What operations will you do with the data? Especially the altitude? If it is just extra info, simply have another column, altitude, of type FLOAT or DOUBLE, depending on the precision needed. If you are computing distances in 3D or trying to do 3D "CONTAINS", then MySQL's SPATIAL won't help. "... order of coordinates ..." -- You must handle this as you store the info. Is JSON really needed? It gets in the way of many operations. If you don't need a Spatial index, then simply have 3 Float or Double columns. "... storing each point individually ..." -- Again, what will you do with the points? If the only 'entity' you have is a 'point', the 3 columns, one row per point seems right. If multiple 'points' are defining a 'polygon', that is another matter. One altitude per polygon A single row per polygon -- one GEOMETRY column for the polygon, one FLOAT for altitude. Each point has a different altitude Now it gets messy. Is "CONTAINS" referring to the projection of the polygon on the surface (of the earth or a plane)? Or something else? You might need two tables -- one table with one row for each polygon stored as a GEOMETRY; the other table has one point per row, perhaps as 3 FLOATs.

As I understood the task. There's an ordinary decart system of coordinates, a rectangle, sides a and b, one of its points, and it'll turn on that point to the corner. The question is how it was previously held, which is the length a and b. A3 -------a------<- A2 | | b | | | -------------- ^ ^ A0 A1 If you can imagine the picture in this way, it's pretty simple. A0of course, stays there. If that's what you need. But in general, a rectangle could have been oriented in the first place. You need to make a clear mission, even for yourself, or you know, "I need to count something!" - "I need some formula."

“Parametrical equation of sphere with centre at point” https://ru.wikipedia.org/wiki/%D0%A1%D1%84%D0%B5%D1%80%D0%B0 Put your limits on the part where the area was cut off. Get the coordinates of camera A0 from two parameters. Make a vector A1 of A0 to point 0.0.0. Make a vector. A2 (x(A0),y(A0),z(A0)+1)♪ A3=Crossing product A1 on A2 will give the enclosure axle. A4=Crossing product A3 on A1 will give another axle. Get three axles for the A1 A3 A4 vector cell, providing orientation to the camera and camera centre A0. Need to normalize.

Don't make it harder.For four points, only two versions of a pair of triangles can be built (this they cross and this is not a surface). Choose any orientation. With a strong enough grid, there's no difference.Also, if you're interested in the visuality, draw the surface from the triangles, and the ribs only show the couple together.

You've gone quite complicated. It's easier.First, we need to build an orthodox coordinate system. You need to find any vector perpendicular to the vector. It's the most unspeakable part of the decision. And in this place, you don't have to be afraid to put the contingent operators - according to the "gea comb" from the topology, you don't have to do it without them.The simplest way to turn the vector into a straight corner is to multiply it into something. But if the multiplied vectors are close to the direction, the accuracy will be affected (and if they are collinears, zero).Therefore, I suggest that three base vectors be considered.i♪ j and kand multiply on the one who's the most unlike. For that. We'll find the coordinate with the smallest mole.♪ May, for certainty, .A. À åå .B. and .C.. So we multiply the vector. i and receive the first of the vectors of the local coordinate system:i1 = n x i = (Ai + Bj + Ck) x i = -Bk + CjNow we'll re-model him to the vector and get the second vector of the base:j1 = n x i1 = (Ai + Bj + Ck) x (-Bk + Cj) = (please open the brackets yourself)Now, the base vectors have to be normalized so that they become single, and the parametric equation of the circumference can be used:r = r0 + R cos φ i1 /i1 + R sin φ j1 /j1

Okay. You've got a little crazy writing, so I'll rephrase:There is a point A(1024, 768) and a point B(600, 500) (500 taken to avoid confusion over the size of the axles). Then the vector AV {600-1024,600-768} = {-424,268} (number less than zero is correct, the vector is directed to the start of coordinates). He'll be perpendicular to the vector N{424,268} or M{-424, 268} (they're just the opposite to take any account of them). Simply put, on X or we change the sign, we get a perpendicular vector.Next: Direct equation (take vector N{424, -268}):x = 424*t + 600;y = -268*t + 500;We'll find the points of intersection with the axles:for Х: У=0 => -268*t+500 = 0 => t = 500/268; coordinates Х=424*(500/268)+600;for Y: x=0 => 424*t+600 = 0 => t = -600/424; coordinates Y=-268*(-600/424)+500;Here=If you lie to be clear, in general form:scr.x = 1024+1; scr.y = 768+1;//take +1 to avoid zeroing;)p.x = 600;p.y = 500;Another variable. k = (scr.x-p.x)/(scr.y-p.y); / / here, so it doesn't work out zero.Then the points are: with axis X: [p.y*k+p.x, 0]with axle Y: [0, p.x/k+p.y]

If a multi-angle is swell, his triangulation is trivial.Non-pollutant multi-angles differ from the swelling of only one characteristic: the presence of angles exceeding the depressed (180 degrees). Since this angle is not possible in the triangle, it is easier to link a triangle with it. For practical purposes (location, moments, centre of gravity, etc.), the processing of algorithms to allow the deductions to be accounted for is not difficult.What's important is that they don't cross. And if the cut-off is in the order of the "inverted" angle, the process of presenting an unpopular multiangle in the form of a swelling multiangle and a lot of triangles is not a problem.If classical triangulation is required, a large number of triangles and many extracts can be formed and each triangle can be expected to cross with each extract. If changes are made, remove the reference triangle and add the result to the end of the triangle list. Repeat cycles are not necessary, but the triangle deduction function is not complex (the triangle has no diagonals) and is written once.

w2=w1*(W+a)/W Same heightP.S. I'm sorry, there's a reply.

Using boys as 3rd editor doesn't make sense. Unity is a game move, not a 3rd editor. The subjects that can be created in a fact student are only necessary for tests.Disposition more than simple: 3d editors are needed for simulations for your discretion: 3d max, Maya or any other.

obtain a normalized vector between points.Calculate the angle of the vector in the Y-Z plane (through the arctangens) - angle XCalculate the angle of the vector in the plane Z-XY - the angle YHere's an example code you can adapt to C+++:function VectorToEuler(const A: TKMVertex3): TKMVertex2; var Dist: Single; begin Dist := Sqrt(Sqr(A.X) + Sqr(A.Y)); // Переводим XY в угол X Result.X := ArcTan2(-A.X / Dist, A.Y / Dist) / Pi * 180; Dist := Sqrt(Sqr(A.X) + Sqr(A.Y) + Sqr(A.Z)); // Переводим Z и XY в угол Y Result.Y := ArcTan2(A.Z / Dist, Sqrt(Sqr(A.X) + Sqr(A.Y)) / Dist) / Pi * 180; end; ** The code is not optimized, I leave it to you.

Here's the code. java.awt.geom.Line2D.linesIntersectHe only refers to one method java.awt.geom.Line2D.relativeCCWpublic static boolean linesIntersect(double x1, double y1, double x2, double y2, double x3, double y3, double x4, double y4) { boolean r1 = relativeCCW(x1, y1, x2, y2, x3, y3) * relativeCCW(x1, y1, x2, y2, x4, y4) &lt;= 0; boolean r2 = relativeCCW(x3, y3, x4, y4, x1, y1) * relativeCCW(x3, y3, x4, y4, x2, y2) &lt;= 0; return r1 &amp;&amp; r2; } public static int relativeCCW(double x1, double y1, double x2, double y2, double px, double py) { x2 -= x1; y2 -= y1; px -= x1; py -= y1; double ccw = px * y2 - py * x2; if (ccw == 0.0) { ccw = px * x2 + py * y2; if (ccw > 0.0) { px -= x2; py -= y2; ccw = px * x2 + py * y2; if (ccw < 0.0) { ccw = 0.0; } } } return (ccw < 0.0) ? -1 : ((ccw > 0.0) ? 1 : 0); }

For exampleatan2(y1 - y0, x1 - x0)

Find the center of gravity.Get a beam to the top.On the beams, find points remote from the centre of gravity in the right proportion of the old peaks.

Here's the C#: points.Concat(new[] { points.First() }) .Pairwise((p1, p2) => new { x = p2.x - p1.x, y = p2.y - p1.y }) .Pairwise((v1, v2) => v1.x * v2.y - v1.y * v2.x) .Select(Math.Sign) .Pairwise((prevsign, nextsign) => (prevsign == nextsign) && prevsign != 0) .All(_ => _); We need to connect. https://www.nuget.org/packages/morelinq/ for Pairwise♪On C++, I hope it's easy.

If I understood the condition correctly, point O should be placed at the intersection of the BAC angle bissectries and the mid-point perpendicular to AB.In general, the AOB equal triangle with the AB base, the OAB angle is half the BAC angle. We need to find a height. There are several ways to make it difficult.

https://ru.wikipedia.org/wiki/%D0%9A%D0%BE%D0%BC%D0%BF%D0%BB%D0%B5%D0%BA%D1%81%D0%BD%D0%BE%D0%B5_%D1%87%D0%B8%D1%81%D0%BB%D0%BE#.D0.90.D0.BB.D0.B3.D0.B5.D0.B1.D1.80.D0.B0.D0.B8.D1.87.D0.B5.D1.81.D0.BA.D0.B0.D1.8F_.D1.84.D0.BE.D1.80.D0.BC.D0.B0 That's what it looks like.(a + ib)(c + id) = (ac - bd) + i(bc + ad) ...где i - мнимая единица, для которой i^2 = -1 ...which is https://ru.wikipedia.org/wiki/%D0%9A%D0%BE%D0%BC%D0%BF%D0%BB%D0%B5%D0%BA%D1%81%D0%BD%D0%BE%D0%B5_%D1%87%D0%B8%D1%81%D0%BB%D0%BE#.D0.93.D0.B5.D0.BE.D0.BC.D0.B5.D1.82.D1.80.D0.B8.D1.87.D0.B5.D1.81.D0.BA.D0.B0.D1.8F_.D0.BC.D0.BE.D0.B4.D0.B5.D0.BB.D1.8C For vectors (in an integrated plane) the single length gives another vector a single length whose argument (the angle with Re+, the horizontal axis right) is the sum of the arguments of the constituents.If they weren't the only ones, the length of the result would be the product of their length.

Your mistake is most likely here:self.x = self.x*cos(ang)+self.y*sin(ang) self.y = -self.x*sin(ang)+self.y*cos(ang) // тут вы обращаетесь к уже измененному self.x Intermediate variables should be introduced:x = self.x y = self.y self.x = x*cos(ang)+y*sin(ang) self.y = -x*sin(ang)+y*cos(ang)

Take the coordinates.x1_1, x1_2 - coordinates of the left and right side of the first rectanglex2_1, x2_2 - coordinates of the left and right side of the second rectangleAnd compare. max(x1_1,x2_1) and min(x1_2,x2_2)♪ If the first value exceeds the second, the rectangles do not intersect (which, I understand, should be checked for y). If not, that's the location on the crossing."I think so."Update That's how 25 years ago it was handled in Turbo Vision:inline void TRect::intersect( const TRect& r ) { a.x = max( a.x, r.a.x ); a.y = max( a.y, r.a.y ); b.x = min( b.x, r.b.x ); b.y = min( b.y, r.b.y ); } inline void TRect::Union( const TRect& r ) { a.x = min( a.x, r.a.x ); a.y = min( a.y, r.a.y ); b.x = max( b.x, r.b.x ); b.y = max( b.y, r.b.y ); } inline Boolean TRect::contains( const TPoint& p ) const { return Boolean( p.x >= a.x && p.x < b.x && p.y >= a.y && p.y < b.y ); }

Recording points (1, ax, ay, az)Read the plane as the vector-designator (D, A, B, C)i j k e 1 ax ay az 1 bx by bz 1 cx cy cz Similar to the two planes that have been reached.

No code algorithm (satisfactory elementary):We have: Two points A♪ B; len distance from point А to required point Cfull_len = |B - A| // длина вектора, соединяющего две точки == длина отрезка C = A + (B - A) * (len / full_len) The stocking and multiplication are obvious operations.