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enter image description here
(B738 AFM, 0-2000 feet elevation, up to 27°C) Example of V1 and Vr differences.

The table shows that the V1 speeds are invariably one or two knots lower than their matching Vr speeds. Given that V1 can equal Vr, what's the limiting factor (probably factors) that puts V1 one or two knots below Vr? Vmbe might be one, but is it the only possible limiting factor here?

Or perhaps the V1 speeds are the same with the matching Vmcg speeds because that's operationally more advantageous in some ways than V1s being the same as the higher Vr speeds?

Or are the V1 speeds the speeds that bring the balanced field lengths, which require the mimimum runway distance?

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$V_R$ is:

FAR 23 (e.g. multi-engine small planes):

$\geq 1.05\;V_{MC}$ (105% of minimum control in flight)

$\geq 1.1\;V_{S_1}$ (110% of V-stall for a specific configuration)

FAR 25 (e.g. transport jets):

$\geq V_1$

$\geq 1.05\;V_{MC}$

Must result in $V_2$ before 35 ft height

Must result in $V_{LOF}$ [when the airplane is rotated at its maximum practicable rate]

$\geq 1.1\;V_{MU}$ (all engines operative; AEO)

$\geq 1.05\;V_{MU}$ (one engine out)

$\geq 1.08\;V_{MU}$ (AEO, geometry limited)

Geometry limited is when there is a greater risk of a tail-strike, e.g., a stretched variant of an airplane (think A321 or 737-900).

So the 1-2 knot disparity in your example can be down to the difference between $V_{MCG}$ and the 105% of $V_{MC}$ or the different $V_{MU}$ speeds. (See below for full text on points 2 and 4.)

To clear any confusion about the Weight:V1 tables such as the one in the question, those tables are not for 1 runway length, those weights are the field limit weight, i.e., each higher weight in the table is a longer runway.

enter image description here
(Twitter) $V_{MU}$ test.


Source: http://wpage.unina.it/fabrnico/DIDATTICA/SPER_VOLO/APPUNTI_3CFU/11.2-TO&LAND.pdf

FAA AC No: 25-7C text:

  1. VR may not be less than V1; however, it can be equal to V1 in some cases.
  2. VR may not be less than 105 percent of the air minimum control speed (VMCA).
  3. VR must be a speed that will allow the airplane to reach V2 at or before reaching a height of 35 ft. above the takeoff surface.
  4. VR must be a speed that will result in liftoff at a speed not less than 110 percent of VMU (unless geometry limited) for the all-engines-operating condition and not less than 105 percent of the VMU (unless geometry limited) determined at the thrust/weight ratio corresponding to the one-engine-inoperative condition for each set of conditions such as weight, altitude, temperature, and configuration when the airplane is rotated at its maximum practicable rate.

(Emphasis mine.)

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  • $\begingroup$ Geometry limited is, more precisely, when tail-strike occurs at pitch less than stall angle of attack. Since A319, with it's much longer gear, is already geometrically limited, I think it's safe to assume all 737 variants are geometrically limited too. $\endgroup$ – Jan Hudec Sep 16 '17 at 19:27
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    $\begingroup$ In that chart, with whatever assumptions go into it, a 180K aircraft can have a V1 of 160. A lighter aircraft would have better stopping performance. So why at 130K is the V1 133 knots instead of fully up to the VR of 135? By your definitions, V1 can = VR, but in this table they never are. If not due to stopping margin (most common cause of reducing V1 below VR), why not? $\endgroup$ – Ralph J Sep 16 '17 at 21:38
  • $\begingroup$ Thank you for your detailed reply but I'm not sure your writing answers my question. It explains why V1 should be above Vmca or Vmu but not why V1 can't be the same as Vr (which was my question). Please correct me if I'm missing your point. $\endgroup$ – lemonincider Sep 17 '17 at 12:29
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    $\begingroup$ @lemonincider - check update. $\endgroup$ – ymb1 Sep 17 '17 at 13:52
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V1 is the maximum speed at which you plan on rejecting takeoff should something go wrong, other than in situations where you have serious doubts about the aircraft's ability to fly e.g. catastrophic fire or structural failure before rotation. It is also tends to vary quite a bit depending on operational conditions. Aircraft manuals will have many tables, charts, and calculation instructions like the one above in order to allow pilots to calculate what their V-speeds will actually be on any given day... though, admittedly, this is usually handled by flight planning software there days.

Generally speaking, V1 is dictated by how much runway you require to stop and how much you have available to you. A lower takeoff weight or a lower density altitude will both serve to bring Vr down, which, ceteris paribus, can bring V1 and Vr closer to each other. A longer runway gives you more room to stop, so it can bring V1 up.

On a balanced field, V1=Vr so you have to option to reject the takeoff right until you're actually rotating. To look at it another way, you have as much time as you could possibly ask for in order to decide not to take off if you encounter a problem during the take-off roll.

As for Vmcg, this is the minimum speed at which you have sufficient control authority on the ground to overcome the adverse yaw you experience with the critical engine inoperative. It would only really come into play to make sure that you don't back yourself into a corner where, if you were to somehow have an engine failure, you also wouldn't have enough airspeed to be able to overcome the asymmetric thrust from the still-operating engine(s), so you wouldn't be able to continue to accelerate and take off safely, but are also going too fast to be able to stop within the available runway distance. Honestly can't say I could think of a situation where something like this could be an issue with anything but the most off-the-wall back country and bush flying these days.

FYI, in case I wasn't clear enough, an engine failure at or above V1 is not, by itself, grounds for rejecting takeoff. That's where your V2 number comes in. You rotate at Vr and climb out at V2. The only time you would reject a takeoff above V1 is if you think that the chances of dying are greater if you get airborne than if you run off the end of the runway.

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  • $\begingroup$ Thank you for your detailed explanation. "which, ceteris paribus, can bring V1 and Vr closer to each other." Could you clarify why this is the case? $\endgroup$ – lemonincider Sep 15 '17 at 23:47
  • $\begingroup$ Lower density alt. helps improve TO performance primarily in 2 ways: 1.) improved engine performance so your aircraft accelerates quicker and 2.) keeping you at a lower TAS throughout so the difference between the relative wind over your wings (and, by extension, the aerodynamic force available to you) and how quickly you're moving over the ground is adjusted in your favor. Stopping distance depends on how quickly you're moving. TO ground roll depends on how quickly you can get to Vr. Lower density altitude can help with both those aspects, though how much varies from one a/c to the other. $\endgroup$ – habu Sep 16 '17 at 0:01
  • $\begingroup$ Lower weight simply means you can both take off at a lower airspeed and need less runway to stop. The interplay of how much this would help reduce V1 vs. Vr as weight decreases also varies from one aircraft to the other and, as your chart shows, may be more academic for certain take-off configurations (e.g. minimal flaps, as in your chart). $\endgroup$ – habu Sep 16 '17 at 0:05
  • $\begingroup$ Wow, ceteris paribus. I had to look that one up. $\endgroup$ – Terry Sep 16 '17 at 1:46
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    $\begingroup$ Balanced field length numbers do not necessarily (or even usually) have V1=Vr.... $\endgroup$ – Lnafziger Sep 16 '17 at 16:06

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