Performance refusal-critical failure speed

[eddychewingum]

Hello,

Maybe I am wrong but shouldn't the acceleration-stop and acceleration-go be swapped?

Bart

[PerfManJ]

No, the graph is correct as shown. If an engine failure (VEF) occurs at low speed, then the distance to stop is shorter than the distance to continue to liftoff. The extreme case would be an engine failure at brake release: the stopping distance would be basically zero, and the distance to continue to liftoff would basically be the 3-engine ferry takeoff distance. As VEF increases, the distance to accelerate to VEF increases and aircraft kinetic energy increases (KE = 1/2 (mass) (velocity)^2) therefore the stopping distance increases. In other words, the faster you go, the longer it takes to stop.

[eddychewingum]

Ok I see but if you loose an engine at Vcef then the speed will decrease to zero if we stop. If you follow the acceleration to stop line the speed is increasing not decreasing...

[PerfManJ]

The Y-axis shows "Total Distance" which is the distance from brake release to stop (accel-stop line) or continue to liftoff (accel-go line). It is not speed vs. the distance along the runway. The scissor chart shows the accel-go and accel-stop distances over a range of VEFs rather than just one so that VCEF can be determined.

For example, if you enter the chart from the X-axis at 100 knots, the distance associated with the accel-go line is the total distance to accel to 100 knots, fail engine, and continue to liftoff. The distance associated with the accel-stop line is the total distance to accel to 100 knots, fail engine, then stop.

You may be confusing the scissor chart with the illustrations at the beginning of the Takeoff Section in SMP 777 which show speed vs. distance along the runway for the different CFL scenarios.