2^12 microseconds is 1000000/2^12 or 1000000/4096. Take that value, divide by 12288 (number of bits in a 1536 byte Ethernet frame) and it’s 0.0000003…, which is how long it takes to transmit 1 bit at 3,000,000 bits per second (or the 3Mps mentioned in the article).
This is more accurate description (from Metcalfe himself):
There is a limit on Ethernet packet sizes to enable sharing on shared cable
back when, reduce latency, provide for packet buffers, specify error
control rates. PARC Ethernet max size was set big enough to carry a whole
disk sector with metadata. That’s all I know.
I don’t understand what the significance of 2^12 us is.
2^12 microseconds is 1000000/2^12 or 1000000/4096. Take that value, divide by 12288 (number of bits in a 1536 byte Ethernet frame) and it’s 0.0000003…, which is how long it takes to transmit 1 bit at 3,000,000 bits per second (or the 3Mps mentioned in the article).
Yes, I understand the calculation. But why does it matter how long this time is? The author does not relate it to the memory bandwidth.
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This is more accurate description (from Metcalfe himself):
See https://blog.benjojo.co.uk/asset/1hhfq2UR8P
FWIW the Alto used 512-byte sectors, so maybe it was some other disk?
In Metcalfe and Boggs (1976) they say the max frame size of 3 Mbit/s PARC Ethernet was about 4000 bits.
The 1500 byte frame size was a feature of 10 Mbit/s DIX Ethernet, along with 48 bit MAC addresses.