I'm trying to understand quaternions better, but I don't know how the math behind FromToRotation works. I tried looking this up but couldn't find any results.
You may know that a rotation can be represented by a quaternion of the following form:
cos (phi / 2)
sin (phi / 2) * axis.x
sin (phi / 2) * axis.y
sin (phi / 2) * axis.z
axis is the rotation axis and phi is the rotation angle. These are the two measures you need to define your quaternion.
There are multiple rotations that map a vector from to another vector to. The shortest rotation is the one where the axis is perpendicular to both of the vectors. Hence, the axis is:
axis = normalize(from x to)
x denotes the cross product.
And the angle is the angle between the two vectors:
phi = acos(dot(from, to) / (norm(from) * norm(to))
norm is the vector norm or vector length.
With these values, you can then calculate the quaternion.
I am building a small device that also uses magnetometer data in order to calculate the compass heading. The LSM9DS0 IMU sensor works great if the heading is calculated as a yaw (if the sensor is on a flat surface).
I have 3D printed a shell in which i am going to assemble all the electronics. My problem is that it is poorly designed and the IMU sensor is not on a flat surface, but it has to stay on 90 degrees. So by this, the Z axis is no more my way to calculate the yaw (or heading), but it changed to Y.
In order to calculate the heading on Z, i was using this formula:
heading.value = atan2((float)dof.my, (float)dof.mx);
if(heading.value < 0) heading.value += 2*PI;
if(heading.value > 2*PI) heading.value -= 2*PI;
heading.value *= 180/PI;
...where my is the magnetometer Y and mx the magnetometer X
Now, I don't know how to calculate the heading based on other axis.
I know this thread hasn't been active for a while, but during my search I came across this publication by NXP which explains the solution really nicely.
In brief:
Align the accelerometer readings (G) and the magnetometer readings (B) so they follow the NED coordinate system with the x-axis pointing forward and the y and z-axis to the right and down, respectively.
Calculate the roll and pitch
// Using atan2 to restrict +/- PI
const roll = Math.atan2(Gy, Gz)
// Using atan to restrict to +/- PI/2
const pitch = Math.atan(-Gx / (Gy * Math.sin(roll) + Gz * Math.cos(roll)))
Calculate the yaw / compass heading (here Vx, Vy, Vz correspond to the Hard-Iron effects that can be calculated separately as discussed in this publication):
// Using atan2 to restring to +/- PI
let yaw = Math.atan2( (Bz-Vz)*Math.sin(roll) - (By-Vy)*Math.cos(roll),
(Bx-Vx)*Math.cos(pitch) + (By-Vy)*Math.sin(pitch)*Math.sin(roll) + (Bz-Vz)*Math.sin(pitch)*Math.cos(roll))
Correct the heading to [0, 2*PI)
if( yaw < 0 ) {
yaw += 2*Math.PI
}
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.
Imagine a dome with its centre in the +z direction. What I want to do is to move that dome's centre to a different axis (e.g. 20 degrees x axis, 20 degrees y axis, 20 degrees z axis). How can I do that ? Any hint/tip helps.
Add more info:
I've been dabbling with rotation matrices in wiki for a while. The problem is, it is not a commutative operation. RxRyRz is not same as RzRyRx. So based on the way I multiple it I get a different final results. For example, I want my final projection to have 20 degrees from the original X axis, 20 degrees from original Y axis and 20 degrees from original Z axis. Based on the matrix, giving alpha, beta, gamma values 20 (or its corresponding radian) does NOT result the intended rotation. Am I missing something? Is there a matrix that I can just put the intended angles and get it at the end ?
Using a rotation matrix is an easy way to rotate a collection of (x,y,z) points. You can calculate a rotation matrix for your case using the equations in the general rotation section. Note that figuring out the angle values to plug into those equations can be tricky. Think of it as rotating about one axis at a time and remember that the order of your rotations (order of multiplications) does matter.
An alternative to the general rotation equations is to calculate a rotation matrix from axis and angle. It may be easier for you to define correct parameters with this method.
Update: After perusing Wikipedia, I found a simple way to calculate rotation axis and angle between two vectors. Just fill in your starting and ending vectors for a and b here:
a = [0.0 0.0 1.0];
b = [0.5 0.5 0.0];
vectorMag = #(x) sqrt(sum(x.^2));
rotAngle = acos(dot(a,b) / (vectorMag(a) * vectorMag(b)))
rotAxis = cross(a,b)
rotAxis =
-0.5 0.5 0
rotAngle =
1.5708