There is a term in multi-engine flying called balanced field length. The balanced field length is the length of the runway required for a twin to operate safely with an engine failure at any point during the take-off phase. This is different then the more common "Takeoff Ground Run Distance" used in single engine flying. Takeoff Ground Run Distance is the minimum distance it takes an airplane (single or twin) to accelerate from 0 mph to when its wheels leave the ground. For the Cessna 150 at gross weight, at a sea level density altitude with zero headwind, the take-off distance is 735 ft. That means that in these conditions we should be able to use runways as small as 735 ft safely - if nothing goes wrong and the pilot does everything right on takeoff. In practice it is usually a good idea to stretch that length a little to leave some margin for error. Of course the more runway the better but endless amounts of runway is not always practical or possible. Therefore we use the takeoff ground-run distance, pulled off of a chart in the Pilot Operating Handbook, as a guideline as to how much runway we need.
Balanced field length is a little different then takeoff ground run distance. The Balanced field length as I mentioned at the beginning takes into account the runway needed to operate a twin engine airplane safely even if it were to have an engine failure at any point during the take-off. That means adding on extra runway length to account for having to stop the airplane if an engine were to fail during the takeoff roll on the ground, or in the case of larger airplanes, the extra length required to continue accelerating to takeoff if there is no longer enough distance to stop.
Larger airplanes use a pre-flight calculated V speed, denoted V1, as their takeoff decision speed. If an engine fails below this speed they are going slow enough and have enough runway to shut down the remaining engines and stop. Above this speed, and they no longer have the space to stop and are therefore committed to taking off, even if they're still on the ground. In a light twin such as ours, we don't use the V1 system because the airplane doesn't have the power to continue a takeoff on only one engine if we're still on the ground. This means that if we are still on the ground with an engine failure, we have no choice but to retard the throttles and stop. We may also even have to re-land and stop on the remaining runway if we have just taken off and have an engine failure before we have attained our best rate of climb speed and have the landing gear up. The landing gear down causes extra drag and our airplane may not be able to climb on only one engine with it down. Therefore our light twin balanced field length takes this into account. Many light twins have Accelerate-Stop charts in their Flight Manuals, which denote how much distance will be used to accelerate to a given airspeed, and then slow to a stop. My Twin Comanche doesn't have a chart like this, so we are forced to estimate it... OR, like I have just spent my entire last night doing, we can develop our OWN accelerate-stop distance charts using two different charts given in our Flight Manual, and a little math.
The chart I have just developed, I should point out, is an unapproved aircraft performance chart, and it has been untested. So I would certainly not wager mine or my passengers lives on it. It did however serve to be good brain excercise, and I do think it would probably fairly accurate if it actually was tested.
So here's what I did. Please any math/physics savey readers who have some suggestions/comments/rebutals, please, by all means, speak up. Anyways, my existing flight manual has two charts I used data from: Takeoff Ground Run Distance, which gives the distance required to accelerate to 80 mph and liftoff, and the Landing Ground Run Distance, which gives the distance required to stop given a touchdown speed of 70 mph. Now we can't really just add these two distances together, because they are in reference to two different speeds, and the resulting distance also wouldn't account for the extra distance we'd need if we were lifted-off, but still accelerating just above the the runway to our best rate of climb speed (Vy) of 112 mph, as is standard takeoff procedure (Accelerate to Vr of 80 mph, rotate and liftoff, accelerate just above the runway to Vy of 112 mph, climb out at that speed and retract the gear). So first of all we need to extrapolate the distances to a higher airspeed. To do this, we have to find our curve of acceleration, which I have charted on a Speed vs Distance graph. Since we know the engines are outputting a constant force, and we know that as speed increases, drag increases to the square of speed, so I think a curve somewhere along the lines of D=V^2 should be a good representative of both acceleration and deceleration seperately, where D equals distance travelled and V equals aircraft speed. I also had to calculate a stretch factor to make the curve fit the data given in the Flight Manual. To do that I used the speed given, divided by the distance given to calculate a stretch factor to make the curve fit the data. I then divided the stretch factor into the calculated speed, giving me this:
Dc = Vc^2/(Vg^2/Dg)
Where:
Dc = Distance calculated
Vc = Speed calculated
Vg = Speed given in charts
Dg = Distance given in charts
Now I can go to the charts in the flight manual to look up the distance needed to accelerate to 80 mph with a given aircraft weight in a given density altitude, to calculate the estimated distance to accelerate to a higher airspeed. For my chart I chose to calculate the distance needed to accelerate to 100 mph. I chose that airspeed because our best rate of climb with a single engine operating is 105 mph, so below that we would have to be able to stop, above 100 mph, we should be able to accelerate the last 5 mph and climb out.
So then I had to go through each density altitude and weight combination listed in the flight manual charts, find the resulting distance to accelerate to 80 mph, and plug that into my formula. Example:
2000 ft density altitude, at gross weight, showed 1500 ft ground roll to 80 mph.
Dc = Vc^2/(Vg^2/Dg)
Dc = 100^2/(80^2/1500)
Dc = 2344 ft required to accelerate to 100 mph.
I then used virtually the exact same method to calculate stopping distance from 100 mph, and added the accelerate and stop distances together, to get my estimated balanced field length.
That would mean that IF (big if) the formula is correct, I would be able to use a runway as short as the calculated balanced field length and I could have an engine failure at any point during the take-off and still have enough runway to safely land and stop below 100 mph or climb away above that speed.
Like I said that was simply a problem I did for fun and there is no guarentee those numbers are correct even IF my understanding of math and physics is correct. So I would never actually try to pass them off as true, and wouldn't recommend myself or anyone else try to use this method.
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I was looking for this topic and find yours very interesting. I will guess that your twin is a propeller driven engine. I worked with tofl on twin jets and everything was done in computers.
ReplyDeleteAgain thanks for your interes in pilots and aviation.
That's correct, my aircraft is a Piper Twin Comanche, which is a small twin engine piston aircraft.
ReplyDeleteIts a 1964 model, which means no computers and a very pathetic Flight Manual!
I appreciate your comment :)
I was doing a writeup on V1 speeds and came across this, and just wanted to add some stuff. I am an engineer and not a pilot.
ReplyDeleteFor a certain aircraft/condition/field/weather/TO thrust/etc./etc. there's going to be some speed V1 at which the aircraft is going fast enough that, if one engine should fail, the remaining engine(s) can provide enough thrust to accelerate to V2 (engine-out takeoff speed) and safely take off in the remaining available runway.
There will also be some speed V1 at which the aircraft is going slow enough that it can come to a full stop without exiting the runway.
A FAR 25 (?) operating aircraft must be able to reach some V1 speed that satisfies *both* conditions - it must be able to, at some arbitrary V1 speed, either RTO or single-engine-out TO.
So you have to have some ACCEL distance (V=0 to V=V1), some GO distance (V=V1 to V=V2), and some STOP distance (V=V1 to V=0). The required field distance must be equal to or greater than both ACCEL+GO *and* ACCEL+STOP. If you do the math you can figure out that when GO=STOP, the required field distance is at a minimum. This requires some specific V1 speed, V1BAL.
Thus, the shortest takeoff distance is the Balanced Field Length.
If the field is longer than the balanced field length, STOP and GO distances may be different, and a range of V1 values are possible. (Some operators allow use of V1MAX or V1MIN.)
For most operations, thrust will be selected so that the balanced field length is equal to the actual field length (minus a safety margin) which reduces engine wear, and means there is only one value of V1 you can use, V1BAL: the V1 such that STOP=GO.
The critical bit I want to correct here is that V1 speed is NOT ONLY the speed at or below which the aircraft has sufficient stopping distance. IT IS ALSO the speed at or above which the aircraft can continue to takeoff with one engine out.
Regarding your more recent posts, I want to extend great sympathy for you over the loss of your Twin Comanche and my happiness to know your superior skill has ensured a "good" landing.
Thanks for the comments and extra info! I think overall we're in agreement though, as I stated in my second paragraph almost exactly what your critical bit of correcting is.
ReplyDeleteMy post is however written from the perspective of flying a light-twin, which aren't mandated by the FAA to have the capability of climbing out on a single engine.
As such in most cases light twins such as a Twin Comanche won't have the power to continue takeoff if its still on the ground regardless of the airspeed its at. This simplifies the options quite a bit and leaves us with only one option on the ground - Accel+Stop, therefore we can't calculate any sort of real V1 airspeed because there's no decision to make - if we're still on the ground we have to reject the takeoff.
Our take-off decision point is at the point where we retract the landing gear, thus committing to the takeoff, and that is done once we're airborne and no longer have enough runway to land and stop again.
I used 100 mph as my "V1" speed in the charts I made because its above the speed at which we rotate and become airborne (90 mph), and very near Vyse (105 mph) which is the speed we need to achieve for best rate of climb with a single engine operating. As such generally when we decide to retract the landing gear we're flying somewhere near that airspeed, so it seemed like a good number to pick for my equivalent "V1" speed to make my charts.
I should point out I'm aware that your pic shows a Canadian reg and thus aren't necessarily beholden to the FAA anyway.
ReplyDeleteI agree probably are in agreement (hah!), it's just that it's so often people don't metion the second bit of the V1 speed - the single-engine out requirement - that I though it was worth extra emphasis and perhaps some discussion as well.
My real correction is if anything rather pedantic - in your third paragraph you don't seem to address the fact that, except in the case of a balanced field takeoff, *there's no one particular V1 speed*. If for example you are using V1MIN as a decision speed, you still (technically) have more than enough distance to stop, because STOP<GO. Likewise, if you're using V1MAX, you still have more than enough distance to to continue takeoff on one engine. But using a V1 other than V1BAL means you're introducing some vagueness into the decision to abort or not - below V1MIN you know you cannot safely take off on one engine, and above V1MAX you cannot safely execute a RTO. This isn't really conducive to a good decision speed - there's an external bias - and is further encouragement to reduce TO thrust to ensure you're using the entire runway and making V1MIN=V1MAX=V1BAL.
I have no qualification to discuss the legalities of such things, only the engineering degrees. :)
Is it absolutely impossible for a Twin Comanche on a single engine to accelerate from Vr to Vyse while on the ground regardless of available runway length? (e.g. tire limitations?) Or is it only that the length required is certain to be far beyond braking distance?
I'm going to go ahead and say "absolutely impossible" doesn't really matter, as clearly you are more than qualified to tell me it is not allowable to continue a takeoff roll once an engine failure has occurred on a light twin.
Note that, if air resistance and non-braking wheel friction is ignored and the aircraft remains on the ground, the distance required to accelerate to a speed V1 is V1^2/(2a), where a is the acceleration due to the engines. The distance required to decelerate from V1 to V=0 is 3*V1^2/(2d), where d is the deceleration due to braking force. The values of a and d will have no particular relation, and d will further vary with airspeed as airspeed alters lift and lift alters weight on wheels. (This can be sort of glossed over on an aircraft with ground spoilers but certainly not on a light twin.)
Adding the element of the aircraft going airborne adds a relationship of kinetic/potential energy and that requires more algebra than I feel like dealing with tonight. :)
Cheers! :D
Ya the Canadian reg doesn't really matter, its an american-built aircraft so its still subject to FAA manufacturing standards, light aircraft just don't have near as strict performance standards that transport category aircraft do, Canadian or American built.
ReplyDeleteI like your explanation of V1min/V1max.
I can't say its absolutely impossible for a Twin Comanche to continue take-off on one engine... I've never tried it lol, but climb performance on one engine with the landing gear and flaps down is close to nil (and even negative on hot days at gross weight), so even if you did manage to get off the runway and into ground effect, you wouldn't be able to climb out of ground effect to clear any obstacles at the end of the runway. Light twins just don't have the power reserves that transport category aircraft do.
Because of that its never a good idea to try to continue a takeoff on one engine in a light twin if you're still on the runway. Your chances of survival are greater to reject the takeoff regardless of how much runway you have left.
So allowable or not, to do so would be foolish and suicidal.
I appreciate your thoughts on the subject, especially the acceleration/deceleration formulas. If I still had the airplane I definitely would have made use of them.
I like your explanation of V1min/V1max.
ReplyDeleteThanks!
Thanks also for explaining light twin operations. I was aware they have substantially reduced power reserves compared to transport category aircraft but I hadn't appreciated that they cannot even achieve positive rate on one engine while dirty.
If calculating ASDR continues to interest you, the equation that relates distance traveled to velocity change is:
x2-x1 = (v2^2-v1^2) /(2a)
Where acceleration is a function of engine force, air drag, and friction due to wheels (mostly rolling resistance) and due to brakes. You can probably ignore rolling friction and assume constant braking force, or at least that's what I'd try to fit the data to first.
If you're considering distance required to lift off, have an engine failure just before gear up, and safely land again, that's going to require a little more analysis because you have to bleed off airspeed and flare.
It's a really fascinating subject and I think I'll be perusing your blog for a while. :)