Triangulation of a location

[reggie]

I'm trying to work out a triangulation problem.

I'm in a big room.
There are 3 reflectors on one wall. I know the distance between the reflectors 1-2 and 2-3.
From any location in the room I can scan the wall with the reflectors.
I know the angle between reflectors 1 and 2.
I know the angle between reflectors 2 and 3.
Where am I relative to reflector 2?

as an example the distance between reflectors 1 and 2 is 10 feet
the distance between reflectors 2 and 3 is 10 feet
the angle between reflectors 1 and 2 is 20 degrees
the angle between reflectors 2 and 3 is 12 degrees

It seems like I should be able to do this but I can't seem to make it work.
Please help.

 

[stapel]

From any location in the room I can scan the wall with the reflectors.

What does it mean to "scan the wall with the reflectors"? Are these swivelling mirrors...?

I can't seem to make it work.

Kindly please reply with a clear listing of what you've tried so far. Thank you!

Eliz.

 

[soroban]

Hello, reggie!

I think I understand.
I have a really looong solution . . . maybe someone can streamline it.

I'm in a big room.
There are 3 reflectors on one wall. I know the distance between the reflectors 1-2 and 2-3.
From any location in the room I can scan the wall with the reflectors.
I know the angle between reflectors 1 and 2.
I know the angle between reflectors 2 and 3.
Where am I relative to reflector 2?

As an example: the distance between reflectors 1 and 2 is 10 feet,
the distance between reflectors 2 and 3 is 10 feet,
the angle between reflectors 1 and 2 is 20 degrees,
the angle between reflectors 2 and 3 is 12 degrees.

      A   10    B      10      C
      o - - - - o - - - -  - - o
                           - *
       *       *         - *
                         *
        *     * x      *
                     *
         *20 * 12  *
                 *
          * *  *
             *
           o
           P

\(\displaystyle \text{The reflectors are at: }A,B,C.\;\;AB = BC = 10\)

\(\displaystyle \text{The observer is at }P\!:\;\;\angle AP\!B = 20^o,\;\angle BPC = 12^o\)

\(\displaystyle \text{Let }x = BP.\)


\(\displaystyle \text{In }\Delta APB\!:\;\;\frac{x}{\sin A} \,=\,\frac{10}{\sin20^o} \quad\Rightarrow\quad x \:=\:\frac{10\sin A}{\sin20^o}\;\;{\bf[1]}\)

\(\displaystyle \text{In }\Delta BPC\!:\;\;\frac{x}{\sin C} \,=\,\frac{10}{\sin12^o} \quad\Rightarrow\quad x \:=\:\frac{10\sin C}{\sin12^o}\;\;{\bf[2]}\)


\(\displaystyle \text{Equate {\bf[1]} and {\bf[2]}: }\;\frac{10\sin A}{\sin20^o} \,=\,\frac{10\sin C}{\sin12^o} \quad\Rightarrow\quad \sin12^o\sin A \:=\:\sin20^o\sin C\;\;{\bf[3]}\)


\(\displaystyle \text{We know: }\:A + C + 32^o \:=\:180^o\quad\Rightarrow\quad C \:=\:148^o-A\)

\(\displaystyle \text{Then {\bf[3]} becomes: }\;\sin12\sin A \:=\:\sin20\sin(148-A)\)

\(\displaystyle \text{and we have: }\;\sin12\sin A \;=\;\sin20(\sin148\cos A - \cos148\sin A)\)

. . \(\displaystyle \sin12\sin A \;=\;\sin20\sin148\cos A - \sin120\cos148\sin A\)

. . \(\displaystyle \sin12\sin A - \sin120\cos148\sin A \;=\;\sin20\sin148\cos A\)

. . \(\displaystyle (\sin12 - \sin120\cos148)\,\sin A \;=\;\sin20\sin148\cos A\)

. . \(\displaystyle \frac{\sin A}{\cos A} \;=\;\frac{\sin20\sin148}{\sin12+\sin20\sin148}\)

. . \(\displaystyle \tan A \:=\:0.4565735189 \quad\Rightarrow\quad A \;\approx\;25^o\)


\(\displaystyle \text{Substitute into {\bf[1]}: }\;x \;=\;\frac{10\sin25^o}{\sin20^o} \;\approx\;12.36\text{ ft}\)

 

[reggie]

soroban,

You've definitely got the picture. (I couldn't figure out how to draw it)

I'm standing at point P with a rotating laser and I'm scanning the wall. The retro-reflectors on the wall bounce back a signal that allows me to determine the angle between them. I want to know how far away I am from the wall and how far, left or right, I am from point B.

I'm building a navigation system for my robot and in the end I want to create a coordinate system with point B as (0,0). I'll have to look it over but I think your analysis might just break my roadblock.

Thanks! any further insight would be appreciated.

Eliz,

I hope the above discussion has made this clearer. I spent the whole day and a couple of blocks of paper drawing triangles and trying to apply the law of sines,the law of cosines, circle geometry and anything else I could think of to the problem. Unfortunately I've just been frustrated with simultanious equations that blow up in my face. I know I'm going to feel stupid when I finally get to the answer. I'm sure it's going to be a lot simplier then I'm making it. I've just been away from this stuff for too long.

 

[soroban]

Hello, Reggie!

Once we know \(\displaystyle \angle A\) and the distance \(\displaystyle x\), we can locate \(\displaystyle P\) with respect to \(\displaystyle B\).

Refer to my diagram . . .

\(\displaystyle \text{Label: }\:\angle A = 25^o,\;BP = x = 12.36\)

\(\displaystyle \text{Let }\theta = \angle CBP\)

\(\displaystyle \theta \text{ is an exterior angle of }\Delta ABP\quad\Rightarrow\quad \theta \:=\:25^o + 20^o \:=\:45^o\)


\(\displaystyle \text{The horizontal displacement of }P\text{ with respect to }B\text{ is: }\;x\cos\theta \;=\;12.36\cos45^o \;\approx\;8.74\)

\(\displaystyle \text{The vertical displacement of }P\text{ with respect to }B\text{ is: }\;x\sin\theta \;=\;12.36\sin45^o \;\approx\;8.74\)


. . . . . . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


\(\displaystyle \text{You probably want a }generalized\text{ solution,}\)

. . . \(\displaystyle \text{where: }\;\begin{Bmatrix}d_1 &=& AB \\ d_2 &=&BC \\ \alpha &=&\angle APB \\ \beta &=& \angle BPC \end{Bmatrix}\)


It's quite messy, but it can be done . . .

 

[reggie]

Thanks for all of your help.

I've gone throught your solution and and I think I understand where my problem was. I believe your proceedure in clear and correct, however I think you made some transposition mistakes in the execution. When you expanded your equations you seemed to change sin20 to sin120 and on the next line you have a minus sin120cos148sinA where I believe it should be +sin120cos148sinA .
It almost all gets corrected on the next line where you equate sinA/cosA = sin20sin148/(sin12+sin20sin148) except that the denominator should be sin12+sin20COS148.
That's makes the TanA=-2.1 = -65deg = 115deg That makes BP=26.6.

Your follow up post works great and I've been able to put it all in a spreadsheet and it all seems to work. I think my robot will now know how to find it's way home.

You've helped me a lot. Thanks again.