure. The following text, admittedly a bit longish,I copiedfrom an amateur radio book. It is about programs giving "distance and bearing" between two points. Just replace "station" by "location" and you're allright. I did not include the actual (Basic) programs. If anyone wants them, let me know. They are from 1985. But it is really the reasoning I thought was interesting.The text is by John Morris, GM4ANB for illuminati. We meet him just after he has explained a basic, sphere-based program. -- snip -- Latitude correction So far it has been assumed that the Earth is a perfect sphere. Of course it is not, being somewhat flattened at the poles and bulging at the equator. It is convenient to assume that the Earth is ellipsoidal; that is a cross section through the equator is circular, but a cross section through a line of longitude is elliptical. This assumption introduces two corrections. The most obvious one is that the distances must now be measured along the boundary of a more or less eccentric ellipse. Surprisingly, this is a relatively small correction. A rather larger error arises because latitude is not quite what most people think it is. Fig 5.2 shows a cross section of the Earth through a line of longitude. The eccentricity is much exaggerated. For a given point P, on the surface of the Earth, the angle ECP is called the central angle. It is this angle which all of the preceding distance and bearing formulae assume is the latitude. In fact the latitude is calculated from the normal to the surface of the Earth at P, giving angle ETP. This arises from the way latitude was historically measured; by the positions of the stars relative to the horizon. The distance a is called the semi-major axis of the ellipse, and b is the semi-minor axis. For a given latitude the central angle can be calculated as follows: Tan(central angle) = (b/a).(b/a).tan(latitude) For the Earth the value of a is about 6378km, and b is about 6356km. The average of these, 6367km, can be used as an average Earth radius, as was done in Programs 5.1 and 5.2. Program 5.3 shows the lines which should be changed or added in Program 5.2 so that central angles are used instead of latitudes. Line 30 sets up a few constants giving the size and shape of the Earth. The new subroutine, in lines 2000-2100, is very similar to the original in Program 5.2, but instead of the latitudes uses the central angles, Nl and N2, in the formulae. The latter are calculated in line 2000 using the above formula and the assumed shape of the Earth. A typical run after the lines of Program 5.3 has been included is shown in Fig 5.3. The results are similar to those from Program 5.2, but not identical. In this version of the subroutine the maximum distance error is reduced from 0.5% to about 0.2%. The residual error arises because the distance calculation itself still assumes a spherical Earth, even though the correct central angles have been used. The azimuth error is extremely small. The bearing from one point to another does not depend on the shape of the Earth, only on the central angles. Once these are known no further correction for the shape of the Earth is needed. Ellipticity correction Program 5.4 shows a replacement subroutine for those in Programs 5.2 and 5.3 which takes account of the ellipticity of the Earth when calculating distance. As may be seen, getting the last few tenths of a percent of accuracy is a messy business. The mathematics behind this routine are rather long, but the steps involved can be summarised as follows: Lines 2000 to 2070 find the central angle and azimuth, as previously. The rest of the routine finds the distance. Lines 2080 to 2170 find the great circle on which the two stations lie. This great circle is in fact an ellipse, its ellipticity depending on its inclination to the equator. If the great circle intersects the equator at a large angle (such as a line of longitude) the ellipticity will be the same as that of the Earth. If the angle is small the ellipticity will be small; in the case of the equator itself it is zero. AL is this angle of intersection. XI and (XI+CA) are the angular distances round the great circle from the equator of the two stations. EP is the square of the eccentricity of the ellipse on which both stations lie. Once the ellipse of interest and the positions of the two stations on it have been identified lines 2180 to 2200 perform the actual distance calculation. The formula in line 2200 is an approximation to an infinite series in EP. As EP will, for the Earth, always be small (its maximum is about 0.007)
RE: Oblate Spheroid correction for computing distances?
Jim, For a simple mental arithmetic answer, I always understood that the English nautical mile (6080 feet when I was at school, about right for the English Channel - but 6076 or so for an International Nautical Mile) was by design very close to 1 minute of latitude, or longitude at the Equator. So 60 n.m. of 6080 feet = 69.09 English statute miles (of 5280 feet) = 1 degree. For metric users, the original definition (or rather the second, after they'd tried the length of a 1-second pendulum at Paris) of the metre was as 1/1000th of the quadrant of longitude through Paris. So taking 90 degrees = 10 million metres gives 1 degree = 111.111 km = 69.04 miles. Thus figures of 69 miles and 111 km are quite close enough for my everyday purposes. Multiply by cos(latitude) to get the length for a degree of longitude at any latitude. So here at 51 N, a degree of longitude is about 43.5 miles. The question of what do you mean by latitude has a bearing on the length of the degree of latitude. It is complicated by both the flattening of the Earth and the effect of the Earth's rotation on the apparent direction of gravity. The apparent direction of gravity affects what one practically measures as horizontal and vertical with level and plumb-line, as opposed to an imaginary vertical passing through the centre of the Earth - if you can decide where you think that is! At this point I'll retire and leave others who are better qualified to explain that and the resulting change in the length of a degree of latitude! Regards Andrew James -Original Message- From: J.Tallman [mailto:[EMAIL PROTECTED] Sent: 03 February 2004 15:58 To: sundial@rrz.uni-koeln.de Subject: Re: Oblate Spheroid correction for computing distances? Hello All, As previously mentioned, the earth is not a perfect sphere, and is distorted by the effects of gravity. So, is it flattened at the poles, or is it elongated at the equator? Is it a combination of both effects? I can envision a stretching effect at the poles, and a bulging effect at the equator, both of which I would think would affect the linear distances between theoretical degrees of latitude. If that is the case then I would think that the only true distances would be found in the mid-latitudes. I would really like to know how to calculate distances using the coordinates, as well. For example, the linear distance of a degree of longitude at the equator is greater than the linear distance that I would find here at 39 N. Every once in a while a sundial customer asks me how far away he can move before his sundial becomes inaccurate. I have always figured about 2 degrees of longitude is an acceptable range, but have no idea how to convert that to linear distance at a given latitude, other than manipulating the mapping sites that can handle coordinates. Best, Jim Tallman Sr. Designer FX Studios 513.829.1888 This message has been scanned for viruses by MailControl, a service from BlackSpider Technologies. This correspondence is confidential and is solely for the intended recipient(s). If you are not the intended recipient, you must not use, disclose, copy, distribute or retain this message or any part of it. If you are not the intended recipient please delete this correspondence from your system and notify the sender immediately. This message has been scanned for viruses by MailControl - www.mailcontrol.com -
Re: Oblate Spheroid correction for computing distances?
Dear Thad As many of us know, we can geometrically compute the distance between two locations (lat, long) and (lat2, long2) assuming that the Earth is a perfect sphere (which of course it isn't). Has anyone seen a correction for this flattening at the poles, or bowing around the equator? As always, Meeus has the answer. The crucial difference is that between geographic latitude and geocentric latitude: The geographic latitude is the apparent altitude of the nearer celestial pole measured above the northern (or southern) horizon. Meeus calls this phi. The geocentric latitude is the angle that a radius from the centre of the Earth to the observer makes with the plane of the Equator. Meeus calls this phi'. The difference is given as: phi - phi' = 692.73 sin(2 phi) - 1.16 sin(4 phi) The constants are arc-seconds. The greatest difference is at a latitude of 45 degrees when the difference is about 11.5 arc-minutes. This translates into about 11.5 nautical miles. This is the about the error where you live! Geographic latitude is what is normally measured and used. This is what is marked on maps. There is an implicit assumption that the plane of the horizon is perpendicular to the local gravitational vector. This means you can use a normal sextant or other instrument that measures relative to the horizon or you can use an instrument that has some kind of spirit-level built in. Beware of massive mountains nearby! Frank King University of Cambridge England -
Re: Oblate Spheroid correction for computing distances?
Hello All, As previously mentioned, the earth is not a perfect sphere, and is distorted by the effects of gravity. So, is it "flattened" at the poles, or is it "elongated" at the equator? Is it a combination of both effects? I can envision a stretching effect at the poles, and a bulging effect at the equator, both of which I would think would affect the linear distances between theoretical degrees of latitude. If that is the case then I would think that the only "true" distances would be found in the mid-latitudes. I would really like to know how to calculate distances using the coordinates, as well.For example, the linear distance of adegree of longitude at the equator is greater than the linear distance that I would find here at 39 N. Every once in a while a sundial customer asks me how far away he can move before his sundial becomes inaccurate. I have always figured about 2 degrees of longitude is an acceptable range, but have no idea how to convert that to linear distance at a given latitude, other than manipulating the mapping sites that can handle coordinates. Best, Jim Tallman Sr. Designer FX Studios 513.829.1888
Re: Oblate Spheroid correction for computing distances?
WGS 84 ellipsoid semi-major (equatorial) axis: 6 378 137.0 m semi-minor (polar) axis: 6 356 752.3142 m (biaxial ellipsoid or just ellipsoid is the preferred (at least in North America) term for spheroid) What kind of differences might you see when comparing great circle routes on an approximating sphere with geodesics on the ellipsoid? As an example, the distance between Washington and L.A. on the sphere is approximately 3711 km. On the ellipsoid, it is 3719 km. Here are the expressions for computing the distance in km for one degree of latitude or longitude on the WGS 84 ellipsoid as a function of latitude, phi: lat = 111.13295 - 0.55982Cos 2 phi + 0.00117Cos 4 phi long = 111.41288 Cos phi - 0.09350 Cos 3 phi + 0.00012 Cos 5 phi See Navigation 101: Basic Navigation with a GPS Receiver http://gauss.gge.unb.ca/papers.pdf/gpsworld.october00.pdf for further details. Navigate is a handy application for computing geodesics on various ellipsoids for PDAs using the Palm OS: http://fermi.jhuapl.edu/navigate/index.html -- Richard Langley Professor of Geodesy and Precision Navigation On Tue, 3 Feb 2004, Thaddeus Weakley wrote: Hello All - Tony's posting reminds me of a question that I have had for a long time. As many of us know, we call geometrically compute the distance between two locations (lat, long) and (lat2, long2) assuming that the Earth is a perfect sphere (which of course it isn't). Has anyone seen a correction for this flattening at the poles, or bowing around the equator? If so, please share. Thanks, Thad Weakley 42.2N, 83.8W - Do you Yahoo!? Yahoo! SiteBuilder - Free web site building tool. Try it! === Richard B. LangleyE-mail: [EMAIL PROTECTED] Geodetic Research Laboratory Web: http://www.unb.ca/GGE/ Dept. of Geodesy and Geomatics EngineeringPhone:+1 506 453-5142 University of New Brunswick Fax: +1 506 453-4943 Fredericton, N.B., Canada E3B 5A3 Fredericton? Where's that? See: http://www.city.fredericton.nb.ca/ === -
