Well I happen to have 5 minutes available
Hypotheses : 4 knots per second during acceleration (or 2m/sē) and +0.4G deceleration
Calculation based on v1 at 120 knots, giving an ASD of X = 1430 m
Then computation for RTO initiated at 125kt : what is speed at length X ?
58kt !!!
Total ASD for rejection at 125kt : 1545m
I am shocked !!
However, 1430m of ASD sounds pretty short, any runway used by commercial aircraft will be longer than that.
Would anyone care to criticize my hypothesis, confirm whether or not the usual ASD (no margins at all) for your A/C will be this short ? (I tried to be representative of a medium jet)
The runway exit speed as a function of rejection speed looks like this :
https://i.gyazo.com/695a1bd56a8a324f...ac6cbf903b.png
There is a beginning to the curve because runway exit speed does not exist if you reject under V1, by definition of V1.
Sharp increase in the beginning, then the steepness of the curve decreases but you're already at high speeds.
There is an end to the curve : the speed you reach at the end of the normal ASD without ever braking.
Next question is how well can an airliner handle a runway exit at higher speeds. Even on grass.