+Zoraima Posted July 7, 2003 Share Posted July 7, 2003 Hi All: I looked in the FAQs first before posting this, and went through several pages of the "Getting Started" but didn't see an answer to my question. I'm wondering how to go about triangulating between points that are far apart, let's say driving distance. If you do the location on a map, you still don't come out with the coordinates, you just have a spot on a map. Thanks all! Susan ----------------------------------------------------------- Even the smallest person can change the course of the future. --Galadriel, "The Lord of the Rings: Fellowship Of the Ring" Quote Link to comment

+Renegade Knight Posted July 7, 2003 Share Posted July 7, 2003 Pythagriams theorom will solve the problem if you draw it out. I could of spelled Pythragiams wrong. Do the math in UTM it's easier. and there are programs out there that will solve the problem. Quote Link to comment

katybird Posted July 7, 2003 Share Posted July 7, 2003 quote:Originally posted by Renegade Knight:Pythagriams theorom will solve the problem if you draw it out. I could of spelled Pythragiams wrong. Do the math in UTM it's easier. and there are programs out there that will solve the problem. Something tells me that, if I even knew what you were talking about, I'd know how to triangulate. Quote Link to comment

+Renegade Knight Posted July 7, 2003 Share Posted July 7, 2003 Well... True enough. A squared plus B squared equils C squared. C is the long leg. It's how to solve the lengh of the missing leg of a right triangle. If you triangluate you can can draw out the right triangles. Then it's a little algebra. Or as they like to make my kids do. Just plug in numbers until you get the right answer. If I could spell it I'd of looked up a link for you. The angled leg can be split into the straight (north and south legs) and those distances once solved add directly to the UTM coordinates of the known point to get the UTM coordinates of the other point. Once you have that you can convert to LL if you need too. Drawing it helps. Quote Link to comment

katybird Posted July 7, 2003 Share Posted July 7, 2003 quote:Originally posted by Renegade Knight:Well... True enough. A squared plus B squared equils C squared. C is the long leg. Drawing it helps. I remember that from school - but I tell you - I have NEVER used that in the 15 years since I last HAD to know it! LOL! Quote Link to comment

+tozainamboku Posted July 7, 2003 Share Posted July 7, 2003 I was FTF this cache which required trangulation to find. Renegade Knight is correct in describing using the Pythagorean Theorem to solve the problem. Here is my try a little further explanation. First, use UTM coordinates as Renegade Knight suggest. UTM coordinates are just the offset in meters - north and east of some point. The earth is divided into segments, each segment has its own origin point. If you have UTM 11S E 385607 N 3771023 it means 385607 meters east and 3771023 meter north of the origin for segment 11. If you are working with points near each other you can drop the first few digits of the northing and easting. (Actually you are translating the origin to a point closer to where you are to make the numbers more managable). Second, using the Pythagorean Theorem, if you have to points on at (x,y) and the other at (x',y') then distance between these point is the square root of the difference (x-x') square plus the difference (y-y') squared. You can write this as: d**2 = (x-x')**2 + (y-y')**2 (I used **2 to represent squaring a number because I can't do superscripts in the forum page). Finally, you need to fill what you know and then solve for the unknowns. In the cache above I knew the UTM coordinates of three other caches and the distance from each of these caches to the cache I was trying to find. d1**2 = (x1-x)**2 + (y1-y)**2 d2**2 = (x2-x)**2 + (y2-y)**2 d3**2 = (x3-x)**2 + (y3-y)**2 After some algebraic manipulation you can get two equations x = [d1**2-d2**2-x1**2+x2**2-y1**2+y2**2-2y(y1-y2)]/[2(x1-x2)] y = [d2**2-d3**2-x2**2+x3**2-y2**2+y3**3-2x(x2-x3)]/[2(y2-y3)] Substitute the right hand side of one of these equation for the variably (x or y) in the other and you can get the formula for x and y of the unknown point. If all this is making your head spin - don't worry. You can use a compass (the kind you use to draw circles, not the kind to find which direction you are heading) and draw circles on a map around each known point with the radius based on the distance. Where the circles intersect is the point you are looking for on the map. 東西南北 -- I found it in the last place I looked. Quote Link to comment

Kerry. Posted July 8, 2003 Share Posted July 8, 2003 That's basically the problem, the intersection of 2 (or more) circles. However one problem with using UTM is in/on and over (further than the overlap) zone boundaries, UTM basically totally useless then. Cheers, Kerry. I never get lost everybody keeps telling me where to go Quote Link to comment

+TotemLake Posted July 8, 2003 Share Posted July 8, 2003 Wow! Thanks for that explanation! That's going into my PDA for future reference. Cheers! TL Quote Link to comment

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