# Triangulation issue

#### theboybatis

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.

Apologies if the picture is a bit small it's in the attachment though

Many thanks

Stephen

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#### romsek

MHF Helper
do you mean $V_x = \dfrac K {R^2_x}$ ?

#### theboybatis

Yes, any solutions?

#### Wilmer

Where a = V1/ V2 and b = V1 / V3

I’d like to find x, y and R1 in terms of a, b and R.
Ok....so why this: Where a = V1/ V2 and b = V1 / V3 ?

#### Wilmer

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?

#### theboybatis

Both what you have put and what I originally stated are correct, thanks for taking an interest though.

Regards

Stephen

#### Wilmer

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?

#### theboybatis

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

#### Wilmer

R1 = x2+ y2
Why isn't that R1 = SQRT(x^2 + y^2)
(and similarly for R2, R3)

#### theboybatis

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.

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