Talk:Web Science/Part1: Foundations of the web/Ethernet/Minimum package length vs maximum cable length

Should the data section be at least 46 or at least 47 bytes long?
In the last video you mentioned that for sending one bit, the data section must be 46 bytes long (3'09"). In this video though, you say "Data section > 46 bytes" (3'10"). I believe the first one (>= 46 bytes) to be correct, but want to clear this up. --103.16.26.85 (discuss) 15:47, 23 October 2013 (UTC)
 * I think at least 46 bytes should be correct --Renepick (discuss • contribs) 22:43, 10 September 2014 (UTC)

Propagation delay
In the calculation it is assumed that signals travel at the speed of light in vacuum which does not fit for wire-/fiber-transmissions. In fiber the speed of light is about 35% slower compared to vacuum (http://en.wikipedia.org/wiki/Speed_of_light). Gluteus (discuss • contribs) 11:27, 24 October 2013 (UTC)


 * The propagation speed depends on the physical medium of the link (that is, fiber optics, twisted-pair copper wire, etc.) and is in the range of $$2*10^8$$ meters/sec for copper wires and $$3*10^8$$ for wireless communication, which is equal to the speed of light (https://en.wikipedia.org/wiki/Transmission_time#Propagation_delay) --oleamm (discuss • contribs) 12:13, 27 October 2013 (UTC)

Check your understanding 2
There is a statement:"The clock frequency has an impact to the maximum distance between two ethernet devices." It means: Do you agree with me? --oleamm (discuss • contribs) 13:21, 27 October 2013 (UTC)
 * if we reduce the clock frequency to 1 bit/sec, hence theoretically we can have the maximum distance between two ethernet devices about 300 000 km.
 * if we increase the clock frequency to 10 MBit/sec => the maximum distance will be reduced to 30 meters; if we want to occupy all wire (1500 meters long) => we should send 50 bits (50 clock ticks * 30 meters/tick = 1500 meters).
 * if 100 MBit/sec => we have only 3 meters (per clock tick); if we want to occupy all wire (1500 meters long) => we should send 64 bytes or 512 bits (minimum frame length standard).
 * Besides the above mentioned fact that the propagation speed is not exactly the speed of light I agree with what you have written here. --Renepick (discuss • contribs) 14:16, 27 October 2013 (UTC)

The length of the cable and the shortest frame
Hello, Rene! You count the length of the wire as 10 ns x 300 mln mps = 3 m for one bit transaction. But when you actually send 1 bit, your frame will be 8 + 14 + 1/8 + 4 = 209 bits (Preamble+MAC+1bit+Checksum). So the actual length of the cable would be 209*3 = 627 m. The second thing: how can the shortest frame be 64 bytes if it is 8 + 14 + 46 + 4 = 72 bytes? --141.26.93.178 (discuss) 11:51, 29 October 2014 (UTC)
 * Let me see if I understand your questions correctly. In the beginning of the video I mean that I would send only one bit of any data ignoring the protocol. So If I send one bit of data in an ethernetframe you're calculation is probably correct. (Though I would suggest to display this in the following way 8Byte (Preamble) + 14Byte (Ethernet Header) + 1Bit (Payload) + 4Byte (Footer)) = 209 Bit. Even if this one Bit was send as data payload via ethernet we would have to make the payload (according to the protocol at least 46 Byte long). Also for the entire legth of the frame the preamble is not considered to be part of the frame. This means we have 14 Bytes (Header) + 46 Bytes (Minimum data payload) + 4 Bytes (Footer) = 64 Bytes (minimum length) which is the stated number. Does this explanation satisfy you? --Renepick (discuss • contribs) 19:47, 30 October 2014 (UTC)
 * Thank you for the answer! --oblomov (discuss • contribs) 15:45, 2 November 2014 (UTC)

Hi
J --1.47.67.160 (discuss) 15:37, 6 January 2017 (UTC)