Timeline for How can V1 and VR (rotate) speeds be calculated?
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| Dec 13, 2017 at 19:35 | history | edited | jwzumwalt | CC BY-SA 3.0 |
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| Dec 13, 2017 at 19:28 | history | edited | jwzumwalt | CC BY-SA 3.0 |
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| Dec 13, 2017 at 19:22 | history | undeleted | jwzumwalt | ||
| Dec 13, 2017 at 19:11 | history | deleted | jwzumwalt | via Vote | |
| S Dec 13, 2017 at 15:09 | history | suggested | CommunityBot | CC BY-SA 3.0 |
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| Dec 13, 2017 at 14:43 | comment | added | Ralph J♦ | @MichaelKjörling Delta A would be the delta in acceleration, just as delta V is the delta in velocity (a.k.a. acceleration). So Delta A would be more like the THIRD derivative of position. If the aircraft has a (more or less) steady acceleration with takeoff thrust set, you'd get a drop in acceleration (and thus a large, negative, delta A) at the point an engine fails. Beyond that, I doubt operators have much use for delta A. Engineers, probably have more. You can see "V" on instruments, you can see "A" on a HUD; delta A, not so much. | |
| Dec 13, 2017 at 14:36 | comment | added | user | @RalphJ To a first order approximation, I would expect "delta-a" to be the second derivative of position, where velocity is the first derivative. Change in position (over time) is velocity; change in velocity (over time) is acceleration (or deceleration). It's also worth keeping in mind that velocity is a vector, while speed is a scalar quantity. | |
| Dec 13, 2017 at 14:12 | comment | added | Ralph J♦ | I'm sorry, but "V1" is undefined for a single engine aircraft, since the ability to lose the critical engine & continue the takeoff only applies when the aircraft has multiple engines. A single might have a refusal speed above which a stop on the remaining runway cannot be completed -- don't abort above this speed for a system failure, perhaps; but if THE engine fails on the runway, you're stopping, no matter what your speed! | |
| Dec 13, 2017 at 13:59 | review | Suggested edits | |||
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| Dec 13, 2017 at 13:18 | comment | added | user | With a sufficiently long runway, then, you might at best have to try really hard simply to keep the plane on the ground long enough to even reach V1. If you can get back down on the runway without turning around, and land safely, then surely by definition you never exceeded V1 because the remaining runway length provided adequate stopping distance? | |
| Dec 13, 2017 at 13:18 | comment | added | user | "A single engine airplane like like a Cessna 150 will have V1 and vr at the stall speed." This doesn't make sense, at least if V1 (as I have understood it) is the speed at or above which takeoff rejection will result in a runway overrun no matter what else you do. That would be a function of speed, breaking capability (including runway surface conditions), and weight. | |
| Dec 13, 2017 at 11:49 | history | edited | jwzumwalt | CC BY-SA 3.0 |
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| Dec 13, 2017 at 11:11 | history | edited | jwzumwalt | CC BY-SA 3.0 |
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| Dec 13, 2017 at 10:56 | history | edited | jwzumwalt | CC BY-SA 3.0 |
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| Dec 13, 2017 at 8:47 | history | answered | jwzumwalt | CC BY-SA 3.0 |