Best margin above a take-off obstacle

Jean Claude

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Reading several times the method of jumping just before passing, leaves me speechless.
Let us leave the argument which consists in giving up as soon as the horizon is hidden, what speed of advance to choose for climb which gives the best margin above an obstacle?
Vy because the best rate of climb?
Vx because the best climb slope?
IMHO, the best is a little slower still, because the forward speed gain is greedy in acceleration distance and it is better to concede a slope a little reduced by the induced drag.

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Vance

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Reading several times the method of jumping just before passing, leaves me speechless.
Let us leave the argument which consists in giving up as soon as the horizon is hidden, what speed of advance to choose for climb which gives the best margin above an obstacle?
Vy because the best rate of climb?
Vx because the best climb slope?
IMHO, the best is a little slower still, because the forward speed gain is greedy in acceleration distance and it is better to concede a slope a little reduced by the induced drag.
The definition I use for Vx is the best angle of climb Jean Claude.

Vx is determined during testing and what every indicated air speed is found to be the best angle defines Vx in the pilot’s operating handbook.

If a slower speed is found to work better to clear an obstacle than that would become the new Vx.

I feel that is the point of Vx and Vy.

If I want to clear an obstacle I use Vx.

If I want to reach some altitude in as short a time as possible I use Vy.

I feel it is useful to remember that Vx and Vy are given at specific conditions and decrease as density altitude increases.

Writing in general terms Vx decreased faster than Vy as density altitude increases and eventually they converge.
 

nomie

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Interesting hypothesis. Giving up some angle of climb in order to start climbing earlier in the take off sequence. Has anyone ever done some testing on this? I guess its possible to also work this out mathematically given the forward acceleration and Vx and <Vx climb rates.
 

Jean Claude

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The definition I use for Vx is the best angle of climb Jean Claude.
Yes. It also my definition, like shows my blue sketch (Best slope)

If I want to reach some altitude in as short a time as possible I use Vy.
I also agree

If I want to clear an obstacle I use Vx.
This is my point of disagreement. You do not take into account the lost run to reach Vx.
A slower speed than Vx allow you to start the climb earlier, after a shorter run.
When the obstacle is lower than 50 ft, then the margin is so better despite a smaller angle[/QUOTE]
 
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Smack

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This is my point of disagreement. You do not take into account the lost run to reach Vx.
A slower speed than Vx allow you to start the climb earlier, after a shorter run.
When the obstacle is lower than 50 ft, then the margin is so better despite a smaller angle[/QUOTE]
[/QUOTE]
Jean Claude, if I interpret your point correctly, what you are advocating is an earlier START to the climb (prior to reaching Vx) and a climb angle that is less than maximum (Vx), but (since you start sooner) gets you to the clearance altitude quicker.

Brian
 

JEFF TIPTON

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Vx allows the highest altitude gain in the shortest horizontal distance.

Vy Allow the highest altitude gain in an amount of time.
 

ventana7

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Yes. It also my definition, like shows my blue sketch (Best slope)


I also agree


This is my point of disagreement. You do not take into account the lost run to reach Vx.
A slower speed than Vx allow you to start the climb earlier, after a shorter run.
When the obstacle is lower than 50 ft, then the margin is so better despite a smaller angle
[/QUOTE]
It seems you are advocating for not accelerating to Vx after liftoff but rather beginning a climb out behind the power curve. Sounds like a recipe for disaster to me. Why are you flying from airfields that have so little margin?

Rob
 

Jean Claude

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Jean Claude, if I interpret your point correctly, what you are advocating is an earlier START to the climb (prior to reaching Vx) and a climb angle that is less than maximum (Vx), but (since you start sooner) gets you to the clearance altitude quicker.
Yes, Brian
This is exactly what I think

It seems you are advocating for not accelerating to Vx after liftoff but rather beginning a climb out behind the power curve. Sounds like a recipe for disaster to me. Why are you flying from airfields that have so little margin?
Rob,
Behind the power curve is defined by a drag greater than the propeller thrust due to a too low forward speed. So it's a downhill flight despite full power
It's not as low a speed as I suggest, obviously.
I'm just saying that the best margin is not obtained with Vx, but slower

Vx allows the highest altitude gain in the shortest horizontal distance.
Yes Jeff,
the shortest horizontal distance with Vx.
However, as the takeoff is slower, you must still accelerate in level until Vx.
How much distance with no climb do you think?
 

Jean Claude

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Then, Vx would already be a recipe for disaster for Ventana7?
 

Mike G

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I watched this same discussion play out on the French forum with the same arguements against JC by the "experts".
JC's math made sense to me so I tried it.
Sorry guys he's right.
JC you're pissing in the wind (or a violin in french) trying to tell these guys anything.
Mike G
 

ventana7

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JC,
Maybe your discussion is academic in nature- and if your goal is to simply understand your gyro better that's great.

But if you are flying out of an airstrip where clearing trees at the end is in question I have to wonder why? If the margin for clearing the trees is so slim that the difference in climbing early at a speed below Vx means you clear the trees and a regular take-off (achieving Vx in ground effect then climbing) is inadequate to clear the trees I have to wonder why.

And for sure if your engine hiccups when you are climbing out at a speed below Vx and you are inside the height velocity envelope you will not have enough time to lower the nose and land safely- I have to wonder why.

I remember watching a low time pilot destroy his gyro with about 15 hours on it at Benson Days doing exactly the maneuver you suggest. He had watched another gyro (that obviously had more power) climb directly away from the runway. He tried it got 30 feet in the air settled back to the runway in a slightly nose high attitude and destroyed his machine. Never had a clue what was happening to him.

Ventana
 
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thomasant

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This is an interesting thread. It's good for all gyro pilots to review the "short field take off" technique.

JC is right according to the technique given in the FAA handbook. "If the prerotator is capable of spinning the rotor in excess of normal flight r.p.m., the stored energy may be used to enhance short-field performance. Once maximum rotor r.p.m. is attained, disengage the rotor drive, release the brakes, and apply power. As airspeed and rotor r.p.m. increase, apply additional power until full power is achieved. While remaining on the ground, accelerate the gyroplane to a speed just prior to VX. At that point, tilt the disk aft and increase the blade pitch to the normal in-flight setting. The climb should be at a speed just under VX until rotor r.p.m. has dropped to normal flight r.p.m. or the obstruction has been cleared. When the obstruction is no longer a factor, increase the airspeed to VY."
 

DavePA11

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If your not going to clear the obstacle, do a 180 and fly out the other way or land or don’t try it. Sportcopter M912 could do 180 on take off after certain altitude, but not sure about other gyros. In certain situations of course. All my flying in SC was In NE so not at high altitudes. I had to do this once to get out of a farm that I landed on in Vermont where it was surrounded by hills with trees. SC M912 is an amazing gyro. Wouldn’t be able to do same flying as I did out here around Denver.
 
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WaspAir

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What happened to yesterday's posts? I had two longish contributions, and there were others, and they're gone now (or at least they don't appear for me).
 
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WaspAir

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I'll try again.
the technique given in the FAA handbook. "If the prerotator is capable of spinning the rotor in excess of normal flight r.p.m., the stored energy may be used to enhance short-field performance. Once maximum rotor r.p.m. is attained, disengage the rotor drive, release the brakes, and apply power. As airspeed and rotor r.p.m. increase, apply additional power until full power is achieved. While remaining on the ground, accelerate the gyroplane to a speed just prior to VX. At that point, tilt the disk aft and increase the blade pitch to the normal in-flight setting. The climb should be at a speed just under VX until rotor r.p.m. has dropped to normal flight r.p.m. or the obstruction has been cleared. When the obstruction is no longer a factor, increase the airspeed to VY."
This technique is not possible in any gyroplane I know about.

First, it requires the ability to change collective pitch. from spin-up to flight. There are only two models sold the to the public that can do this, the J-2 and the A&S18A. I have substantial experience with both, and neither can permit this technique.

For the McCulloch J-2, one can pre-spin to 125% of flight rpm before starting the take-off roll. However, when you disengage the rotor drive, you also automatically go to full flight pitch at the same time (the same control disengages the transmission, releases tension on the belt drive clutch, and lifts the swash plate to flight collective position.). The take-off run is done at full throttle, while the rotor rpm decays (that's why you spin to 125%, so that you have a margin to allow for that decay). Anything less than full throttle lengthens the run and costs more rpm, seriously impeding take-off performance. There is no additional power to be added, and the blades are already in flight pitch, so this procedure is impossible.

For the A&S18A, you can prespin to 150% of flight rpm to store energy before any take-off roll. After disengaging the rotor drive, the blades must be kept in flat pitch during the run to minimize rpm loss (again, there is no increase in rpm possible). The run is done at full throttle, so there is no power to be added (again, less than full throttle will eat distance and cost rpm, so it's a stupid idea). When you choose to leave the ground, you DO NOT tilt the disc aft (the manual suggests putting the stick aft/left but that's only an inch of stick displacement to anticipate pitching/rolling after leaving the ground). You push a button on the throttle that releases the blades to take-off pitch (about double normal flight pitch). As you rise, coning causes the blade pitch to reduce to the normal flight value through pitch-cone coupling via the delta-three hinging. The departure from the ground is in a level attitude. For short fields, typically one pushes the button to leave the ground at about 45 (Vx is 50) but immediately establishes Vx to hold until the obstacle is cleared (the manual says "climb at best angle"). Rotor rpm will be down to the flight value in the blink of an eye (you will be nowhere near the obstacle when that happens) so that bit about holding just under Vx until one or the other is nonsense.

I think the author of this piece read the 18A manual, but didn't understand it, and has never seen the aircraft in operation, much less served as PIC.
 
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WaspAir

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I checked the POH data for the J-2 and the A&S18A (Standard Airworthiness models for which there is a reliable H-V diagram and for which I have the manuals) to see how much margin there is between the edge of the avoid region on the H-V curve and Vx.

In the 18A, between 25 and 50 feet AGL, the edge of the H-V avoid region is at 45 mph (it slopes back rapidly to slower speeds as you gain altitude, but 50 feet is the classic obstacle clearance scenario). Vx is 50, leaving only a 5 mph gap. Acceleration from 45 to 50 is very rapid with 180 hp through a constant speed prop and uses minimal distance. I am skeptical that the sacrifice of climb angle from using a lower speed will be compensated by avoiding the distance consumed in going from 45 to 50. (You can leave the ground as slow as zero airspeed in an 18A but you need significant airpseed to sustain a climb and stay H-V safe.)

In the J-2, the H-V edge at 50 feet AGL (chosen for comparable numbers with the 18A; the curve is a more complicated shape) is 55 mph while best angle is 62 for a 7 mph gap. In my experience, the climb rate below Vx is extremely poor. One can leave the ground crisply at only 45, but if you want to gain any altitude, you need to accelerate in ground effect before starting to climb out.
 
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thomasant

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Yes, the posts have disappeared.
I feel that the information in the FAA manual is a bit mixed up. The reason I quoted the FAA manual regarding the short field technique is to bring to notice the inapplicable portions in today's context. The large majority of gyroplanes today are single and two place tandems with semi-rigid rotor systems.

Regarding the climb speed below Vx, I believe it is because of the difference in Thrust produced vs Horse power produced for a certain speed in relation to the Power curve. Perhaps the experts can explain this further.
 

XXavier

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My post disappeared too...

But it's easy to reconstruct, as it was very short:


Vx is always less than Vy, that is the airspeed for best climb. I don't know how how Vx may be computed for a given aircraft, but concerning Vy, Richard von Mises, in his 'Theory of Flight (Dover Books, still in print...) links it to the curves for power required & power available. He was writing about FW aircraft, but I believe that this may be true for all heavier-than-air aircraft...

Captura de pantalla 2020-02-08 a las 23.31.15.png
 
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ventana7

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I’ll try to rewrite my posts too.

An F18 can rotate to vertical on takeoff, climb at 44,000 fpm AND be accelerating the entire time. It has a thrust to weight ratio so gigantic it can do this,. We’ve all seen aerobatic planes climb vertically at fullpower while decelerating the whole time until they do a wing over or tail slide. Those planes don’t posses enough excess thrust to make a vertical climb AND accelerate. For most of those aerobatic planes there is likely some angle of climb they could perform AND also accelerate at the same time.
If your gyro needs X amount of power to accelerate from breaking ground to reaching Vx in ground effect, then it will need X+ amount of power to both climb AND accelerate simoutanesly. How much is that + amount of power needed? And at what angle of climb? Obviously the steeper the climb angle the more excess power you need to also be accelerating.

With the F18 and the aerobatic plane they can both climb to a safe altitude and safely perform tests to determine their flight envelope.

Since gyros don’t rotate at a fixed airspeed and controlled rotor rpm, I can’t think of any safe way to duplicate a gyro takeoff and <Vx climb at a safe altitude while already in flight.

If you decide to test this during actual takeoff with your gyro as JC and Mike have you will be able to determine your flight envelope when you exceed it by finding your gyro sinking to the runway unable to trade your remaining altitude for adequate acceleration to allow a flare. You will be on the wrong end of the height velocity envelope.

If you begin a climb and find yourself not accelerating you will have just a second or two to react, and the movement from climb attitude to nose over to gain airspeed will be extremely steep.

And remember all this is so you can clear a 50’ tree by 5 more feet than normal procedures would.
If your margin is so close WHY ARE YOU TAKING OFF FROM AN UNSUITABLE AIRFIELD.
 

ventana7

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I’ll try to rewrite my posts too.

An F18 can rotate to vertical on takeoff, climb at 44,000 fpm AND be accelerating the entire time. It has a thrust to weight ratio so gigantic it can do this,. We’ve all seen aerobatic planes climb vertically at fullpower while decelerating the whole time until they do a wing over or tail slide. Those planes don’t posses enough excess thrust to make a vertical climb AND accelerate. For most of those aerobatic planes there is likely some angle of climb they could perform AND also accelerate at the same time.
If your gyro needs X amount of power to accelerate from breaking ground to reaching Vx in ground effect, then it will need X+ amount of power to both climb AND accelerate simoutanesly. How much is that + amount of power needed? And at what angle of climb? Obviously the steeper the climb angle the more excess power you need to also be accelerating.

With the F18 and the aerobatic plane they can both climb to a safe altitude and safely perform tests to determine their flight envelope.

Since gyros don’t rotate at a fixed airspeed and controlled rotor rpm, I can’t think of any safe way to duplicate a gyro takeoff and <Vx climb at a safe altitude while already in flight.

If you decide to test this during actual takeoff with your gyro as JC and Mike have you will be able to determine your flight envelope when you exceed it by finding your gyro sinking to the runway unable to trade your remaining altitude for adequate acceleration to allow a flare. You will be on the wrong end of the height velocity envelope.

If you begin a climb and find yourself not accelerating you will have just a second or two to react, and the movement from climb attitude to nose over to gain airspeed will be extremely steep.

And remember all this is so you can clear a 50’ tree by 5 more feet than normal procedures would.
If your margin is so close WHY ARE YOU TAKING OFF FROM AN UNSUITABLE AIRFIELD.
$.
 
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