07-11-2021, 03:13 AM
You have 4 mechanics working on each tyre at the same time. They are working concurrently.
The perfect pitstop is 45 seconds if none of them make mistake.
If 1 mechanic has a delay that costs him 9 seconds and the other 3 mechanics make no mistakes then the pit time will be 54 seconds (45+9).
If All mechanics have each delay of 9 seconds on their tyre, then it will still be the same 54 second pitstop.
It's not going to be 4 x 9 = 36 seconds extra.
The times above are assuming there is no damage to fix too.
The perfect pitstop is 45 seconds if none of them make mistake.
If 1 mechanic has a delay that costs him 9 seconds and the other 3 mechanics make no mistakes then the pit time will be 54 seconds (45+9).
If All mechanics have each delay of 9 seconds on their tyre, then it will still be the same 54 second pitstop.
It's not going to be 4 x 9 = 36 seconds extra.
Bumble, post: 30393, member: 353 Wrote:TireTime=(20.0,45.0,9.0,1.0) taken from BT24 .hdv - 9.0 is the random delayThe 1.0 at the end means concurrent.
The times above are assuming there is no damage to fix too.