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The statement made in the question is mentioned in one of the renowned publications.

My contradictions which I have already ruled out as a potential solution:

  • As Density decreases Vmcg decreases thus V1 is not limited by Vmcg.

  • As Density decreases Vmbe decreases as there's insufficient cooling of the brakes, thus V1 should ideally be limited by Vmbe.

  • At lower density the aircraft needs to be at a higher groundspeed to achieve the same CAS, but the ASDA would be limiting factor as it would require a greater stopping distance, this would again limit V1.

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    $\begingroup$ "in one of the renowned publications" -> why don't you say which one? $\endgroup$ – Bianfable Dec 21 '20 at 15:06
  • $\begingroup$ Do you disagree with the statement ? $\endgroup$ – Darshan Patil Dec 21 '20 at 15:10
  • $\begingroup$ Please provide more of the circumstance surrounding the statement "V1 increase with decreasing density". Are we talking of the FLL, Field Length Limit Weight all the time? Let us assume a balanced field for simplicity. The statement can become correct if we take an actual weight less than the FLL then if we start calculating for lower densities, at some point this same actual lower weight will become the FLL weight. As the density is lowered, at each increment, the Required FL will increase and it is probable that the V1 will increase along with it. $\endgroup$ – skipper44 Dec 21 '20 at 16:24
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I guess the statement is about V1 from the balanced field length, which is where the accelerate-go distance is equal to the accelerate-stop distance:

Balanced Field Length
(image source: boldmethod > Engine Failure On Takeoff: Do You Stop Or Go?)

How do these curves change when the density decreases? Lower density means less thrust, therefore accelerating will be slower and uses more distance. This affects both curves equally for the acceleration to V1, but afterwards only the accelerate-go distance:

  • Let us call the force accelerating the aircraft initially $ F = 2 \times T $ (thrust from two engines).
  • In the accelerate-go case, you would only accelerate with a force of $ F = T $ after the engine failure.
  • In the accelerate-stop case, you brake with a certain force $ F = F_\text{brake} $.
  • Now, at lower density, the thrust $ T $ will decrease, but the brake force $ F_\text{brake} $ will be unaffected.
  • The accelerate-go case is therefore affected during the entire time by the lower value of $ T $.
  • The accelerate-stop case is only affected during the accelerate part, but afterwards the brake force is still the same.

Therefore, both curves will move up in the diagram, but the accelerate-go distance will move up more. This implies that their intersection (the balanced field length) will occur further to the right, resulting in a higher value for V1.

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  • $\begingroup$ If both curves move upwards, wouldn't the net V1 decrease ? $\endgroup$ – Darshan Patil Dec 21 '20 at 15:32
  • $\begingroup$ But why would you say that accelerate go distance would, increase more that accelerate stop distance? Is there something I'm missing ? Is it a parametric thing? $\endgroup$ – Darshan Patil Dec 21 '20 at 15:39
  • $\begingroup$ That edit makes it much more understandable, appreciate your effort and time ! Good day @Bianfable $\endgroup$ – Darshan Patil Dec 21 '20 at 15:50
  • $\begingroup$ 'Now, at lower density, the thrust T will decrease, but the brake force Fbrake will be unaffected.' May I ask how this is possible? With reduced density, the maximum brake energy speed reduces. The brakes becomes more inefficient. $\endgroup$ – Anas Maaz Dec 21 '20 at 17:13
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    $\begingroup$ @AnasMaaz I know. I'm not suggesting brake temperature makes no difference. I'm talking about the effect of different densities on the brakes. And AFAIK air density has essentially no effect on the force the brakes can deliver, but maybe I'm wrong about that. $\endgroup$ – Bianfable Dec 21 '20 at 20:00
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We all know that V1 is limited by both the Vmcg (minimum control speed-ground) and the Vmbe (maximum brake energy). It is also affected by Vr (rotation speed). But that is for some other question. Firstly, V1 cannot be lower than Vmcg because if an engine fails below V1, the air flow over the vertical stabilizer and the rudder is not sufficient enough to control the aircraft. On the other hand, V1 cannot be higher than Vmbe because, if you try to stop the aircraft at a higher speed than Vmbe, the brakes can fail and in worst case catch fire.

If the density is lower, the Vmcg reduces because less thrust is produced by the engines. So, in an engine failure the amount of force that needs to be counteracted by the rudder against the live engine reduces. As, you have mentioned thus with reduced densities, Vmcg does not limit the V1.

A reduction in density reduces the value of Vmbe as well. So, if we consider a balanced field, the value of V1 will be higher than Vmbe. As we have discussed before, this is not an ideal situation. To ensure a safe take off hence, the value of V1 is reduced until it becomes less or equal to Vmbe. This will increase the TODR (Take off distance required) and reduce the ASDR (Accelerate stop distance required). Why does this happen? Any time, the V1 is reduced, the ASDA decreases and the TODR increases. The ASDA decreases because the lower the V1, the earlier it is assumed an engine will fail, so the less the aircraft is accelerated. Therefore, the lesser the acceleration, the shorter the aircraft can be stopped. The TODR increases because a reduced V1 just like before results in a reduced acceleration. It simply takes more distance to take off with one engine operative than with two.

To answer your question. At a lower density, the reduction in Vmbe means, you will have to reduce the value of the V1 to that of Vmbe to get the best field performance. The result is an increased TODR and a decreased ASDA. The total field length required increases because it is limited by the TODR.

enter image description here

The dark dotted line is the ideal V1 which occurs at the balanced field (ASDA = TODR). Here, the value of Vmbe is reduced due to a decrease in density. So, if you reduce the V1 to that of Vmbe, the TODR increases and the ASDA decreases. The yellow line is the total field length and it increases to account for the TODR.

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    $\begingroup$ If I understand you correctly, you are saying that V1 must be reduced down to Vmbe, if balanced field length V1 is larger at that density. That makes total sense to me. But what about the case where the balanced field length V1 is still below Vmbe despite the lower density? $\endgroup$ – Bianfable Dec 21 '20 at 18:28
  • $\begingroup$ What do you mean by V1 going below Vmbe exactly? You can reduce it, but that will increase field length as TODR increases. If you are asking what if the balanced V1 is below Vmbe then that is great. That is what we want. With V1 below Vmbe the aircraft can be safely stopped before V1. Moreover, with a balanced field the total field required reduces. $\endgroup$ – Anas Maaz Dec 21 '20 at 19:15
  • $\begingroup$ The whole reason why we bring V1 to the same value of Vmbe is solely to ensure that V1 is equal to Vmbe (V1 should always be less than or equal to Vmbe). If Vmbe is higher than V1 you need not change a thing. $\endgroup$ – Anas Maaz Dec 21 '20 at 19:54
  • $\begingroup$ Yes, that is exactly what makes perfect sense to me. But the original question by the OP was: why does V1 increase with decreasing density? So, what happens here if you are not limited by Vmbe? $\endgroup$ – Bianfable Dec 21 '20 at 20:04
  • $\begingroup$ It doesn't have to. V1 is a flexible value. It can be varied between Vmcg and Vmbe. (Vmcg <= V1 <= Vmbe). So, a reduced density reduces both Vmcg and Vmbe. So, for a given airport with a given pressure altitude and density, a reduced density will result in a reduced V1. $\endgroup$ – Anas Maaz Dec 21 '20 at 20:27

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