I wish to use PostGIS to select all the points within a polygon, but this question is about defining the actual polygon.
I'm looking to define a polygon that is based on a great circle, specified by two points on the earths surface defined by latitude and longitude coordinates. The polygon that I'm after should be defined by a width left and right of the the center line (the center line being the line made by the great circle)
The resulting shape would be a long curved rectangular shape.
The purpose being to select all the points within x distance of the great circle line.
I think you are confused about the kind of data you are dealing with, if you use a equidistant projection you could use something as simple as this:
ST_DWithIn(ST_MakeLine(point1, point2),distanceInSRIDunits)
There is an old discussion in the postgis mail list that will be useful to you.
Related
I am using Autodesk.Viewing.Markupscore extension to draw a polygon in 3D view in top view (orthographic)
I want to see only those elements which are inside the polygon
For which I am using the cutplanes concept because bounding box logic is taking too long to process.
So Please suggest a way to apply cut planes along these lines.
Try programmatically cutplanes with an array of vector4s specifying the four sides say:
vectorArray.push(new THREE.Vector4 (0.9999999999999998, 0, 0, 0.06520926952362059)) //x,y,z,w
viewer.setCutPlanes(vectorArray)
See documentation here
I have a single point and a set of shapes. I need to know if the point is contained within the compound shape of those shapes. That is, where all of the shapes intersect.
But that is the easy part.
If the point is outside the compound shape I need to find the position within that compound shape that is closest to the point.
These shapes can be of the type:
square
circle
ring (circle with another circle cut out of the center)
inverse circle (basically just the circular hole and a never ending fill outside that hole, or to the end of the canvas is there must be a limit to its size)
part of circle (as in a pie chart)
part of ring (as above but
line
The example below has an inverted circle (the biggest circle with grey surrounding it), a ring (topleft) a square and a line.
If we don't consider the line, then the orange part is the shape to constrain to. If the line is taken into account then the saturated orange part of the line is the shape to constrain to.
The black small dots represent the points that need to be constrained. The blue dots represent the desired result. (a 1, b 2 etc.)
Point "f" has no corresponding constrained result, since it is already in the orange area.
For the purpose of this example, only point "e" is constrained to the line, all others are constrained to the orange orange area.
If none of the shapes would intersect, then the point cannot be constrained. If the constraint would consist of two lines that cross eachother, then every point would be constrained to the same position (the exact position where the lines cross).
I have found methods that come close to this, but none that I can combine to produce the above functionality.
Some similar questions that I found:
Points within a semi circle
What algorithm can I use to determine points within a semi-circle?
Point closest to MovieClip
Flash: Closest point to MovieClip
Closest point through Minkowski Sum (this will work if I can convert the compound shape to polygons)
http://www.codezealot.org/archives/153
Select edge of polygon closest to point (similar to above)
For a point in an irregular polygon, what is the most efficient way to select the edge closest to the point?
PS: I noticed that the orange area may actually come across as yellow on some screens. It's the colored area in any case.
This isn't much of an answer, but it's a bit too long to fit into a comment ...
It's tempting to think, and therefore to advise you, to find the nearest point in each of the shapes to the point of interest, and to find the nearest of those nearest points.
BUT
The area you are interested in is constructed by union, intersection and difference of other areas and there will, therefore, be no general relationship between the closest points of the original shapes and the closest point of the combined shape. If you understand what I mean. For example, while the closest point of A union B is the closest of the set {closest point of A, closest point of B}, the closest point of A intersection B is not a simple function of that same set; at least not for the general case.
I suggest, therefore, that you are going to have to compute the (complex) shape which represents the area of interest and use one of the algorithms you've already discovered to find the closest point to your point of interest.
I look forward to someone much better versed in computational geometry proving me wrong.
Let's call I the intersection of all the shapes, C the contour of I, p the point you want to constrain and r the result point. We have:
If p is in I, then r = p
If p is not in I, then r is in C. So r is the nearest point in C to p.
So I think what you should do is the following:
If p is inside of all the shapes, return p.
Compute the contour C of the intersection of all the shapes, it is defined by a list of parts (segments, arcs, ...).
Find the nearest point to p in every part of C (computed in 2.) and return the nearest point among them to p.
I've discussed this question at length with my brother, and together we came to conclude that any resulting point will always lie on either the point where two shapes intersect, or where a shape intersects with the line from that shape perpendicular to the original point.
In the case of a circular shape constraint, the perpendicular line equals the line to its center. In the case of a line shape constraint, the perpendicular line is (of course) the line perpendicular to itself. In the case of a rectangle, the perpendicular line is the line perpendicular to the closest edge.
(And the same, theoretically, for complex polygon constraints.)
So a new approach (that I'll have to test still) will be to:
calculate all intersecting (with a shape constraint or with the perpendicular line from the original point to the shape constraint) points
keep only those that are valid: that lie within (comply with) all constraints
select the one closest to the original point
If this works, then one more optimization could be to determine first, which intersecting points are nearest and check if they are valid, and then work outward away from the original point until a valid one is found.
If this does not work, I will have another look at the polygon clipping method. For that approach I've come across this useful post:
Compute union of two arbitrary shapes
where clipping complex polygons is made much easier through http://code.google.com/p/gpcas/
The method holds true for all the cases (all points and their results) above, and also for a number of other scenarios that we tested (on paper).
I will try a live version tomorrow at work.
This is my firs excursion on the HTML5 canvas, I have working knowledge of jQuery and Javascript.
I'm trying to create a "spinning globe" effect with it.
The idea is to have a circle and meridians "spinning" on it, to give the effect of a rotating globe.
I've drawn the circle and now I'm trying to create lines that start from the right (following the curve of the circle), move towards the centre straightnening up (in the middle they are straight) and follow the inverse curvature on the left, ending with the circle.
I'm trying to do this with the HTML5 canvas and jQuery but I'm not sure of where to start... should I create an arc and then try to animate it?
I'm even wondering if the canvas is the right tool or if I should use anything else.
Any suggestion is welcome!
Sebastian
You could use a quadratic bezier curve, which is basically just a curve with a start point, an end point, and a "control point" in the middle, which is what you would want to change as the globe spins. In this case, all of your lines would start and end at the north and south poles, respectively, of your "globe". For example, to make one of these lines:
// start drawing a line
canvas.beginPath();
// move the the top of your globe
canvas.moveTo(0,0);
/* draw a curve with the control point specified by the first two args,
* end point by the second two:
* (in your case, the control point would be in the middle of the globe) */
canvas.quadraticCurveTo(control_point_x, control_point_y, 0, 50);
// finish drawing, stroke and end
canvas.stroke();
canvas.closePath();
You would also have to take in to account how you will clear the lines after each frame, of course.
See: The Canvas element API, Complex Paths
This is what I got, didn't have the time to proceed any further: http://jsfiddle.net/Z6h3Z/
I use bezier curves where the two control points are in a sort of oval arc centered at the poles.
What I got stuck at is the distribution of points along the arc to look more realistic.
Short question: Given a point P and a line segment L, how do I find the point (or points) on L that are exactly X distance from P, if it guaranteed that there is such a point?
The longer way to ask this question is with an image. Given two circles, one static and one dynamic, if you move the dynamic one towards the static one in a straight line, it's pretty easy to determine the point of contact (see 1, the green dot).
Now, if you move the dynamic circle towards the static circle at an angle, determining the point of contact is much more difficult, (see 2, the purple dot). That part I already have done. What I want to do is, after determining the point of contact, decrease the angle and determine the new point of contact (see 3, 4, the red dot).
In #4, you can see the angle is decreased by less than half, and the new point of contact is half-way between the straight-line point and the original point. In #7, you can see the angle is bisected, but the new point of contact moves much farther than half way back towards the straight-line point.
In my case, I always want to decrease the angle to 5/6ths its original value, but the original angle and distance between the circles are variable. The circles are all the same radius. The actual data I need after decreasing the angle is the vector between the new center of the dynamic circle and the static circle, that is, the blue line in 3, 4, 6, and 7, if that makes the calculation any easier.
So far, I know I have to move the dynamic circle along the line that the purple circle is a center of, towards the center of the static circle. Then the circle has to move directly back towards the original position of the dynamic circle. The hard part is knowing exactly how far back it has to move so that it's just touching the other circle.
To answer your short question, if you are on the Cartesian plane, then find the equation of the line L is sitting on (given the two endpoints of L, this is simple). Find the equation of the perpendicular to said line, which passes through P (this is done by taking the negative inverse of the slope, plugging in P's x and y values, and solving for the intercept). Then find the point where the two perpendicular lines intersect by using their equations as a single system of equations (with x's and y's equal). Then find the distance between the point of intersection and the point P, which is one leg of a triangle. Finally, with that distance and the distance X you are given, use Pythagorean theorem to find the distance of the other leg of the triangle. Now the point you are looking for is a point on L, and also on the line on which L sits. So using the distance you just obtained, the intersection point you had found before, and the equation of L's line, you can find the desired point's coordinates. There can only be a maximum of 2 such points, so all you have to test for is whether the coordinates of the points found are actually on L, or beyond L but still on its line. Sorry for the long answer and if you wanted a geometric explanation rather than an algebraic one.
Draw a circle with the same centre as the stationary circle and the radius of the sum of both radii. There are two intersections with the translation line of the moving circle's centre. The place of the moving circle's center at the time of contact is the closer of those two intersections.
Let the ends of your segment be A and B, and the center of your stationary circle be C. Let the radius of both circles be r. Let the center of the moving circle at the moment of collision be D. We have a triangle ACD, of which we know: the distance AC, because it is constant, the angle DAC, because that's what you are changing, and the distance CD, which is exactly 2r. Theoretically, two sides and angle should let you get all the rest of a triangle...
I'm using OpenMap and I'm reading a ShapeFile using com.bbn.openmap.layer.shape.ShapeFile. The bounding box is read in as lat/long points, for example 39.583642,-104.895486. The bounding box is a lower-left point and an upper-right point which represents where the points are contained. The "points," which are named "radians" in OpenMap, are in a different format, which looks like this: [0.69086486, -1.8307719, 0.6908546, -1.8307716, 0.6908518, -1.8307717, 0.69085056, -1.8307722, 0.69084936, -1.8307728, 0.6908477, -1.8307738, 0.69084626, -1.8307749, 0.69084185, -1.8307792].
How do I convert the points like "0.69086486, -1.8307719" into x,y coordinates that are usable in normal graphics?
I believe all that's needed here is some kind of conversion, because bringing the points into Excel and graphing them creates a line whose curve matches the curve of the road at the given location (lat/long). However, the axises need to be adjusted manually and I have no reference as how to adjust the axises, since the given bounding box appears to be in a different format than the given points.
The ESRI Shapefile technical description doesn't seem to mention this (http://www.esri.com/library/whitepapers/pdfs/shapefile.pdf).
0.69086486, -1.8307719 is a latitude and a longitude in radians.
First, convert to degrees (multiply by (180/pi)), then you will have common units between your bounding box and your coordinates.
Then you can plot all of it in a local frame with the following :
x = (longitude-longitude0)*(6378137*pi/180)*cos(latitude0*pi/180)
y = (latitude-latitude0)*(6378137*pi/180)
(latitude0, longitude0) are the coordinates of a reference point (e.g. the lower-left corner of the bounding box)
units are degrees for angles and meters for distances
Edit -- explanation :
This is an orthographic projection of the Earth considered as a sphere whose radius is 6378137.0 m (semi-major axis of the WGS84 ellipsoid), centered on the point (lat0, lon0)
In OpenMap, there are a number of ways to convert from radians to decimal degrees:
Length.DECIMAL_DEGREE.fromRadians(radVal);
Math.toDegrees(radVal) // Standard java library
For an array, you can use ProjMath.arrayDegToRad(double[] radvals);
Be careful with that last one, it does the conversion in place. So if you grab a lat/lon array from an OMPoly, make a copy of it first before converting it. Otherwise, you'll mess up the internal coordinates of the OMPoly, which it expects to be in radians.