projection

Perspective Projection: determine the 2D screen coordinates (x,y) of points in 3D space (x,y,z)

自古美人都是妖i 提交于 2019-12-20 09:49:08
问题 I wish to determine the 2D screen coordinates (x,y) of points in 3D space (x,y,z). The points I wish to project are real-world points represented by GPS coordinates and elevation above sea level. For example: Point (Lat:49.291882, Long:-123.131676, Height: 14m) The camera position and height can also be determined as a x,y,z point. I also have the heading of the camera (compass degrees), its degree of tilt (above/below horizon) and the roll (around the z axis). I have no experience of 3D

projecting Tango 3D point to screen Google Project Tango

丶灬走出姿态 提交于 2019-12-20 06:31:29
问题 Ptoject Tango provides a point cloud, how can you get the position in pixels of a 3D point in the point cloud in meters? I tried using the projection matrix but I get very small values (0.5,1.3 etc) instead of say 1234,324 (in pixels). I include the code I have tried //Get the current rotation matrix Matrix4 projMatrix = mRenderer.getCurrentCamera().getProjectionMatrix(); //Get all the points in the pointcloud and store them as 3D points FloatBuffer pointsBuffer = mPointCloudManager

D3 albersUsa projection funtion return null

会有一股神秘感。 提交于 2019-12-20 06:13:50
问题 I'm pretty new to D3 and I'm trying to set point on a map. I'm confused as I created a projection with this code: var projection = d3.geo .albersUsa() .scale(500) .translate([el.clientWidth / 2, el.clientHeight / 2]); I use this projection to draw a map and it works fine. But then whenever I call projection([10, 20]) it returns null whichever values I'm passing in. What is my error? 回答1: From the documentation # projection(location) […] May return null if the specified location has no defined

Orthographic projection Python

不羁岁月 提交于 2019-12-19 11:54:08
问题 I use orthographic projection to plot maps. I use this programm: from mpl_toolkits.basemap import Basemap import numpy as np import matplotlib.pyplot as plt import os, sys from sys import argv import pylab from mpl_toolkits.basemap import Basemap, shiftgrid from matplotlib import mpl from matplotlib import rcParams import matplotlib.pyplot as plt import matplotlib.mlab as mlab import matplotlib.patches as patches import matplotlib.path as path import matplotlib.dates as dt from numpy import

What is “linear projection” in convolutional neural network

独自空忆成欢 提交于 2019-12-19 09:04:06
问题 I am reading through Residual learning, and I have a question. What is "linear projection" mentioned in 3.2? Looks pretty simple once got this but could not get the idea... I am basically not a computer science person, so I would very appreciate if someone provide me a simple example. 回答1: First up, it's important to understand what x , y and F are and why they need any projection at all. I'll try explain in simple terms, but basic understanding of ConvNets is required. x is an input data

Android augmented reality app convert List location from Point of Screen

不问归期 提交于 2019-12-19 08:27:33
问题 I have a problem and I can't find a solution. I have made an Android augmented reality app. In my app I have a list of points of interest, every point of interest as a location(longitude, latitude, altitude). I know my position and I have the azimuth (of camera). Now my question is: how can I draw those points (gps points) on the screen(x, y points)? 回答1: i couldn't got what you are exactly trying do draw,but i assume you are trying to draw those poi on camera screen something like this: .

square to trapezoid

强颜欢笑 提交于 2019-12-18 21:20:35
问题 I know that transforming a square into a trapezoid is a linear transformation, and can be done using the projective matrix, but I'm having a little trouble figuring out how to construct the matrix. Using the projective matrix to translate, scale, rotates, and shear is straightforward. Is there a simple projective matrix which will transform a square to a trapezoid? 回答1: a,b,c,d are the four corners of your 2D square. a,b,c,d are expressed in homogeneous coordinate and so they are 3x1 matrices

Projecting into KeyValuePair via EF / Linq

半腔热情 提交于 2019-12-18 18:49:14
问题 I'm trying to load a list of KeyValuePairs from an EF / Linq query like this: return (from o in context.myTable select new KeyValuePair<int, string>(o.columnA, o.columnB)).ToList(); My problem is that this results in the error "Only parameterless constructors and initializers are supported in LINQ to Entities." Is there an easy way around this? I know I could create a custom class for this instead of using KeyValuePair but that does seem like re-inventing the wheel. 回答1: Select only columnA

Projecting into KeyValuePair via EF / Linq

十年热恋 提交于 2019-12-18 18:49:00
问题 I'm trying to load a list of KeyValuePairs from an EF / Linq query like this: return (from o in context.myTable select new KeyValuePair<int, string>(o.columnA, o.columnB)).ToList(); My problem is that this results in the error "Only parameterless constructors and initializers are supported in LINQ to Entities." Is there an easy way around this? I know I could create a custom class for this instead of using KeyValuePair but that does seem like re-inventing the wheel. 回答1: Select only columnA

Formula for a orthogonal projection matrix?

青春壹個敷衍的年華 提交于 2019-12-18 12:31:29
问题 I've been looking around a bit and can't seem to find just what I"m looking for. I've found "canonical formulas," but what's the best way to use these? Do I have to scale every single vertex down? Or is there a better way? A formula would really help me out, but I'm also looking for an explanation about the near and far z planes relative the viewer's position 回答1: Here is a reasonable source that derives an orthogonal project matrix: Consider a few points: First, in eye space, your camera is