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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

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

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.

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??

=> If you have four RX points, then you will have six possible hyperbolae.

How do you know this?

thanks for your response

dTOA12

dTOA23

dTOA34

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?

The asymptotes

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

why is this?

(0,0) vs (10,10) and (0,10) vs (10,0)

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.?

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.

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.

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?

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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.