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While checking Google's Universal sentence encoder paper, I found that they mention that using a similarity based on angular distance performs better than raw cosine similarity. More specifically, they used the following formula for the similarity:
The distance between those two lines is known as the angular distance. The distance between the two points is called the angular distance Such a distance is expressed in degrees.
$\begingroup$ According to ["Metrics for 3D rotations: Comparison and Analysis"][1] which compares a number of metrics, you might want to use the geodesic distance of the normed quaternions, since these lie on the unit 3-sphere (the euclidean distance is not appropriate there). It so happens that it is exactly the formula for $\theta$ that ...
The goal is to find angular distance between any two points. My only thought so far is to draw a line between a central point and all the lines from the central point to all the others and see where these lines intersect the circle.
In this context, "linear distance" appears to mean "arc length"[*], while "angular distance" is presumably $\Delta\theta$, the angular difference measured at the center of the circle.
$\begingroup$ The cosine distance is not even a distance for vectors of uni length because it violates the triangle inequality. Also, the question was not about the cosine distance but about the angular distance d_a(x,y) = \theta(x,y) which would also violate the identity of indiscernibles, but the triangle inequality should hold. $\endgroup$
This form makes it fairly transparent how azimuthal symmetry allows you to automatically eliminate some of the angular dependencies in certain integration problems. Another advantage of this form is that you now have at least two variables, namely $\phi$ and $\phi'$ , that appear in the equation only once, which can make finding series ...
The angular displacement of Earth is {eq}0.5 radians {/eq}. Example Problem 2 - Rotating Disk Through what angular displacement must a disk of radius 30 cm rotate so that a point on the outer edge ...
The angular distance looks to be calculated by the dot product of the vectors. It has been about $10$ years since linear algebra so I am very rusty on this but I think I can get a handle on this one hopefully.
Angular Distance Between Quaternions. Ask Question Asked 5 years, 2 months ago. Modified 3 years, 2 months ...