# Laser triangulation / Range finder

1. Dec 11, 2015

### Anton Alice

Hello.
On a german wiki article I have found the operation principle of a range finder:

https://de.wikipedia.org/wiki/Abstandsmessung_(optisch)#Lasertriangulation

Here are two pictures for illustration:

The Laser, the CCD sensor, and the lens are in a fixed relation to each other, inside a chassis.
Now it is said, that the scattering of the laser on the object surface is projected by the lens onto a point on the CCD. Then is is said, that in order to measure the distance of the object one has to either move the apparatus, or the object itself (which is illustrated by DZ in the second picture).

I don't understand why this is necessary. I could have measured the distance to the object without moving anything, because I know the orientation and distance of the laser relative to the lens and CCD. And by the position of the image point on the CCD chip I can determine the angle to the object.
Nothing more is needed...

2. Dec 12, 2015

### Staff: Mentor

Where? The German wikipedia article does not claim that.

3. Dec 16, 2015

### Anton Alice

Well, not explicitly, but the formula relies on the fact, that DZ is not zero.
The question is, why do I derive a formular depending on DZ, if it is also possible to measure the distance much easier? I mean, the position of the ray spot on the CCD sensor can be absolutely related to an angle, right?

4. Dec 16, 2015

### Staff: Mentor

The formulas there use differences relative to some (known, fixed) reference distance which leads to a known point at the camera.
It is.

5. Dec 17, 2015

### Anton Alice

First: The first picture and the second picture are somewhat different: With "reference distance" you mean the x0 from the first picture, i guess.
In the second picture there is no known reference distance. There is a DZ-travel in space, and a corresponding dz-travel on the CCD. so both the positions x and x+DZ are unknown. The distance is then so to speak calculated by the derivative: " If DZ creates a dz with such a rate, then the distance x must be such and such".

Why do I need a reference distance at all? What is the advantage of this method compared to a simple triangulation by 2 know angles and 1 know distance between laser and CCD?

6. Dec 17, 2015

### Staff: Mentor

Nothing moves. The illustration just shows two different possible positions of the object, and as you can see from the different response in the cameras, the sensor can distinguish between them (=it can measure the distance because different distances lead to different measurement results).