I try to setting camera position of pptk, with this method:
xyz = pptk.rand(100, 3)
v = pptk.viewer(xyz)
v.set(lookat=[0,0,0])
this working well, the camera position change.
But when I left-clik on it, to move the point cloud, the camera position return to the original position
Do you have an idea how to maintain the camera position
Thank you
Please send the look-at coordinates as a tuple (0,0,0) instead of a list [0,0,0].
v.set(lookat=(0,0,0))
I found this bug too. I'm currently working around it by first using the mouse wheel to move in/out on the point cloud before clicking; then the look at position doesn't change.
Related
I truly want to know. How do i get the mouse position to a part in roblox? Like, i want to get the mouse position, on screen, aligned with a part. Let's say the part position is 10,10,10 and i want my mouse to point at that exact position, how can i do that? (I just need to get the position, ex: 1093x899, 1789x305. Didn't understand? Here, i simply want to know the position, on which my mouse would say that there is the 'part' that i want my mouse to point at. Any help is nice!
Take a look at the docs for Camera:WorldToScreenPoint.
I've adapted the code sample from there:
local camera = workspace.CurrentCamera
-- get the position of the part you care about
local targetPart = game.Workspace.YOUR_PART_NAME
local worldPoint = targetPart.Position
-- get the screen coordinates
local vector, onScreen = camera:WorldToScreenPoint(worldPoint)
if onScreen then
local screenPoint = Vector2.new(vector.X, vector.Y)
print(screenPoint)
else
print("the part isn't on your screen right now")
end
You'll need to hook this up to a loop of some kind to repeated get the coordinates.
I am trying to get the color of a pixel on my screen using node.js. I want it to be returned in RGB format, e.g. (255, 0, 0). My current solution is to use screenshot-desktop to screenshot my entire screen in JPG format, decode it to get the raw pixel data, and get the color of a given pixel. However, this lags out my entire computer for 1-2 seconds as it is taking the screenshot. This is unusable as I would like to do this multiple times per second. So my question is: How can I get the color of a given pixel on the screen, without taking a full screenshot?
I am using Linux with X11. There is an X11 library for node.js, so I asssume I should use that to get the pixel color, I'm just not sure how. If you could show me how to do it in C then I can easily use node.js to do the same thing.
Thanks!
Oh my gosh I just figured it out after posting this. I was using robotjs for reading the mouse position and I totally forgot it can do screen stuff too! So, the solution would be to do
var robot = require('robotjs');
var color = robot.getPixelColor(x, y);
X11 solution using x11 node library ( I am the author ):
query windows tree with QueryTree starting at the root window
get every child geometry using GetGeometry request
if your point is not inside any child, use current window id and get 1x1 pixmap from the current image: GetImage(format, currentWindow, x, y, 1, 1, planeMask) ( 2 for format and 0xffffffff for plane mask should work ). Make sure you calculate relative x y position as you travers windows tree.
if child window covers your point query children for that window and repeat again. Note that QueryTree returns windows in bottom to top stacking order so make sure you pick last one covering your point
Once you have 1x1 pixmap from the topmost window under your point - the buffer should contain only color bytes for your image, RGB order and bit mask might depend on red_mask, green_mask, blue_mask from display.screen[0].depths[visual].
If you cache "topmost window" between requests and only start from root when no match anymore the above solution might be much more performant then the one using robotjs ( although much more low level and complicated ) Good luck!
I'm actually developing a game in Swift SpriteKit.
I set the position of a SKSpriteNode:
skPart1.position = CGPointMake(0, 100)
So the node should start at the very left window-edge.
|-------------------------|
|-------------------------|
|-------------------------|
|-------------------------|
|==========|--------------|
|==========|--------------|
|==========|--------------|
But in reality half of my SKSpriteNode is outside of the screen:
|-------------------------|
|-------------------------|
|-------------------------|
|-------------------------|
==========|--------------------|
==========|--------------------|
==========|--------------------|
I've read the same problem on stackoverflow, but the only solution provided there was, to set the scaleMode to AspectFit.
I've figured out, that it works with SKShapeNode. But why not in SKSpriteNode?
And I've alreay tried that.
How can I fix that?
SKSpriteNode has an anchorPoint property that defaults to the center of the image, which is why half of the sprite is off the left side of the screen.
You can adjust the anchorPoint so that it behaves like your SKShapeNode with the anchor in the bottom left. Try this :
skPart1.anchorPoint = CGPointMake(0,0);
I know you have accepted an answer, however this is the ideal way to handle the situation and it's why the property exists.
I think this actually depends on what you're using as your SKShapeNode. If you're using a rectangle, then the point you give it will be the lower left corner of the rectangle. But if you use a SKShapeNode circle, it'll drop the circle centered on that point you give it, and you'll see very similar behavior to the SKSpriteNode.
The SKSpriteNode is using the center of the image as the point it places your sprite at, and so when you're placing your node at (0, 100), exactly half of it is being draw to the left of the screen.
If you want your sprite to be drawn as far left as possible, but completely on the screen, you should be able to accomplish this by offsetting for one half of the sprite's width.
skPart1.position = CGPoint(x: skPart1.size.width / 2, y: 100)
Following Quote from this source:
http://www.cambridgeincolour.com/tutorials/image-projections.htm
Equirectangular image projections map the latitude and longitude
coordinates of a spherical globe directly onto horizontal and vertical
coordinates of a grid, where this grid is roughly twice as wide as it
is tall.
I have a 13312 px width and 6656 pixel height Panorama picture. It's a equirectangular projection of a room and have a 2:1 ratio.
I use following formular to calculate the xPosition:
var xPosition = ( panorama.width / 360 ) * azimuth
Azimuth = Phi = Heading = Angle to the left or right
How do I project this now on a 1366x768px browser screen?
I think my results are wrong, because it's not on the point where it should be.. it could be because the sphere has a distortion on the left and right:
Is there any formular to calculate the position with attention to the distortion and scale it to fit on the browser screen? I looked many (MANY) sources to find a solution for this, but they always just say that equirectangular are just lat and long.. they don't consider the distortion.
Last question: To find a special solution, I tryed to put a plane on the circle and expanded the line which shows the alpha angle. I though with Phytagoras I could find the position.. but this didn't worked either.. maybe I did something wrong? Is this the way even possible or am I doing it wrong?
edit
THIS is what I'm actually looking for: http://othree.github.io/360-panorama/three-2d/
The black grid in the background. What is the name of this? For what do I have to google or look for? When you start the 2D Panorama, if you want to get the coordinations of the top right corner of the window, what do you have to do?
The whole calculation problem was about to create a Google Streetview similiar view from a 2:1 equirectangular image. We already found a solution for this with a great help from Martin Matysiak (https://github.com/marmat | Google).
It's been a while so I can't give a direct answer to what the main solution is, but I can provide a URL to an AddOn Martin wrote for adding the custom Markers that we actually were trying to make.
You can follow https://github.com/marmat/google-maps-api-addons and look for yourself. In the end it helped a lot to solve the main problem and let us continue with our main Framework for Google Business Tours.
If you follow the link in the threejs demo you included, it would take you to the source code.
particularly look at:
https://github.com/mrdoob/three.js/blob/dev/examples/webgl_panorama_equirectangular.html
and
https://github.com/mrdoob/three.js/blob/dev/src/geometries/SphereBufferGeometry.js
not sure if there is distortion though. The distortion comes from the fact that the texture is mapped to the sphere, and the sphere is rendered in 3D (openGL).
Have to implement a Directx9 project that involves zoom towards the cursor like Google maps with the mouse scroll wheel
(similar to this implementation by Phrogz).
Need the math and the variables required for the same.
Solved this problem using below steps
Decide per scroll movement, call it Z-SHIFT, in Z-direction towards the target point
such that the camera should travel to target in fixed scrolls(SCROLL_COUNT)
Calculate the distance to travel in X and Y directions, say DIST_X and DIST_Y
Movement per scroll in X-direction and Y-direction will be calculated as
X-SHIFT = DIST_X/SCROLL_COUNT
Y-SHIFT = DIST_Y/SCROLL_COUNT
Z-SHIFT = Pre decided suitable value
We have mathematical equation to guide the coordinates of the camera per scroll which when placed in the code provides the required zoom to cursor effect.