dborsje Posted October 13, 2003 Share Posted October 13, 2003 Dear all, I am involved in a demining project and I have a question I can't solve myself. Suppose I have 5 waypoints: a, b, c, d and e. I need to calculate the area in square metres, delimited by these 5 waypoints. Any help would be greatly appreciated. Dick Borsje, Angola Dick Borsje Quote Link to comment
+planetrobert Posted October 13, 2003 Share Posted October 13, 2003 well how i would do thot is... take your points... 1) find the distances between them around the perimeter of the area they enclose 2) assuming the 5 points are in the shape of a true pentagon take the distance from the point(top) of the pentagon te each of the 2 base points. now you have divided the 5 points into 3 triangles which are easy to figure out. Now where did I set my GPS??? planetrobert.net Quote Link to comment
+hikemeister Posted October 13, 2003 Share Posted October 13, 2003 PS -- the area of a triangle is 0.5 bh where b = base length, and h = height. SirRalanN -- good practical answer to that guy's guestion! Quote Link to comment
+parkrrrr Posted October 13, 2003 Share Posted October 13, 2003 Other formulas for the area of a triangle, including one that works if all you have is the lengths of the three sides and another that works if you have the (planar, so use UTM if it's a relatively small area) coordinates of the vertices, can be found here. Quote Link to comment
+planetrobert Posted October 13, 2003 Share Posted October 13, 2003 whew, glad it made sence, i was half asleep when typing that post Now where did I set my GPS??? planetrobert.net Quote Link to comment
+blindleader Posted October 13, 2003 Share Posted October 13, 2003 Assuming you have rectangular coordinates for each of the points - The area of a polygon whose vertices are P1, P2, .., Pn is given by the expression K = [(x1y2 + x2y3 + x3y4 + ... + xny1) - (x2y1 + x3y2 + x4y3 + ... + x1yn)]/2 I'm afraid the chances of getting a useful answer using the technique given by SirRalanN are just about zero. A geocaching forum is not the first place I would go to ask a math question.The math forum would be a much better place. The answers to nearly any math problem you can think of are already there. A quick search on "irregular polygon" turns up a number of pages, including the one containing the above formula. Quote Link to comment
+briansnat Posted October 13, 2003 Share Posted October 13, 2003 If you have an eTrex Vista, you can just walk the perimeter and it will calculate the area for you. "You can't make a man by standing a sheep on his hind legs. But by standing a flock of sheep in that position, you can make a crowd of men" - Max Beerbohm Quote Link to comment
+parkrrrr Posted October 13, 2003 Share Posted October 13, 2003 quote:Originally posted by blindleader:A geocaching forum is not the first place I would go to ask a math question.http://mathforum.org/ would be a much better place. The answers to nearly any math problem you can think of are already there. Not to be overly snarky, but if that's the quality of the information to be found there, it's best forgotten about. That formula is only useful for convex polygons, and they didn't bother to mention that tiny little restriction. It makes you wonder how many of the other formulae listed there have similarly limited applicability. The notion of subdividing the polygon into its constituent triangles works for any polygon and has been used by GIS people for many years to great effect. (Provided you can figure out which triangles to use; that's the part where it helps to plot the polygon on paper.) [Edit: I see now that SirRalanN's method also assumes a convex polygon, but that assumption was unnecessary. You just have to pick three triangles whose intersection is empty and whose union is equivalent to the original pentagon. In the case of a convex pentagon, SirRalanN's triangles fulfill that criterion.] Quote Link to comment
Dan Vull Posted October 13, 2003 Share Posted October 13, 2003 Utilizing a bit of cogo would produce the area without all of that nonsense. Heck send me the coords of the points and I'll comp it out as fast as I can type in the coord sets. Quote Link to comment
+parkrrrr Posted October 14, 2003 Share Posted October 14, 2003 Blindleader was kind enough to write me privately and point out that, as usual, the truth lies somewhere in the middle. It turns out that both mathforum and mathworld got the constraints wrong. The determinant formula works for any simple polygon, which is to say any polygon without self-intersections, holes, or discontinuities. So, since the area in question is probably a simple polygon, the determinant formula should work just fine (again, assuming that the area the polygon covers is small enough to be approximately planar.) I apologize for any confusion I might have caused, and I apologize for my prejudice toward mathforum. I'll go soak my head now. Quote Link to comment
+blindleader Posted October 14, 2003 Share Posted October 14, 2003 I wouldn't go so far as to say either site got the constraints wrong. mathworld doesn't say that the formula applies only to convex polygons. In fact it doesn't say anything about concave polygons. It presents the formula in a more general presentation about convex polygons. Neither site mentions the assumption that we are talking about simple polygons. Such an assumption in talking about elementary plane geometry is reasonable, unless you're of the school that thinks Betrand Russell did the world a service by taking I-don't-know-how-many-dozen-pages to prove that 1+1=2. I hope vtpaddler recognizes that his cogo program uses "all of that nonsense" (the determinant formula) to calculate areas. Land surveyors always assume simple polygons and a flat Earth too. Now about this here snarkiness that's going around... Quote Link to comment
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