How can you calculate the effective transfer rate of canbus?
It's just about the payload vs overhead ratio.
In the example they give, which looks fairly accurate, you have 64 data bits out of 130. Which is the extreme worst case with 19 stuffing bits, which is highly unlikely.
It appears that whoever made the article considered identifier part of the payload - rightly so, since it is often used for higher layer purposes.
Then you have 64 data bits + 11 identifier bits = 75 bits payload. 75/130 = 57.69%.
So at 1Mbps, the effective transfer rate will be 577kbps. They have for some reason truncated the decimals instead of rounding them.