'three.js How to render a simple white dot/point/pixel
I'm using THREE.WebGLRenderer and I would like to draw a few same-sized white dots at specific positions in 3D space.
Should I use sprites, calculate the 2D screen coordinates and use SpriteMaterial.useScreenCoordinate?
Should I simply recalculate the size of the sprites using the distance of them to the camera?
Can I use SpriteMaterial.scaleByViewport or SpriteMaterial.sizeAttenuation? Is there any documentation for this?
Is there something like GL_POINTS? It would be nice to just define 1 vertex and get a colored pixel at that position. Should I experiment with PointCloud?
Thanks for any hints!
Edit: All points should have the same size on the screen.
Solution 1:[1]
Using .sizeAttenuation and a single-vertex PointCloud works, but it feels a bit… overengineered:
var dotGeometry = new THREE.Geometry();
dotGeometry.vertices.push(new THREE.Vector3( 0, 0, 0));
var dotMaterial = new THREE.PointsMaterial( { size: 1, sizeAttenuation: false } );
var dot = new THREE.Points( dotGeometry, dotMaterial );
scene.add( dot );
Solution 2:[2]
For r125
The excerpt is taken from threejs official example. After some modification here how made it to work.
var dotGeometry = new BufferGeometry();
dotGeometry.setAttribute( 'position', new Float32BufferAttribute( [0,0,0], 3 ) );
var dotMaterial = new PointsMaterial( { size: 0.1, color: 0x00ff00 } );
var dot = new Points( dotGeometry, dotMaterial );
scene.add( dot );
Solution 3:[3]
Yet another update: The interface for attribute has changed somewhat:
For r139
const dotGeometry = new THREE.BufferGeometry();
dotGeometry.setAttribute('position', new THREE.BufferAttribute(new Float32Array([0,0,0]), 3));
const dotMaterial = new THREE.PointsMaterial({ size: 0.1, color: 0xff0000 });
const dot = new THREE.Points(dotGeometry, dotMaterial);
scene.add(dot);
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Bemmu |
| Solution 2 | e-motiv |
| Solution 3 | Berthur |
