trigonometry

问题 When comparing two vectors it is simple to calculate the angle between them, but in R it is noticeably harder to calculate the angle between a vector and a matrix of vectors efficiently. Say you have a 2D vector A=(2, 0) and then a matrix B={(1,3), (-2,4), (-3,-3), (1,-4)}. I am interested in working out the smallest angle between A and the vectors in B. If I try to use min(acos( sum(a%*%b) / ( sqrt(sum(a %*% a)) * sqrt(sum(b %*% b)) ) )) it fails as they are non-conformable arguments. Is

问题 When comparing two vectors it is simple to calculate the angle between them, but in R it is noticeably harder to calculate the angle between a vector and a matrix of vectors efficiently. Say you have a 2D vector A=(2, 0) and then a matrix B={(1,3), (-2,4), (-3,-3), (1,-4)}. I am interested in working out the smallest angle between A and the vectors in B. If I try to use min(acos( sum(a%*%b) / ( sqrt(sum(a %*% a)) * sqrt(sum(b %*% b)) ) )) it fails as they are non-conformable arguments. Is

问题 With the algorithm below, when the projection plane is tangent to the equator (the center line of the equirectangular image), projected image looks rectilinear. But when the projection plane is tilted, (py0 != panorama.height/2), lines are warped. The two last "lines" in the algorithm below needs to be "rectified", in order to adjust px and/or py when the center line of the destination plane is not at the same level than the center line of the equirectangular image. // u,v,w : // Normalized

问题 I'm working with two unit vectors but not sure how to calculate this. I need it so that if they point in the same direction the answer is 1, opposite directions the answer is 0, perpendicular (either up or down) the answer is 0.5, etc. Examples: For two vectors (1,0) and (-1,0) (so, opposite vectors), the answer I get is 0. For two vectors (1,0) and (1/sqrt(2),1/sqrt(2)) (so, the unit vector pointing at a 45 degree angle) I get 0.25. For two vectors (0,1) and (-1,0) (so, perpendicular vectors

问题 I'm working in a 3D environment (a 100x100x100 virtual cube) where each point is defined by its x, y and z position. We have two points: A with position x:20 y:35 z:40 and B x:50 y:40 z:85. Using this algo (https://gis.stackexchange.com/questions/108547/how-to-calculate-distance-azimuth-and-dip-from-two-xyz-coordinates) I'm able to find the distance, the azimuth and the elevation of B seen from A. let ax = 20; let ay = 35; let az = 40; let bx = 50; let by = 40; let bz = 85; let dx = bx-ax;

问题 I want to write a c++ program that will calculate collision between sphere and plane. The rule is that the angle of the falling object equals to angle of reflection. What do I have for sphere: //sphere coordinates and radius float x; float y; float z; float r; //sphere velocity vector projections float vx; float vy; float vz; Plane is described by plane equation coefficients: float A; float B; float C; float D; With sphere-plane collision detection I have no problem. But how to find velocity

问题 i have a func to compare angles.. if the angles si between the range -5/+5deg from the starting heading this should return Color .green otherwise should return .orange. i simplify the project so it can be easy to use in swift Playground. here my fun: func convertToPositiveAngle(_ angle : Angle) -> Angle { var toRet = Angle(degrees: fmod(angle.degrees, 360)) if toRet.degrees < 0 { toRet.degrees += 360.0; } return toRet } let headingStart = 3.0 let currentHeading = 3.0 func centro(startHeading

问题 This has been asked before however I've found no satisfactory answers so I'm going to try and makes sure this is on topic and generates some good responses. This question should not be on maths.stackexchange because it's about the Javascript code required to complete the function I started below. Imagine the following triangle. Each point A, B and C have x and y coordinates and we know what these coordinates are for Ax, Ay, Cx and Cy. We also know the length of a, b and c and the angle of A,

问题 I use maxima's tex output and would like to change the way it outputs sin(x). It currently does: tex(sin(x)); $$\sin x$$ But I would like to have brackets around the x, but not always; for instance, if i use this: sin(x/2) already has \left( and \right) tex(sin(x/2)); \begin{equation} \sin \left({{x}\over{2}}\right) \end{equation} is this possible, maybe with the texput function? 回答1: Well, if you are not adverse to putting in a little Lisp code: (%i1) :lisp (setf (get '%sin 'tex) nil) (%i1)

问题 Hello I want to ratate image with only using trigonometric functions. Assume "q" as a "theta" y'=x*sin(q) + y(-tan(q/2))*sin(q)+y x'=x+2 y (-tan(q/2)) + x sin(q) (-tan^2(q/2)) + y x' and y' is going to be our new x and y values. I am quite new in image proccesiing but i wrote a code which i think is kind of right. In this example "theta" is considered "30 degrees" but i will develop the code and it will work for any other degrees... So basically, I want from u to find my misake or if the code