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Point in Spherical Polygon using Python [duplicate]

Say I have an arbitrary set of latitude and longitude pairs representing points on some simple, closed curve. In Cartesian space I could easily calculate the area enclosed by such a curve using Green's Theorem. What is the analogous approach to calculating the area on the surface of a sphere? I guess what I am after is (even some approximation of) the algorithm behind Matlab's areaint function.

There several ways to do this.
1) Integrate the contributions from latitudinal strips. Here the area of each strip will be (Rcos(A)(B1-B0))(RdA), where A is the latitude, B1 and B0 are the starting and ending longitudes, and all angles are in radians.
2) Break the surface into spherical triangles, and calculate the area using Girard's Theorem, and add these up.
3) As suggested here by James Schek, in GIS work they use an area preserving projection onto a flat space and calculate the area in there.
From the description of your data, in sounds like the first method might be the easiest. (Of course, there may be other easier methods I don't know of.)
Edit – comparing these two methods:
On first inspection, it may seem that the spherical triangle approach is easiest, but, in general, this is not the case. The problem is that one not only needs to break the region up into triangles, but into spherical triangles, that is, triangles whose sides are great circle arcs. For example, latitudinal boundaries don't qualify, so these boundaries need to be broken up into edges that better approximate great circle arcs. And this becomes more difficult to do for arbitrary edges where the great circles require specific combinations of spherical angles. Consider, for example, how one would break up a middle band around a sphere, say all the area between lat 0 and 45deg into spherical triangles.
In the end, if one is to do this properly with similar errors for each method, method 2 will give fewer triangles, but they will be harder to determine. Method 1 gives more strips, but they are trivial to determine. Therefore, I suggest method 1 as the better approach.

I rewrote the MATLAB's "areaint" function in java, which has exactly the same result.
"areaint" calculates the "suface per unit", so I multiplied the answer by Earth's Surface Area (5.10072e14 sq m).
private double area(ArrayList<Double> lats,ArrayList<Double> lons)
{
double sum=0;
double prevcolat=0;
double prevaz=0;
double colat0=0;
double az0=0;
for (int i=0;i<lats.size();i++)
{
double colat=2*Math.atan2(Math.sqrt(Math.pow(Math.sin(lats.get(i)*Math.PI/180/2), 2)+ Math.cos(lats.get(i)*Math.PI/180)*Math.pow(Math.sin(lons.get(i)*Math.PI/180/2), 2)),Math.sqrt(1- Math.pow(Math.sin(lats.get(i)*Math.PI/180/2), 2)- Math.cos(lats.get(i)*Math.PI/180)*Math.pow(Math.sin(lons.get(i)*Math.PI/180/2), 2)));
double az=0;
if (lats.get(i)>=90)
{
az=0;
}
else if (lats.get(i)<=-90)
{
az=Math.PI;
}
else
{
az=Math.atan2(Math.cos(lats.get(i)*Math.PI/180) * Math.sin(lons.get(i)*Math.PI/180),Math.sin(lats.get(i)*Math.PI/180))% (2*Math.PI);
}
if(i==0)
{
colat0=colat;
az0=az;
}
if(i>0 && i<lats.size())
{
sum=sum+(1-Math.cos(prevcolat + (colat-prevcolat)/2))*Math.PI*((Math.abs(az-prevaz)/Math.PI)-2*Math.ceil(((Math.abs(az-prevaz)/Math.PI)-1)/2))* Math.signum(az-prevaz);
}
prevcolat=colat;
prevaz=az;
}
sum=sum+(1-Math.cos(prevcolat + (colat0-prevcolat)/2))*(az0-prevaz);
return 5.10072E14* Math.min(Math.abs(sum)/4/Math.PI,1-Math.abs(sum)/4/Math.PI);
}

You mention "geography" in one of your tags so I can only assume you are after the area of a polygon on the surface of a geoid. Normally, this is done using a projected coordinate system rather than a geographic coordinate system (i.e. lon/lat). If you were to do it in lon/lat, then I would assume the unit-of-measure returned would be percent of sphere surface.
If you want to do this with a more "GIS" flavor, then you need to select an unit-of-measure for your area and find an appropriate projection that preserves area (not all do). Since you are talking about calculating an arbitrary polygon, I would use something like a Lambert Azimuthal Equal Area projection. Set the origin/center of the projection to be the center of your polygon, project the polygon to the new coordinate system, then calculate the area using standard planar techniques.
If you needed to do many polygons in a geographic area, there are likely other projections that will work (or will be close enough). UTM, for example, is an excellent approximation if all of your polygons are clustered around a single meridian.
I am not sure if any of this has anything to do with how Matlab's areaint function works.

I don't know anything about Matlab's function, but here we go. Consider splitting your spherical polygon into spherical triangles, say by drawing diagonals from a vertex. The surface area of a spherical triangle is given by
R^2 * ( A + B + C - \pi)
where R is the radius of the sphere, and A, B, and C are the interior angles of the triangle (in radians). The quantity in the parentheses is known as the "spherical excess".
Your n-sided polygon will be split into n-2 triangles. Summing over all the triangles, extracting the common factor of R^2, and bringing all of the \pi together, the area of your polygon is
R^2 * ( S - (n-2)\pi )
where S is the angle sum of your polygon. The quantity in parentheses is again the spherical excess of the polygon.
[edit] This is true whether or not the polygon is convex. All that matters is that it can be dissected into triangles.
You can determine the angles from a bit of vector math. Suppose you have three vertices A,B,C and are interested in the angle at B. We must therefore find two tangent vectors (their magnitudes are irrelevant) to the sphere from point B along the great circle segments (the polygon edges). Let's work it out for BA. The great circle lies in the plane defined by OA and OB, where O is the center of the sphere, so it should be perpendicular to the normal vector OA x OB. It should also be perpendicular to OB since it's tangent there. Such a vector is therefore given by OB x (OA x OB). You can use the right-hand rule to verify that this is in the appropriate direction. Note also that this simplifies to OA * (OB.OB) - OB * (OB.OA) = OA * |OB| - OB * (OB.OA).
You can then use the good ol' dot product to find the angle between sides: BA'.BC' = |BA'|*|BC'|*cos(B), where BA' and BC' are the tangent vectors from B along sides to A and C.
[edited to be clear that these are tangent vectors, not literal between the points]

Here is a Python 3 implementation, loosely inspired by the above answers:
def polygon_area(lats, lons, algorithm = 0, radius = 6378137):
"""
Computes area of spherical polygon, assuming spherical Earth.
Returns result in ratio of the sphere's area if the radius is specified.
Otherwise, in the units of provided radius.
lats and lons are in degrees.
"""
from numpy import arctan2, cos, sin, sqrt, pi, power, append, diff, deg2rad
lats = np.deg2rad(lats)
lons = np.deg2rad(lons)
# Line integral based on Green's Theorem, assumes spherical Earth
#close polygon
if lats[0]!=lats[-1]:
lats = append(lats, lats[0])
lons = append(lons, lons[0])
#colatitudes relative to (0,0)
a = sin(lats/2)**2 + cos(lats)* sin(lons/2)**2
colat = 2*arctan2( sqrt(a), sqrt(1-a) )
#azimuths relative to (0,0)
az = arctan2(cos(lats) * sin(lons), sin(lats)) % (2*pi)
# Calculate diffs
# daz = diff(az) % (2*pi)
daz = diff(az)
daz = (daz + pi) % (2 * pi) - pi
deltas=diff(colat)/2
colat=colat[0:-1]+deltas
# Perform integral
integrands = (1-cos(colat)) * daz
# Integrate
area = abs(sum(integrands))/(4*pi)
area = min(area,1-area)
if radius is not None: #return in units of radius
return area * 4*pi*radius**2
else: #return in ratio of sphere total area
return area
Please find a somewhat more explicit version (and with many more references and TODOs...) here.

You could also have a look at this code of the spherical_geometry package: Here and here. It does provide two different methods for calculating the area of a spherical polygon.

Theory behind Wolfenstein-style 3D rendering

I'm currently working on a project about 3D rendering, and I'm trying to make simplistic program that can display a simple 3D room (static shading, no player movement, only rotation) with pygame
So far I've worked through the theory:
Start with a list of coordinates for the X and Z of each "Node"
Nodes are kept in an order which forms a closed loop, so that a pair of nodes will form either side of a wall
The height of the wall is determined when it is rendered, being relative to distance from the camera
Walls are rendered using painter's algorithm, so closer objects are drawn on top of further ones
For shading "fake contrast", which brightens/darkens walls based on the gradient between it's two nodes
While it seems simple enough, the process behind translating the 3D coordinates into 2D points on the screen is proving the difficult for me to understand.
Googling this topic has so far only yeilded these equations:
screenX = (worldX/worldZ)
screenY = (worldY/worldZ)
Which seem flawed to me, as you would get a divide by zero error if any Z coordinate is 0.
So if anyone could help explain this, I'd be really greatful.

Well the
screenX = (worldX/worldZ)
screenY = (worldY/worldZ)
is not the whole stuff that is just the perspective division by z and it is not meant for DOOM or Wolfenstein techniques.
Well in Doom there is only single angle of viewing (you can turn left/right but cannot look up/down only duck or jump which is not the same). So we need to know our player position and direction px,py,pz,pangle. The z is needed only if you want to implement also z axis movement/looking...
If you are looking in a straight line (Red) all the object that cross that line in the 3D are projected to single x coordinate in the player screen...
So if we are looking at some direction (red) any object/point crossing/touching this red line will be place at the center of screen (in x axis). What is left from it will be rendered on the left and similarly whats on right will be rendered on the right too...
With perspective we need to define how large viewing angle we got...
This limits our view so any point touches the green line will be projected on the edge of view (in x axis). From this we can compute screen x coordinate sx of any point (x,y,z) directly:
// angle of point relative to player direction
sx = point_ang - pangle;
if (sx<-M_PI) sx+=2.0*M_PI;
if (sx>+M_PI) sx-=2.0*M_PI;
// scale to pixels
sx = screen_size_x/2 + sx*screen_size_x/FOVx
where screen_size_x is resolution of our view area and point ang is angle of point x,y,z relative to origin px,py,pz. You can compute it like this:
point_ang = atan2(y-py,x-px)
but if you truly do a DOOM ray-casting then you already got this angle.
Now we need to compute the screen y coordinate sy which is dependent on the distance from player and wall size. We can exploit triangle similarity.
so:
sy = screen_size_y/2 (+/-) wall_height*focal_length/distance
Where focal length is the distance at which wall with 100% height will cover exactly the whole screen in y axis. As you can see we dividing by distance which might be zero. Such state must be avoided so you need to make sure your rays will be evaluated at the next cell if standing directly on cell boundary. Also we need to select the focal length so square wall will be projected as square.
Here a piece of code from mine Doom engine (putted all together):
double divide(double x,double y)
{
if ((y>=-1e-30)&&(y<=+1e-30)) return 0.0;
return x/y;
}
bool Doom3D::cell2screen(int &sx,int &sy,double x,double y,double z)
{
double a,l;
// x,y relative to player
x-=plrx;
y-=plry;
// convert z from [cell] to units
z*=_Doom3D_cell_size;
// angle -> sx
a=atan2(y,x)-plra;
if (a<-pi) a+=pi2;
if (a>+pi) a-=pi2;
sx=double(sxs2)*(1.0+(2.0*a/view_ang));
// perpendicular distance -> sy
l=sqrt((x*x)+(y*y))*cos(a);
sy=sys2+divide((double(plrz+_Doom3D_cell_size)-z-z)*wall,l);
// in front of player?
return (fabs(a)<=0.5*pi);
}
where:
_Doom3D_cell_size=100; // [units] cell cube size
view_ang=60.0*deg; // FOVx
focus=0.25; // [cells] view focal length (uncorrected)
wall=double(sxs)*(1.25+(0.288*a)+(2.04*a*a))*focus/double(_Doom3D_cell_size); // [px] projected wall size ratio size = height*wall/distance
sxs,sys = screen resolution
sxs2,sys2 = screen half resolution
pi=M_PI, pi2=2.0*M_PI
Do not forget to use perpendicular distances (multiplied by cos(a) as I did) otherwise serious fish-eye effect will occur. For more info see:
Ray Casting with different height size

Draw ellipses around points

I'm trying to draw ellipses around points of a group on a graph, with matplotlib. I would like to obtain something like this:
A dataset for a group (the red one for example) could look like this:
[[-23.88315146 -3.26328266] # first point
[-25.94906669 -1.47440904] # second point
[-26.52423229 -4.84947907]] # third point
I can easily draw the points on a graph, but I encounter problems to draw the ellipses.
The ellipses have diameters of 2 * standard deviation, and its center has the coordinates (x_mean, y_mean). The width of one ellipse equals the x standard deviation * 2. Its height equals the y standard deviation * 2.
However, I don't know how to calculate the angle of the ellipses (you can see on the picture the ellipses are not perfectly vertical).
Do you have an idea about how to do that ?
Note:
This question is a simplification of LDA problem (Linear Discriminant Analysis). I'm trying to simplify the problem to its most basic expression.

This is a well-studied problem. First take the convex hull of the set of points
you wish to enclose. Then perform computations as described in the literature.
I provide two sources below.
"Smallest Enclosing Ellipses--An Exact and Generic Implementation in C++" (abstract link).
Charles F. Van Loan. "Using the Ellipse to Fit and Enclose Data Points."
(PDF download).

This has a lot more to do with mathematics than programming ;)
Since you already have the dimensions and only want to find the angle, here is what I would do (based on my instinct):
Try to find the line that best fits the given set of points (trendline), this is also called Linear Regression. There are several methods to do this but the Least Squares method is a relatively easy one (see below).
Once you found the best fitting line, you could use the slope as your angle.
Least Squares Linear Regression
The least squares linear regression method is used to find the slope of the trendline, exactly what we want.
Here is a video explaining how it works
Let's assume you have a data set: data = [(x1, y1), (x2, y2), ...]
Using the least square method, your slope would be:
# I see in your example that you already have x_mean and y_mean
# No need to calculate them again, skip the two following lines
# and use your values in the rest of the example
avg_x = sum(element[0] for element in data)/len(data)
avg_y = sum(element[1] for element in data)/len(data)
x_diff = [element[0] - avg_x for element in data]
y_diff = [element[1] - avg_y for element in data]
x_diff_squared = [element**2 for element in x_diff]
slope = sum(x * y for x,y in zip(x_diff, y_diff)) / sum(x_diff_squared)
Once you have that, you are almost done. The slope is equal to the tangent of the angle slope = tan(angle)
Use python's math module angle = math.atan(slope) this will return the angle in radians. If you want it in degrees you have to convert it using math.degrees(angle)
Combine this with the dimensions and position you already have and you got yourself an ellipse ;)
This is how I would solve this particular problem, but there are probably a thousand different methods that would have worked too
and may eventually be better (and more complex) than what I propose.

I wrote a simple function to implement Mathieu David's solution. I'm sure there are many ways to do this, but this worked for my application.
def get_ellipse_params(self, points):
''' Calculate the parameters needed to graph an ellipse around a cluster of points in 2D.
Calculate the height, width and angle of an ellipse to enclose the points in a cluster.
Calculate the width by finding the maximum distance between the x-coordinates of points
in the cluster, and the height by finding the maximum distance between the y-coordinates
in the cluster. Multiple both by a scale factor to give padding around the points when
constructing the ellipse. Calculate the angle by taking the inverse tangent of the
gradient of the regression line. Note that tangent solutions repeat every 180 degrees,
and so to ensure the correct solution has been found for plotting, add a correction
factor of +/- 90 degrees if the magnitude of the angle exceeds 45 degrees.
Args:
points (ndarray): The points in a cluster to enclose with an ellipse, containing n
ndarray elements representing each point, each with d elements
representing the coordinates for the point.
Returns:
width (float): The width of the ellipse.
height (float): The height of the ellipse.
angle (float): The angle of the ellipse in degrees.
'''
if points.ndim == 1:
width, height, angle = 0.1, 0.1, 0
return width, height, angle
else:
SCALE = 2.5
width = np.amax(points[:,0]) - np.amin(points[:,0])
height = np.amax(points[:,1]) - np.amin(points[:,1])
# Calculate angle
x_reg, y_reg = [[p[0]] for p in points], [[p[1]] for p in points]
grad = LinearRegression().fit(x_reg, y_reg).coef_[0][0]
angle = np.degrees(np.arctan(grad))
# Account for multiple solutions of arctan
if angle < -45: angle += 90
elif angle > 45: angle -= 90
return width*SCALE, height*SCALE, angle

Matplotlib Axes3d, line intersect with 2d object

I am trying to do this but in 3d and using a 2d circle instead of a box.
I have a line starting between the two points [ (0,0,0), (3,4,5) ] and I want to see if it intersects through
circle = Circle((2, 1), 0.5)
ax.add_patch(circle)
art3d.pathpatch_2d_to_3d(circle, z=1, zdir="x")
Is it possible to test for a path intersect on a 2d object plotted on 3d axis? From the linked example above, I want to do path.intersects_circle where I define a circle as:
I have had a look through the Bbox documentation and it seems that I can't use this method for a circle?

This sounds more like an algebraic problem than related to matplotlib.
This is how I understand your question:
you have a circle at (x=2,y=1) with a radius of r=0.5
this circle is located in a plane at a constant z=1
1.) You need to determine where your vector pierces the plane which is parallel to the x,y-plane and at z=1. For the vector you specify in your question this intersection is at:
x = 3./(2.**0.5)
y = 4./(2.**0.5)
z = 1.
2.) You need to determine if this intersection falls into the part of the plane covered by the circle. The maximum y-coordinate your circle reaches is 1.5 - the y-coordinate of the intersection is already larger. Hence your straight line does not pierce the circle.
All this being said, I would recommend implementing an algebraic check based on the intersection with the plane and determining if this intersection is part of the circle. And only then using matplotlib.

Radius of a cube

This may be a silly question, but I basically need to find the radius of a cube in MAya.
Or another explanation would be I need to find the absolute center point, to the axis of choice example Y, so it would find out the radius of the Y direction if specified, or the distance from the center point to the furthest X direction, this would be the "radius".
Would love the help!

By analogy in 2 dimensions - or think of this as the center slice of the sphere in the cube.