I've got two points between which im drawing a line (x1,y1 and x2,y2) but i need to know the coordinates of x3,y3 which is gapSize away from point x2,y2. Any ideas on how to solve this problem (the program is written in objective-c if that is helpful at all)?
You can simply calculate the angle in radians as
double rads = atan2(y2 - y1, x2 - x1);
Then you get the coordinates as follows:
double x3 = x2 + gapSize * cos(rads);
double y3 = y2 + gapSize * sin(rads);
Is this what you meant?
Compute the distance between P1 and P2: d=sqrt( (y2-y1)^2 + (x2-x1)^2)
Then x2 = (d*x1 + gapSize*x3) / (d+gapSize)
So x3 = (x2 * (d+gapSize) - d*x1) / gapSize
Similarly, y3 = (y2 * (d+gapSize) - d*y1) / gapSize
Sorry for the math. I didn't try to code it but it sounds right. I hope this helps.
There are many ways to do this. Simplest (to me) is the following. I'll write it in terms of mathematics since I can't even spell C.
Thus, we wish to find the point C = {x3,y3}, given points A = {x1,y1} and B = {x2,y2}.
The distance between the points is
d = ||B-A|| = sqrt((x2-x1)^2 + (y2-y1)^2)
A unit vector that points along the line is given by
V = (B - A)/d = {(x2 - x1)/d, (y2-y1)/d}
A new point that lies a distance of gapSize away from B, in the direction of that unit vector is
C = B + V*gapSize = {x2 + gapSize*(x2 - x1)/d, y2 + gapSize*(y2 - y1)/d}
Related
I implemented both the naive line drawing algorithm and bresenham algorithm. When I run the program with a 1000 lines, the naive line drawing algorithm is faster than the bresenham algorithm. Could anyone explain why?
Here is my code for both methods
def simpleLine(x1, y1, x2, y2):
dy = y2-y1;
dx = x2-x1;
x = x1
m = dy/dx;
b = y1-m*x1;
if(x1>x2):
x1,x2 = x2,x1
x=x1
while(x<=x2):
y=m*x+b;
PutPixle(win,x,round(y));
x=x+1
'
def BresenhamLine(x1, y1, x2, y2):
dx = abs(x2 - x1)
dy = abs(y2 - y1)
p = 2 * dy - dx
duady = 2 * dy
duadydx = 2 * (dy - dx)
x = x1
y = y1
xend = x2
if(x1 > x2):
x, y,xend = x2, y2,x1
while(x < xend):
PutPixle(win,x,y)
x =x+1
if(p<0):
p = p + 2*dy
else:
y = y-1 if y1>y2 else y+1
p = p+2*(dy-dx)
Bresenham's algorithm was invented for languages and machines with different performance characteristics than your python environment. In particular, low-level languages on systems where floating point math is much more expensive than integer math and branches.
In Python, your simple version is faster even though it uses floating point and rounding, because Python is slow and it executes fewer python operations per pixel. Any difference in speed between single integer or floating point operations is dwarfed by the cost of just doing python stuff.
I have a line with two points in latitude and longitude
A: 3.222895, 101.719751
B: 3.227511, 101.724318
and 1 point
C: 3.224972, 101.722932
How can I calculate minimum distance between point C and a line consists of point A and B?
It will be convenient if you can provide the calculation and objective-c code too. The distance is around 89 meters (using ruler in Google Earth).
Thanks to mimi and this great article http://www.movable-type.co.uk/scripts/latlong.html but they don't give the whole picture. Here is a detail one. All this points are collected using Google Earth using Placemark to mark the locations. Make sure lat/long are set to decimal degrees in Preferences.
lat A = 3.222895
lon A = 101.719751
lat B = 3.222895
lon B = 101.719751
lat C = 3.224972
lon C = 101.722932
Earth radius, R = 6371
1. First you have to find the bearing from A to C and A to B.
Bearing formula
bearingAC = atan2( sin(Δλ)*cos(φ₂), cos(φ₁)*sin(φ₂) − sin(φ₁)*cos(φ₂)*cos(Δλ) )
bearingAB = atan2( sin(Δλ)*cos(φ₂), cos(φ₁)*sin(φ₂) − sin(φ₁)*cos(φ₂)*cos(Δλ) )
φ is latitude, λ is longitude, R is earth radius
2. Find A to C distance using spherical law of cosines
distanceAC = acos( sin(φ₁)*sin(φ₂) + cos(φ₁)*cos(φ₂)*cos(Δλ) )*R
3. Find cross-track distance
distance = asin(sin(distanceAC/ R) * sin(bearingAC − bearing AB)) * R
Objective-C code
double lat1 = 3.227511;
double lon1 = 101.724318;
double lat2 = 3.222895;
double lon2 = 101.719751;
double lat3 = 3.224972;
double lon3 = 101.722932;
double y = sin(lon3 - lon1) * cos(lat3);
double x = cos(lat1) * sin(lat3) - sin(lat1) * cos(lat3) * cos(lat3 - lat1);
double bearing1 = radiansToDegrees(atan2(y, x));
bearing1 = 360 - ((bearing1 + 360) % 360);
double y2 = sin(lon2 - lon1) * cos(lat2);
double x2 = cos(lat1) * sin(lat2) - sin(lat1) * cos(lat2) * cos(lat2 - lat1);
double bearing2 = radiansToDegrees(atan2(y2, x2));
bearing2 = 360 - ((bearing2 + 360) % 360);
double lat1Rads = degreesToRadians(lat1);
double lat3Rads = degreesToRadians(lat3);
double dLon = degreesToRadians(lon3 - lon1);
double distanceAC = acos(sin(lat1Rads) * sin(lat3Rads)+cos(lat1Rads)*cos(lat3Rads)*cos(dLon)) * 6371;
double min_distance = fabs(asin(sin(distanceAC/6371)*sin(degreesToRadians(bearing1)-degreesToRadians(bearing2))) * 6371);
NSLog(#"bearing 1: %g", bearing1);
NSLog(#"bearing 2: %g", bearing2);
NSLog(#"distance AC: %g", distanceAC);
NSLog(#"min distance: %g", min_distance);
Actually there's a library for this. You can find it here https://github.com/100grams/CoreLocationUtils
Calculate bearing for each: C to A , and C to B:
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();
dLon= lon2-lon1;
Calculate cross-track distance:
var dXt = Math.asin(Math.sin(distance_CB/R)*Math.sin(bearing_CA-bearing_CB)) * R;
R is the radius of earth, dXt is the minimum distance you wanted to calculate.
Code to carry out this calculation is posted at here.
This implements an accurate solution in terms of ellipsoidal geodesics.
For the basic geodesic calculations, you can use
GeographicLib or the port of these algorithms to C which are included in version 4.9.0 of PROJ.4. This C interface is documented here.
Here's the result of compiling and running intercept.cpp:
$ echo 3.222895 101.719751 3.227511 101.724318 3.224972 101.722932 | ./intercept
Initial guess 3.225203 101.7220345
Increment 0.0003349040566247297 0.0003313413822354505
Increment -4.440892098500626e-16 0
Increment 0 0
...
Final result 3.225537904056624 101.7223658413822
Azimuth to A1 -135.1593040635131
Azimuth to A2 44.84069593652217
Azimuth to B1 134.8406959363608
Distance to line is 88.743m:
$ echo 3.224972 101.722932 3.225537904056624 101.7223658413822 | GeodSolve -i
-45.15927221 -45.15930407 88.743
See post here:
https://stackoverflow.com/a/33343505/4083623
For distance up to a few thousands meters I would simplify the issue from sphere to plane.
Then, the issue is pretty simply as a easy triangle calculation can be used:
We have points A and B and look for a distance X to line AB. Then:
Location a;
Location b;
Location x;
double ax = a.distanceTo(x);
double alfa = (Math.abs(a.bearingTo(b) - a.bearingTo(x))) / 180
* Math.PI;
double distance = Math.sin(alfa) * ax;
If you know how to calculate the distance of two points, get the distances between each two points, you get AB, AC, and BC. You want to know the closest distance between point C and line AB.
First get the value of P
P=（AB+BC+AC)/2
Using P, you need to get S
S=SQRT(（P(P-AC)（P-AB)(P-AC))
SQRT means square root. Then you get what you want by
2*S/AB
I've got two points between which im drawing a line (x1,y1 and x2,y2) but i need to know the coordinates of x3,y3 which is gapSize away from point x2,y2. Any ideas on how to solve this problem (the program is written in objective-c if that is helpful at all)?
You can simply calculate the angle in radians as
double rads = atan2(y2 - y1, x2 - x1);
Then you get the coordinates as follows:
double x3 = x2 + gapSize * cos(rads);
double y3 = y2 + gapSize * sin(rads);
Is this what you meant?
Compute the distance between P1 and P2: d=sqrt( (y2-y1)^2 + (x2-x1)^2)
Then x2 = (d*x1 + gapSize*x3) / (d+gapSize)
So x3 = (x2 * (d+gapSize) - d*x1) / gapSize
Similarly, y3 = (y2 * (d+gapSize) - d*y1) / gapSize
Sorry for the math. I didn't try to code it but it sounds right. I hope this helps.
There are many ways to do this. Simplest (to me) is the following. I'll write it in terms of mathematics since I can't even spell C.
Thus, we wish to find the point C = {x3,y3}, given points A = {x1,y1} and B = {x2,y2}.
The distance between the points is
d = ||B-A|| = sqrt((x2-x1)^2 + (y2-y1)^2)
A unit vector that points along the line is given by
V = (B - A)/d = {(x2 - x1)/d, (y2-y1)/d}
A new point that lies a distance of gapSize away from B, in the direction of that unit vector is
C = B + V*gapSize = {x2 + gapSize*(x2 - x1)/d, y2 + gapSize*(y2 - y1)/d}
I'm looking for a smooth way to calculate the distance between two GPS Points, so I get the result like: "You have to go x meters up and y meters to the left - so I can work with a 2d-coordinate system, where I have my position as (0,0) and the other positions is showing the distance in (x, y) in meters from my position.
My idea was to calculate the distance between the points using the haversine formula. (This returns my hypotenuse)
In addition to that, I'm calculating the bearing between this two points. This is my alpha.
With this two values, I wanted to use basic trigonometry functions to resolve my problem.
So I tried to calculate:catheti_1 = sin(alpha) * hypotenuse, catheti_2 = cos(alpha) * hypotenuse.
Maybe I'm doing something wrong, but my results are useless at the moment.
So my question is: How can I calculate the distance in x and y direction between two GPS points?
I'm calculating alpha in the following procedure:
public static double bearingTo(GPSBean point1, GPSBean point2) {
double lat1 = Math.toRadians(point1.latitude);
double lat2 = Math.toRadians(point2.latitude);
double lon1 = Math.toRadians(point1.longitude);
double lon2 = Math.toRadians(point2.longitude);
double deltaLong = lon2 - lon1;
double y = Math.sin(deltaLong) * Math.cos(lat2);
double x = Math.cos(lat1) * Math.sin(lat2) - Math.sin(lat1)
* Math.cos(lat2) * Math.cos(deltaLong);
double bearing = Math.atan2(y, x);
return (Math.toDegrees(bearing) + 360) % 360;
}
I just implemented your code, using approximate coordinates of NYC and Boston as reference points, and implementing the Haversine formula as found at http://www.movable-type.co.uk/scripts/latlong.html (which you didn't show):
long1 = -71.02; lat1 = 42.33;
long2 = -73.94; lat2 = 40.66;
lat1 *=pi/180;
lat2 *=pi/180;
long1*=pi/180;
long2*=pi/180;
dlong = (long2 - long1);
dlat = (lat2 - lat1);
// Haversine formula:
R = 6371;
a = sin(dlat/2)*sin(dlat/2) + cos(lat1)*cos(lat2)*sin(dlong/2)*sin(dlong/2)
c = 2 * atan2( sqrt(a), sqrt(1-a) );
d = R * c;
When I run this code, I get d = 306, which agrees with the answer from the above site.
For the bearing I get 52 deg - again, close to what the site gave.
Without seeing the rest of your code it's hard to know why your answer is different.
Note: when the two points are close together, you could make all kinds of approximations, but this code should still work - the formula has good numerical stability because it's using the sin of the difference between longitudes, latitudes (rather than the difference of the sin).
Addendum:
Using your code for x, y (in your question), I get sensible values for the distance - agreeing with the "proper" answer to within 120 m (which isn't bad since one is a straight line approximation and the other follows the curvature of the earth). So I think your code is basically OK now you fixed the typo.
Use Haversine formula to Calculate distance (in km) between two points specified by latitude/longitude (in numeric degrees)
from: Haversine formula - R. W. Sinnott, "Virtues of the Haversine"
Sky and Telescope, vol 68, no 2, 1984
http://www.census.gov/cgi-bin/geo/gisfaq?Q5.1
Example usage from form:
result.value = LatLon.distHaversine(lat1.value.parseDeg(), long1.value.parseDeg(), * lat2.value.parseDeg(), long2.value.parseDeg());
Javascript :
LatLon.distHaversine = function(lat1, lon1, lat2, lon2) {
var R = 6371; // earth's mean radius in km
var dLat = (lat2-lat1).toRad();
var dLon = (lon2-lon1).toRad();
lat1 = lat1.toRad(), lat2 = lat2.toRad();
var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(lat1) * Math.cos(lat2) * Math.sin(dLon/2) * Math.sin(dLon/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c;
return d;
}
If anybody is interested to have a simpler formula that anyone can understand. Here is mine, it works for Sweden, but you can adapt it to work anywhere by making a more general formula for calculation of longfactor.
Hope you can understand even if it is written in an odd language.
<gpsDist lat1,long1,lat2,long2> all parameters in 1/100000 degree.
Example: <getDist 5950928,1327120,5958505,1302241> => 16303
Same at https://gps-coordinates.org/distance-between-coordinates.php => 16.35 KM.
<var $latFactor,1.112>
<function getDist,
-<var $longFactor,<calc 0.638 - ($1/100000-55)*0.0171,3>>
-<var $latDist,<calc ($3-$1)*$latFactor>>
-<var $longDist,<calc ($4-$2)*$longFactor>>
-<sqrt $latDist*$latDist + $longDist*$longDist>
->
/Bertil Friman
Given vertices V1 (x1,y1,z1), V2 (x2,y2,z2), V3 (x3,y3,z3) of a triangle T, I have to find z coordinate of a point by it's x,y coordinate if I know that (x,y) lies within projection of triangle Tp (x1,y1), (x2,y2), (x3,y3).
Actually, triangle plane in 3D is defined by equation: Ax+By+Cz+D=0, and I can find z = (D-Ax-By)/C
The problem is that A, B, C, D are too expensive to calculate in run-time:
A = y1(z2-z3) + y2(z3-z1) + y3(z1-z2)
B = z1(x2-x3) + z2(x3-x1) + z3(x1-x2)
C = x1(y2-y3) + x2(y3-y1) + x3(y1-y2)
D = -x1(y2*z3 – y3*z2) – x2(y3*z1 – y1*z3) – x3 (y1*z2 – y2*z1)
Is it possible to calculate A, B, C, D using, say, opengl shaders? Are there optimized algorithms to find plane coefficients?
The technique is called Barycentric coordinates but the wiki page is pretty hard to follow -
See http://www.alecjacobson.com/weblog/?p=1596
float calcY(vec3 p1, vec3 p2, vec3 p3, float x, float z) {
float det = (p2.z - p3.z) * (p1.x - p3.x) + (p3.x - p2.x) * (p1.z - p3.z);
float l1 = ((p2.z - p3.z) * (x - p3.x) + (p3.x - p2.x) * (z - p3.z)) / det;
float l2 = ((p3.z - p1.z) * (x - p3.x) + (p1.x - p3.x) * (z - p3.z)) / det;
float l3 = 1.0f - l1 - l2;
return l1 * p1.y + l2 * p2.y + l3 * p3.y;
}
Code from http://www.gamedev.net/topic/597393-getting-the-height-of-a-point-on-a-triangle/ - carefull about computer graphics vs maths use of Y Z
ps. I Don't know of any faster version using shaders. One quick dirty+solution is to render the triangle using colors based on the height of the vertices and pick the pixel color at your X,Y - in practice this never ends up being much faster on a desktop machine, don't know about opengl-es