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Balanced Field Length again, sorry

GravityHater

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The definition given me is 'the runway length where AG distance equals AS distance"

Is this a misnomer?

It seems to me that if AG EVER equals AS to any precision, then it would be a phenomenal coincidence.

Essentially, one is the distance it takes to stop the airplane after accelerating to Vef, the other is the distance to climb to rch after vef.

I can't see how they would ever be the same, unless a human manipulated something to make them equal.

Thanks for helping a newb

((( )))*

*this post protected by the above 'sarcasm and angry-wit shield', mfg by Dow Corning.
 

Cardinal

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I'm not a fan of that definition. The runway length obivously has no bearing on what distance the aircraft will consume during a reject or while staggering into the air on one donkey.

A clearer, if non-technical way to say it might be:

"You have enough runway to do either". Of A/G and A/S distances, one will be more limiting, but you as an operator don't care which one it is.
 

A Squared

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It seems to me that if AG EVER equals AS to any precision, then it would be a phenomenal coincidence.

Yes they are equal, and no it is not a coincidence. V1 is selected so as to make them equal. There's a reason for that. When they are equal, that is the shortest runway on which you can accelerate, lose an engine and go; or accelerate, lose an engine and stop. Think about it, if you picked a very high V1, say V2-1 knot, your distance to accelerate to v1, then go, would be very short, only slightly longer than a normal takeoff, however, your accelerate-stop distance would be very long, because you are stopping from a very high speed.

On the other hand, if you picked a very low V1, say 0.5 knots, your distance to continue the takeoff on one engine would be very long, it would be essentially a entire takeoff on one engine. However the accelerate-stop distance would be extremely short, just a few feet.

Obviously either of these extremes, very low v1 and very high v1, would require a lot of runway. The shortest required runway distance will be when v1 is chosen so that AS and AG distances are equal.

THe goal isn't for these distances be balanced, per se. THe goal is that at all times durng the takeoff, you have a course of action which will not result in the airplane crashing if an engine fails at any point. before v-1, stop, after v-1, go. The balanced field is merely the minimum runway for which you always have a safe course of action. Any shorter than a balanced field and there will be a window during which an engine failure will leave you with no safe option, you don't have enough runway to continue, and you don't have enough runway to stop.
 

91100 100 set

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Yes, very good explanation. I've staggered through understanding that one in a coherent fashion for awhile. It seems sometimes my brain gets it and sometimes it doesn't, depending on the time of day, my energy level, the price of beads in China, etc.
 

surplus1

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A Squared,

Nice effort. Now take it to the next step and explain why the field is sometimes intentionally unbalanced.
 

A Squared

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surplus1 said:
A Squared,

Nice effort. Now take it to the next step and explain why the field is sometimes intentionally unbalanced.

Not quite sure what you're getting at, but I'll take a stab: On a slippery runway you may not be able to maintain directional control at airspeeds below VMC, due to the fact that you no longer have the help of effective nosewheel steering. so regardless of what speed gives you equal AG and AS distances, you'd establish V1 at or above VMC on a slippery runway.
 
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typhoonpilot

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surplus1 said:
A Squared,

Nice effort. Now take it to the next step and explain why the field is sometimes intentionally unbalanced.

I'll take a stab Surplus.

An unbalanced field is where clearway is factored into the calculation. You can carry more weight off an unbalanced field (than a balanced field), by reducing the V1. Unfortunately this calculation is very complex. You need to factor in Take off run (all engines), Take off run (with engine failure at Vef), Take off Distance (All engines), Take off distance (with engine failure at Vef), brake energy limits, and Accelerate stop.

The main benefit of using balanced field length in flight planning boils down to simplicity at a small weight penalty.

TP
 

CFIse

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GravityHater said:
Essentially, one is the distance it takes to stop the airplane after accelerating to Vef, the other is the distance to climb to rch after vef.

And jut to check your terminology, which may have been confusing you.

Vref is reference landing speed - and varies with weight but not runway length.

The speed you're thinking of, and that others have used, is V1, and that can be varied to either balance the field or for other reasons.
 

EatSleepFly

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And just to check your reading skills: :)

He said Vef, not Vref.

Vef means the speed at which the critical engine is assumed to fail during takeoff.

Not to be confused with V1...
 
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Yank McCobb

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Yep. Every time this subject is discussed, someone posts a reference to "Vef" and then someone else comes on and "corrects" them by explaining the difference between "Vref" and "V1". Every time, without fail.:)
 
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