This is from your link and supports my (And Jims) statements, it also answers my actual question, so thanks for helping me find an answer, even if it was on accident.
Quoted from Page 33 -
"The maximum clearing time (time it takes for breakers to completely open) decreases as current increases. This is because of the blow-apart contact design which utilizes the magnetic field built-up around the contacts. As current increases the magnetic field strength increases, which aids in opening the contacts."
Sorry,Zog,but there is one more detail to be taken into account by you.There is short time delay over load region that occurs just before the INST region in the breaker characteristic.The two were clubbed together and simply called instantaneous trip component in that link.Study the following lines in the same page 33.
This circuit breaker has an adjustable instantaneous trip point from 900 A to 2000 A, which
is 4.5 to 10 times the 200 A trip unit rating.
If the trip point adjustment is set to minimum (900 A), and a fault current of 900 amps or greater occurs, the breaker will trip within 1 cycle (16.8 ms). If the trip point setting is set to maximum (2000 A), and a fault current of 900 amps occurs, the breaker will trip between approximately 12 and 55 seconds. A greater fault current will cause the breaker to trip faster.
See the underlined sentence.If the fault current is above 900A for the INST setting of 900A,you can not determine the exact trip time for each value of fault current above 900A from the
same characteristic of the breaker.But if it is less than 900A,trip time may be determined from the breaker characteristic for such values of fault current.The same holds for each value of trip setting from 900A t0 2000A.
Am I clear?