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 ΒΈ 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 ΒΈ 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??

The asymptotes

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

This one is on us!

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

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.

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