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rotate hyperbola for multilateration

Posted on 2009-03-30
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Last Modified: 2012-05-06
Please find the attached tutorial
page 3;

To plot a rotated hyperbola, it is neces-
sary to do a coordinate transformation. The
transformation used to plot the hyperbola cor-
responding to a particular baseline takes points
calculated in an x, y coordinate system in which
the baseline is vertically upward (in which the
hyperbolae are easy to calculate), and rotates
the points clockwise by an angle ¸ so that the y22
axis is correctly oriented with the actual base-
line in the networks x, y coordinate system. It
then offsets the resulting points away from be-
ing centered around (0,0) to the center (x0, y0)
of the actual baseline. From a carefully drawn
sketch of the geometry of the (22) to (2) trans-
formation, determine the matrix expression for
onverting (x22, y22) values to (x2, y2) values, and
confirm that it is the same as used in the trans-
form program
You need to find the matrix co-
efficients so that:
x2      a11  a12      y22
y2  =  a21  a22 =  x22

Can someone please explain this to me... why do we need to rotate the hyperbola?

Your help would be really appreciated... thanks in advance.
toa.pdf
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Question by:jtiernan2008
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18 Comments
 
LVL 18

Expert Comment

by:WaterStreet
ID: 24021989
"why do we need to rotate the hyperbola?"

I took a quick look at the attached text.

It's been a long time since I've done this kind calculation, but it seems that the purpose in rotating the hyperbola is to be able to calculate the Time Of Arrival (TOA) at different points in a 3D space.  Let's see what others say.


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LVL 27

Expert Comment

by:d-glitch
ID: 24022086
If you have four RX points, then you will have six possible hyperbolae.

If you pick your point conveniently (at the corners of square/rectangle), then four of
your hyperbolae will be aligned with the axes, and will not require rotation.
These four may be adequate to solve the position problem.

The hyberbolae associated with the RX pairs on the diagonals do have to be rotated.
But they may not be necessary.

What are the requirements for the program you are writing?
Do you need to plot the hyperbolae??

0
 
LVL 1

Author Comment

by:jtiernan2008
ID: 24022103
why is it easier to calculate when the baseline is upward?
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LVL 1

Author Comment

by:jtiernan2008
ID: 24022155
This is all part of the writing up part of the report and theory and I also need to understand the concept in for the presentation if asked.

=> If you have four RX points, then you will have six possible hyperbolae.
How do you know this?

thanks for your response
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LVL 1

Author Comment

by:jtiernan2008
ID: 24022666
If you have four RX points do you not have 3 possible hyperbolas that is 1 hyperbola per two readers?

dTOA12
dTOA23
dTOA34

0
 
LVL 27

Expert Comment

by:d-glitch
ID: 24022966
One per pair:  12  13  14  23  24  34
0
 
LVL 1

Author Comment

by:jtiernan2008
ID: 24023628
From the tutorial it states 'why is it easier to calculate when the baseline is upward? '
why is this?

=> he hyberbolae associated with the RX pairs on the diagonals do have to be rotated.
what are the diagonals, what does this mean?
0
 
LVL 18

Expert Comment

by:WaterStreet
ID: 24024156
"What are the diagonals?"

The asymptotes

See the diagram about 1/3 down in http://home.windstream.net/okrebs/page63.html


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LVL 1

Author Comment

by:jtiernan2008
ID: 24024293
=>so the hyperbolas associated with the RX pairs on the assymptotes do have to be rotated
why is this?
0
 
LVL 27

Accepted Solution

by:
d-glitch earned 2000 total points
ID: 24024389
If you go back to your earlier picture and think about the basics

    http://www.experts-exchange.com/images/122524/pic.png


Suppose you have RX1 at (0,10)  and RX2 at (0,0).  The DTOA (t1-t2) is 4 ns.
You have no other information.

Can you sketch the locus of possible TX positions?  
Can you use a compass to construct 3-5 points of the curve?
It will be half of a hyperbola.  Do you understand why it only half?

Can you get the equation of the hyperbola using the distance formula and algebra?

Can you get the equation into standard form?
Maybe something like    (y-a)² - (x-b)² = c²

===========================================
Now suppose you have RX2 at (0,0)  and RX3 at (10,10).  The DTOA (t2-t3) is 9 ns.
You should still be able to sketch the curve and derive an equation for it.
But you won't be able to get into a standard form because is not aligned with the axes.

Whether or not this matters depends on the problem you are trying to solve and
the technique you want to use.
 
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LVL 27

Expert Comment

by:d-glitch
ID: 24024399
Sorry the diagonals I was referring to are formed by RX's on opposite corners the square:  

     (0,0) vs (10,10)    and      (0,10) vs (10,0)
0
 
LVL 1

Author Comment

by:jtiernan2008
ID: 24024538
ok I understand... so if aligned on axis this is fine because you can get it into standard form.
If not alligned with axis we need to arrange rotation...

So if the readers are on the co-ordinates as per the drawing (0,0) (10,10) (0,10) and (10,0). these are alligned with the co axes ....

but when are they not alligned with the axes... is this when we introduce a z axes?

by diagonals, do you mean the hyperbola formed between (0,0) and (0,10) etc.?
0
 
LVL 27

Expert Comment

by:d-glitch
ID: 24029545
You have to take the RX's in pairs.
If the line through pair of RX's is aligned with either axis, the the hyperbola will be also.

So   [(0,0)  (0,10)]     [(0,0)  (10,0)]     [(10,0)  (10,10)]    [(0,10)  (10,10)]
give aligned hyperbolae.

And   [(0,0)  (10,10)]     [(0,10)  (10,0)]   give rotated ones.
The lines through [(0,0)  (10,10)]   and   [(0,10)  (10,0)]   are the diagonals of the square RX layout.

This is the sort of rotation (in 2D) that they are talking about in the tutorial you listed
at the top of this question.  Look at Figures 5 and 6.

For 3D, there is another type of rotation required.  You have to rotate the hyperbola
around its axis.
0
 
LVL 1

Author Comment

by:jtiernan2008
ID: 24031532
So if I have all the 4 readers on the floor this is in 2d but if I raise one of them to the roof this will be in 3d.
0
 
LVL 27

Expert Comment

by:d-glitch
ID: 24031806
You can have all the readers on the floor, and still let the TX move in 3D.
You can solve for the 3D position of TX using a variety of methods.

But if all the RX's are in a plane, you will not be able to tell the difference
between (X,Y,+Z) and (X,Y,-Z).

If all the RX are at ground level, and the TX can not go below ground, then you
don't have to worry about the ambiguity.
0
 
LVL 1

Author Comment

by:jtiernan2008
ID: 24032054
Ah now I understand that's interesting... so basically we never need to worry about the ambiguity because we do not go below ground so it will always be +Z.

Finally

=> if all the RX's are in a plane

From http://en.wikipedia.org/wiki/Plane_(mathematics) I understand this as the RX's on the floor in one plane or RX's on the wall another plane... am I understanding this correctly?
0
 
LVL 27

Expert Comment

by:d-glitch
ID: 24032213
Three points (or three RX's) define a plane unless they happen to lie on a line.

Walls, floors, ceilings, ramps, and slanted roofs are all planes.

Three RX's on the floor define a horizontal plane.

Two RX's on the floor, and one not on the floor define a plane too.
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LVL 1

Author Closing Comment

by:jtiernan2008
ID: 31564471
thanks again for your help
0

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