Triangulation issue

Sep 2014
6
0
Brighton
Hello,

Could any one help me with following question;

From Vx =K/R2 where Vx isthe Voltage at sensor x, R is the separation & K is a constant.
I get to
R1 = x2+ y2
aR1 = (R –x)2 + y2
bR1 = x2+ (R – y)2
Where a = V1/ V2 and b = V1 / V3
I’d like to find x, y and R1 in terms of a, b andR.


Sensor pic2.png
Apologies if the picture is a bit small it's in the attachment though

Many thanks

Stephen
 

Attachments

romsek

MHF Helper
Nov 2013
6,725
3,029
California
do you mean $V_x = \dfrac K {R^2_x}$ ?
 
Dec 2007
3,184
558
Ottawa, Canada
R1 = x^2 + y^2
aR1 = (R – x)^2 + y^2
bR1 = x^2+ (R – y)^2

For handling ease: u = R1, v = R
u = x^2 + y^2
au = (v - x)^2 + y^2
bu = x^2 + (v - y)^2

Using the above 3 equations, you are trying for this:
x = {expression using a, b, v as only variables}

Is that correct?
 
Sep 2014
6
0
Brighton
Both what you have put and what I originally stated are correct, thanks for taking an interest though.

Regards

Stephen
 
Dec 2007
3,184
558
Ottawa, Canada
What d'heck are x and y coming from?
The coordinates of a point that moves along the hypotenuse (Sensor1 at origin)?

Anyway, getting x in terms of a,b,v seems possible,
but the equation will be uglier that MacBeth's 7 witches combined...

And how sure are you that the 3 equations you ended up with are correct?
 
Sep 2014
6
0
Brighton
X and Y are the position of the point source or subject, this is shown in the attached pictures.

I'm about as sure as I can be about the equations, if you translate distance of the point source from sensor x, i.e. Rx into variables x and y and known value R using Pythagoras I think that's what yuo get.

Regards

Stephen
 
Sep 2014
6
0
Brighton
Sorry, it should be
R1^2 = x^2 + y^2
aR1^2 = (R – x)^2 + y^2
bR1^2 = x^2 + (R – y)^2

Apologies, as you can probably tell I'm new to this.