3d

问题 Imagine a client asked you to design a piece of software that would satisfy some pretty rough specs as follows: 1: it will be marketed towards the fashion industry 2: the users will be people that "design clothes and stuff" (there is likely a specific term for it but it doesn't come to mind) 3: For various reasons, it would be useful to be able to quickly prototype designs and see how they look on models. To make it more useful, it should be rendering 3D models so that the designers can look

问题 Imagine a client asked you to design a piece of software that would satisfy some pretty rough specs as follows: 1: it will be marketed towards the fashion industry 2: the users will be people that "design clothes and stuff" (there is likely a specific term for it but it doesn't come to mind) 3: For various reasons, it would be useful to be able to quickly prototype designs and see how they look on models. To make it more useful, it should be rendering 3D models so that the designers can look

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 It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. Closed 7 years ago . I'm trying to plot 3D points onto a 2D surface (I'm currently working with SDL.NET as my 'game engine'). The surface will be 800x400 pixels in size and the 3d coordinates will range from -4000 to 4000. My view

问题 I am using SceneForm Android SDK to render 3D Model in Android APP. To display the 3D Model I am using this code below which also supports Rotating and Zooming feature in 3D Model private fun renderLocalObject(position: Int) { ModelRenderable.builder() .setSource(appCompatActivity, Uri.parse(models.get(position))) .setRegistryId(models.get(position)) .build() .thenAccept { modelRenderable: ModelRenderable -> addNodeToScene(modelRenderable) } .exceptionally { throwable: Throwable? -> var

问题 I am using the R programming language. I am following a tutorial on data visualization over here: https://plotly.com/r/3d-surface-plots/ I created my own data and made a 3D plot: library(plotly) set.seed(123) #generate data a = rnorm(100,10,10) b = rnorm(100,5,5) c = rnorm(100,5,10) d = data.frame(a,b,c) #3d plot fig <- plot_ly(z = ~as.matrix(d)) fig <- fig %>% add_surface() #view plot fig As seen here, there is a point on this 3D plot where "y = 97". I am not sure how this is possible,

问题 I am trying to create a cylinder in 3D space. I get a start and endpoint in 3D and to place the vertices, I need to create 2 circles for the bottom and the top of the cylinder. I thought about making a vector u with coordinates (1,0,0) and calculate a direction vector dir = end - start. Now I make the cross product to get a vector w = u x dir: this way I am 90 degrees away from the dir vector with w. The problem is, how do I calculate the next position w' with an arbitrary angle between w and

问题 I tried to understand the mathematics behind the projection matrix and I found this page. The matrix from this page: I found this matrix is similar to the matrix of Xna. I understood how they got to m33 and m44 columns of the matrix, but how they got to m11 and m22? And I didn't understand why I have to give the aspect ratio if I already give the field of view angle. The last thing I want to know is: what is the equation for transfaring from ndc space to window space and how they get to this

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 wanted to ask you if anyone knew how to check if a point was inside a given triangle, on a reference system in space. I am aware that speaking of 2d systems I can obtain this point with the following procedures: To determine if a given point v lies within a given triangle, consider a single vertex, denoted v0, v1 and v2 are the vectors from the other two vertices v0. Expressing the vector from v0 to v in terms of v1 and v2 then gives v = v0 + av1 + bv2 where a, b are constant. Solve for a,