Cheap and accurate indoor triangulation.

SLK001

Joined Nov 29, 2011
1,549
I want to model self driving cars. since GPS is not accurate, i was wondering if you know of any tutorials to build an accurate triangulation setup or kits i could buy for less than $100? IT should be accurate within a few centimeters
This has been a holy grail for years. Sorry, but it ain't gonna happen.
 

BR-549

Joined Sep 22, 2013
4,928
The last time I looked, 30 kHz had a wavelength of 10,000 meters. 33 kHz should be just shy of 10,000 meters.

For a 1 cm wavelength, the frequency should be 30 GHz.
 

Sensacell

Joined Jun 19, 2012
3,432
Aren't we talking about the wavelength of sound now, not EM radiation?

This is one of those things where the solution could make you very rich.
So many applications could use this solution, but it's a doozy.
 

BR-549

Joined Sep 22, 2013
4,928
Is the idea just to plot the location in the house? The car doesn't need to know how far away an obstacle is, right?

And how big is the car? Why do you need cm resolution?
 

WBahn

Joined Mar 31, 2012
29,979
The last time I looked, 30 kHz had a wavelength of 10,000 meters. 33 kHz should be just shy of 10,000 meters.

For a 1 cm wavelength, the frequency should be 30 GHz.
So you are saying that the speed of sound and the speed of light are the same?
 

WBahn

Joined Mar 31, 2012
29,979
Why not a grid of hooded IR detectors in the ceiling? With an IR transmitter on car, pointed straight up?
And how many such detectors would you need in order to determine the toy car's position to within a couple of centimeters in, say, a 12' x 12' room?
 

AnalogKid

Joined Aug 1, 2013
10,987
No, with a single 360 degree hemispherical emitter and three wide-angle detectors at the periphery.

Three points determine a plane, and the intersections of three spherical coordinate vectors determine uniquely any point in the plane, including points outside the closed space determined by the detector locations. To make it easier, put one detector in each of the room's four corners with 90 degree fields of view. Now, all valid locations are within the plane's internal boundaries. You don't need to know the incident angle of the signal at each detector, only the relative signal strength. Start with the dimensions of the detector rectangle, add a butt-load of high school trig, and shake well.

ak
 
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WBahn

Joined Mar 31, 2012
29,979
I agree. My question was specifically targeted at the suggested approach that was offered, which sounds like it's trying to make an optical touchscreen in which the light from the emitter is hitting just one (or a small number) of detectors. I'd still like to see BR-549 estimate the number of detectors that this approach would require. After that, the narrowness of the IR beam would need to be considered.
 

BR-549

Joined Sep 22, 2013
4,928
Oh, I don't know. It's easy to suggest things when one has an unlimited budget. And my idea would fail when going under the coffee table.

Perhaps a tera Hz emitter and 8 detectors at the 8 corners of the house.
 

WBahn

Joined Mar 31, 2012
29,979
No, with a single 360 degree hemispherical emitter and three wide-angle detectors at the periphery.

Three points determine a plane, and the intersections of three spherical coordinate vectors determine uniquely any point in the plane, including points outside the closed space determined by the detector locations. To make it easier, put one detector in each of the room's four corners with 90 degree fields of view. Now, all valid locations are within the plane's internal boundaries. You don't need to know the incident angle of the signal at each detector, only the relative signal strength. Start with the dimensions of the detector rectangle, add a butt-load of high school trig, and shake well.

ak
I think this is going to run in a huge problem. In using the relative signal strength, you are assuming two things that are very poor assumptions -- that every emitter is very uniform over every possible orientation relative to any of the emitters, and that every detector is very uniform in its sensitivity from every possible incident path. In addition, you have reflections that are going to be seen by the detectors that will corrupt the signal-strength measurements. Since you are looking centimeter scale resolution, your differential signal-strength measurements need to be sensitive to about the resolution commensurate with the IR beam spreading out over the course of one more centimeter of travel.
 
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