[GIS] How to properly utilize DWITHIN Spatial Filter in OpenLayers

distancefilteropenlayers-2spatial-index

Given the following code:

var points = [];                                
points.push(new OpenLayers.Geometry.Point(-110, 38));
points.push(new OpenLayers.Geometry.Point(-108, 40));
var lineString = new OpenLayers.Geometry.LineString(points);
var filteredPoint = new OpenLayers.Geometry.Point(-108.25, 39.78);
var newFilter = new OpenLayers.Filter.Spatial({
        type: OpenLayers.Filter.Spatial.DWITHIN,
        distanceUnits: 'm',
        distance: 92600,
        value: lineString
}); //92600 meters, or 50 nautical miles                                  
alert(newFilter.evaluate(filteredPoint));

When the alert is fired, it results in false. I'm a beginner to OpenLayers, looking for a way in Javascript to determine whether or not a point is within a certain distance, either side of a line. I cannot use a bounding box for this feature, because I want the user to be able to pick any two points and a distance from the line, so an MBR isn't available here. filteredPoint is less than 20nm from the 2nd point in the linestring, so my assumption is that this alert would return true, but alas it is false.
Meanwhile, given the following code, in addition to that above:

var bounds = new OpenLayers.Bounds();
bounds.extend(new OpenLayers.LonLat(-110, 38));
bounds.extend(new OpenLayers.LonLat(-108, 40));
bounds.toBBOX();
var newFilter2 = new OpenLayers.Filter.Spatial({
        type: OpenLayers.Filter.Spatial.BBOX,
        value: bounds       
}); 
alert(newFilter2.evaluate(filteredPoint));

That code gives a result of true inside of this bounding box.
I would think that the two bits of code are trying to find points in the block below represented by the asterisk. Am I wrong in interpreting DWITHIN as such?


----------                     /
|      * |                 *--/
|  bbox  |                   /
|        |                  /
----------                 /

Any tips or guidance in the right direction would be greatly appreciated. My overall goal is to determine how many points out of ~3700 points I have fall within a distance of the line.

Best Answer

What i can see from Openlayers source code only BBOX and INTERSECTS filters only implemented. Have a look at the evaluate() method.

You could use WFS getFeature call with CQL filters to filter the features at server or if it is only display purpose, SLD could be implemented with DWITHIN filter. (http://trac.osgeo.org/openlayers/ticket/1543)

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[GIS] Finding the minimum and maximum distance between an observer and a lat/lon bounding box

The "bounding box" can have several interpretations. The two most natural are (a) the sides are meridians and the top and bottom are circles of constant latitude and (b) the sides are still meridians, but now the top and bottom are geodesics (which therefore do not necessarily have constant latitude). Fortunately, the solution method is the same in both interpretations, because in all cases the sides are portions of circles. This reduces the problem to optimizing the distance between a fixed point x and an arc of a circle (in three dimensions). Do this for each of the four sides and retain the best results.

A simple way to represent the arc of a circle is to choose coordinates for x in which the circle's center is at the origin, the circle is determined by two vectors e and f of equal length (say, r) and at right angles to each other, and the arc is given by all points of the form cos(q) * e + sin(q) * f for angles q ranging from q0 through q1. Critical points for the optima occur either at the endpoints of the arc or where the tangent vector is perpendicular to the direction vector. The latter tells us the inner product of x - (cos(q) * e + sin(q) * f) with (-sin(q) * e + cos(q) * f) is zero. This easily reduces to the equation

tan(q) = -xe / xf

where xe is the inner product of x with e and xf is the inner product of x with f. There will usually be two solutions (differing by 180 degrees). Use ATan (or its equivalent) to find them. Any of these solutions that lies in the interval [q0, q1] is a critical point.

In this fashion you collect (in most cases) up to two critical points per side plus the four vertices of the bounding box: that gives between 4 and 12 distance calculations to make in order to find the largest and smallest distance. These are 3D (Pythagorean) distances: compute them using the Pythagorean formula, but don't bother to take square roots when comparing; just take square roots for the largest and smallest distances.

^{Portions of the four circles determined by the sides of this bounding box are shown in gray. The arcs of latitude are in red and the arcs of the meridians are in blue.}

There is a notable exception to this nice situation: x might lie on, or extremely close to, the axis of the circle. In this case, every point along the circle (and therefore every point along the arc) has the same distance to x. Make sure to check for this situation. It's easy to detect: both xf and xe are zero (or extremely small compared to x itself).

It remains, then, to describe how to find e, f, q0, and q1. That's easy:

For a meridian at longitude lambda, the circle's center is already at (0,0,0), so don't change the coordinates of x (which presumably is referenced to a geocentric system). Take e = (Rcos(lambda), Rsin(lambda), 0) and f = (0,0,R) where R is the earth's radius (in a spherical model). Here, q0 will be the minimum latitude among the bounding box coordinates and q1 will be the maximum latitude.
For a circle of latitude phi, take the center to be (0, 0, Rsin(phi)). Subtract these coordinates from x. In these new coordinates, we can let e = (Rcos(phi), 0, 0) and f = (0, R*cos(phi), 0). Now q0 is the westernmost longitude and q1 is the easternmost longitude in the bounding box.
Equations for e, f, q0, and q1 are more complicated for arbitrary geodesics (case (b) in the first paragraph). Because it is not apparent these are needed, I won't derive them here.

In summary,

Only simple calculations are needed to establish the 3D coordinate representation of x, e, f and to find q0 and q1. (This is all done four times, once for each side of the bounding box.)
Four arctangent calculations will determine between 0 and 8 critical points. At this time, you can inspect xe and xf (four times) to check whether any of the sides of the bounding box are equidistant from x. (You will have to decide how such a situation is to be handled).
Squared pythagorean distances (sums of squares) have to be determined between x and the four corners of the bounding box plus these other critical points. Take the square roots of the min and max of these distances.

[GIS] Problems with features, popups and strategy with OpenLayers

you can catch your opened(active) popup with this way:

map.popups[a]

you can change so update its content:

map.popups[a].contentHTML = 'Test Content';
map.popups[a].redraw();

or

map.popups[a].setContentHTML = 'Test Content'

you can show or hide your popup:

map.popups[a].show();
map.popups[a].hide();

you can update position of your popup:

map.popups[a].lonlat.lon = 10;
map.popups[a].lonlat.lat = 10;
map.popups[a].updatePosition();

for openlayers cluster you should check out Cluster Strategy Example here. Only one thing you should do is that define fillColor or externalGraphic for your cluster.

          var style = new OpenLayers.Style({
                pointRadius: "${radius}",
                fillColor: "${getColor}",
                fillOpacity: 0.8,
                strokeColor: "#cc6633",
                strokeWidth: 2,
                strokeOpacity: 0.8
            }, {
                context: {
                    radius: function(feature) {
                        return Math.min(feature.attributes.count, 7) + 3;
                    },
                    getColor: function(feature) {
                        return // Your color selection logic;
                    }
                }
            });

As the last question, i dont understand very well what you want? but try to use removeFeatures instead of destroyFeatures....

i hope it helps you...

Best Answer

Related Solutions

[GIS] Finding the minimum and maximum distance between an observer and a lat/lon bounding box

[GIS] Problems with features, popups and strategy with OpenLayers

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