+Enchanted Shadow Posted January 2, 2006 Share Posted January 2, 2006 I have a specific need at the moment, which has led to a general question as well. To start with the specific - I need to calculate the exact coordinates of a point halfway between two other points. I've tried searching through Google for help with this (since the board's Search function is disabled at the moment), but the few answers I found were not written very clearly. Since this involves somewhat complex calculations (Great Circle Formula, etc...), clarity is vital. Does anyone have a good reference for the mathematics needed for this type of calculation? Now, that being asked, that leads me to a more general question... Are there any good references or programs for GPS coordinate based mathematics in general? For example: 1. Finding the coordinates of the midpoint between two points 2. Finding the coordinates of a point equidistant from three other points. 3. Finding the coordinates of the intersecting point of three circles, where the circles are of *differing* radiuses centered on three seperate sets of coordinates. 4. Etc... I must admit, I was a little surprised to find so few resources along these lines... But then again, it's always possible that I hadn't yet been working with the right search terms... Any help that can be provided (either with my specific or general question), would be appreciated! Quote Link to comment

+WeightMan Posted January 2, 2006 Share Posted January 2, 2006 If the points you are talking about are fairly close together, like a few miles or so, you really can assume a flat earth and make your calculations much easier. If the points are far apart then you will have to go the great circle calculations. There will be some error by assuming the flat earth, but for short distances the error should not be a problem given only three digit accuracy for the coords and the 20 foot or so errors on the GPSr. For the short distances I would just average the coords. Quote Link to comment

+Sonoran Privateers Posted January 2, 2006 Share Posted January 2, 2006 You could just avoid the math. First place both points into Google Earth as pleces. Then, use the measuring tool to measure the distance between them (if you don't already know it), and then find half way....put in a push pin. Zoom WAY in, and read the coordinates from the bottom of the map window that follow you're mouse arrow. I know, the math would be much cooler, but this gets it done for those of us that are to far out of math class to be useful. Quote Link to comment

+Timber_Wolf Posted January 3, 2006 Share Posted January 3, 2006 We have at work an program called PCMap. With this Program i can tell you everything you need. Quote Link to comment

+fizzymagic Posted January 3, 2006 Share Posted January 3, 2006 (edited) To start with the specific - I need to calculate the exact coordinates of a point halfway between two other points. Depends on what you mean by "halfway between." Is it along a geodesic? Or some other path? Assuming a geodesic, it's quite easy. Just calculate the distance and azimuths between the two points. Then project from the first point at the same forward azimuth but exactly half the distance. You can check your result by doing the projection from the second point using the reverse azimuth. Of course, there are actually 2 points that can be called "exactly halfway between" any other two -- the other one is on the other side of the Earth! As for the rest of your questions, the answer is "it depends." For short distances, UTM is usually accurate enough to solve these problems using plane geometry; for longer distances, the math gets quite complex. Edited January 3, 2006 by fizzymagic Quote Link to comment

+Lone Duck Posted January 3, 2006 Share Posted January 3, 2006 I think that any software mapping program that interfaces with a GPS can do this, and possibly better than Google Earth. Most of them have a built in distance tool, so you can simply divide that to get the mid point. Quote Link to comment

+Sonoran Privateers Posted January 3, 2006 Share Posted January 3, 2006 I think that any software mapping program that interfaces with a GPS can do this, and possibly better than Google Earth. Most of them have a built in distance tool, so you can simply divide that to get the mid point. I agree about the other software doing it better than GE. I just reference GE because it is free, and that way I don't have to assume anyone has the same software I have. Quote Link to comment

+Timber_Wolf Posted January 3, 2006 Share Posted January 3, 2006 I think that any software mapping program that interfaces with a GPS can do this, and possibly better than Google Earth. Most of them have a built in distance tool, so you can simply divide that to get the mid point. Yes maybe every program can do it. But PCMap in the Version we have is for goverment use only. For training i gave our guys some Cachedata so they can get used to this program. And it works pretty well for Caches. Quote Link to comment

+Sonoran Privateers Posted January 3, 2006 Share Posted January 3, 2006 I think that any software mapping program that interfaces with a GPS can do this, and possibly better than Google Earth. Most of them have a built in distance tool, so you can simply divide that to get the mid point. Yes maybe every program can do it. But PCMap in the Version we have is for goverment use only. For training i gave our guys some Cachedata so they can get used to this program. And it works pretty well for Caches. If it's just for the gubmint, I'm confused as to how it is to be useful to the OP Quote Link to comment

+Pasha Posted January 3, 2006 Share Posted January 3, 2006 I guess I don't understand the question. If I have two points: 47 31.289/122 36.968 and 47 31.396/122 36.907 (both waypoints of one of my caches), isn't the point between them going to be 47 31.3425/122 36.9375 At least close enough for our purposes? I can see error getting significant if one point is New York and one is Australia, but for a cache? Quote Link to comment

+fizzymagic Posted January 3, 2006 Share Posted January 3, 2006 I guess I don't understand the question. If I have two points: 47 31.289/122 36.968 and 47 31.396/122 36.907 (both waypoints of one of my caches), isn't the point between them going to be 47 31.3425/122 36.9375 At least close enough for our purposes? I can see error getting significant if one point is New York and one is Australia, but for a cache? Well, your waypoints are so close together that a linear approximation is quite good. But they don't have to be all that far apart before it doesn't work so well. Let's take two waypoints 20 miles apart: N 47 31.289 W 122 36.968 N 47 34.259, W 122 11.691 The distance is 20 miles, forward azimuth 80 degrees. Using your method, the midpoint would be: N 47 32.774 W 122 24.330 but the actual midpoint is: N 47 32.786, W 122 24.336 Your method gives a result that is off by almost 80 feet. Enough to make finding a cache very hard. Doing the calculation using UTM: First waypoint was: 10T E 528899 N 5263189 Second waypoint was: 10T E 560557 N 5268933 Average is: 10T E 544728 N 5266061, or N 47 32.786, W 122 24.335 which is only 4 feet off. Moral of the story: For very short distances (< 1 mile) a linear approximation works fine. For intermediate distances, UTM works pretty well (but remember that it can't be used to get azimuth angles). For long distances, only the hard math will do. Quote Link to comment

+Nellsnake Posted January 3, 2006 Share Posted January 3, 2006 1. Finding the coordinates of the midpoint between two points 2. Finding the coordinates of a point equidistant from three other points. 3. Finding the coordinates of the intersecting point of three circles, where the circles are of *differing* radiuses centered on three seperate sets of coordinates. 4. Etc... Any help that can be provided (either with my specific or general question), would be appreciated! The calculations you are asking about are fairly straightforward to do by hand or with a PC if you use the UTM (Cartesian) coordinate system. As fizzymagic said, this is the way to go if the points are within several miles of each other and the curvature of the earth can be neglected. 1. The midpoint on a line between two points (x1,y1) and (x2,y2) would just be ((x1+x2)/2, (y1+y2)/2). 2. Finding a point that is equidistant from three different points is the same thing as finding the center of a virtual circle whose circumference contains all three points (A, B, and C). One way to do this is to find the equations of the perpendicular bisectors of the lines connecting the three points (lines AB, AC, and BC). The perpendicular bisecting line is the locus of all points that are equidistant from the two points in question. If you then take any two of the three perpendicular bisecting lines and solve them simultaneously, you will find their intersection point. This is the unique point that is equidistant from all three points (i.e., the exact center of the circle specified by those three points). You can solve it for the other two combinations as well to confirm that you get the same answer. To find the equation of the perpendicular bisector of any two points, you: A. Derive the equation for the line segment connecting the two points by fitting the x,y coordinates to the equation: y = mx + b. This is fairly straightforward algebra. Actually, you will only need to solve for the slope m of the line (the y intercept b is not needed for any further calculations). B. Find the midpoint of the line segment (see #1 above). C. The perpendicular bisector is the line (y = m'x + b') that goes through this midpoint with slope m' equal to -1/m (i.e., the negative of the reciprocal of the slope calculated above). You have x and y values and a slope for that line, so you can solve for b'. 3. To find the coordinates of the intersecting point of three circles of differing radii, it is easiest to first calculate the two intersection points of two of the three circles and then figure out which of the two points lies on the circumference of the third circle. To calculate the intersection points of two circles, you can refer to the following webpage, for example: Intersection of Two Circles Quote Link to comment

robertlipe Posted January 3, 2006 Share Posted January 3, 2006 The math is much simpler if you assume the world if flat, but that way of thinking pretty much is out of vogue now unless you're considering only very small pieces of the planet. The "teach a man to fish" department mandates that I share the link to http://williams.best.vwh.net/avform.htm as it's a concise list of all kind of things that Fizzymagic knows well enough to sing to the tune of "Pinball Wizard" but make my head hurt. Quote Link to comment

+CharlieP Posted January 3, 2006 Share Posted January 3, 2006 Although it may not be scientific enough for your needs, I have found a few caches that required determining intersecting circles by using the map page "measure distance" function. The procedure involves determining approximately where the intersection will occur, and then placing a few waypoints near the intersection at the correct radial distance from each reference point. The intersection becomes apparent and I place a waypoint there and check the distances and adjust as necessary. It works, and can be done in the field with just the GPS. Quote Link to comment

+Pasha Posted January 3, 2006 Share Posted January 3, 2006 The "teach a man to fish" department mandates that I share the link to http://williams.best.vwh.net/avform.htm as it's a concise list of all kind of things that Fizzymagic knows well enough to sing to the tune of "Pinball Wizard" but make my head hurt. Thanks Robert. That's what I was asking, I guess... why isn't the straight-line method correct? I understand we're dealing with a sphere. Is the problem that the distance between longitudinal degrees is not a constant? I'll eagerly peruse your link and see if I can find the answer. Quote Link to comment

robertlipe Posted January 3, 2006 Share Posted January 3, 2006 why isn't the straight-line method correct? I understand we're dealing with a sphere. Is the problem that the distance between longitudinal degrees is not a constant? I'll eagerly peruse your link and see if I can find the answer. Fizzymagic summed it up pretty well, though you might not have recognized the "catch'. Pythagorean theorem (and other triangle/straight line math) on coords works reasonably well when the distance is short enough that the planet can be figured flat or you're near the equator where a degree of long and a degree of lat cover the same linear distance. Move that same problem closer to the poles where the ratio of the distance covered by the two coords gets out of whack and the problem becomes harder. To be clear, Fizzymagic (and others) understand this better than I do. I'm secure enough in my geekdom to readily admit that. By profession, he's closer to a math guy than I am. But I (like many others) don't need to be able to recite this stuff once it's explained reasonably well to me; I just need a place to find the explanations and the formulae every now and then. In my case, that education came in the form of Chapter 3 in http://mappinghacks.com (and shoot me now if the moderators delete that as a commercial link to the O Reilly book by Erle, Gibson, and Walsh). This particular issue is covered in hack #27. I'm not really a mapping guy It wasn't until I saw the pictures in that book before I "got it". Maps are flat. The Planet isn't. To make the planet appear flat on a map, you have to scrunch it. Scrunching it means that the lines (or traiangles) you draw on the scrunched imaged are distoraged when compared to the non-flat version of the planet. Of course, as Mr. Magic alluded to, the world isn't really a sphere, either, but it's closer. It was that book that also helped me realize that the various countries didn't prefer different map projections just to be pig-headed. A map that renders Chile (a long, skinny country nearish to the equator) looks terrible for Greenland (a much squattier country that's mostly in the arctic circle). ding! Quote Link to comment

+BigWhiteTruck Posted January 3, 2006 Share Posted January 3, 2006 Is the problem that the distance between longitudinal degrees is not a constant? Yup. Quote Link to comment

+ParrotRobAndCeCe Posted January 3, 2006 Share Posted January 3, 2006 To start with the specific - I need to calculate the exact coordinates of a point halfway between two other points. Depends on what you mean by "halfway between." Is it along a geodesic? Or some other path? Assuming a geodesic, it's quite easy. Just calculate the distance and azimuths between the two points. Then project from the first point at the same forward azimuth but exactly half the distance. You can check your result by doing the projection from the second point using the reverse azimuth. Of course, there are actually 2 points that can be called "exactly halfway between" any other two -- the other one is on the other side of the Earth! Well, if you want to go down that path, there are actually an infinite number of points, all of which lie on the great circle that intersects the great circle arc between the two points perpendicularly and medially. Quote Link to comment

+ParrotRobAndCeCe Posted January 3, 2006 Share Posted January 3, 2006 (edited) Deleted duplicate post resulting from the PHP/MySQL timeout feature. Edited January 3, 2006 by ParrotRob Quote Link to comment

+ParrotRobAndCeCe Posted January 3, 2006 Share Posted January 3, 2006 (edited) Deleted duplicate post resulting from the PHP/MySQL timeout feature. Edited January 3, 2006 by ParrotRob Quote Link to comment

+WeightMan Posted January 3, 2006 Share Posted January 3, 2006 (edited) How can there be more than two points? A great circle is a line (admittedly on a sphere). That line is one point wide. There is one point midway between the two points on the shorter segment and one point midway between the two points on the longer segment. Where might there be other points on that great circle and equally distant between the two points. As far as I can see any other points equidistant would not be on the great circle. Editted to remove modified quote. Edited January 3, 2006 by WeightMan Quote Link to comment

+ParrotRobAndCeCe Posted January 3, 2006 Share Posted January 3, 2006 How can there be more than two points? A great circle is a line (admittedly on a sphere). That line is one point wide. There is one point midway between the two points on the shorter segment and one point midway between the two points on the longer segment. Where might there be other points on that great circle and equally distant between the two points. As far as I can see any other points equidistant would not be on the great circle. Editted to remove modified quote. Consider the following two points on the earth's surface: Point A: N 10°0'0", W 90°0'0" Point B: S 10°0'0", W 90°0'0" The great circle that bisects the arc between these two points is the equator. ANY point along the equator is equidistant between these two points. In fact, the infinite set of points that are equidistant between those two points all lie along a plane that extends through and beyond the earth as well, it's just that most of those points are rather inaccessible Quote Link to comment

+fizzymagic Posted January 3, 2006 Share Posted January 3, 2006 Of course, there are actually 2 points that can be called "exactly halfway between" any other two -- the other one is on the other side of the Earth! Well, if you want to go down that path, there are actually an infinite number of points, all of which lie on the great circle that intersects the great circle arc between the two points perpendicularly and medially. I suppose, although that is greatly stretching the definition of "halfway between." I would say that for any point to be "between" two endpoints, it has to lie on the geodesic connecting the endpoints. Given that definition, there are only two such points unless the endpoints are exactly opposite each other. Quote Link to comment

+Moose Mob Posted January 3, 2006 Share Posted January 3, 2006 Of course, there are actually 2 points that can be called "exactly halfway between" any other two -- the other one is on the other side of the Earth! Well, if you want to go down that path, there are actually an infinite number of points, all of which lie on the great circle that intersects the great circle arc between the two points perpendicularly and medially. I suppose, although that is greatly stretching the definition of "halfway between." I would say that for any point to be "between" two endpoints, it has to lie on the geodesic connecting the endpoints. Given that definition, there are only two such points unless the endpoints are exactly opposite each other. On the surface of a sphere, would there not be an infinite set of point equidistant from two points? Granted, it would not be the same [straight] line. Quote Link to comment

+sept1c_tank Posted January 3, 2006 Share Posted January 3, 2006 And those points would scribe another (great circle) that crosses the original line perpendicularly. But only two of the points on that circle would be halfway between the original two...unless you consider points below the surface of the earth. I think? Quote Link to comment

+WeightMan Posted January 3, 2006 Share Posted January 3, 2006 And those points would scribe another (great circle) that crosses the original line perpendicularly. But only two of the points on that circle would be halfway between the original two...unless you consider points below the surface of the earth. I think? Precisely. Midpoints are equidistant, but not all equidistant points are midpoints. We are only considering the surface of a sphere here. Actually it's not a sphere, but let's not quibble over that point. It is close enough. Quote Link to comment

+Markwell Posted January 3, 2006 Share Posted January 3, 2006 The other thing to keep in mind with all of this heady math is that it sounds like the original poster is trying to solve a specific cache: To start with the specific - I need to calculate the exact coordinates of a point halfway between two other points. While it may be correct to talk about Great Circle and geodesics and UTM, etc. But the real question is HOW DID THE HIDER FIGURE IT OUT. My guess: Unless they are known for doing something really funky with math, you could just average the two points and that's what they intended. Quote Link to comment

+Pasha Posted January 3, 2006 Share Posted January 3, 2006 That's what I said. Seriously, I much appreciate the links and comments people have provided here. Interesting stuff. Quote Link to comment

+Vinny & Sue Team Posted January 4, 2006 Share Posted January 4, 2006 To start with the specific - I need to calculate the exact coordinates of a point halfway between two other points. Depends on what you mean by "halfway between." Is it along a geodesic? Or some other path? Assuming a geodesic, it's quite easy. Just calculate the distance and azimuths between the two points. Then project from the first point at the same forward azimuth but exactly half the distance. You can check your result by doing the projection from the second point using the reverse azimuth. Of course, there are actually 2 points that can be called "exactly halfway between" any other two -- the other one is on the other side of the Earth! Well, if you want to go down that path, there are actually an infinite number of points, all of which lie on the great circle that intersects the great circle arc between the two points perpendicularly and medially. Thank you for taking the time to write this. I was so tempted to say the same thing! Topology rules! Quote Link to comment

+Enchanted Shadow Posted January 4, 2006 Author Share Posted January 4, 2006 Thank you, everyone, for your help and advice! To clarify, I am looking for the mathematics that will yield the greatest precision. I don't like methods that end up rounding, or making approximations. So, none of the "assume the world is flat" methods will work for me. Robert Lipe's link has been the most helpful so far. With it, I was able to grab formula to calculate the exact distance between two points, as well as the exact coordinates of a third point at any percentage of distance between two other points. I'm in the process of making a central set of geocaching tools in a VBA Excel spreadsheet, and I've implemented what I've found so far. (And this way, I never have to do them by hand again! ) But that being said, of the examples I gave in my original post, the second and third items are still unresolved. i.e.: ----------------- 2. Finding the coordinates of a point equidistant from three other points. 3. Finding the coordinates of the intersecting point of three circles, where the circles are of *differing* radiuses centered on three seperate sets of coordinates. ----------------- I'm not sure if the formula I need are actually on the Aviation Formulary page. If they are, I'm not immediately discerning them. Does anyone know if they're there? Or if not, then where I might find the formula in question? I'm continuing to search for them, but coordinate geometry ON a sphere is tricky to find, needless to say. p.s. Thanks again for all your help!!!!! Quote Link to comment

+WaldenRun Posted January 4, 2006 Share Posted January 4, 2006 The other thing to keep in mind with all of this heady math is that it sounds like the original poster is trying to solve a specific cache: To start with the specific - I need to calculate the exact coordinates of a point halfway between two other points. While it may be correct to talk about Great Circle and geodesics and UTM, etc. But the real question is HOW DID THE HIDER FIGURE IT OUT. My guess: Unless they are known for doing something really funky with math, you could just average the two points and that's what they intended. Great point that MW makes. I have a cache that requires a midpoint to be solved. I used UTM calcs to hide the cache, but did not say that on the cache page. I provided enough of a spoiler that you could find the cache even if you averaged from the ddmm.mmm format. -WR Quote Link to comment

+fizzymagic Posted January 4, 2006 Share Posted January 4, 2006 Robert Lipe's link has been the most helpful so far. With it, I was able to grab formula to calculate the exact distance between two points, as well as the exact coordinates of a third point at any percentage of distance between two other points. Robertlipe's link to the aviation formulary is quite helpful, but if you think there is a formula for the exact distance between two points, you misunderstand it. There is, in fact, no such closed-form formula. My guess is that you are using the spherical-Earth approximation. Good enough over short distances. There is a first-order ellipsoidal approximation that works pretty well for points that are both in the same hemisphere of latitude where you replace the spherical radius with the average radius of the two points; that is also a closed-form expression, though I don't think it appears on the aviation formulary page. In any case, I'd be very, very careful about claiming "exact distances" unless you've studied the problem very carefully! For the other things, I still recommend using UTM coordinates for relatively short distances (up to a few tens of miles). With those, the solutions to your other problems are just plane geometry. There are, for example, closed-form solutions for the point equidistant from three other points and the intersection of three circles. I'm not going to give those here since I have a puzzle cache that requires figuring it out, though! Quote Link to comment

+Enchanted Shadow Posted January 4, 2006 Author Share Posted January 4, 2006 Robertlipe's link to the aviation formulary is quite helpful, but if you think there is a formula for the exact distance between two points, you misunderstand it. There is, in fact, no such closed-form formula. My guess is that you are using the spherical-Earth approximation. Good enough over short distances. There is a first-order ellipsoidal approximation that works pretty well for points that are both in the same hemisphere of latitude where you replace the spherical radius with the average radius of the two points; that is also a closed-form expression, though I don't think it appears on the aviation formulary page. Really? The section on that page I took it from is the following: ----------------------------------- Distance between points The great circle distance d between two points with coordinates {lat1,lon1} and {lat2,lon2} is given by: d=acos(sin(lat1)*sin(lat2)+cos(lat1)*cos(lat2)*cos(lon1-lon2)) A mathematically equivalent formula, which is less subject to rounding error for short distances is: d=2*asin(sqrt((sin((lat1-lat2)/2))^2 + cos(lat1)*cos(lat2)*(sin((lon1-lon2)/2))^2)) ----------------------------------- Is that not what it claims to be? Quote Link to comment

+fizzymagic Posted January 4, 2006 Share Posted January 4, 2006 (edited) Robertlipe's link to the aviation formulary is quite helpful, but if you think there is a formula for the exact distance between two points, you misunderstand it. Really? The section on that page I took it from is the following: The great circle distance d between two points with coordinates {lat1,lon1} and {lat2,lon2} is given by: ... Is that not what it claims to be? It's exactly what it claims to be: the great-circle distance. Great circles are on spheres; these formulae approximate the shape of the Earth as a sphere. Unfortunately, the Earth is not a sphere, so the results you get will not be exactly correct. Here's an example: consider the distance between two caches, Dokey Oakey in California (N 37° 18.945 W 122° 09.904) and Windy Run in Virginia (N 38° 54.297 W 077° 05.814) The great-circle distance is about 2426 miles, while the true WGS84 distance is about 2433 miles. That's a difference of 7 miles, or about 0.3%. That's a pretty significant error! Edited January 4, 2006 by fizzymagic Quote Link to comment

+Enchanted Shadow Posted January 5, 2006 Author Share Posted January 5, 2006 It's exactly what it claims to be: the great-circle distance. Great circles are on spheres; these formulae approximate the shape of the Earth as a sphere. Unfortunately, the Earth is not a sphere, so the results you get will not be exactly correct. Here's an example: consider the distance between two caches, Dokey Oakey in California (N 37° 18.945 W 122° 09.904) and Windy Run in Virginia (N 38° 54.297 W 077° 05.814) The great-circle distance is about 2426 miles, while the true WGS84 distance is about 2433 miles. That's a difference of 7 miles, or about 0.3%. That's a pretty significant error! That doesn't mean there isn't a way to work it out mathematically, it just means that I can't use that particular formula. If I use Vincenty's Algorithm (which allows for the Earth being an ellipsoid, and allows you to set the Major Axis and Flattening you wish to use), I come up with the WGS84 distance. In fact, that's what Ed Williams uses for his Online Great Circle Calculator (http://williams.best.vwh.net/gccalc.htm). Quote Link to comment

+fizzymagic Posted January 5, 2006 Share Posted January 5, 2006 That doesn't mean there isn't a way to work it out mathematically, it just means that I can't use that particular formula. Absolutely right. Vincenty's method is fast and very accurate; it's what I use, too. I called it the "WGS84 distance" to avoid unnecessary complication. Download my little utility application GeoCalc and play with it. I also have a VBA function that calculates distances using Vincenty's method if you want. But since Vincenty's method is not a formula, you can't do some of the things you'd like to do with it directly. For example, to find the point at the intersection of three circles you have to use an iterative method. And those get kind of tricky, especially trying to do in Excel! Quote Link to comment

+Enchanted Shadow Posted January 5, 2006 Author Share Posted January 5, 2006 Absolutely right. Vincenty's method is fast and very accurate; it's what I use, too. I called it the "WGS84 distance" to avoid unnecessary complication. Download my little utility application GeoCalc and play with it. I also have a VBA function that calculates distances using Vincenty's method if you want. Ed's sample spreadsheet uses Vincenty's Algorithm, so I can probably reverse engineer that. But, it's certainly going to be a chore! But having it coded a different way will probably help me to understand it's implementation better, so I'd certainly be interested in it. I didn't initially see it on your website, though. Am I going blind prematurely? But since Vincenty's method is not a formula, you can't do some of the things you'd like to do with it directly. For example, to find the point at the intersection of three circles you have to use an iterative method. And those get kind of tricky, especially trying to do in Excel! As I discovered, Ed's spreadsheet has that function. So, it's more RE work for me! Quote Link to comment

+Chance Encounter Posted January 5, 2006 Share Posted January 5, 2006 One of our prominent Houston-area cachers (Wayne of the Dreamcachers) has developed a handy set of freeware applications titled DreamcachersWare. There are two versions: one for Windows, and one for PocketPC. Both provide extremely accurate calculations that have become invaluable for those of us dedicated to navigational puzzle caches. From the readme file: The little program, GeoTriang, solves the problem of finding a third point, given two points and bearings. This is the triangulation problem which was so eloquently solved by Geowyz! in his “-23- SUR” geocache. The solution is done with planar coordinates, and is therefore only accurate for distances of up to about 10 miles. Triangulate, a similar sounding program, will find the common intersection of three circles. It is used to solve problems such as those encountered in “Trog’s Travel” by Team Troglodyte or “Be the GPSr” by LostGuys. Since the given circles might not all intersect at exactly the same spot, an error term is calculated in addition to the averaged position. These calculations are also done with planar coordinates. The program, Vincenty, does projections on the WGS84 spheroid. It uses the formulae derived by T. Vincenty, Survey Review XXII,176, April 1975. A couple of words of documentation are necessary. If the “Distance” field is blank or zero, the program computes distance and bearings from the two given points. If the “Distance” field contains a valid number, the coordinates of Point 2 are computed using Point 1 as the starting point and the distance and forward azimuth for the projection. Snellius solves Captain J type problems which involve a Snellius construction. It also uses planar coordinates. The default values solve “Lost Treasure IV”. Link: http://web.hgcs.org/forum/index.php?topic=118.0 Quote Link to comment

+Enchanted Shadow Posted January 5, 2006 Author Share Posted January 5, 2006 One of our prominent Houston-area cachers (Wayne of the Dreamcachers) has developed a handy set of freeware applications titled DreamcachersWare. There are two versions: one for Windows, and one for PocketPC. Both provide extremely accurate calculations that have become invaluable for those of us dedicated to navigational puzzle caches. Thank you for the suggestion. But planar coordinates (which most of that suite seems to use) isn't accurate enough for what I'm looking for. Quote Link to comment

Zurcher Posted January 14, 2006 Share Posted January 14, 2006 Does anyone know of a neat , easy on line calculator , like Fizzymagic.... that can figure out the intersection of two arcs on this earth. Example: take lat/lon of los angeles and newyork.....lets say its 3000 miles and give the two , unequal lenght radii....one 2000 other 1300 miles and get lat lon of the 2 points of intersection. no azmuth or degrees... [ I know..they intersect in 2 places....] Z Quote Link to comment

+blunoz Posted January 15, 2006 Share Posted January 15, 2006 Unfortunately, it's not a cheap or readily accessible solution, but if you know anyone who has ESRI's ArcGIS software, it can do all the things you are talking about. It's a very powerful Geographic Information System software package that we use at my work. Of note - I just finished taking an online class in ArcGIS where they provided us a copy of the ArcGIS software and a "student" license to use it for the 3 month duration of the class. It was a really good class. A lot of major universities like Penn State are offering graduate certificates in GIS now. Anyway, like I said, I know it's not cheap, but if you know anyone in your local county planning office, you might ask if you could sit and use their computer for a half hour or so. (It was actually a friend of mine who works in a county planning office that introduced me to GIS). Kevin/Blunoz Quote Link to comment

+the hermit crabs Posted January 15, 2006 Share Posted January 15, 2006 Does anyone know of a neat , easy on line calculator , like Fizzymagic....that can figure out the intersection of two arcs on this earth. Example: take lat/lon of los angeles and newyork.....lets say its 3000 miles and give the two , unequal lenght radii....one 2000 other 1300 miles and get lat lon of the 2 points of intersection. no azmuth or degrees... [ I know..they intersect in 2 places....] Z Compsys21 will do that (with its "Circle/Circle" function). (Thanks to blindleader for including that link in this post.) Quote Link to comment

+Cardinal Red Posted January 15, 2006 Share Posted January 15, 2006 Using Fizzy's Numbers Two waypoints 20 miles apart with a forward azimuth of 80 degrees: N 47 31.289 W 122 36.968 N 47 34.259, W 122 11.691 Midpoint = N 47 32.786, W 122 24.336 Went to this Website: Great Circle Calculator by Ed Williams Calculated Distance = 20.000114691134492 Statute Miles Calculated Azimuth = 79.995855 degrees Dropped down to the second calculator and entered a distance of 10.0005735 Statute Miles @ 79.995855 degrees Calculated Midpoint = N 47 32.7856, W 122 24.3354 Quote Link to comment

Zurcher Posted January 16, 2006 Share Posted January 16, 2006 Great big thanks to the "HERMIT" for the tip...re using Compsys21, works like a charm! Once I realized I need to work with nautical miles !! Z Quote Link to comment

+EScout Posted January 16, 2006 Share Posted January 16, 2006 The first problem is the easiest, especially if you have a Meridian or Sportrak with their excellent projection feature. Use the first waypoint as the starting point, go down to the coordinates and input the coords of the second point. The distance and bearing (in precision of onehundredths of the distance unit you choose, and one hundredth of a degree, or in mils) between the two will show. Reduce the distance by half and you will have the new midpoint that you can save as a waypoint. Another way to do this in the field, is to use the PalmOS program: "Navigate". Also very quick to input and calculate, this uses the great circle calculation as does your GPSr. The second problem, finding a point equal distance from 3 other points, can also be done in the field without a map, just with your GPSr. Draw this out and you will see that the points are on a circle with the desired point in the middle. Conect all the points. You will see equal sides and triangles with equal angles. Remember what this means for triangles and how the 3 angles add up to 180d or 3200 mils. Enter the points in the route function to get the bearings from each to the other. It helps to have a GPSr that uses mils for more precision. You will do a projection for your final answer. Do a projection from each of the 3 points for a final check. I have done this in the field and it works. Quote Link to comment

+the hermit crabs Posted January 16, 2006 Share Posted January 16, 2006 Great big thanks to the "HERMIT" for the tip...re using Compsys21, works like a charm! Once I realized I need to work with nautical miles !! Z Glad it worked for you -- and you can change it from nautical miles to statute miles (or feet, or meters) by selecting Units -> Distance Units from the list of options at the top). Quote Link to comment

+AtoZ Posted January 16, 2006 Share Posted January 16, 2006 FIRST USE UTM not Lat/Long and it is very easy. UGH people make it so hard. cheers Quote Link to comment

+Walkaboutlost Posted March 16, 2006 Share Posted March 16, 2006 Need help. Is there a program or help to find a projected waypoint if Two other waypoints are given and their respective distance from the required unknown waypoint is known. ie. First known waypoint co ords are 200 meters from unknown Second waypoint co ords are 737 meters from the unknown no bearings are given. Quote Link to comment

+jon & miki Posted March 16, 2006 Share Posted March 16, 2006 Need help. Is there a program or help to find a projected waypoint if Two other waypoints are given and their respective distance from the required unknown waypoint is known. ie. First known waypoint co ords are 200 meters from unknown Second waypoint co ords are 737 meters from the unknown no bearings are given. The quickest way is to use a map drawn to scale and draw two circles around the known points. Some of the mapping programs will let you draw circles on the maps, which should make the process somewhat easier. As you probably already figured out, in most cases there will be two places where the two circles intersect. One of those two intersections will be your location. If you can get a third known waypoint and distance, you should be able to narrow it down to just one of the two points. Quote Link to comment

+BillsBayou Posted March 16, 2006 Share Posted March 16, 2006 Use Haversines. Convert to radians first. Bx = cos(lat2).cos(Δlong) By = cos(lat2).sin(Δlong) lat3 = atan2(sin(lat1) + sin(lat2), √((cos(lat1)+Bx)² + By²)) lon3 = lon1 + atan2(By, cos(lat1)+Bx) Δlong is the difference between the two longitudes Or you can just punch in the coordinates here: http://www.movable-type.co.uk/scripts/LatLong.html and hit the "Calculate Midpoint" button about 1/3 the way down the page. Quote Link to comment

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