I am trying to create a Circle Geometry in MySQL using the co-ordinates of the center and a radius. I searched everywhere...all i could find in the MySQL doc on the site were for polygons. May be i am looking in the wrong place. can anybody help me with an appropriate SQL that can help me create a table that stores this Circle geometry as one of the columns in the table?
Also, i am not even sure if there is a way to do so in MySQL?..The version i am using is MySQL 5.6.
Thanks in advance.
As of MySQL v5.6.1, you can use Buffer(g, d):
Returns a geometry that represents all points whose distance from the geometry value g is less than or equal to a distance of d.
Obviously, in your case g should be the point at the centre of the circle and d should be its radius.
There are two Parts:
A.For given tested points you have to check their relation with given circle.
B.You want to generate points on circumference of given circle.
A.Yes, First of all take the distance between your given point(test Point) and the centre of circle point. Both of these points are defined in Latitude and longitude. Distance formula between two points(x1,y1) and (x2,y2) is distance d= sqrt[ (x2-x1)^2 + (y2-y1)^2 ].
Now,
If this distance is less than radius of circle then your tested point is inside your circle.
If this distance is Greater than radius then tested point is outside the circle.
If this calculated distance is equal to radius of circle then this tested point is on your circle i.e. on the circumference of your circle.
B. In a circle the total angle theta is 360 degree or 2*Pi in radians.
For given Circle whose centre is (x1, y1) and radius is r.
x = x1 + r * cos(theta)
y = y1 + r * sin(theta)
where, theta is running from Zero to 2*Pi and Pi is 3.1415.
Depending upon how you do it. Example: if you wants 10 points on circle, then increment=(2*Pi-Zero)/10.
fist theta is zero, then theta is Zero+increment, then theta is Zero +increment+increment i.e. 2* increment and then zero + 3*increment and then so on. unless you get theta equal to 2*Pi.
For all above thetas calculate x and y. These all x and y coordinate points are on the circumference of the circle.
Related
I am working on code to find the points on the circumference of a circle. I have the center point of the circle and the radius and I need to draw a circle around it. This will help me define the boundary. Please help me with formula for finding the these points on the circumference.
For a circle with origin (j, k) and radius r:
x(t) = r cos(t) + j
y(t) = r sin(t) + k
where you need to run this equation for t taking values within the range from 0 to 360, then you will get your x and y each on the boundary of the circle.
For more information:
http://www.math.com/tables/geometry/circles.htm
Lets say I have a lat lng coordinate and I want to place that at the center of a square that is 10km wide and then get the minimum lat/lng and maximum lat/lng.
Is there an easy way to do this that already exists?
If it doesn't need to be exact, it is pretty easy:
For the latitude, 1 km is 0.009 degrees (follows from the original definition of meter). Since your square is 5 km around the center, you just need to add and subtract 0.045 degrees from the center point.
For the longitude, it is slightly more complicated: Divide the above value with the cosine of the latitude.
In code:
lat_min = lat_center - 0.045;
lat_max = lat_center + 0.045;
long_min = long_center - (0.045 / Math.cos(lat_center*Math.PI/180);
long_max = long_center + (0.045 / Math.cos(lat_center*Math.PI/180);
(Math.PI/180 is needed to convert from degrees to radians).
Beware: Does not work around the poles.
How is the square oriented? Parallel to the equator? If so, then just do a bearing of 45 deg, 5km * sqrt(2) distance from your lat/lon to get the upper right corner. Similar for the bottom left, use a bearing of 225 deg.
See Destination point given distance and bearing from start point at http://www.movable-type.co.uk/scripts/latlong.html
In Google maps, the closer one gets to the pole, the more strechted out the map gets and sp each pixel of map represents less movment (asymtotically to 0 at the north pole)
I'm looking for a formula to connect the width of a pixel in degrees to the latitute (i.e. the real world distance represented by a pixel on the map). I have some data points here for zoom level 12 (IIRC)
Lat Width
0 0.703107352
4.214943141 0.701522096
11.86735091 0.688949038
21.28937436 0.656590105
30.14512718 0.60989762
35.46066995 0.574739011
39.90973623 0.541457085
41.5085773 0.528679228
44.08758503 0.507194173
47.04018214 0.481321842
48.45835188 0.468430215
51.17934298 0.442887842
63.23362741 0.318394373
72.81607372 0.208953319
80.05804956 0.122131316
90 0
The reason for doing this is I want to input lat/lng pairs and sort out exactly what pixel they would be located with respect to 0,0
I might be wrong but are you sure thos points are the pixel height? They seem to be a cosine which would be the pixel width not the height.
After a little trigonometry the pixel height adjusts to the formula:
where R is the earth radius, phi is the latitude and h is the height of a pixel in the equator.
This formula does not adjust to your points, that's why I asked if it was the width instead.
Anyway if you want so much precision that you cannot use the approximation in the previous answer you should also consider the R variable with the latitude and even with that I don't think you'll get the exact result.
Update:
Then the formula would be a cosine. If you want to take the variable radius of the earth the formula would be:
where R is the radius of the earth and d(0) is your pixel width at the equator. You may use this formula for R assuming the eearth to be an ellipsoid:
with a = 6378.1 (equator) and b = 6356.8 (poles)
While I am not sure what "height of a pixel" means, the plot of data (shown below) seems to fit the equation
y = a + bx + cx^2 + dx^3 where y = height, x = latitude
with coefficients
a = 7.0240278979641990E-01
b = 3.7784208874521786E-04
c = -1.2602864112736206E-04
d = 3.8304225582846095E-07
The general approach to find the equation is to first plot the data, then hypothesize the type of function, and then do a regression to find the coefficients.
I'm receiving data from GPS and store them in MySQL database. I have following columns:
User ID (int)
Latitude (double)
Longitude (double)
Horizontal Accuracy (double)
Horizontal accuracy is radius around Lat/Long, so my user with equivalent probability can be in any point of this area.
I need to find out probability that two users was intersecting. But I also have vision area, which is 30 meters. If horizontal accuracy would be 0 I could just measure area of intersection of two circles that have radius of 30 meters around lat/long. But in my case that's not possible because horizontal accuracy could be in range from 5 to 3000. Usually it's more than my vision area.
I think I can measure area of intersection of two cones where inner circle of this cone will have radius of horizontal accuracy + 30 meters and outer circle will have radius of horizontal accuracy. But this solution seems to be little bit complicated.
I want to hear some thoughts about that and other possible solution.
I've checked MySQL Spatial extension and as far I can see it can't do such calculations for me.
Thanks.
I worked on just such a problem as you are describing. How I approached it was to convert the Lat/Long (world coordinates) into X/Y (Cartesian coordinates) then I applied the Pythagorean Theorem a^2 + b^2 = c^2 to solve the problem.
First you need to convert the Lat/Long Coordinates.
To get X you Multiply the Radius by the cosine (cos) of the angle (NOTE: this angle has to be expressed as radians).
To get Y you do the same as above but use the sine function (sin).
To convert degrees to radials Multiply the angle by the quantity of PI (Approx. 3.14159...) / 180.
Radians = Angle * (PI / 180);
To solve for the c^2 "C Squared" c = SQRT (a*a + b*b);
For more information on Degrees to Radians: http://www.mathwarehouse.com/trigonometry/radians/convert-degee-to-radians.php
For more information on: Converting Lat/Long to X/Y coordinates: http://www.mathsisfun.com/polar-cartesian-coordinates.html
I usually find the information that I need for this kind of problem by asking a question on ask.com.
All the best.
Allan
I've got two LatLon (latitude-longitude) objects which represent two locations on the surface of the globe. I want to find the angle (in radians) between the center of the earth and these two LatLon objects.
I'm going to use this angle and the radius of the earth to calculate the arc length between the two locations (I figure this will give better precision than using simple Pythagoras, and be faster than computing the great circle distance).
I already have code to give me the Pythagorean distance and the great circle distance.
Using something like this - how to calculate the angle between two vectors
I thought this at first (after some calc on paper) is this Pythagorean thing?
angle_between_radian = sqrt(deltaLA^2 + deltaLO^2)*PI /180
edit: delta = delta>180?360-delta:delta
We working on sphere then above must wrong ^^. But this link may help:Calculate distance, bearing and more between Latitude/Longitude points.