How Far Do the Birds Fly?Date: 08/12/2002 at 17:16:24 From: John Vance Subject: Radius of the Earth Dear Dr Math, I am a racing pigeon enthusiast, and one problem I wanted to solve for myself and others in the sport is how to calculate the survey distance our birds fly from point A to point B, where A is the release point and B is the home loft of a returning bird. We have the latitude and longitude of each home loft and the release points that we race from, and up until now have used a surveying service to calculate these distances for us. With the advent of the Internet and Web sites, more and more information is available to the curious and I have gone on a quest to create an area on my site where other flyers can figure out the distance between two points. From your site, I have been able to program a Web page to do the calculations, and that is when I came to the problem I now face. The formula I use when written in expressions my server software can understand is as follows: mylength = ACOS((COS(a1)*COS(b1)*COS(a2)*COS(b2)) + (COS(a1)*SIN(b1) *COS(a2)*SIN(b2)) + (SIN(a1)*SIN(a2)))*r the value "r" however is the great unknown. I live at 48.8236666667 -122.5888333333 Since the radius of the earth is ever-changing from equator to pole, and since I am somewhat near the halfway mark in Latitude, I chose to use the volumetric radius of the earth as given by NASA (3958.76 miles) since I thought that this radius would most closely approximate my local conditions. When I read this entry in the Dr. Math archives: Using Longitude and Latitude to Determine Distance http://mathforum.com/library/drmath/view/51711.html Where Nat Keller uses a value for "r" of 3963.1 miles it only further points out that few measurements would agree, since the likelihood that two calculations from two different solutions would use the same radius values is small. We all agree on the formula, just not the appropriate radius values. I think there is a 22-mile difference between the radius of the earth at the equator and at the poles. Is there a formula that derives a close approximation of the average radius between two latitude readings? For example, if the two latitude readings I was using were 48.8236666667 and 44.9386544, is there a formula I could use to calculate an average value for "r"? Thank you, John Vance Date: 08/13/2002 at 09:42:57 From: Doctor Rick Subject: Re: Radius of the Earth Hi, John. Recently I have been learning a little about working with ellipsoids - a closer approximation to the shape of the earth, with its slight polar flattening. Maybe I've only learned enough to be dangerous, but here goes! I think the correct radius to use is the radius of curvature of the earth in the direction of interest, at the point of interest. Over a long course the radius of curvature would change, so this can be only an approximation. However, we're only talking about a maximum error of +-11/3960 = 0.28% or so even when we pretend the earth is a sphere, so we aren't going to get too large an error by pretending it's an ellipsoid but the radius of curvature doesn't change, unless the course is, say, 1000 miles or more. The radius of curvature at a point on the ellipsoid depends on the direction of travel. Two radii are defined: M, the radius of curvature in the meridian (north-south), and N, the radius of curvature in the prime vertical (east-west). They are M = a(1 - e^2)/(1 - e^2*sin^2(phi))^3/2 N = a/sqrt(1 - e^2*sin^2(phi)) where a is the radius of the earth at the equator, e is the eccentricity of the earth (e = ), and phi is the latitude. (Specifically it's the geodetic latitude; there are several ways to define latitude on an ellipsoid, but this one is what we measure astronomically.) The radius of curvature along a course at bearing beta is expressed in terms of the two principal radii of curvature: R = M*N/(N*cos^2(beta)+M*sin^2(beta)) The mean radius at a given latitude - that is, averaging over all directions at a given point - is sqrt(MN) = a*sqrt(1-e^2)/(1 - e^2*sin^2(phi)) I got these formulas from the following site: Geodesy Course (The Hong Kong Polytechnic University) http://www.lsgi.polyu.edu.hk/Cyber-Class/geodesy/index.htm From the same Web site, the values of a and e for the formulas I gave are: a = 6378137 m = 3963.191 miles e = 0.08181919084 - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 08/13/2002 at 12:15:43 From: John Vance Subject: Radius of the Earth Thank you for the help. sqrt(MN) = a*sqrt(1-e^2)/(1 - e^2*sin^2(phi)) I have tested this formula in an Excel spreadsheet and only get a correct answer at phi=0 degrees, and then it is the polar radius 6356752.314 (shouldn't the polar radius be at phi=90 degrees?). I believe that I am not processing the sin(phi) properly. Below are my values. a=6378137 meters e^2=0.00669438 1-e^2=0.99330562 sqrt(1-e^2)=0.996647189 a*sqrt(1-e^2)=6356752.314 When phi=45 degrees the below values are what I get: sin(phi)=0.707106781 sin^2(phi)=0.5 or sin(phi)*sin(phi)=0.5 1-e^2*sin^2(phi)=0.99665281 If I set phi to 90 degrees, I get Sqrt(MN)=6399593.626 or 21456.62576 meters in excess of the equatorial radius. What am I doing wrong? Thanks. Date: 08/13/2002 at 13:53:50 From: Doctor Rick Subject: Re: Radius of the Earth Hi, John. I think your only error is in your expectations. Remember that we are calculating the radius of curvature, and this has nothing necessarily to do with the distance from the center of the earth. The earth is flattened at the poles. The flatter a curve (or a surface containing a curve), the greater its radius of curvature - the radius of the circle that most closely approximates the curve. So the radius of curvature at the poles will be greater than the equatorial radius, and indeed we find that it is a^2/b. It's the same in all directions, due to the symmetry at the poles, so the mean radius of curvature is also a^2/b. At the equator, it's easy to see that the radius of curvature in an east-west direction is a, because the equator *is* a circle of this radius - no approximation is required. In the north-south direction we'll get a smaller radius of curvature; somehow it works out that the mean is exactly b. The geodetic latitude is the complement of the angle between a line perpendicular to the surface and the polar axis. This perpendicular line does not necessarily go through the center of the earth. If a north-south path on the ellipsoid is short enough, it can be approximated by a circle of radius equal to the radius of curvature. Thus the arc length is approximately the radius of curvature times the difference in geodetic latitudes of the endpoints. This is as much as I understand so far regarding the calculation of arc lengths on an ellipsoid, but I think it is all you need. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 08/13/2002 at 18:10:40 From: John Vance Subject: Radius of the Earth Thanks so much for the help. My quick fix solutition to the problem is as follows: a=equatorial radius b=polar radius f=(a-b)/a (flattening) R(phi)=a-(a*f*sin(phi)) at phi=90 degrees sin(90)=1 R(phi)=a-(a*f) R(phi)=a-(a*(a-b)/a) R(phi)=a-(a-b) R(phi)=b at phi=0 degrees sin(0)=0 R(phi)=a-(a*f*sin(phi)) R(phi)=a at phi=45 sin(45)=0.707106781 R(phi)=a-(a*f*sin(phi)) R(phi)=6363015.744 meters or 3953.79 statute miles Whereas, (a+b)/2 = 6367444.657 meters or 3956.55 staute miles which is the result if phi were equal to 30 degrees. Not the perfect model but closer than using "a". Now if I was calculating the distance of two points, one at 43 degrees latitude and the other at 48 degrees latitude, I could approximate the average radius over the distance as: phi=Lat(ave)=(43+48)/2 phi=45.5 and plug this value into: R(phi)=a-(a*f*sin(phi)) One of several questions remaining would be if radians would be a better appoximation for phi. I am not really all that good at trig to know when radians apply. Your comments are welcome as to this solution as generalized as it is. Thank you, John Vance Date: 08/14/2002 at 08:30:34 From: Doctor Rick Subject: Re: Radius of the Earth Hi, John. I take it that you don't want to make use of the radius of curvature at all, and instead you are going with the approach of interpolating between a and b. Perhaps the following web page will convince you that the radius of curvature is really the way to go: GIS FAQ Question 5.1 http://www.census.gov/cgi-bin/geo/gisfaq?Q5.1 This page has a table of radius of curvature versus latitude (just over halfway down the page; r = mean radius of curvature, R' = my M). You will see there that the radius of curvature is GREATEST at the poles and LEAST at the equator, contrary to your calculation based on distance from the center of the earth. In reference to radians, it's not a matter of which gives a better approximation for phi; that would be like saying that meters give a better approximation for a distance than feet. Radians are just a different unit for measuring angles: 1 radian = 180/pi degrees, or 57.295779 degrees. The angle MUST be in radians, however, before you multiply it by the radius (of curvature) to get an arc length. If you use the angle in degrees, you won't just have a worse approximation, you'll be off by a factor of 57.295779 - not even close! I hope this helps. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 08/14/2002 at 14:54:25 From: John Vance Subject: Radius of the Earth Hello Dr Math, My assumption was that delta R (change in the length of the radius from the surface to the center of the earth) would be greater moving away from the equator then moving away from the pole and that is why I used a sin function. For example, delta R from 90 degrees to 89 degrees would be less than from 0 degrees to 1 degree. If I have this backward, then possibly I should be using the cosine. You may wonder why I even care about this. We race our birds from distances of 75 miles out to 500 and 600 miles. Each bird carries a band with an RFID chip that holds a unique ID number. Each loft has an antenna that reads the chip upon the bird's return and logs the ID number and the time of arrival to 1/10th of a second. Through GPS we know the Lat and Long of the release point and each loft. We calculate the distance, divide that by the time of flight and derive a velocity for each bird in the race. The velocity is printed on our race results to 1/1000th of a ypm. Some lofts are 70 - 80 miles farther than other lofts in the race so the better we are able to calculate the straight line distance the better we are able to declare the race winner based on the ypm criteria. It is true that the birds do not travel in a straight line, but we must calculate distances as though they do. As it is now, we pay for a surveyor to calculate our distances from each loft to each release point. I want to develop a Web site that would do the calculations for free. Also, we don't know what value the surveyor is using for r so I must be able to justify why my value for r is as accurate as is possible under the circumstances. Sorry it has taken so much of your time to get this far. I have learned a lot from all your input. Thank you! One more thing, and this is possibly why I was confused about the Radii of Curvature: latitude...........r...................R'..................N.......... 00 degrees . 6357 km (3950 mi) . 6336 km (3937 mi) . 6378 km (3963 mi) 15 degrees . 6360 km (3952 mi) . 6340 km (3940 mi) . 6379 km (3964 mi) 30 degrees . 6367 km (3957 mi) . 6352 km (3947 mi) . 6383 km (3966 mi) 45 degrees . 6378 km (3963 mi) . 6367 km (3957 mi) . 6389 km (3970 mi) 60 degrees . 6388 km (3970 mi) . 6383 km (3966 mi) . 6394 km (3973 mi) 75 degrees . 6396 km (3974 mi) . 6395 km (3974 mi) . 6398 km (3975 mi) 90 degrees . 6399 km (3976 mi) . 6399 km (3976 mi) . 6399 km (3976 mi) Would this mean that if I am calculating distance near Lat 45, I should assume the radius of the earth is 6378 km, and if near the equator that the radius is 6399 km? Thanks again for all the time you have spent, John V Date: 08/15/2002 at 13:42:19 From: Doctor Rick Subject: Re: Radius of the Earth Hi, John. If you don't want to include the bearing in the calculation, then the average radius of curvature is the best you can do; that means the figures you quote. I gave you a formula to calculate the average radius of curvature of the ellipsoid exactly, rather than choosing an arbitrary means of interpolation as you have been doing. You see from the table, though, that there is still a variation in the actual radius of curvature at lat 45 degrees of plus or minus 11 km from the average radius of curvature. That is, if the course is north-south, the radius of curvature is 11 km less than the average, and if the course is east-west, the radius of curvature is 11 km greater than the average. You'll be better off if you can use the formula I gave for the radius of curvature as a function of both latitude and bearing. If you can calculate the bearing between the two points (and I can show you formulas for this if you don't have one), you can then use the formula to get a more accurate radius for use in the distance formula. In response to your previous comments about the need for this precision: I don't know the speed of a pigeon in a 600-mile race, but I'd estimate that you're measuring the time to a relative precision of about one millionth (10^-6) and printing the velocity to a relative precision of a millionth or even better. For the velocity to have an accuracy of this magnitude, you'd need to calculate the distance to an accuracy of plus or minus a yard or so. I doubt that you can get anywhere near that accuracy. Over 600 miles the radius of curvature could change by about 4 miles if it's a north-south course, and I would expect the sphere-based distance formula to show errors from other sources as well over such a distance. I suspect you'd need a much more complicated numerical algorithm to get the accuracy you seek. This goes beyond my basic knowledge of geodesy and ellipsoids. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 08/21/2002 at 14:11:21 From: John Vance Subject: Radius of the Earth I suppose at this point I am incredibly buried in the terminology of this solution. If I have the Lat and Lon of the release point and the home loft, what exactly should I do to arrive at the most accurate distance calculation? For example, if the release point is at (I have converted to decimal notation): Lat: 43.801216 Lon: -123.048701 and the home loft at: Lat: 48.823666 Lon: -122.588833 What would I do according to the instructions you have been giving me? I must admit that I think I do not understand how to properly identify the vars and plug them into a formula to give the proper distance based on the Lat - Lon - direction of flight. This would probably clear up my misunderstanding if I saw how you calculated the results. Thank you, John Vance Date: 08/21/2002 at 15:15:15 From: Doctor Rick Subject: Re: Radius of the Earth Hi, John. Here is one way to get an approximation. First find the heading from point 1, using this formula (from the Aviation Formulary on the Web): heading = mod(atan2(-sin(lon1-lon2)*cos(lat2), cos(lat1)*sin(lat2) -sin(lat1)*cos(lat2)*cos(lon1-lon2)), 2*pi) I get 0.060276182 radians, or 3.453570825 degrees east of north. (If your atan2 function takes arguments (x,y) like Excel rather that (y,x) like C, then switch the arguments.) Now we can use this heading in the calculation of the radius of curvature: M = a(1 - e^2)/(1 - e^2*sin^2(phi))^3/2 N = a/sqrt(1 - e^2*sin^2(phi)) R = M*N/(N*cos^2(beta)+M*sin^2(beta)) I get: M = 3937.942347 miles N = 3963.621319 miles R = 3938.03493 miles Because the route is nearly north-south, the radius along the route is close to the radius of curvature in the meridian (north-south). You have the formula for the distance; use the radius R. If you want even better accuracy, you can calculate the latitude and longitude of the midpoint of the route, using the formula here: Latitude and Longitude of a Point Halfway between Two Points http://mathforum.org/library/drmath/view/51822.html Then calculate the heading at the midpoint, using the formula I used above, and use the latitude and heading at the midpoint in the calculation of R. It's still an approximation, but it's about as good as I could do. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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