I'm working on a program on hololens2 research mode on unity. Hololens give us a depth image that is distance from depth sensor to object in front, for every pixel.
What I do is for every pixel I project pixel to image plane, then backproject it according to depth distance captured by depth sensor and it gives me the xyz in depth sensor coordinate frame. now it is needed to transform this coordinate to global coordinate system. to do so I get camera coordinate from unity by cam_pose = Camera.main.transform and in the other hand saved depth sensor extrinsic matrix.
From these two matrices I create a depth_to_world = cam_pose # inv(extrinsic). Now for every xyz on depth I perform global_xyz = depth_to_world # xyz to get point in real world. Problem is it return a point with 10-15 cm error. What am I doing wrong? (code is in python)
x = self.us[Depth_i, Depth_j] # projection from pixels to image plane
y = self.vs[Depth_i, Depth_j] # projection from pixels to image plane
D = distance_img[Depth_i, Depth_j] #distance_img is depth image
distance = 1000*float(D) / np.sqrt(x * x + y * y + 1) #distance according to spherical image plane D is in millimeter
depth_to_world = cam_pose # np.linalg.inv(Constants.camera_extrinsic)
X = (np.array([x * distance, y * distance, 1.0 * distance, 1])).reshape(4, 1)
point = (depth_to_world # X )[0:3, 0]
I got it! according to (https://github.com/petergu684/HoloLens2-ResearchMode-Unity) first I passed unity world origin to a winrt plugin, and depth_to_world was depth_to_world = inv(extrinsic) * cam_pose witch cam_pose is given by TryLocateAtTimeStamp. And other point is that unity coordinate is left handed (surprisingly!) so we should multiply a (-1) to z. (z <- -z)
my depth_to_world transformation was near but not correct.
We have these logarithmic spirals which are circling around the centre of the coordinate system:
x = ebθ cos(θ)
y = ebθ sin(θ)
where the ebθ is the distance between the point (which is on the spiral) and the centre; and the θ is the angle between the line connecting the point and the origin and the axis x.
Consider a spiral where the angle is θ ϵ <0,10π> and the parameter is b=0.1. By thickening points on the spirals (and the angle θ) calculate the circumference with the relative precision better than 1%. Draw the spiral!
I'm preparing for a (MATLAB) test and I'm stuck with this exercise. Please help, any hint is appreciated.
Start by computing a list of x,y for your range of theta and value of b. For more accurate results, have your theta increment in smaller steps (I chose 5000 arbitrarily). Then, its simply computing the distance for each pair of consecutive points and summing them up.
t = linspace(0,10*pi,5000);
b = 0.1;
x = exp(b*t).*cos(t);
y = exp(b*t).*sin(t);
result = sum(sqrt((x(2:end) - x(1:end-1)).^2 + (y(2:end)-y(1:end-1)).^2))
I'm trying to estimate a position based on signal strength received from 4 Wi-Fi Access Points. I measure the signal strength from 4 access points located in each corner of a square room with 100 square meters (10x10). I recorded the signal strengths in a known position (x, y) = (9.5, 1.5) using an Android phone. Now I want to check how accurate can a multilateration method be under the circumstances.
Using MATLAB, I applied a formula to calculate distance using the signal strength. The following MATLAB function shows the application of the formula:
function [ d_vect ] = distance( RSS )
% Calculate distance from signal strength
result = (27.55 - (20 * log10(2400)) + abs(RSS)) / 20;
d_vect = power(10, result);
end
The input RSS is a vector with the four signal strengths measured in the test point (x,y) = (9.5, 1.5). The RSS vector looks like this:
RSS =
-57.6000
-60.4000
-44.7000
-54.4000
and the resultant vector with all the estimated distances to each access points looks like this:
d_vect =
7.5386
10.4061
1.7072
5.2154
Now I want to estimate my position based on these distances and the access points position in order to find the error between the estimated position and the known position (9.5, 1.5). I want to find the intersection area (In order to estimate a position) between four circles where each access point is the center of one of the circles and the distance is the radius of the circle.
I want to find the grey area as shown in this image :
http://www.biologycorner.com/resources/venn4.gif
If you want an alternative way of estimating the location without estimating the intersection of circles you can use trilateration. It is a common technique in navigation (e.g. GPS) to estimate a position given a set of distance measurements.
Also, if you wanted the area because you also need an estimate of the uncertainty of the position I would recommend solving the trilateration problem using least squares which will easily give you an estimate of the parameters involved and an error propagation to yield an uncertainty of the location.
I found an answear that solved perfectly the question. It is explained in detail in this link:
https://gis.stackexchange.com/questions/40660/trilateration-algorithm-for-n-amount-of-points
I also developed some MATLAB code for the problem. Here it goes:
Estimate distances from the Access Points:
function [ d_vect ] = distance( RSS )
result = (27.55 - (20 * log10(2400)) + abs(RSS)) / 20;
d_vect = power(10, result);
end
The trilateration function:
function [] = trilat( X, d, real1, real2 )
cla
circles(X(1), X(5), d(1), 'edgecolor', [0 0 0],'facecolor', 'none','linewidth',4); %AP1 - black
circles(X(2), X(6), d(2), 'edgecolor', [0 1 0],'facecolor', 'none','linewidth',4); %AP2 - green
circles(X(3), X(7), d(3), 'edgecolor', [0 1 1],'facecolor', 'none','linewidth',4); %AP3 - cyan
circles(X(4), X(8), d(4), 'edgecolor', [1 1 0],'facecolor', 'none','linewidth',4); %AP4 - yellow
axis([0 10 0 10])
hold on
tbl = table(X, d);
d = d.^2;
weights = d.^(-1);
weights = transpose(weights);
beta0 = [5, 5];
modelfun = #(b,X)(abs(b(1)-X(:,1)).^2+abs(b(2)-X(:,2)).^2).^(1/2);
mdl = fitnlm(tbl,modelfun,beta0, 'Weights', weights);
b = mdl.Coefficients{1:2,{'Estimate'}}
scatter(b(1), b(2), 70, [0 0 1], 'filled')
scatter(real1, real2, 70, [1 0 0], 'filled')
hold off
end
Where,
X: matrix with APs coordinates
d: distance estimation vector
real1: real position x
real2: real position y
If you have three sets of measurements with (x,y) coordinates of location and corresponding signal strength. such as:
m1 = (x1,y1,s1)
m2 = (x2,y2,s2)
m3 = (x3,y3,s3)
Then you can calculate distances between each of the point locations:
d12 = Sqrt((x1 - x2)^2 + (y1 - y2)^2)
d13 = Sqrt((x1 - x3)^2 + (y1 - y3)^2)
d23 = Sqrt((x2 - x3)^2 + (y2 - y3)^2)
Now consider that each signal strength measurement signifies an emitter for that signal, that comes from a location somewhere at a distance. That distance would be a radius from the location where the signal strength was measured, because one would not know at this point the direction from where the signal came from. Also, the weaker the signal... the larger the radius. In other words, the signal strength measurement would be inversely proportional to the radius. The smaller the signal strength the larger the radius, and vice versa. So, calculate the proportional, although not yet accurate, radius's of our three points:
r1 = 1/s1
r2 = 1/s2
r3 = 1/s3
So now, at each point pair, set apart by their distance we can calculate a constant (C) where the radius's from each location will just touch one another. For example, for the point pair 1 & 2:
Ca * r1 + Ca * r2 = d12
... solving for the constant Ca:
Ca = d12 / (r1 + r2)
... and we can do this for the other two pairs, as well.
Cb = d13 / (r1 + r3)
Cc = d23 / (r2 + r3)
All right... select the largest C constant, either Ca, Cb, or Cc. Then, use the parametric equation for a circle to find where the coordinates meet. I will explain.
The parametric equation for a circle is:
x = radius * Cos(theta)
y = radius * Sin(theta)
If Ca was the largest constant found, then you would compare points 1 & 2, such as:
Ca * r1 * Cos(theta1) == Ca * r2 * Cos(theta2) &&
Ca * r1 * Sin(theta1) == Ca * r2 * Sin(theta2)
... iterating theta1 and theta2 from 0 to 360 degrees, for both circles. You might write code like:
for theta1 in 0 ..< 360 {
for theta2 in 0 ..< 360 {
if( abs(Ca*r1*cos(theta1) - Ca*r2*cos(theta2)) < 0.01 && abs(Ca*r1*sin(theta1) - Ca*r2*sin(theta2)) < 0.01 ) {
print("point is: (", Ca*r1*cos(theta1), Ca*r1*sin(theta1),")")
}
}
}
Depending on what your tolerance was for a match, you wouldn't have to do too many iterations around the circumferences of each signal radius to determine an estimate for the location of the signal source.
So basically you need to intersect 4 circles. There can be many approaches to it, and there are two that will generate the exact intersection area.
First approach is to start with one circle, intersect it with the second circle, then intersect the resulting area with the third circle and so on. that is, on each step you know current intersection area, and you intersect it with a new circle. The intersection area will always be a region bounded by circle arcs, so to intersect it with a new circle you walk along the boundary of the area and check whether each bounding arc intersects with a new circle. If it does, then you leave only the part of the arc that lies inside a new circle, remember that you should continue with an arc from a new circle, and continue traversing the boundary until you find the next intersection.
Another approach that seems to result in a worse time complexity, but in your case of 4 circles this will not be important, is to find all the intersection points of two circles and choose only those points that are of interest for you, that is which lie inside all other circles. These points will be the corners of your area, and then it is rather easy to reconstruct the area. After googling a bit, I have even found a live demo of this approach.
I am in need of an idea! I want to model the vascular network on the eye in 3D. I have made statistics on the branching behaviour in relation to vessel diameter, length etc. What I am stuck at right now is the visualization:
The eye is approximated as a sphere E with center in origo C = [0, 0, 0] and a radius r.
What I want to achieve is that based on the following input parameters, it should be able to draw a segment on the surface/perimeter of E:
Input:
Cartesian position of previous segment ending: P_0 = [x_0, y_0, z_0]
Segment length: L
Segment diameter: d
Desired angle relative to the previous segment: a (1)
Output:
Cartesian position of resulting segment ending: P_1 = [x_1, y_1, z_1]
What I do now, is the following:
From P_0, generate a sphere with radius L, representing all the points we could possibly draw to with the correct length. This set is called pool.
Limit pool to only include points with a distance to C between r*0.95 and r, so only the points around the perimeter of the eye are included.
Select only the point that would generate a relative angle (2) closest to the desired angle a.
The problem is, that whatever angle a I desire, is actually not what is measured by the dot product. Say I want an angle at 0 (i.e. that the new segment is following the same direction as the previous`, what I actually get is an angle around 30 degrees because of the curvature of the sphere. I guess what I want is more the 2D angle when looking from an angle orthogonal from the sphere to the branching point. Please take a look at the screenshots below for a visualization.
Any ideas?
(1) The reason for this is, that the child node with the greatest diameter is usually follows the path of the previous segment, whereas smaller child nodes tend to angle differently.
(2) Calculated by acos(dot(v1/norm(v1), v2/norm(v2)))
Screenshots explaining the problem:
Yellow line: previous segment
Red line: "new" segment to one of the points (not neccesarily the correct one)
Blue x'es: Pool (text=angle in radians)
I will restate the problem with my own notation:
Given two points P and Q on the surface of a sphere centered at C with radius r, find a new point T such that the angle of the turn from PQ to QT is A and the length of QT is L.
Because the segments are small in relation to the sphere, we will use a locally-planar approximation of the sphere at the pivot point Q. (If this isn't an okay assumption, you need to be more explicit in your question.)
You can then compute T as follows.
// First compute an aligned orthonormal basis {U,V,W}.
// - {U,V} should be a basis for the plane tangent at Q.
// - W should be normal to the plane tangent at Q.
// - U should be in the direction PQ in the plane tangent at Q
W = normalize(Q - C)
U = normalize(Q - P)
U = normalize(U - W * dotprod(W, U))
V = normalize(crossprod(W, U))
// Next compute the next point S in the plane tangent at Q.
// In a regular plane, the parametric equation of a unit circle
// centered at the origin is:
// f(A) = (cos A, sin A) = (1,0) cos A + (0,1) sin A
// We just do the same thing, but with the {U,V} basis instead
// of the standard basis {(1,0),(0,1)}.
S = Q + L * (U cos A + V sin A)
// Finally project S onto the sphere, obtaining the segment QT.
T = C + r * normalize(S - C)
I want to plot a latitude and longitude using matlab. Using that latitude and longitude as center of the circle, I want to plot a circle of radius 5 Nm.
r = 5/60;
nseg = 100;
x = 25.01;
y = 55.01;
theta = 0 : (2 * pi / nseg) : (2 * pi);
pline_x = r * cos(theta) + x;
pline_y = r * sin(theta) + y;
hold all
geoshow(pline_x, pline_y)
geoshow(x, y)
The circle does not look of what I expected.
Drawing a circle on earth is more complex that it looks like.
Drawing a line or a poly line is simple, because the vertices are defined.
Not so on circle.
a circle is defined by all points having the same distance from center (in meters! not in degrees!!!)
Unfortuantley lat and lon coordinates have not the same scale.
(The distance between two degrees of latidtude is always approx. 111.3 km, while for longitude this is only true at the equator. At the poles the distance between two longitudes approach zero. In Europe the factor is about 0.6. (cos(48deg))
There are two solution, the first is more universal, usefull for nearly all problems.
convert spherical coordinate (of circle center) to cartesian plane with unit = 1m, using a transformation (e.g equidistant transformation, also called equirectangular transf., this transformation works with the cos(centerLat) compensation factor)
calculate points (e.g circle points) in x,y plane using school mathematics.
transform all (x,y) points back to spherical (lat, lon) coordinates, using the inverse transformation of point 1.
Other solution
1. write a function which draws an ellipse in defined rectangle (all cartesian x,y)
2. define bounding of the circle to draw:
2a: calculate north-south diameter of circle/ in degrees: this a bit tricky: the distance is define in meters, you need a transformation to get the latitudeSpan: one degrees of lat is approx 111.3 km (eart circumence / 360.0): With this meters_per_degree value calc the N-S disatcne in degrees.
2b: calculate E-W span in degrees: now more tricky: calculate like 2a, but now divide by cos(centerLatitude) to compensate that E-W distances need more degrees when moving north to have the same meters.
Now draw ellipseInRectangle using N-S and E_W span for heigh and width.
But a circle on a sphere looks on the projected monitor display (or paper) only like a circle in the center of the projection. This shows:
Tissot's Error Ellipse