foggycoder
Super Member
The following has been paraphrased slightly from the Vail website:
Your computer software transmits morse in the form of packets of data.
Each transmission (packet) consists of:
? Timestamp (milliseconds since 1 Jan 1970, 00:00:00 in Reykjavík), so your computer clock needs to be synchronised to an internet time server
? Transmission Duration (milliseconds)
Just like a radio repeater, a CW Repeater broadcasts everything it receives (Vail is a CW Repeater). When it receives those packets of morse data, it broadcasts them to whoever is listening, including your computer.
When your computer gets back the exact same thing it sent, it compares the Current Time to the Timestamp in the packet. This is the Round-Trip Time: the time it takes for a packet to get from your computer to the CW Repeater and back.
The Receive Delay is the difference between the time the packet was sent and the time it was received.
This is the point at which I start to get confused...
When your computer gets a packet it didn't send, it adds the Receive Delay to the Timestamp, and schedules to play the tones and silences in the packet at that time.
By adding the maximum Round-Trip Time to the longest recent Transmission Duration (the length of a dah, hopefully), your computer can make a guess about how much time needs to be added to a received Timestamp, in order to have it play back in the future at the time it comes in. This time is just an estimate. If you're communicating with somebody with a higher Round-Trip Time than you have, you'll need to raise your Receive Delay to account for it.
If a packet arrived so late that it can't be played in time, an error will occur. In technical terms: the Timestamp of the packet plus the Receive Delay is less than the Current Time. It can't be scheduled to play, because we can't go back in time.
This could be happening for three reasons:
1. You (the person getting the error) need a larger Receive Delay
2. The receiving computer's clock is in the future (running fast)
3. The sending computer's clock is in the past (running slow)
I'm guessing that each packet follows a different path across the internet and therefore the packet stream may arrive in a different order in which it was sent - hence the need for time parameters (to sort them back into transmission order).
I have two questions:
[list type=decimal]
[*]Can anyone provide a clearer explanation?
[*]Might this be an explanation of G0KZZ's experience of glitches and latency on CWCOM?
[/list]
Your computer software transmits morse in the form of packets of data.
Each transmission (packet) consists of:
? Timestamp (milliseconds since 1 Jan 1970, 00:00:00 in Reykjavík), so your computer clock needs to be synchronised to an internet time server
? Transmission Duration (milliseconds)
Just like a radio repeater, a CW Repeater broadcasts everything it receives (Vail is a CW Repeater). When it receives those packets of morse data, it broadcasts them to whoever is listening, including your computer.
When your computer gets back the exact same thing it sent, it compares the Current Time to the Timestamp in the packet. This is the Round-Trip Time: the time it takes for a packet to get from your computer to the CW Repeater and back.
The Receive Delay is the difference between the time the packet was sent and the time it was received.
This is the point at which I start to get confused...
When your computer gets a packet it didn't send, it adds the Receive Delay to the Timestamp, and schedules to play the tones and silences in the packet at that time.
By adding the maximum Round-Trip Time to the longest recent Transmission Duration (the length of a dah, hopefully), your computer can make a guess about how much time needs to be added to a received Timestamp, in order to have it play back in the future at the time it comes in. This time is just an estimate. If you're communicating with somebody with a higher Round-Trip Time than you have, you'll need to raise your Receive Delay to account for it.
If a packet arrived so late that it can't be played in time, an error will occur. In technical terms: the Timestamp of the packet plus the Receive Delay is less than the Current Time. It can't be scheduled to play, because we can't go back in time.
This could be happening for three reasons:
1. You (the person getting the error) need a larger Receive Delay
2. The receiving computer's clock is in the future (running fast)
3. The sending computer's clock is in the past (running slow)
I'm guessing that each packet follows a different path across the internet and therefore the packet stream may arrive in a different order in which it was sent - hence the need for time parameters (to sort them back into transmission order).
I have two questions:
[list type=decimal]
[*]Can anyone provide a clearer explanation?
[*]Might this be an explanation of G0KZZ's experience of glitches and latency on CWCOM?
[/list]