I am new to Power BI and DAX, so I hope you can help me.
I have two tables without any relationship:
Table A contains lat/lng and date of tracked positions.
Table B contains lat/lng and names of all stadiums.
I want to find the closest stadium near the tracked position. Also if possible I want to validate, if the position was in a specific radius of that stadium.
Any help greatly appreciated.
Here's one possible approach:
First, calculate the minimal distance using the Haversine function.
Add this as a calculated column to your Tracked table.
Nearest =
MINX(Stadiums,
ROUND(2 * 3959 *
ASIN(SQRT(
SIN((Stadiums[Lat] - Tracked[Lat]) * PI()/360)^2 +
COS(Tracked[Lat] * PI()/180) * COS(Stadiums[Lat] * PI()/180) *
SIN((Stadiums[Lon] - Tracked[Lon]) * PI()/360)^2)), 1))
In this formula, 3959 is the radius of the Earth in miles.
We can now match up distances to find the nearest stadium:
Stadium = CALCULATE(MAX(Stadiums[Stadium]),
FILTER(Stadiums,
ROUND(2 * 3959 *
ASIN(SQRT(
SIN((Stadiums[Lat] - Tracked[Lat]) * PI()/360)^2 +
COS(Tracked[Lat] * PI()/180) * COS(Stadiums[Lat] * PI()/180) *
SIN((Stadiums[Lon] - Tracked[Lon]) * PI()/360)^2)), 1)
= Tracked[Nearest]))
Note: I rounded the values to avoid not matching from possible floating point errors. This may or may not be necessary.
Related
I have two locations and I have the formula to get the distance between them but I want to calculate the travel time between them without using google maps or any API calling. Is there any formula to do that in flutter?
I think the real travel time depends on a lot of factors, for example the road you take or the speed limit.
Whitout some complex elaboration, which are usually perfomed by a backend, I don't think it is possible to calculate it.
Anyway, since you have got the distance, you could simply estimate it by giving a supposed speed and then using basic physics formulas.
time = distance / speed.
You cannot find the travel distance between two locations without using the Google API.
However you can get the air distance in meters using this formula in flutter.
double calculateDistance(
double lat1, double lon1, double lat2, double lon2) {
var p = 0.017453292519943295;
var c = cos;
var a = 0.5 -
c((lat2 - lat1) * p) / 2 +
c(lat1 * p) * c(lat2 * p) * (1 - c((lon2 - lon1) * p)) / 2;
var temp = 12742 * asin(sqrt(a));
return (temp * 1000);
}
I am using postgres and postgis.
I have Posts which have a geometry, with an attribute visible_within_m which is how many meters from that point the Post should be shown in results.
I can find Posts within some random radius of some random points by doing ST_DWithin(geometry, ST_SetSRID(ST_Point(a, b), 4326), 10000)
However, I want to know how many Posts are visible with a radius of some random point.
How can I look up how many Posts are visible within a radius of some arbitrary point?
Is there a better way to do this?
You can calculate the distance between each point and the center of your circle. If the distance is grater than the radius then it is outside otherwise it's inside.
const EARTH_RADIUS = 6371000;
const toRad = function(num){return num*Math.PI/180};
var calculateDistance =
function(lat1, lon1, lat2, lon2){
var dLat = toRad(lat2 - lat1);
var dLon = toRad(lon2 - lon1);
var a = Math.sin(dLat / 2) * Math.sin(dLat / 2) + Math.cos(toRad(lat1)) *
Math.cos(toRad(lat2)) * Math.sin(dLon / 2) * Math.sin(dLon / 2);
var distance = EARTH_RADIUS * 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
return distance;
}
Instead of using a constant value for the distance, use the value stored in visible_within_m
SELECT * FROM mytable
WHERE ST_DWithin(geometry, ST_SetSRID(ST_Point(a, b), 4326), visible_within_m);
On a side note, st_dwithin with geometries uses the distance unit of the projection, so for 4326 it is a (meaningless) distance in degrees, not in meters.
I've looked around and seen mention of the haversine formula to determine distance between two coordinates (lat1, lng1) and (lat2, lng2).
I've implemented this code:
function haversineGreatCircleDistance(
$latitudeFrom, $longitudeFrom, $latitudeTo, $longitudeTo, $earthRadius = 6371000)
{
// convert from degrees to radians
$latFrom = deg2rad($latitudeFrom);
$lonFrom = deg2rad($longitudeFrom);
$latTo = deg2rad($latitudeTo);
$lonTo = deg2rad($longitudeTo);
$latDelta = $latTo - $latFrom;
$lonDelta = $lonTo - $lonFrom;
$angle = 2 * asin(sqrt(pow(sin($latDelta / 2), 2) +
cos($latFrom) * cos($latTo) * pow(sin($lonDelta / 2), 2)));
return $angle * $earthRadius;
}
And am trying to determine:
1) what units this is returning? (goal being in feet)
2) is this equation written the right way?
For example what should be the distance between these two points?
(32.8940695525,-96.7926336453) and (33.0642604502, -96.8064332754)?
I'm getting 18968.0903312 from the formula above.
Thanks!
1) what units this is returning? (goal being in feet)
Whatever units in which you supply the Earth's radius.
2) is this equation written the right way?
Test it. You can compare your results with an existing Haversine formula implementation, like this one.
I have a long list of locations in excel format and I need to calculate driving distance between these locations using lat&long. Is there any macro I can use.
Cheers
from : link
this is too simple
Miles:
=ACOS(COS(RADIANS(90-Lat1)) * COS(RADIANS(90-Lat2)) + SIN(RADIANS(90-Lat1)) *
SIN(RADIANS(90-Lat2)) * COS(RADIANS(Long1-Long2))) * 3959
Kilometers:
=ACOS(COS(RADIANS(90-Lat1)) * COS(RADIANS(90-Lat2)) + SIN(RADIANS(90-Lat1)) *
SIN(RADIANS(90-Lat2)) * COS(RADIANS(Long1-Long2))) * 6371
I have 2 coordinates and would like to do something seemingly straightforward. I want to figure out, given:
1) Coordinate A
2) Course provided by Core Location
3) Coordinate B
the following:
1) Distance between A and B (can currently be done using distanceFromLocation) so ok on that one.
2) The course that should be taken to get from A to B (different from course currently traveling)
Is there a simple way to accomplish this, any third party or built in API?
Apple doesn't seem to provide this but I could be wrong.
Thanks,
~Arash
EDIT:
Thanks for the fast responses, I believe there may have been some confusion, I am looking to get the course (bearing from point a to point b in degrees so that 0 degrees = north, 90 degrees = east, similar to the course value return by CLLocation. Not trying to compute actual turn by turn directions.
I have some code on github that does that. Take a look at headingInRadians here. It is based on the Spherical Law of Cosines. I derived the code from the algorithm on this page.
/*-------------------------------------------------------------------------
* Given two lat/lon points on earth, calculates the heading
* from lat1/lon1 to lat2/lon2.
*
* lat/lon params in radians
* result in radians
*-------------------------------------------------------------------------*/
double headingInRadians(double lat1, double lon1, double lat2, double lon2)
{
//-------------------------------------------------------------------------
// Algorithm found at http://www.movable-type.co.uk/scripts/latlong.html
//
// Spherical Law of Cosines
//
// Formula: θ = atan2( sin(Δlong) * cos(lat2),
// cos(lat1) * sin(lat2) − sin(lat1) * cos(lat2) * cos(Δlong) )
// JavaScript:
//
// var y = Math.sin(dLon) * Math.cos(lat2);
// var x = Math.cos(lat1) * Math.sin(lat2) - Math.sin(lat1) * Math.cos(lat2) * Math.cos(dLon);
// var brng = Math.atan2(y, x).toDeg();
//-------------------------------------------------------------------------
double dLon = lon2 - lon1;
double y = sin(dLon) * cos(lat2);
double x = cos(lat1) * sin(lat2) - sin(lat1) * cos(lat2) * cos(dLon);
return atan2(y, x);
}
See How to get angle between two POI?
Depending on how much work you want to put in this one, I would suggest looking at Tree Traversal Algorithms (check the column on the right), things like A* alpha star, that you can use to find your find from one point to another, even if obstacles are in-between.
If I understand you correctly, you have the current location and you have some other location. You want to find the distance (as the crow flies) between the two points, and to find a walking path between the points.
To answer your first question, distanceFromLocation will find the distance across the earth's surface between 2 points, that is it follows the curvature of the earth, but it will give you the distance as the crow flies. So I think you're right about that.
The second question is a much harder. What you want to do is something called path-finding. Path finding, require's not only a search algorithm that will decide on the path, but you also need data about the possible paths. That is to say, if you want to find a path through the streets, the computer has to know how the streets are connected to each other. Furthermore, if you're trying to make a pathfinder that takes account for traffic and the time differences between taking two different possible paths, you will need a whole lot more data. It is for this reason that we usually leave these kinds of tasks up to big companies, with lots of resources, like Google, and Yahoo.
However, If you're still interested in doing it, check this out
http://www.youtube.com/watch?v=DoamZwkEDK0