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Posted on 2009-04-12
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please find attached the tutorial if the screenshot does not make sense

ok... this is from Matlab and shows the intersection of the hyperbolas at the point (4,8) to show the location of TX while f1, f2, f3, f4 are RX's.

This gives a location ok for x,y or x,y,0

but what about if I wish to locate a point at x,y,z.

this is done with hyperboloids intersecting?

Can someone please explain what would be required and the concept behind it?

screenshot.jpg
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Question by:jtiernan2008
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Author Comment

ID: 24126197
tutorial attached below;
toa.pdf
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Expert Comment

ID: 24127862
"this is done with hyperboloids intersecting?"
yes
The concepts are the same as in the 2D case. There is a relation between distance and time. The details are in the tutorial. I know, it takes som,e study and practice. But the tutorial is 12 pages long. One cannot expand on everything in it without typing 24 pages.
Now if you have a specific question perhaps it can be answered in fewer pages.
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Author Comment

ID: 24129659
I dont see anything about it in the tutorial? I just refers to it at the start.

"The manner in which the arrival time mea-
surements locate the source can be determined
graphically from the fact that the differences inhe arrival times at a pair of stations i,j con-
strain the source to lie on a hyperboloid of rev-
olution about the baseline between the two sta-
tions."

"to lie on a hyperboloid of revolution" - what does this mean?

Do you know any reading material on this like the pdf that shows how to plot and find the intersection of these hyperboloids?
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Author Comment

ID: 24129792
I understand the 2D version (x,y) fine as explained in the tutorial not the 3D version (x,y,z)
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Accepted Solution

yuk99 earned 2000 total points
ID: 24130391
Hyperboloid with baseline along z can be described by equation:
z^2/a^2 - x^2/b^2-y^2/c^2=1
Hyperboloid of rotation is hyperboloid with b=c. So at z>a its intersection with xy plane will be a circle.
Remember when you derived hyperbola playing with circles of different radii with the same difference? With 3D you have one more degree of freedom. So you have to do the same, but now with spheres with centers located on z axis.
Also note that rotation of hyperboloid in 3D is more complex then what you did in 2D.
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Author Comment

ID: 24130403
Thanks for the great explanation... I appreciate it and understand what you are saying...
Do you know any reading material on this?
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Expert Comment

ID: 24130548
For TOA in 3D I don't know.
For mathematical concepts of hyperboloid, just google it. You will find a lot of material, including wikipedia, mathworld, etc.
Rotation in 3D and rotation matrices are also easy to find there.
Good luck!
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