I have a very basic question. I am analyzing a movie. In the movie I am fitting a curve to define an object and computing the first derivative of a curve at it's starting point. The first derivative is changing in each frame. I want to measire the first derivative in the from of angle given in degrees or radians. Is there a faster way to do it in MATLAB.
I know its a very simple question, but if someone can explain me the concept then that would be very helpful. Thanks
The derivative of the curve is defined as dy/dx the ratio between the change in X direction and Y direction. This ratio is also the tan of the angle between the curve and the X axis. Therefore, if you want the angle (in radians) all you need it compute atan of the derivative.
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My math is too rusty to figure this out. I want to derive the onscreen angle (the angle as seen on the 2d screen) of a 3d vector.
Given the x and y rotation of a vector (z rotation is zero and doesn't mstter), what does the angle on screen look like?
We know when y is zero and x is positive, the angle is 90. When y is zero and x is negative the angle is -90. When y is 90, for any value of x, the angle is 180. When y is -90, for any value of x, the angle is 0.
So what the formula here so I can derive the angle for the other values of x and y rotation?
The problem, as stated, doesn't make sense. If you're holding z to zero rotation, you've converted a 3D problem to 2D already. Also, it seems the angle you're measuring is from the y-axis which is fine but will change the ultimate formula. Normally, the angle is measured from the x-axis and trigometric functions will assume that. Finally, if using Cartesian coordinates, holding y constant will not keep the angle constant (and from the system you described for x, the angle would be in the range from -90 to 90 - but exclusive of the end points).
The arctangent function mentioned above assumes an angle measured from the x-axis.
Angle can be calculated using the inverse tangent of the y/x ratio. On unity3d coordinated system (left-handed) you can get the angle by,
angle = Mathf.Rad2Deg * Mathf.Atan(y/x);
Your question is what will a 3-d vector look like.
(edit after posted added perspective info)
If you are looking at it isometrically from the z-axis, it would not matter what the z value of the vector is.
(Assuming a starting point of 0,0,0)
1,1,2 looks the same as 1,1,3.
all x,y,z1 looks the same as any x,y,z2 for any values of z1 and z2
You could create the illusion that something is coming "out of the page" by drawing higher values of z bigger. It would not change the angle, but it would be a visual hint of the z value.
Lastly, you can use Dinal24's method. You would apply the same technique twice, once for x/y, and then again with the z.
This page may be helpful: http://www.mathopenref.com/trigprobslantangle.html
Rather than code this up yourself, try to find a library that already does it, like https://processing.org/reference/PVector.html
I'm trying to calculate the angle between 2 geographic (Latitude,Longitude) points in MATLAB. The points are:
(-65.226,125.5) and (-65.236,125.433).
I used the MATLAB function, azimuth, as:
azimuth(-65.226,125.5,-65.236,125.433)
I convert the result to radians, and plotting this using quiver, I get the following plot:
I want the red vector to point from the top right dot to the bottom left dot.
The points are at fairly high latitude (~65S), and the separation of the points is low (about 0.1 degrees). Thus, I can't really understand how the curvature of the earth could affect the azimuth prediction that much..
Does anyone have any experience with azimuth in MATLAB, or have a better suggestion to calculating the angle between the coordinate pairs?
Thanks!
Here you can detailed information and formulae on how to find angle between two latitude-longitude points.
As shown in image, there is a binary polygonal image. I want to find the principal direction in the image with respect to X-axis. I have shown the principal direction and X-axis with blue line. This can be done using PCA but my problem is such a small rectangle will have around 1000 pixels and I have to find Principal directions for around 100 polygons (polygon can be of arbitrary shape).
One approach that I have thought is:
Project that rectangle onto a line which is oriented at degrees at an interval (say) 5 degrees. The projection which has the maximum variance is the desired projection axis, and thus that is the desired angle. But this also falls under a greedy approach and thus will take time. Is there a smarter approach?
Also, if anybody could explain the exact procedure to do this using PCA, it would be helpful. I know the steps:
1. Take the covariance matrix.
2. Get the top eigenvector corresponding to largest eigenvalue -> that will be the principal direction.
But I am confused in the following statement which I often read everywhere:
A column vector: [0.5 0.5] is the first principal component and it gives the direction of the maximum variance. So can do I exactly calculate the angle by which I should rotate the data so that it will become parallel to X-axis.
Compute the eigenvector associated with the highest eigen value. Call that v. Normalize v. v = v/norm(v);
Compute angle between that and the horizontal direction: angle=acos(sum(v.*[1,0]))
Rotate by -angle, transformation matrix T = [cos(-angle) -sin(-angle); sin(-angle) cos(-angle)], multiply all points by that matrix. Do that for all polygons.
I am trying to implement the generalized Hough transform in matlab. The algorithm requires the gradient direction at each point in the shape. How can I measure phi as shown in the figure below?
The normal to the curve [x(t), y(t)] is [-dy(t)/dt, dx(t)/dt]. Thus, with x being the x-coordinates and y the y-coordinates, the normals are
[-diff(y(:)),diff(x(:))]
and the angle phi is
atan2(diff(x(:)),-diff(y(:)))
I have been trying to find the difference between the 2 but to no luck minus this
The primary diff erence between
the two representations is that a quaternion’s axis of rotation is scaled
by the sine of the half angle of rotation, and instead of storing the angle in the
fourth component of the vector, we store the cosine of the half angle.
I have no idea what
sine of the half angle of rotation
or
cosine of the half angle
means?
Quaternios and Axis-angle are both 4D representations of 3D rotations/orientations and both have pro's and cons.
Axis-angle: represents the rotation by its angle a and the rotation axis n. For example, a rotation of 180 degrees around the Y-Axis would be represented as a = 180, n= {0,1,0}. The representation is very intuitive, but for actually applying the rotation, another representation is required, such as a quaternion or rotation matrix.
Quaternion: represents a rotation by a 4D vector. Requires more math and is less intuitive, but is a much more powerful representation. Quaternions are easily interpolated (blending) and it is easy to apply them on 3D point. These formula's can easily be found on the web. Given a rotation of a radians about a normalized axis n, the quaternion 4D vector will be {cos a/2, (sin a/2) n_x, (sin a/2) n_y, (sin a/2) n_z}. That's where the sine and cosine of the half angle come from.
It means that if you, for example, want to make a 180deg rotation around the Z axis (0,0,1), then the quaternion's real part will be cos(180deg/2)=0, and its imaginary part will be sin(180deg/2)*(0,0,1)=(0,0,1). That's q=0+0i+0j+1k. 90-degree rotation will give you q=cos(90deg/2)+sin(90deg/2)*(0i+0j+1k)=sqrt(2)/2+0i+0j+sqrt(2)/2*k, and so on.
OTOH, if you're asking what sine and cosine are, check if your languange provides sin() and cos() functions (their arguments will probably be in radians, though), and check out http://en.wikipedia.org/wiki/Sine.