Find location using only distance and range?
- by pinnacler
Triangulation works by checking your angle to three KNOWN targets.
"I know the that's the Lighthouse of Alexandria, it's located here (X,Y) on a map, and it's to my right at 90 degrees." Repeat 2 more times for different targets and angles.
Trilateration works by checking your distance from three KNOWN targets.
"I know the that's the Lighthouse of Alexandria, it's located here (X,Y) on a map, and I'm 100 meters away from that." Repeat 2 more times for different targets and ranges.
But both of those methods rely on knowing WHAT you're looking at.
Say you're in a forest and you can't differentiate between trees, but you know where key trees are. These trees have been hand picked as "landmarks."
You have a robot moving through that forest slowly.
Do you know of any ways to determine location based solely off of angle and range, exploiting geometry between landmarks? Note, you will see other trees as well, so you won't know which trees are key trees. Ignore the fact that a target may be occluded. Our pre-algorithm takes care of that.
1) If this exists, what's it called? I can't find anything.
2) What do you think the odds are of having two identical location 'hits?' I imagine it's fairly rare.
3) If there are two identical location 'hits,' how can I determine my exact location after I move the robot next. (I assume the chances of having 2 occurrences of EXACT angles in a row, after I reposition the robot, would be statistically impossible, barring a forest growing in rows like corn). Would I just calculate the position again and hope for the best? Or would I somehow incorporate my previous position estimate into my next guess?
If this exists, I'd like to read about it, and if not, develop it as a side project. I just don't have time to reinvent the wheel right now, nor have the time to implement this from scratch. So if it doesn't exist, I'll have to figure out another way to localize the robot since that's not the aim of this research, if it does, lets hope it's semi-easy.
