### Network

W

Jan 1, 1970
0
How accurate could something like this be:

Three antennae on ten foot towers are set up about 100 yards apart.

A handheld unit, no larger than about eight inches square and an inch
or so thick, uses signals from those to determine its position within
the triangle (the unit will usually be held fairly parallel to the
ground, rarely at greater than a 45 degree angle).

Now it can't just figure out it's basic spot but also what angle it is
to a baseline -- that is, it isn't enough to say that it's 83 feet
from tower #1, 212 feet from tower #2, etc., it also has to read out
that it is 18.42 degrees to the line between towers 1 and 2.

In other words, on a map with the towers plotted, it must be possible
to determine exactly where the unit is _and_ where it is pointing.

'Absolute' accuracy is not as important as repeatable and relational
accuracy -- that is, it can read 83 feet from tower #1 when it is
actually 82.4 feet, as long as it always reads the same thing at that
point and it reads 41.5 feet from that tower when it is exactly half
as far away (when it is actually 41.2 feet away). ((Does that make
sense?))

The complete unit has to be no more than $1500, and the towers with transmitters no more than$500 each (these do not have to be
weatherproof -- they will be taken down after each day's use -- pipe
towers whose 'feet' slip into pipes driven into the ground are
probably best).

What accuracy could be reasonably reached?
Within a yard, within a foot, within an 1/8 of an inch?

Perhaps more importantly, what accuracy in the direction it's pointing
could be reached? (i.e. if there were two such units, and you plotted
their readings on a map in a computer, could you determine whether
laser pointers attached to them will cross 250 yards outside the
triangle?)

J

#### Jasen Betts

Jan 1, 1970
0
How accurate could something like this be:

Three antennae on ten foot towers are set up about 100 yards apart.

A handheld unit, no larger than about eight inches square and an inch
or so thick, uses signals from those to determine its position within
the triangle (the unit will usually be held fairly parallel to the
ground, rarely at greater than a 45 degree angle).
Now it can't just figure out it's basic spot but also what angle it is
to a baseline -- that is, it isn't enough to say that it's 83 feet
from tower #1, 212 feet from tower #2, etc., it also has to read out
that it is 18.42 degrees to the line between towers 1 and 2.

it needs to knows how far off the ground it is, or it needs a fourth
transmitter.
In other words, on a map with the towers plotted, it must be possible
to determine exactly where the unit is _and_ where it is pointing.

'Absolute' accuracy is not as important as repeatable and relational
accuracy -- that is, it can read 83 feet from tower #1 when it is
actually 82.4 feet, as long as it always reads the same thing at that
point and it reads 41.5 feet from that tower when it is exactly half
as far away (when it is actually 41.2 feet away). ((Does that make
sense?))

The complete unit has to be no more than $1500, and the towers with transmitters no more than$500 each (these do not have to be
weatherproof -- they will be taken down after each day's use -- pipe
towers whose 'feet' slip into pipes driven into the ground are
probably best).

What accuracy could be reasonably reached?
Within a yard, within a foot, within an 1/8 of an inch?

Perhaps more importantly, what accuracy in the direction it's pointing
could be reached? (i.e. if there were two such units, and you plotted
their readings on a map in a computer, could you determine whether
laser pointers attached to them will cross 250 yards outside the
triangle?)

that depends how close they are to each other.

differential GPS will give you centimetre precision (or better) with just
one tower. - should be uner $1000 for the tower. angle is harder to do, I know there are inertial angle sensors I have no idea of their cost, precision or operating requirements. Bye. Jasen W #### Wildepad Jan 1, 1970 0 it needs to knows how far off the ground it is, or it needs a fourth transmitter. Thanks for that info. Unfortunately, ground height will be changing, so I guess a fourth tower will be necessary. that depends how close they are to each other. The readings will be in clumps -- 80 to 100 possible readings within a foot or two of each other, then another 80-100 a few yards away, then another . . . and they all hae to relate to one another. Basically, the entire interior of the triangle (now a square?) will be covered. differential GPS will give you centimetre precision (or better) with just one tower. - should be uner$1000 for the tower.

angle is harder to do, I know there are inertial angle sensors
I have no idea of their cost, precision or operating requirements.

Sigh! The angle is the really important part -- position is
meaningless unless the exact direction is included.
Bye.
Jasen

Thanks!

R

#### Richard

Jan 1, 1970
0
Sigh! The angle is the really important part -- position is
meaningless unless the exact direction is included.

If you know the position (lat/lon) of the fixed location and of the point
you are located, you can compute the angle.

1. Determine delta lat and delta lon between the two points (it helps to
read latitude and longitude as decimals, or convert them to decimals)
2. Find atan(delta lat/delta lon) to get the angle from the fixed point.
If using Excel, this will be the angle in radians, so convert to degrees by
multiplying by 180/pi.

Making a triangle, delta lat is the y-axis distance from the fixed point to
the new location; delta lon is the y-axis difference. Thus, you can see that
Pythagoras comes to aid once again.

I hope this helps (and that my trig is still accurate).

Richard

W

Jan 1, 1970
0
If you know the position (lat/lon) of the fixed location and of the point
you are located, you can compute the angle.

I need to compute the angle of the device relative to a line, not what
angle it is from its position to another point.

This device (#1) would be attached to a device (#2) which is acquiring
other data. #1's readings would be used to orient #2's data when
constructing a 3-D map. Since it is perfectly possible for what #2 is
measuring to have great discontinuities, and even be rotated at
strange angles from sampling position to sampling position, it would
fall to the readings from #1 to provide an accurate reference not only
of position within a triangle but also of the data's angle relative to
a baseline.

R

#### Richard

Jan 1, 1970
0
I need to compute the angle of the device relative to a line, not what
angle it is from its position to another point.

This device (#1) would be attached to a device (#2) which is acquiring
other data. #1's readings would be used to orient #2's data when
constructing a 3-D map. Since it is perfectly possible for what #2 is
measuring to have great discontinuities, and even be rotated at
strange angles from sampling position to sampling position, it would
fall to the readings from #1 to provide an accurate reference not only
of position within a triangle but also of the data's angle relative to
a baseline.

From what I understand, you wnat to know the direction your device is
pointing.

How about a 2- or 3-axis compass co-located with your roving positioning
unit (GPS)? The GPS would provide your position and the compass would
provide the direction you are pointing your device.

See http://www.pnicorp.com/. I have used their TCM series units. Great for
pitch and roll info (which you probably don't need), as well as direction.

Richard

W

Jan 1, 1970
0
From what I understand, you wnat to know the direction your device is
pointing.

How about a 2- or 3-axis compass co-located with your roving positioning
unit (GPS)? The GPS would provide your position and the compass would
provide the direction you are pointing your device.

The problem with those is resolution.

iirc, the affordable ones only provide about 2 degrees of accuracy.
For this app, I need an order of magnitude better than that.

E

#### ehsjr

Jan 1, 1970
0
How accurate could something like this be:

Three antennae on ten foot towers are set up about 100 yards apart.

A handheld unit, no larger than about eight inches square and an inch
or so thick, uses signals from those to determine its position within
the triangle (the unit will usually be held fairly parallel to the
ground, rarely at greater than a 45 degree angle).

Now it can't just figure out it's basic spot but also what angle it is
to a baseline -- that is, it isn't enough to say that it's 83 feet
from tower #1, 212 feet from tower #2, etc., it also has to read out
that it is 18.42 degrees to the line between towers 1 and 2.

In other words, on a map with the towers plotted, it must be possible
to determine exactly where the unit is _and_ where it is pointing.

'Absolute' accuracy is not as important as repeatable and relational
accuracy -- that is, it can read 83 feet from tower #1 when it is
actually 82.4 feet, as long as it always reads the same thing at that
point and it reads 41.5 feet from that tower when it is exactly half
as far away (when it is actually 41.2 feet away). ((Does that make
sense?))

The complete unit has to be no more than $1500, and the towers with transmitters no more than$500 each (these do not have to be
weatherproof -- they will be taken down after each day's use -- pipe
towers whose 'feet' slip into pipes driven into the ground are
probably best).

What accuracy could be reasonably reached?
Within a yard, within a foot, within an 1/8 of an inch?

Perhaps more importantly, what accuracy in the direction it's pointing
could be reached? (i.e. if there were two such units, and you plotted
their readings on a map in a computer, could you determine whether
laser pointers attached to them will cross 250 yards outside the
triangle?)

the GPS? A laser pointer and protractor can give you the
relative angle from the handheld unit to each of the antennas.
But your description of the angle needed is incomplete. A point
can't have an angle with respect to a line. For example:

HandHeld_Unit

A-------------------------------------------------B

What is the angle of the hand held to the line defined
by points A and B?

You need to define a second line in order to find an angle.

Ed

P

#### Peter Bennett

Jan 1, 1970
0
the GPS? A laser pointer and protractor can give you the
relative angle from the handheld unit to each of the antennas.
But your description of the angle needed is incomplete. A point
can't have an angle with respect to a line. For example:

HandHeld_Unit

A-------------------------------------------------B

What is the angle of the hand held to the line defined
by points A and B?

You need to define a second line in order to find an angle.

Ed

I think you misunderstand the angle that he wants.

He has a line drawn on the handheld device (or perhaps a laser beam
projecting from it), and wants to know the angle between that line (or
the laser beam) and your A/B line.

To get the orientation of the device, he will need a compass or
something of similar function in the device.

W

Jan 1, 1970
0
the GPS? A laser pointer and protractor can give you the
relative angle from the handheld unit to each of the antennas.

This has to be accurate to about 1/4 degree, and the reading has to be
taken within about the second that the unit can be held steady.
But your description of the angle needed is incomplete. A point
can't have an angle with respect to a line. For example:

HandHeld_Unit

A-------------------------------------------------B

What is the angle of the hand held to the line defined
by points A and B?

You need to define a second line in order to find an angle.

The handheld unit isn't a mathematical point, it's an eight inch
square. As such, it has a front edge which is the 'second line'.

Thanks for responding!

W

Jan 1, 1970
0
To get the orientation of the device, he will need a compass or
something of similar function in the device.

I hope not. I was thinking about two receivers in the device, their
antennae at opposite corners of the unit. The difference in their
location should (I hope) provide the angle in relation to a baseline.

Orientation to something outside the triangle (or square) is
unnecessary (for the mapping process, it doesn't matter which way is
north).

Thanks for responding!

R

#### Richard

Jan 1, 1970
0
Now it can't just figure out it's basic spot but also what angle it is
to a baseline -- that is, it isn't enough to say that it's 83 feet
from tower #1, 212 feet from tower #2, etc., it also has to read out
that it is 18.42 degrees to the line between towers 1 and 2.

In other words, on a map with the towers plotted, it must be possible
to determine exactly where the unit is _and_ where it is pointing.

'Absolute' accuracy is not as important as repeatable and relational
accuracy -- that is, it can read 83 feet from tower #1 when it is
actually 82.4 feet, as long as it always reads the same thing at that
point and it reads 41.5 feet from that tower when it is exactly half
as far away (when it is actually 41.2 feet away). ((Does that make
sense?))

The complete unit has to be no more than $1500, and the towers with transmitters no more than$500 each

After several days, we have determined that you really want angular
measurament accuracy, but \$700 is too expensive (TCM2.5 compass) to achieve
0.1 degree accuracy because it busts the budget for the entire system.

You state you want to determine the angle to the straight line that connects
two towers. You do, of course, realize that every point on that line is at a
determine the angle that a beam will intersect the line? If so, a compass
tied to each unit still may be your best bet.

Other alternatives (some may border on the ridiculous):
- Make the "pointers" long enough so there is a definite position
difference between the from and back; mount a GPS unit on each end and
compute the angle between their positions (expensive).
- Mount one or more accelerometers in the units (at the end closest to the
beam output) then, starting with the units pointing in a known direction
(you need to know what that direction is), the data from the
accelerometer(s) could be used to determine speed and direction of movement.
Compute angular difference from the starting point. Using one accelerometer
for horizontal movement and one for vertical movement, you will be able to
determine both the geographical point of intersection and the altitude of
intersection.

The above would require some way to obtain and process the data. The units
could be hardwired to a controller unit or, that being impractical, you
could use a wireless system (Bluetooth, etc.).
Perhaps more importantly, what accuracy in the direction it's pointing
could be reached? (i.e. if there were two such units, and you plotted
their readings on a map in a computer, could you determine whether
laser pointers attached to them will cross 250 yards outside the
triangle?)

Assuming the line runs east-west, if your two units are inside the triangle
(south of the line) and the beams bisect the line at any angle between 0.1
and 89.9 degrees for the left-most unit and 359.9 and 269.9 degrees for the
right-most unit, you can be sure the beams will intersect at some point if
the units are at the same height and vertically pointing at the same angle.
The intersection point will be determined by the positions of the two roving
units and the angle they are pointing. Distance between the units and
distance to the line both come into play. Trig will, once again, be useful
in determining where they will intersect.

If I still don't understand what you are trying to do, maybe you could be a
bit more specific about what you are trying to accomplish and seeing if the
folks in this group can give you other ideas of how to do it.

Accuracy can be expensive. Don't get discouraged, just research as many
alternatives as possible.

Good luck, no matter how you end up approaching this.

Richard
Old guy; perpetual student

R

#### Rich Grise

Jan 1, 1970
0
How accurate could something like this be:

Three antennae on ten foot towers are set up about 100 yards apart.

A handheld unit, no larger than about eight inches square and an inch
or so thick, uses signals from those to determine its position within
the triangle (the unit will usually be held fairly parallel to the
ground, rarely at greater than a 45 degree angle).

Now it can't just figure out it's basic spot but also what angle it is
to a baseline -- that is, it isn't enough to say that it's 83 feet
from tower #1, 212 feet from tower #2, etc., it also has to read out
that it is 18.42 degrees to the line between towers 1 and 2.

In other words, on a map with the towers plotted, it must be possible
to determine exactly where the unit is _and_ where it is pointing.

Put two loop antennas on the handheld, set the transmitters at the
towers to three different frequencies, and have three receivers
in the box, which can resolve the relative phase of the two loop
antennas' signals (quadrature detector?); this will give you relative
angles to the three towers, then just trig it out.

With just radio, there's no way to determine distance, unless you're
using some kind of radar. But angles are simple, with two loops. (at
90 degrees to each other of course.)

Good Luck!
Rich

W

Jan 1, 1970
0
Accuracy can be expensive. Don't get discouraged, just research as many
alternatives as possible.

The problem now looks like neither radio nor sound would be accurate
enough -- neither one can go much below 3cm at any cost, which is far
too coarse for positioning, let alone determining the angle.

Maybe that's why no one else is trying it?

Thanks for the input!

E

#### ehsjr

Jan 1, 1970
0
This has to be accurate to about 1/4 degree, and the reading has to be
taken within about the second that the unit can be held steady.

Throw in the towel. You need accuracy of +/- 15 minutes and the
unit will be held by a human, not mounted on a pivot and locked
in place, even if momentarily? Seems like a non-starter. I think
you need to re-define the problem, or the approach. Heck, GPS
won't give you that kind of accuracy, forgetting the other issues.
Ed

J

#### Jasen Betts

Jan 1, 1970
0
the GPS? A laser pointer and protractor can give you the
relative angle from the handheld unit to each of the antennas.
But your description of the angle needed is incomplete. A point
can't have an angle with respect to a line. For example:

He wants to know which way the handheld unit is pointing WRT some baseline.
HandHeld_Unit

A-------------------------------------------------B

What is the angle of the hand held to the line defined
by points A and B?

they look parallel to me

J

#### Jasen Betts

Jan 1, 1970
0
Put two loop antennas on the handheld, set the transmitters at the
towers to three different frequencies, and have three receivers
in the box, which can resolve the relative phase of the two loop
antennas' signals (quadrature detector?); this will give you relative
angles to the three towers, then just trig it out.

With just radio, there's no way to determine distance, unless you're
using some kind of radar. But angles are simple, with two loops. (at
90 degrees to each other of course.)

the only problem I see with this is that the operator holding the device

E

#### ehsjr

Jan 1, 1970
0
Jasen said:
He wants to know which way the handheld unit is pointing WRT some baseline.

Yup. And you can't solve that problem until "handheld unit" is
given some detail. For example, which way is the handheld unit
shown below pointing with respect to the baseline below it?

------
| Hand |
| Held |
| Unit |
------

-------------------------------------------- baseline

As it turns out, the closer you question him, the more detail
comes out. Originally, the handheld unit was "about eight inches
square". Now he says it is a square which removes some ambiguity,
and the second line is the front edge of the square, removing the
remqaining ambiguity. And he adds that he needs accuracy to

they look parallel to me

Really? They look perpendicular to me. Let me draw and extend
the edge line for you:
-------------
|HandHeld_Unit|
|-------------
|
|
A-------------------------------------------------B

The details are important! His original description
was mathematically equivalent to a point and a line.
We can assume anything we want - and be completely
wrong - until he specifies the ambiguity out of the
question. Hell, his reference line on the handheld unit
could have been either diagonal of (what turned out to be)
a square.

Ed

W

Jan 1, 1970
0
Originally, the handheld unit was "about eight inches
square". Now he says it is a square which removes some ambiguity,

Um, what part of "square" is missing from "eight inches square"?
and the second line is the front edge of the square, removing the
remqaining ambiguity.

What difference could that possibly make? Pick a straight line --
front edge, side, diagonal, whatever -- as long as you don't change
what is defining the orientation of the unit between readings, how the
'angle of the unit' changes relative to the baseline at consecutive
And he adds that he needs accuracy to

I was asking what accuracy is possible. Adding that I need 1/4 degree
expresses my disappointment at there not being anything readily
available.
Really? They look perpendicular to me. Let me draw and extend
the edge line for you:
-------------
|HandHeld_Unit|
|-------------
|
|
A-------------------------------------------------B

You added four different lines, completely changing the situation!

The orientation of the text, which is how you originally represented
the unit, is parallel to the A-B line.

But if you want to consider that the orientation of the unit is
determined by the side, fine -- it can't possibly matter since such
information is only used when determining how the angle has changed
The details are important! His original description
was mathematically equivalent to a point and a line.

Um, no, it wasn't. I described a three dimensional object -- width and
length (eight inches square) and height.

W

Jan 1, 1970
0
Throw in the towel. You need accuracy of +/- 15 minutes and the
unit will be held by a human, not mounted on a pivot and locked
in place, even if momentarily? Seems like a non-starter.

Just because something is handheld does not prevent it from being held
solidly up against something when readings are taken.

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