# Homework Help: Trilateration with a Difference!

1. Mar 28, 2014

### Sparkypedia

Good Evening all,

Let me start by introducing myself, my name is Alex and I am in the final stages of an Electronic Engineering degree. This is not directly 'homework'. It will take me a little while to set the scene so feel free to get a cup of tea!

One of my modules (Wireless Sensor Networks) has a project that involves using 3 wireless motes and a base station. For those that have not come across this, it is simply 3 wireless sensor platforms that relay their respective data to a hub (base-station) that is connected to a computer serial port. The project was meant to be fairly straightforward; each mote has a series of embedded sensors. I was meant to pick one and then relay each motes data back to the computer and display it in a homemade GUI. I have already accomplished all of this quite successfully. (I chose to use the embedded light dependent resistors (LDR's) as my sensors.

For some brownie points (this is not in the projects spec) I thought it would be interesting if I could track the co-ordinates of a target moving around within the sensor grid. The idea being that I turn all the lights off and then send someone with a torch walking around room. The mote closest to the target would give the highest sensor reading.

The problem with this is I cannot use normal trilateration, as I do not know the distance between each mote and the target. All I have to go on is the motes LDR readings which can vary depending on the light source anywhere between readings of 3 and 1300.

Known Variables

Mote No.1,2,3 X,Y coordinates

Unknown Variables

Target's X,Y Coordinates

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Originally I assumed that the triangle of motes was isosceles, from that point I could work out the distances between each mote. From this using the LDR readings I could then take a ratio approach and work out the distances between each mote and the target (between 2 points not 3). Using this method I could build up 6 sets of coordinates each representing the outer limits of the targets potential positions. Using these I could plot curved lines with the aim of identifying the point of intersection.

However I have realised that I cannot assume that the triangle is isosceles... As I said I am a soon to be electronic engineer not a mathematician, this problem is going into uncharted territories. I love challenges but I would appreciate a point in the right direction. Thanks for reading this, if nothing else it has been interesting learning about the differences between triangulation and trilateration!

Cheers,

Alex

2. Mar 28, 2014

### Staff: Mentor

(It's still schoolwork, so per the PF rules, I moved it to the HH forums)

So you are wanting to locate the position of a person with a torch/flashlight based on 3 LDRs spaced out in the room? There is the variable of which way the torch is being pointed as well, then, which would mess up any algorithm. The person could be standing right in front of one sensor but with his back to it, and pointing the torch directly at the sensor furthest from him. Am I understanding the setup correctly?

3. Mar 28, 2014

### Sparkypedia

Fair enough, I thought I might not get away with putting it here ;)

You have good understanding of the setup, I had not considered the issue of the direction of the torch. This does not need to be pinpoint accuracy though, I just want to have a rough idea of the persons position.

Thanks

4. Mar 28, 2014

### Staff: Mentor

If the light source were not directional, and were always visible to all 3 LDRs (assuming they have high enough gain), then your intensity algorithm would work a lot better. Can you mount a vertical light bulb on an RC vehicle and guide it around the room? That could work, or at least would be worth trying out...

5. Mar 28, 2014

### Sparkypedia

Yes I could do that, I have an RC car that I can use. Its a good suggestion thanks

6. Mar 28, 2014

### Sparkypedia

If anyone has any ideas for the algorithms them selves it would be very helpful. My method as described above only works with isosceles triangles and to be honest its a bit haphazard.

7. Mar 29, 2014

### Sparkypedia

Solved

After thinking about it for a little while I solved the problem using ratios and a little maths, see the equations below and the attached images for more details. Thanks for your suggestions.

L1=M2x-M1x

∴700cm-50cm=650cm

xt=((L1/(Luminosity_1+Luminosity_2)×Luminosity_2)+M1x)

∴((650cm/(100+90)×90)+50cm)=357.89cm

L2=M3y-M1y

∴600cm-50cm=550cm

yt=((L2/(Luminosity_1+Luminosity_2+Lumonisty_3)×Luminosity_3)+M1y)

∴((550cm/(100+90+20)×20)+50cm)=102.38cm

∴Target (xt,yt)=(357.89,102.38)

This method has 2 limitations:

• Mote no.3’s ‘x’ coordinate must lie between mote no.1 and mote no.2 ‘x’ coordinates.
• Mote no.1 and mote no.2’s ‘y’ coordinate must be the same.

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Last edited: Mar 29, 2014