# Locate Cellphone Using Triangulation

• Dec 5th 2012, 04:18 PM
meghan112
Locate Cellphone Using Triangulation
Hi Everyone...

For a homework problem I need to locate a cell phone using triangulation. Three cell towers form an equilateral triangle with 4.1 mile sides. The transmitting cell phone is somewhere inside the the triangle formed by the towers. The transmit power of the phone is unknown.

The received powers measured at the towers are:
at tower A: 0.00822 microwatts
at tower B: 0.03442 microwatts
at tower C: 0.00972 microwatts.

Find the location of the phone using only this information. The answer should be in the form of x and y coordinates for the phone and each of the three towers.

THANK YOU
• Dec 5th 2012, 09:02 PM
MarkFL
Re: Locate Cellphone Using Triangulation
Using the inverse square law of radiation, what can we say about the phone's distances from the 3 towers?
• Dec 5th 2012, 09:14 PM
meghan112
Re: Locate Cellphone Using Triangulation
I've been trying to use the inverse square law to set up some ratios.

Intensity = Power / Area
Intensity = P / 4 pi (r squared)

Using that formula, I've made three equations by substituting the P for the three different powers given and ra, rb, rb, for the radius. Am I right to think that they are all set to the same "I" because they are receiving power from the same source (the cellphone)?
• Dec 5th 2012, 09:16 PM
meghan112
Re: Locate Cellphone Using Triangulation
correction: rc
• Dec 5th 2012, 09:38 PM
MarkFL
Re: Locate Cellphone Using Triangulation
I think it is the (transmit) power which is the same. The intensity is the received power.

I would let $\displaystyle k=\frac{P\times10^{11}}{4\pi}$ which gives the intensities in hundreds of nanowatts :

$\displaystyle \frac{k}{r_A^2}=822$

$\displaystyle \frac{k}{r_B^2}=3442$

$\displaystyle \frac{k}{r_C^2}=972$
• Dec 5th 2012, 10:23 PM
meghan112
Re: Locate Cellphone Using Triangulation
Oh, that makes more sense! Thank you.

I'm still confused about where to go from this point. Once I have substituted all the given information, I still have too many unknown in order to solve (4 variables and 3 equations). What am I missing here...

Thank you for your help. Very very much appreciated.
• Dec 5th 2012, 10:27 PM
meghan112
Re: Locate Cellphone Using Triangulation
Also, I've switched the given numbers from the original problem on my homework assignment. Realizing now that may cause problems for anyone trying to solve this as I've written it here.
• Dec 5th 2012, 11:09 PM
MarkFL
Re: Locate Cellphone Using Triangulation
I think I would use coordinate geometry. I would put tower A at the origin, tower B in the first quadrant and tower C on the positive x-axis. We know their coordinates. I would next write the two larger distances in terms of the smallest using the ratios previously found. Then, using the distance formula (with a variable point inside the triangle to represent the location of the phone), we will have 3 equation in 3 unknowns.