Store circle in MySQL GeoSpatial Database - mysql

I want to be able to store a circle using a fixed point g and radius d, then get those values back when retrieving the information.
The only way I've found to use those arguments to create a geographic object is to use buffer which produces a polygon:
https://dev.mysql.com/doc/refman/5.6/en/spatial-operator-functions.html#function_buffer
SELECT ASTEXT( BUFFER ( POINT( 10, 10 ), 5 ) );
| POLYGON((10 0,9.50932325672582 0.012045437948275506,9.019828596704395 0.048152733278032045,8.532695255446383 0.10823490035219052,8.049096779838717 0.1921471959676957,7.570198200967361 0.2996874680545609,7.097153227455378 0.4305966426779104,6.631101466077799 0.5845593481697922,6.173165676349102 0.7612046748871322,5.724449065697179 0.9601070687655664,5.286032631740024 1.1807873565164506,4.858972558067784 1.4227138999972784,4.444297669803978 1.6853038769745474,4.043006955075667 1.9679246851935517,3.6560671583635447 2.2698954663726303,3.2844104515298165 2.5904887464504087,2.9289321881345254 2.9289321881345254,2.5904887464504087 3.2844104515298165,2.2698954663726303 3.6560671583635447,1.9679246851935517 4.043006955075667,1.6853038769745474 4.444297669803978,1.4227138999972784 4.858972558067784,1.1807873565164506 5.286032631740024,0.9601070687655664 5.724449065697179,0.7612046748871322 6.173165676349102,0.5845593481697922 6.631101466077799,0.4305966426779104 7.097153227455378,0.2996874680545609 7.570198200967361,0.1921471959676957 8.049096779838717,0.10823490035219052 8.532695255446383,0.048152733278032045 9.019828596704395,0.012045437948275506 9.50932325672582,0 10,0.048152733278032045 10.980171403295605,0.10823490035219052 11.467304744553617,0.1921471959676957 11.950903220161283,0.2996874680545609 12.429801799032639,0.4305966426779104 12.902846772544622,0.5845593481697922 13.368898533922202,0.7612046748871322 13.826834323650898,0.9601070687655664 14.27555093430282,1.1807873565164506 14.713967368259976,1.4227138999972784 15.141027441932216,1.6853038769745474 15.555702330196022,1.9679246851935517 15.956993044924333,2.2698954663726303 16.343932841636455,2.5904887464504087 16.715589548470184,2.9289321881345254 17.071067811865476,3.2844104515298165 17.409511253549592,3.6560671583635447 17.73010453362737,4.043006955075667 18.03207531480645,4.444297669803978 18.314696123025453,4.858972558067784 18.577286100002723,5.286032631740024 18.81921264348355,5.724449065697179 19.039892931234434,6.173165676349102 19.238795325112868,6.631101466077799 19.41544065183021,7.097153227455378 19.569403357322088,7.570198200967361 19.70031253194544,8.049096779838717 19.807852804032304,8.532695255446383 19.89176509964781,9.019828596704395 19.95184726672197,9.50932325672582 19.987954562051726,10 20,10.49067674327418 19.987954562051726,10.980171403295605 19.95184726672197,11.467304744553617 19.89176509964781,11.950903220161283 19.807852804032304,12.429801799032639 19.70031253194544,12.902846772544622 19.569403357322088,13.368898533922202 19.41544065183021,13.826834323650898 19.238795325112868,14.27555093430282 19.039892931234434,14.713967368259976 18.81921264348355,15.141027441932216 18.577286100002723,15.555702330196022 18.314696123025453,15.956993044924333 18.03207531480645,16.343932841636455 17.73010453362737,16.715589548470184 17.409511253549592,17.071067811865476 17.071067811865476,17.409511253549592 16.715589548470184,17.73010453362737 16.343932841636455,18.03207531480645 15.956993044924333,18.314696123025453 15.555702330196022,18.577286100002723 15.141027441932216,18.81921264348355 14.713967368259976,19.039892931234434 14.27555093430282,19.238795325112868 13.826834323650898,19.41544065183021 13.368898533922202,19.569403357322088 12.902846772544622,19.70031253194544 12.429801799032639,19.807852804032304 11.950903220161283,19.89176509964781 11.467304744553617,19.95184726672197 10.980171403295605,19.987954562051726 10.49067674327418,20 10,19.95184726672197 9.019828596704395,19.89176509964781 8.532695255446383,19.807852804032304 8.049096779838717,19.70031253194544 7.570198200967361,19.569403357322088 7.097153227455378,19.41544065183021 6.631101466077799,19.238795325112868 6.173165676349102,19.039892931234434 5.724449065697179,18.81921264348355 5.286032631740024,18.577286100002723 4.858972558067784,18.314696123025453 4.444297669803978,18.03207531480645 4.043006955075667,17.73010453362737 3.6560671583635447,17.409511253549592 3.2844104515298165,17.071067811865476 2.9289321881345254,16.715589548470184 2.5904887464504087,16.343932841636455 2.2698954663726303,15.956993044924333 1.9679246851935517,15.555702330196022 1.6853038769745474,15.141027441932216 1.4227138999972784,14.713967368259976 1.1807873565164506,14.27555093430282 0.9601070687655664,13.826834323650898 0.7612046748871322,13.368898533922202 0.5845593481697922,12.902846772544622 0.4305966426779104,12.429801799032639 0.2996874680545609,11.950903220161283 0.1921471959676957,11.467304744553617 0.10823490035219052,10.980171403295605 0.048152733278032045,10.49067674327418 0.012045437948275506,10 0)) |
1 row in set (0.00 sec)
My problem with this is that I cannot retrieve the point and radius when selecting this row in the future, instead I get the polygon back.
Is there not a better way to store a circle for use with MySQL GeoSpatial Extensions?


Spatial databases are not great at storing curves; rather they tend to be approximated by lots of straight segments, so you're not actually retaining a circle. Support for curved geometries is improving but it's still not really there.
I would probably store the geometry as you have, as well as the radius as a floating point value.
Alternatively, if you only want to store the geometry, you can obtain the centre of the polygon ("circle") with the Centroid function, and then get the radius by converting the polygon to a line and determining the distance between the centroid and the line. There are other ways to determine this distance, too. the caveat is that because this is only an approximation of a circle, the distance between the centre of the circle and its edge is different when measured between the centre and one of the vertices, and when measured from the centre and an edge between two vertices. So if you do not store the radius independently of the geometry, ideally you should measure distance between a vertex and the centre, which will be equal to the original radius (buffer distance). Practically the difference will be rather small so long as your "circle" has dense vertices.

Related

Convert Autodesk Viewer Units to Inches

I am using the viewer with the Edit2D library and am trying to convert the length between two x and y points into real measurements.
For example, after a shape is drawn using the polygon tool, I want to get the length of the first edge.
I get the drawn shape and the first two points on the event shown below, get 2 points, and get the distance between them. It seems they are in Autodesk Units or something. Is there an easy way to convert the units to feet or inches?
I have found
Edit2DExtension.defaultContext.unitHandler.fromDisplayUnits()
as well as
Edit2DExtension.defaultContext.unitHandler.toDisplayUnits()
and also
Autodesk.Viewing.Private.convertUnits().
I've tried all three, but am unsure how to use them and haven't found any good results with them yet.
There may be a way to do it through Edit2d but I haven't found a way yet and there is next to no documentation I can find on this library.
beforeEdit2DAction(event) {
console.log('After Shape has been drawn -> ', event);
let shape = event.action.shape;
let pointA = shape._loops[0][0]; // Value: {x: 21.393766403198242, y: 20.934386880096092}
let pointB = shape._loops[0][1]; // Value: {x: 25.082155227661133, y: 20.934386880096092}
// Distance between 2 points (Assuming Autodesk units)
let length = Autodesk.Edit2D.Math2D.distance2D(pointA, pointB); // 3.6883888244628906
// Need to convert to real world units (preferably ft or inches)
}
The real length is 29.5 FEET
Any ideas, or comments are welcome! Thanks
Edit: Trying Petr's suggestion here's what it returned:

That's an interested question. The "unit handler" keeps track of two types of units:
layer units (Edit2DExtension.defaultContext.unitHandler.config.layerUnits, can be inch for example)
display units (Edit2DExtension.defaultContext.unitHandler.config.displayUnits)
These two properties control how the actual lengths and areas are displayed. For example, the unit handler's toDisplayUnits method is implemented like so:
toDisplayUnits(fromUnits, value) {
this.updateConfig();
return Autodesk.Viewing.Private.convertUnits(fromUnits, this.config.displayUnits, this.config.scaleFactor, value);
}
With that, configuring fromUnits and displayUnits (and scale) properly should give you the real measurements you need.

Mapping Nebraska school districts with D3 v4 - base layer not showing

I am having trouble mapping Nebraska school districts in D3 (v4). (See bl.ock here.) I can map Nebraska counties no problem, but the same code modified for school districts--and pointing to a school district TopoJSON file--gives me a blank page.
Here's how I created the JSON, based on Mike Bostock's excellent instructions :
curl "https://www2.census.gov/geo/tiger/GENZ2017/shp/cb_2017_31_unsd_500k.zip" -o cb_2017_31_unsd_500k.zip
unzip -o cb_2017_31_unsd_500k.zip
shp2json cb_2017_31_unsd_500k.shp -o ne_district.json
ndjson-split "d.features" < ne_district.json > ne_district.ndjson
ndjson-map "d.id = d.properties.GEOID, d" < ne_district.ndjson > ne_district-id.ndjson
geo2topo -n districts=ne_district-id.ndjson > ne_district-id-topo.json
And here's my projection:
var projection = d3.geoConicConformal()
.parallels([40, 43])
.rotate([100, 0])
.scale(8000);
Thanks for your help and apologies in advance for anything important I left out!

The issue is you haven't finished setting your projection parameters. You have rotate the map, which is how you should center a conic projection along the x axis. But you haven't centered the map on the y axis, it is centered on the equator. You
For a conical projection, you can do this one of three ways:
Center the map on a central latitude : projection.center([0,y])
You don't need to use .center with an x value because the map is already centered on the x by rotation, rotation and centering are cumulative
Rotate the map to a central latitude and longitude: projection.rotate([-x,-y])
On a conical projection the rotation on the meridian does not warp the map (generally), we rotate by the negative as we move the earth under us. This option does slightly distort the map relative to the other options - this may be preferrable.
Use the projection translation to center the map
The easiest way is to translate the result while automatically scaling (though you can do this manually too) with projection.fitSize or projection.fitExtent. These methods modify projection.scale and projection.translate. As with centering with .center, you need to keep your rotation - otherwise you'll get an odd tilt to the map.
These methods set translate and scale to appropriate values so that your map area contains the desired features:
var featureCollection = topojson.feature(ne, ne.objects.districts);
projection.fitSize([width,height],featureCollection);
These methods must take objects, not arrays, so we use the featureCollection, not the features as an array
Both methods take an array specifying the size to stretch a provided geojson object over:
projection.fitSize([mapwidth,mapheight],geojsonObject)
projection.fitExtent([[left,top],[right,bottom]],geojsonObject)
Here's an updated gist using fitSize.

game - how can I drag objects (cars with numbers) into targets (start line) AS3.0.?

I am having this problem, where I have several cars, numbers and letters, and need to put 5 cars in the starting places. -random order is ok.
I' having trouble finding in AS3 a way so that the EndX and EndY of each object can be in the starting lines and be considered right no matter the order!
I'm having trouble putting the code here so, heres a titanpad with the code:
this is the code:
being (um, dois, tres, quatro) the movieclip instance name for each numbered car.
https://titanpad.com/42vtnCbvLu

First of all, you could probably benefit by using the distance between two points formula and seeing if the distance is less than a certain value rather than checking in all 4 directions manually:
Math.abs(Math.sqrt((x2-x1)^2 + (y2-y1)^2))
Let the position of the car be (x1,y1) and the start position (x2,y2).
This formula will give you the distance between the two points in any direction, and you could test maybe whether this value is less than your offset.
As for the cars in any order part, I'm interpreting that you have your cars and you want the user to drag them to one of 5 spots, a bit like this:
spot1
spot2
spot3
spot4
spot5
All with respective coordinates. My suggestion would be to have a boolean flag for whether each spot is occupied that stops the program checking whether a car is put there after it has been taken once.
Once all these flags are true, then you can proceed.
Hope this helps.

AS3 Polar coordinate from Cartesian coordinate

I'm playing with body animation in AS3. I did a body with all parts (excluding fingers) and make a XML with the "skeleton". The XML got the instances of each part and the place of the articulation of the next part. I make it work with cardinal coordinates (x,y) and the body moves when I rotate a part and recalculate all the links again (each part in each articulation).
However, this will demand some calculation each little modification of the body, so now I'm optimizing it. As for de design x,y is easier, so when the body instance is created, the class re-build the XML converting coordinates to Polar system (r,t), like this ("Quadro" is the node with coordinates):
dx = Quadro.#x;
dy = Quadro.#y;
Quadro.#r = Math.sqrt(Math.pow(dx,2) + Math.pow(dy,2));
Quadro.#t = (dy>0)? Math.asin(dx/Quadro.#r) : Math.acos(dy/Quadro.#r);
I did some changes to make it work but at list one quadrant is always wrong! In this case, the upper left is wrong. The neck and the head should be in this place and they are in upper right (mirrored).
Any tips for a right conversion in AS3?

Try to use this:
Quadro.#t=Math.atan2(dy,dx);
From Wikipedia:
The Cartesian coordinates x and y can be converted to polar
coordinates r and φ with r ≥ 0 and φ in the interval (−π, π] by:

How does this work in computing the depth map?

From this site: http://www.catalinzima.com/?page_id=14
I've always been confused about how the depth map is calculated.
The vertex shader function calculates position as follows:
VertexShaderOutput VertexShaderFunction(VertexShaderInput input)
{
VertexShaderOutput output;
float4 worldPosition = mul(input.Position, World);
float4 viewPosition = mul(worldPosition, View);
output.Position = mul(viewPosition, Projection);
output.TexCoord = input.TexCoord; //pass the texture coordinates further
output.Normal =mul(input.Normal,World); //get normal into world space
output.Depth.x = output.Position.z;
output.Depth.y = output.Position.w;
return output;
}
What are output.Position.z and output.Position.w? I'm not sure as to the maths behind this.
And in the pixel shader there is this line: output.Depth = input.Depth.x / input.Depth.y;
So output.Depth is output.Position.z / outputPOsition.w? Why do we do this?
Finally in the point light shader (http://www.catalinzima.com/?page_id=55) to convert this output to be a position the code is:
//read depth
float depthVal = tex2D(depthSampler,texCoord).r;
//compute screen-space position
float4 position;
position.xy = input.ScreenPosition.xy;
position.z = depthVal;
position.w = 1.0f;
//transform to world space
position = mul(position, InvertViewProjection);
position /= position.w;
again I don't understand this. I sort of see why we use InvertViewProjection as we multiply by the view projection previously, but the whole z and now w being made to equal 1, after which the whole position is divided by w confuses me quite a bit.

To understand this completely, you'll need to understand how the algebra that underpins 3D transforms works. SO does not really help (or I don't know how to use it) to do matrix math, so it'll have to be without fancy formulaes. Here is some high level explanation though:
If you look closely, you'll notice that all transformations that happen to a vertex position (from model to world to view to clip coordinates) happens to be using 4D vectors. That's right. 4D. Why, when we live in a 3D world ? Because in that 4D representation, all the transformations we usually want to do to vertices are expressible as a matrix multiplication. This is not the case if we stay in 3D representation. And matrix multiplications are what a GPU is good at.
What does a vertex in 3D correspond to in 4D ? This is where it gets interesting. The (x, y, z) point corresponds to the line (a.x, a.y, a.z, a). We can grab any point on this line to do the math we need, and we usually pick the easiest one, a=1 (that way, we don't have to do any multiplication, just set w=1).
So that answers pretty much all the math you're looking at. To project a 3D point in 4D we set w=1, to get back a component from a 4D vector, that we want to compare against our standard sizes in 3D, we have to divide that component by w.
This coordinate system, if you want to dive deeper, is called homogeneous coordinates.