Sorcerer wrote:
"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

Yes, your description on your web site, such as it is, does at
least say that there is also an arrival time measured by the
receiver. Have you noticed that in the discussion above you
said it is not necessary to know the arrival time?

- Randy

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.


| 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.

It is several days since you first posted this nonsense,
so it isn't a blunder made in one posting because you are drunk.
So unless you have been drunk for a week,
you must have lost your mind completely.

This is way beyond what a sober, sane person can claim.

Paul

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

Sorcerer wrote:
"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

Yes, your description on your web site, such as it is, does at
least say that there is also an arrival time measured by the
receiver. Have you noticed that in the discussion above you
said it is not necessary to know the arrival time?

- Randy

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

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

Sorcerer wrote:
"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

Yes, your description on your web site, such as it is, does at
least say that there is also an arrival time measured by the
receiver. Have you noticed that in the discussion above you
said it is not necessary to know the arrival time?

- Randy

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

And that's why all the receivers on Earth in the footprint
of these three satellites will determine their position to
be 1 km from satellite C? :-)

This must be your greatest blunder of all, Androcles.
And that's GREAT!

Paul, the amused stupid ****wit ****head

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

Sorcerer wrote:
"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

Yes, your description on your web site, such as it is, does at
least say that there is also an arrival time measured by the
receiver. Have you noticed that in the discussion above you
said it is not necessary to know the arrival time?

- Randy

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

So on your web site you say the arrival time at the
receiver is needed.

In this thread you say it isn't.

They can't both be right. Which statement of yours is
correct and which is wrong?

- Randy

"Paul B. Andersen" wrote in message
...
| Sorcerer wrote:
| "Randy Poe" wrote in message
| ups.com...
|
| Sorcerer wrote:
| "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
|
| Yes, your description on your web site, such as it is, does at
| least say that there is also an arrival time measured by the
| receiver. Have you noticed that in the discussion above you
| said it is not necessary to know the arrival time?
|
| - Randy
|
| Hint which Blind Brandy Poe will not understand: For every
| receiver in the footprint of these three satellites, the
| transmit times are the same.
|
| And that's why all the receivers on Earth in the footprint
| of these three satellites will determine their position to
| be 1 km from satellite C? :-)

No.

| This must be your greatest blunder of all, Androcles.
| And that's GREAT!
|
I said "No", didn't I?
So there is no blunder except yours, and that's ... NORMAL!

| Paul, the amused stupid ****wit ****head

At least you know your status, I don't need to re-assert it.
Blind Brandy Poe still doesn't know that much.

Androcles

Sorcerer wrote:
"Paul B. Andersen" wrote in message
...
| Sorcerer wrote:
| "Randy Poe" wrote in message
| ups.com...
|
| Sorcerer wrote:
| "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
|
| Yes, your description on your web site, such as it is, does at
| least say that there is also an arrival time measured by the
| receiver. Have you noticed that in the discussion above you
| said it is not necessary to know the arrival time?
|
| - Randy
|
| Hint which Blind Brandy Poe will not understand: For every
| receiver in the footprint of these three satellites, the
| transmit times are the same.
|
| And that's why all the receivers on Earth in the footprint
| of these three satellites will determine their position to
| be 1 km from satellite C? :-)

No.

So what will a receiver at Earth determine its position to be
after it has received these 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


| This must be your greatest blunder of all, Androcles.
| And that's GREAT!
|
I said "No", didn't I?

So you did.
Sobered up?
Realized your blunder?
Blushing?

So there is no blunder except yours, and that's ... NORMAL!

If "no" is the correct answer, then "yes" was a blunder.
Wasn't it?
Or was it?
It was, wasn't it?

Paul B. Andersen wrote:
| 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. :-)

Androcles responded:
| Yes.

Paul, the amused stupid ****wit ****head loving to rub it in

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

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

Sorcerer wrote:
"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

Yes, your description on your web site, such as it is, does at
least say that there is also an arrival time measured by the
receiver. Have you noticed that in the discussion above you
said it is not necessary to know the arrival time?

- Randy

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

| So on your web site you say the arrival time at the
| receiver is needed.

True.

| In this thread you say it isn't.

False.

| They can't both be right. Which statement of yours is
| correct and which is wrong?

The part that was snipped explains it. Have a nice snip.
Oh, and read the thread title.
Does it say "GPS AGAIN! (for Tusseladd)"?
Does it say "GPS AGAIN! (for Blind Poe)"?
No.
So **** off, ineducable ****head.

What is says above is:
"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 = ?"

That doesn't mean satellite C was at 30° 00' 00.00"N, 00° 00' 00.00"E
at time 0.0, does it?
No.
It gives ct(AC) = 26,000 km
It gives ct(BC) = 26,000 km
It gives ct(CC) = 0 km

It doesn't get any easier, Blind Poe.
Hint which Blind Brandy Poe will not understand: For every
receiver in the footprint of these three satellites, the
transmit times are the same.
Hint for a ****head: the receiver is at C which is equidistant
from A and B, and C *DOES* have a precise clock.
When you've understood that (if you ever do) we can go on
to step 2.

Androcles

Sorcerer wrote:
"Randy Poe" wrote in message
oups.com...

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

Sorcerer wrote:
"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

Yes, your description on your web site, such as it is, does at
least say that there is also an arrival time measured by the
receiver. Have you noticed that in the discussion above you
said it is not necessary to know the arrival time?

- Randy

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

| So on your web site you say the arrival time at the
| receiver is needed.

True.

| In this thread you say it isn't.

False.

Look up. Read this:

| 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.

| They can't both be right. Which statement of yours is
| correct and which is wrong?

The part that was snipped explains it.

No snip. Every word lovingly preserved. Look up. See the quote.

Have a nice snip.
Oh, and read the thread title.
Does it say "GPS AGAIN! (for Tusseladd)"?
Does it say "GPS AGAIN! (for Blind Poe)"?
No.
So **** off, ineducable ****head.

What is says above is:
"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 = ?"

That doesn't mean satellite C was at 30° 00' 00.00"N, 00° 00' 00.00"E
at time 0.0, does it?
No.
It gives ct(AC) = 26,000 km
It gives ct(BC) = 26,000 km
It gives ct(CC) = 0 km

Two different receivers, at different places in the footprint,
both receive all three of the pieces of information above.

"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.

This is what those three satellites transmit at those
times, this is what any receiver down below receives. Right?

So how do those two receivers conclude they are in different
positions?


It doesn't get any easier, Blind Poe.
Hint which Blind Brandy Poe will not understand: For every
receiver in the footprint of these three satellites, the
transmit times are the same.
Hint for a ****head: the receiver is at C which is equidistant
from A and B, and C *DOES* have a precise clock.

You mean "Sat_C" refers to a ground receiver in the footprint
of Sat_A and Sat_B, and "Sat_C" is not a satellite?

- Randy

Sorcerer wrote:
"Randy Poe" wrote in message
oups.com...

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

Sorcerer wrote:
"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

Yes, your description on your web site, such as it is, does at
least say that there is also an arrival time measured by the
receiver. Have you noticed that in the discussion above you
said it is not necessary to know the arrival time?

- Randy

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

| So on your web site you say the arrival time at the
| receiver is needed.

True.

| In this thread you say it isn't.

False.

| They can't both be right. Which statement of yours is
| correct and which is wrong?

The part that was snipped explains it. Have a nice snip.
Oh, and read the thread title.
Does it say "GPS AGAIN! (for Tusseladd)"?
Does it say "GPS AGAIN! (for Blind Poe)"?
No.
So **** off, ineducable ****head.

What is says above is:
"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 = ?"

That doesn't mean satellite C was at 30° 00' 00.00"N, 00° 00' 00.00"E
at time 0.0, does it?
No.
It gives ct(AC) = 26,000 km
It gives ct(BC) = 26,000 km
It gives ct(CC) = 0 km

It doesn't get any easier, Blind Poe.
Hint which Blind Brandy Poe will not understand: For every
receiver in the footprint of these three satellites, the
transmit times are the same.
Hint for a ****head: the receiver is at C which is equidistant
from A and B, and C *DOES* have a precise clock.

But you said it doesn't have to know what time the signals from
A and B arrived.

| 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.

So without that information, how can it know it is equidistant
from A and B?

When you've understood that (if you ever do) we can go on
to step 2.

I'm afraid I still don't understand why you don't need to know
arrival time of the signal, or how two receivers can draw different
conclusions about their positions from exactly the same
pieces of information.

Please, feel free to amplify.

- Randy