Henri Wilson wrote:
On Sun, 05 Nov 2006 03:39:49 GMT, "Sorcerer"
wrote:

For simplicity and given:
all sats are at altitude r = 26,000 km
all sats have a reliable clock
the receiver has a broken clock, do not use.

Sat_A says it's at 90° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
Sat_B says it's at 30° 00' 00.00"S, 00° 00' 00.00"E, at time 0.0
Sat_C says it's at 30° 00' 00.00"N, 00° 00' 00.00"E, at time
0.000086726665272239762249227670859584 seconds.

That time gives ct = ?
It doesn't get any easier, ****** Wilson.
A B and C are three points around a hexagon (in a plane)
Teenage trigonometry, a fifteen-year-old could do it.

I would have to be a clever fifteen year old. :-)

I think you left out some vital information.

Indeed he did.
I guess the missing vital information is that all the three
signals are received simultaneously.

Sat_A says it's at 40° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
Sat_B says it's at 10° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
Sat_C says it's at 30° 00' 00.00"N, 00° 00' 00.00"E,
at time 0.0000033356409720092216249702950330609 seconds

Q. Where's the receiver (two possible positions)?
Answer:

C reads 3.3 microseconds, ct = 1000 meters, so the reciever
is somewhere on a sphere of radius 1000 metres centred on C.

If the three signals are received simultaneously,
this doesn't make sense.
Since the signal from C is transmitted 3.3 us later than
the signals from A and B, but are received simultaneously,
the distance to C must be 1 km less than the distance
to A and B.

BUT...
A and B read t = 0, so the receiver has to be equidistance from
each, we are limited to a circle.
The two positions was a trick question.

If the three signals are received simultaneously,
it is correct that the receiver will be equidistant from A and B,
that is on a plane 25 degrees N to the equatorial plane and perpendicular
to the zero meridian. Note that C is 5 degrees above this plane.

So the receiver is on a curve in this plane where the distance
to C is 1 km less than the distance to A and B.
This curve is certainly not a circle 1 km from C!
It isn't trivial to calculate this locus, but it will be
the intersection between a hyperboloid and a plane.

Paul

"Paul B. Andersen" wrote in message
...
| Henri Wilson wrote:
| On Sun, 05 Nov 2006 03:39:49 GMT, "Sorcerer"

| wrote:
|
| For simplicity and given:
| all sats are at altitude r = 26,000 km
| all sats have a reliable clock
| the receiver has a broken clock, do not use.
|
| Sat_A says it's at 90° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
| Sat_B says it's at 30° 00' 00.00"S, 00° 00' 00.00"E, at time 0.0
| Sat_C says it's at 30° 00' 00.00"N, 00° 00' 00.00"E, at time
| 0.000086726665272239762249227670859584 seconds.
|
| That time gives ct = ?
| It doesn't get any easier, ****** Wilson.
| A B and C are three points around a hexagon (in a plane)
| Teenage trigonometry, a fifteen-year-old could do it.
|
| I would have to be a clever fifteen year old. :-)

But you are a stupid 60-year-old.
Most 15-year-olds can multiply c by t.


| I think you left out some vital information.
|
| Indeed he did.
| I guess the missing vital information is that all the three
| signals are received simultaneously.


Wrong guess, Tusseladd. The calculation is made after all the data
is in, however long that takes. Should the receiver move between
signals there will be error, but no 15-year-old would consider that
as significant.


|
| Sat_A says it's at 40° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
| Sat_B says it's at 10° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
| Sat_C says it's at 30° 00' 00.00"N, 00° 00' 00.00"E,
| at time 0.0000033356409720092216249702950330609 seconds
|
| Q. Where's the receiver (two possible positions)?
| Answer:
|
| C reads 3.3 microseconds, ct = 1000 meters, so the reciever
| is somewhere on a sphere of radius 1000 metres centred on C.
|
| If the three signals are received simultaneously,
| this doesn't make sense.
| Since the signal from C is transmitted 3.3 us later than
| the signals from A and B, but are received simultaneously,
| the distance to C must be 1 km less than the distance
| to A and B.

HAHAHAHA!
My clock says it's 1:00 o'clock.
A remote clock says its 12:00 o'clock.
I reckon the remote clock is one light-hour away.
Another remote clock says it's 12:00. The remote clocks are
2 * 0.866 light-hours apart. All clocks are working correctly and
were synchronized when together.
How can that be, confused Tusseladd?



|
| BUT...
| A and B read t = 0, so the receiver has to be equidistance from
| each, we are limited to a circle.
| The two positions was a trick question.
|
| If the three signals are received simultaneously,
| it is correct that the receiver will be equidistant from A and B,
| that is on a plane 25 degrees N to the equatorial plane and perpendicular
| to the zero meridian. Note that C is 5 degrees above this plane.

(40-10)/2 + 10 = 25.
Ok, my typo. I'd intended B at 20 N to make it easy, but since Henri
didn't attempt the problem... shrug


| So the receiver is on a curve in this plane where the distance
| to C is 1 km less than the distance to A and B.
| This curve is certainly not a circle 1 km from C!

| It isn't trivial to calculate this locus, but it will be
| the intersection between a hyperboloid and a plane.
|

The receiver is 1 km from C and equidistant from A and B.
A, B and C are known, and all are on a plane.
The receiver cannot be North or South of C or it will
not be equidistant from A and B.
It can be East or West of C by up to 1000 km.
Therefore it lies on a circle centred on C.
We are not limited to any one algorithm.
True that we need more data, either from a fourth
satellite or a second reading. You snipped that, though.
That's what makes you a ****head troll.
Androcles

On Sun, 05 Nov 2006 23:46:55 GMT, "Sorcerer"
wrote:


"Henri Wilson" HW@.. wrote in message
.. .
| On Sun, 05 Nov 2006 03:39:49 GMT, "Sorcerer"

|
| Q. How could a fourth satellite resolve the position on the circle?
| A. "Four bits of information are required" IN 3D,
| surfaces have nothing to do with it. A tetrahedron has 4 points.
|
| Q. Without a fourth satellite, how could a second reading
| one minute later resolve the two positions?
| A. "Four bits of information are required" IN 3D
| but it's implicit to assume the receiver didn't move and the satellites
| did. A second reading a minute later give the fourth "bit" of
| information.
|
| Q. Why would you care about the second position anyway?
| A. Because you know in practice your receiver is well below
| the GPS constellation, you are not out in space, your altitude
| is limited to the top of Everest and sea level.
|
| You failed, you didn't attempt one question.
|
| If you described your experiment in an intelligible way I might be able to
make
| something out of this gobbledegook.

It was not an experiment, ****head. It was a test to see if you know
how far ct is. You failed.

I haven't the faintest idea of what you are talking about and I'm sure nobody
else has either.

| | If you can give the right answer you'll understand GPS.
| | If you can give the algorithm and know radio and computer
| | electronics you can build your own receiver.
| | Being a mere physicist you'd be lost, go no further than
| | Sydney or rely on engineers, navigators and road signs.
| | You couldn't find your way by the stars, which is all GPS really is
| | and used by Cook to find Botany Bay and his way home again.


HW.
www.users.bigpond.com/hewn/index.htm

The difference between a preacher and a used car salesman is that the latter's product doesn't fall apart till AFTER the sale.

On Mon, 06 Nov 2006 21:32:50 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Sun, 05 Nov 2006 03:39:49 GMT, "Sorcerer"
wrote:


That time gives ct = ?
It doesn't get any easier, ****** Wilson.
A B and C are three points around a hexagon (in a plane)
Teenage trigonometry, a fifteen-year-old could do it.

I would have to be a clever fifteen year old. :-)

Norwegian fifteen year old are usually iodine deficient.....hence not very
bright...


I think you left out some vital information.

Indeed he did.]]

For once I agree with you.

I guess the missing vital information is that all the three
signals are received simultaneously.

That's not all he left out.


Sat_A says it's at 40° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
Sat_B says it's at 10° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
Sat_C says it's at 30° 00' 00.00"N, 00° 00' 00.00"E,
at time 0.0000033356409720092216249702950330609 seconds

Q. Where's the receiver (two possible positions)?
Answer:

C reads 3.3 microseconds, ct = 1000 meters, so the reciever
is somewhere on a sphere of radius 1000 metres centred on C.

If the three signals are received simultaneously,
this doesn't make sense.
Since the signal from C is transmitted 3.3 us later than
the signals from A and B, but are received simultaneously,
the distance to C must be 1 km less than the distance
to A and B.

don't try to work it out. He was obviously drunk...or just raving..

BUT...
A and B read t = 0, so the receiver has to be equidistance from
each, we are limited to a circle.
The two positions was a trick question.

If the three signals are received simultaneously,
it is correct that the receiver will be equidistant from A and B,
that is on a plane 25 degrees N to the equatorial plane and perpendicular
to the zero meridian. Note that C is 5 degrees above this plane.

So the receiver is on a curve in this plane where the distance
to C is 1 km less than the distance to A and B.
This curve is certainly not a circle 1 km from C!
It isn't trivial to calculate this locus, but it will be
the intersection between a hyperboloid and a plane.

You could be right ..but it matters not..


Paul


HW.
www.users.bigpond.com/hewn/index.htm

The difference between a preacher and a used car salesman is that the latter's product doesn't fall apart till AFTER the sale.

Sorcerer wrote:
"Paul B. Andersen" wrote in message
...
| Henri Wilson wrote:
| On Sun, 05 Nov 2006 03:39:49 GMT, "Sorcerer"

| wrote:
|
| For simplicity and given:
| all sats are at altitude r = 26,000 km
| all sats have a reliable clock
| the receiver has a broken clock, do not use.
|
| Sat_A says it's at 90° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
| Sat_B says it's at 30° 00' 00.00"S, 00° 00' 00.00"E, at time 0.0
| Sat_C says it's at 30° 00' 00.00"N, 00° 00' 00.00"E, at time
| 0.000086726665272239762249227670859584 seconds.
|
| That time gives ct = ?
| It doesn't get any easier, ****** Wilson.
| A B and C are three points around a hexagon (in a plane)
| Teenage trigonometry, a fifteen-year-old could do it.
|
| I would have to be a clever fifteen year old. :-)

But you are a stupid 60-year-old.
Most 15-year-olds can multiply c by t.


| I think you left out some vital information.
|
| Indeed he did.
| I guess the missing vital information is that all the three
| signals are received simultaneously.


Wrong guess, Tusseladd. The calculation is made after all the data
is in, however long that takes. Should the receiver move between
signals there will be error, but no 15-year-old would consider that
as significant.

I see.
So it doesn't matter at all when the receiver received the signals.

So we have:
| For simplicity and given:
| all sats are at altitude r = 26,000 km
| Sat_A says it's at 40° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
| Sat_B says it's at 10° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
| Sat_C says it's at 30° 00' 00.00"N, 00° 00' 00.00"E,
| at time 0.0000033356409720092216249702950330609 seconds

And from that, with no information about when the receiver
received the signals, you claim it can calculate it's position
to be 1 km from C and equidistant from A and B. :-)

This must be an all time high, Androcles! :-)

BTW, what would the signals received by a receiver on the ground be?
Wouldn't he receive the signals:
| Sat_A says it's at 40° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
| Sat_B says it's at 10° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
| Sat_C says it's at 30° 00' 00.00"N, 00° 00' 00.00"E,
| at time 0.0000033356409720092216249702950330609 seconds

Would the content of the signals change en route? :-)

Paul

| Q. Where's the receiver (two possible positions)?
| Answer:
|
| C reads 3.3 microseconds, ct = 1000 meters, so the reciever
| is somewhere on a sphere of radius 1000 metres centred on C.
|
| If the three signals are received simultaneously,
| this doesn't make sense.
| Since the signal from C is transmitted 3.3 us later than
| the signals from A and B, but are received simultaneously,
| the distance to C must be 1 km less than the distance
| to A and B.

HAHAHAHA!
My clock says it's 1:00 o'clock.
A remote clock says its 12:00 o'clock.
I reckon the remote clock is one light-hour away.
Another remote clock says it's 12:00. The remote clocks are
2 * 0.866 light-hours apart. All clocks are working correctly and
were synchronized when together.
How can that be, confused Tusseladd?



|
| BUT...
| A and B read t = 0, so the receiver has to be equidistance from
| each, we are limited to a circle.
| The two positions was a trick question.
|
| If the three signals are received simultaneously,
| it is correct that the receiver will be equidistant from A and B,
| that is on a plane 25 degrees N to the equatorial plane and perpendicular
| to the zero meridian. Note that C is 5 degrees above this plane.

(40-10)/2 + 10 = 25.
Ok, my typo. I'd intended B at 20 N to make it easy, but since Henri
didn't attempt the problem... shrug


| So the receiver is on a curve in this plane where the distance
| to C is 1 km less than the distance to A and B.
| This curve is certainly not a circle 1 km from C!

| It isn't trivial to calculate this locus, but it will be
| the intersection between a hyperboloid and a plane.
|

The receiver is 1 km from C and equidistant from A and B.
A, B and C are known, and all are on a plane.
The receiver cannot be North or South of C or it will
not be equidistant from A and B.
It can be East or West of C by up to 1000 km.
Therefore it lies on a circle centred on C.
We are not limited to any one algorithm.
True that we need more data, either from a fourth
satellite or a second reading. You snipped that, though.
That's what makes you a ****head troll.
Androcles

Henri Wilson wrote:
On Mon, 06 Nov 2006 21:32:50 +0100, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Sun, 05 Nov 2006 03:39:49 GMT, "Sorcerer"
wrote:


That time gives ct = ?
It doesn't get any easier, ****** Wilson.
A B and C are three points around a hexagon (in a plane)
Teenage trigonometry, a fifteen-year-old could do it.
I would have to be a clever fifteen year old. :-)

Norwegian fifteen year old are usually iodine deficient.....hence not very
bright...

A comment which becomes you, Henri.

I think you left out some vital information.
Indeed he did.]]

For once I agree with you.

I guess the missing vital information is that all the three
signals are received simultaneously.

That's not all he left out.

With that information, it would make sense. It would be possible
to determine a curve on which the receiver would have to be.

But Androcles says that the signals didn't have to be received
simultaneously, quite the contrary, he says that it doesn't matter
at all when the receiver receives the signal.

Then nothing makes sense.

Because two satellites simultaneously are transmitting their
position and time, a receiver - without regard to when it receives
the signals - has to be equidistant to the satellites ????
And because another satellite 3.3 us later transmits a signal
with its position and time, the receiver has to be positioned
1 km from the latter satellite ????

He must have lost the last remnants of his wits.

Beware, Henri.
You seem to be heading in the same direction.

Paul

"Henri Wilson" HW@.. wrote in message
...
| On Sun, 05 Nov 2006 23:46:55 GMT, "Sorcerer"

| wrote:
|
|
| "Henri Wilson" HW@.. wrote in message
| .. .
| | On Sun, 05 Nov 2006 03:39:49 GMT, "Sorcerer"
|
| |
| | Q. How could a fourth satellite resolve the position on the circle?
| | A. "Four bits of information are required" IN 3D,
| | surfaces have nothing to do with it. A tetrahedron has 4 points.
| |
| | Q. Without a fourth satellite, how could a second reading
| | one minute later resolve the two positions?
| | A. "Four bits of information are required" IN 3D
| | but it's implicit to assume the receiver didn't move and the
satellites
| | did. A second reading a minute later give the fourth "bit" of
| | information.
| |
| | Q. Why would you care about the second position anyway?
| | A. Because you know in practice your receiver is well below
| | the GPS constellation, you are not out in space, your altitude
| | is limited to the top of Everest and sea level.
| |
| | You failed, you didn't attempt one question.
| |
| | If you described your experiment in an intelligible way I might be able
to
| make
| | something out of this gobbledegook.
|
| It was not an experiment, ****head. It was a test to see if you know
| how far ct is. You failed.
|
| I haven't the faintest idea

Of course you don't. You are totally ****in' clueless.

Androcles


of what you are talking about and I'm sure nobody
| else has either.
|
| | | If you can give the right answer you'll understand GPS.
| | | If you can give the algorithm and know radio and computer
| | | electronics you can build your own receiver.
| | | Being a mere physicist you'd be lost, go no further than
| | | Sydney or rely on engineers, navigators and road signs.
| | | You couldn't find your way by the stars, which is all GPS really is
| | | and used by Cook to find Botany Bay and his way home again.
|
|
| HW.
| www.users.bigpond.com/hewn/index.htm
|
| The difference between a preacher and a used car salesman is that the
latter's product doesn't fall apart till AFTER the sale.

"Paul B. Andersen" wrote in message
...
| Sorcerer wrote:
| "Paul B. Andersen" wrote in message
| ...
| | Henri Wilson wrote:
| | On Sun, 05 Nov 2006 03:39:49 GMT, "Sorcerer"
|
| | wrote:
| |
| | For simplicity and given:
| | all sats are at altitude r = 26,000 km
| | all sats have a reliable clock
| | the receiver has a broken clock, do not use.
| |
| | Sat_A says it's at 90° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
| | Sat_B says it's at 30° 00' 00.00"S, 00° 00' 00.00"E, at time 0.0
| | Sat_C says it's at 30° 00' 00.00"N, 00° 00' 00.00"E, at time
| | 0.000086726665272239762249227670859584 seconds.
| |
| | That time gives ct = ?
| | It doesn't get any easier, ****** Wilson.
| | A B and C are three points around a hexagon (in a plane)
| | Teenage trigonometry, a fifteen-year-old could do it.
| |
| | I would have to be a clever fifteen year old. :-)
|
| But you are a stupid 60-year-old.
| Most 15-year-olds can multiply c by t.
|
|
| | I think you left out some vital information.
| |
| | Indeed he did.
| | I guess the missing vital information is that all the three
| | signals are received simultaneously.
|
|
| Wrong guess, Tusseladd. The calculation is made after all the data
| is in, however long that takes. Should the receiver move between
| signals there will be error, but no 15-year-old would consider that
| as significant.
|
| I see.
| So it doesn't matter at all when the receiver received the signals.

The sat's have to be above the horizon, an hour before or after
doesn't work shrug.

|
| So we have:
| | For simplicity and given:
| | all sats are at altitude r = 26,000 km
| | Sat_A says it's at 40° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
| | Sat_B says it's at 10° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
| | Sat_C says it's at 30° 00' 00.00"N, 00° 00' 00.00"E,
| | at time 0.0000033356409720092216249702950330609 seconds
|
| And from that, with no information about when the receiver
| received the signals, you claim it can calculate it's position
| to be 1 km from C and equidistant from A and B. :-)

Yes.


| This must be an all time high, Androcles! :-)

You mean you can't do it?
This must be another time low, Tusseladd. Start blushing now,
get a head start.

|
| BTW, what would the signals received by a receiver on the ground be?
| Wouldn't he receive the signals:
|| Sat_A says it's at 40° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
|| Sat_B says it's at 10° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
|| Sat_C says it's at 30° 00' 00.00"N, 00° 00' 00.00"E,
|| at time 0.0000033356409720092216249702950330609 seconds
|
Did you snip something to say, ****head?
Put it back and we'll continue, ****wit.
Androcles.

|
| | Q. Where's the receiver (two possible positions)?
| | Answer:
| |
| | C reads 3.3 microseconds, ct = 1000 meters, so the reciever
| | is somewhere on a sphere of radius 1000 metres centred on C.
| |
| | If the three signals are received simultaneously,
| | this doesn't make sense.
| | Since the signal from C is transmitted 3.3 us later than
| | the signals from A and B, but are received simultaneously,
| | the distance to C must be 1 km less than the distance
| | to A and B.
|
| HAHAHAHA!
| My clock says it's 1:00 o'clock.
| A remote clock says its 12:00 o'clock.
| I reckon the remote clock is one light-hour away.
| Another remote clock says it's 12:00. The remote clocks are
| 2 * 0.866 light-hours apart. All clocks are working correctly and
| were synchronized when together.
| How can that be, confused Tusseladd?
|
|
|
| |
| | BUT...
| | A and B read t = 0, so the receiver has to be equidistance from
| | each, we are limited to a circle.
| | The two positions was a trick question.
| |
| | If the three signals are received simultaneously,
| | it is correct that the receiver will be equidistant from A and B,
| | that is on a plane 25 degrees N to the equatorial plane and
perpendicular
| | to the zero meridian. Note that C is 5 degrees above this plane.
|
| (40-10)/2 + 10 = 25.
| Ok, my typo. I'd intended B at 20 N to make it easy, but since Henri
| didn't attempt the problem... shrug
|
|
| | So the receiver is on a curve in this plane where the distance
| | to C is 1 km less than the distance to A and B.
| | This curve is certainly not a circle 1 km from C!
|
| | It isn't trivial to calculate this locus, but it will be
| | the intersection between a hyperboloid and a plane.
| |
|
| The receiver is 1 km from C and equidistant from A and B.
| A, B and C are known, and all are on a plane.
| The receiver cannot be North or South of C or it will
| not be equidistant from A and B.
| It can be East or West of C by up to 1000 km.
| Therefore it lies on a circle centred on C.
| We are not limited to any one algorithm.
| True that we need more data, either from a fourth
| satellite or a second reading. You snipped that, though.
| That's what makes you a ****head troll.
| Androcles
|
|

Sorcerer wrote:
"Paul B. Andersen" wrote in message
...
| Sorcerer wrote:
| "Paul B. Andersen" wrote in message
| ...
| | Henri Wilson wrote:
| | On Sun, 05 Nov 2006 03:39:49 GMT, "Sorcerer"
|
| | wrote:
| |
| | For simplicity and given:
| | all sats are at altitude r = 26,000 km
| | all sats have a reliable clock
| | the receiver has a broken clock, do not use.
| |
| | Sat_A says it's at 90° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
| | Sat_B says it's at 30° 00' 00.00"S, 00° 00' 00.00"E, at time 0.0
| | Sat_C says it's at 30° 00' 00.00"N, 00° 00' 00.00"E, at time
| | 0.000086726665272239762249227670859584 seconds.
| |
| | That time gives ct = ?
| | It doesn't get any easier, ****** Wilson.
| | A B and C are three points around a hexagon (in a plane)
| | Teenage trigonometry, a fifteen-year-old could do it.
| |
| | I would have to be a clever fifteen year old. :-)
|
| But you are a stupid 60-year-old.
| Most 15-year-olds can multiply c by t.
|
|
| | I think you left out some vital information.
| |
| | Indeed he did.
| | I guess the missing vital information is that all the three
| | signals are received simultaneously.
|
|
| Wrong guess, Tusseladd. The calculation is made after all the data
| is in, however long that takes. Should the receiver move between
| signals there will be error, but no 15-year-old would consider that
| as significant.
|
| I see.
| So it doesn't matter at all when the receiver received the signals.

The sat's have to be above the horizon, an hour before or after
doesn't work shrug.

|
| So we have:
| | For simplicity and given:
| | all sats are at altitude r = 26,000 km
| | Sat_A says it's at 40° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
| | Sat_B says it's at 10° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
| | Sat_C says it's at 30° 00' 00.00"N, 00° 00' 00.00"E,
| | at time 0.0000033356409720092216249702950330609 seconds
|
| And from that, with no information about when the receiver
| received the signals, you claim it can calculate it's position
| to be 1 km from C and equidistant from A and B. :-)

Yes.

Hint which Androcles will not understand: For every
receiver in the footprint of these three satellites, the
transmit times are the same.

Will all those receivers determine that they are in the
same location? Is every receiver listening to this signal
from C located 1 km from C?

- Randy

"Randy Poe" wrote in message
ups.com...

Sorcerer wrote:
"Paul B. Andersen" wrote in message
...
| Sorcerer wrote:
| "Paul B. Andersen" wrote in message
| ...
| | Henri Wilson wrote:
| | On Sun, 05 Nov 2006 03:39:49 GMT, "Sorcerer"
|
| | wrote:
| |
| | For simplicity and given:
| | all sats are at altitude r = 26,000 km
| | all sats have a reliable clock
| | the receiver has a broken clock, do not use.
| |
| | Sat_A says it's at 90° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
| | Sat_B says it's at 30° 00' 00.00"S, 00° 00' 00.00"E, at time
0.0
| | Sat_C says it's at 30° 00' 00.00"N, 00° 00' 00.00"E, at time
| | 0.000086726665272239762249227670859584 seconds.
| |
| | That time gives ct = ?
| | It doesn't get any easier, ****** Wilson.
| | A B and C are three points around a hexagon (in a plane)
| | Teenage trigonometry, a fifteen-year-old could do it.
| |
| | I would have to be a clever fifteen year old. :-)
|
| But you are a stupid 60-year-old.
| Most 15-year-olds can multiply c by t.
|
|
| | I think you left out some vital information.
| |
| | Indeed he did.
| | I guess the missing vital information is that all the three
| | signals are received simultaneously.
|
|
| Wrong guess, Tusseladd. The calculation is made after all the data
| is in, however long that takes. Should the receiver move between
| signals there will be error, but no 15-year-old would consider that
| as significant.
|
| I see.
| So it doesn't matter at all when the receiver received the signals.

The sat's have to be above the horizon, an hour before or after
doesn't work shrug.

|
| So we have:
| | For simplicity and given:
| | all sats are at altitude r = 26,000 km
| | Sat_A says it's at 40° 00' 00.00"N, 00° 00' 00.00"E, at time 0.0
| | Sat_B says it's at 10° 00' 00.00"N, 00° 00' 00.00"E, at time
0.0
| | Sat_C says it's at 30° 00' 00.00"N, 00° 00' 00.00"E,
| | at time 0.0000033356409720092216249702950330609 seconds
|
| And from that, with no information about when the receiver
| received the signals, you claim it can calculate it's position
| to be 1 km from C and equidistant from A and B. :-)

Yes.

Hint which Androcles will not understand: For every
receiver in the footprint of these three satellites, the
transmit times are the same.

Will all those receivers determine that they are in the
same location? Is every receiver listening to this signal
from C located 1 km from C?

Hint for Blind Ignorant Stupid Brandy Poe:
http://www.androcles01.pwp.blueyonder.co.uk/GPS.htm