Fatal - Magni M24 Orion 05-TD, Guillestre, near Mont-Dauphin - St-Crépin Airfield, Hautes, France 08 MAR 2022

Abid

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But that's just it, isn't it? Trike controls are SO different, there's little chance of airplane guys reverting to airplane habits.
As you say, gyro controls basically seem the same, but they often need to be used very differently. IMO, gyro CFIs really need to make extra sure their airplane students really understand what's going on in a gyro, and not just with a head nod.

Well yes. I had never thought about it that way but I see your point.
 

ferranrosello

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I attempted to comment on your post#19; it did not show the quote as i thought it should/would.
You are the one that stated you can demonstrate a 3000' rate of climb and would loose rrpm; not me.
I am under the impression that as the density altitude increases (air molecules become less dense) the rotor rpm will increase to compensate for the same all up weight; and the engine produces less power and the propeller produces less thrust for the same engine rpm. I cannot attest to higher altitudes, I rarely fly above 1500' agl and even in the summer months don't get to DA of 5000'.
Allow me to ask this another way, Please.
On your next flight, have the goal of maintaining a constant airspeed while S&L and climbing and descending. Maintain the IAS during the entire phase of the test; and report your findings regarding RRPM, please. Thank You.
When teaching aerodynamic principles, in order to explain what is going on in very simple terms, we say that the lift equals the weight in all non-accelerated flying conditions. That includes climbs.

However, this is not true. A steady climb is a non-accelerated flying condition, however the load factor is always under 1 g.

The power available in our gyrocopters is too low to let a climb with a steep angle. Our climbing angles are so small that there is nearly no difference in the g load. But if you have power enough to climb in a steep angle, the lift required will be much lower. This is because the engine thrust is partially compensating the aircraft weight, and the g load decreases.

An extreme case is a fighter in a vertical flight path. The lift would be 0.

Remember that what keep our rotors turning is lift. Less lift implies less rotor rpm. I’m sure that with a rate of climb of 3000 fpm at 60 mph the rotor rpm drop will be noticeable. But with a rate of climb of 500-1000 fpm there is no difference.

Of course, in a descent the g load is increased, and the rotor rpm will increase at the very same IAS.
 

WaspAir

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I have some scepticism about a sustained 3000 fpm climb rate.

What pitch angle (with respect to the ground) is assumed here? Not the climb angle (about 35 degrees for that claimed airspeed and climb rate), but the direction of the thrust line?

What weight is assumed for aircraft and pilot?

What total propeller thrust is assumed at that airspeed?

It should be a simple matter to determine the proportion of lift carried by the rotor and that carried by the prop in a steady state climb (after any initial zoom is complete) if some of this information is known.
 

ferranrosello

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Actually, the g load is decreased in a straight path descent, just like a climb. But I agree with everything else in the post.
Aaron, usually we descend with reduced power in order to avoid too much air speed. However, in a steep descend (without reaching 90º) the engine thrust is partially adding to aircraft weight, and lift required to compensate it will be higher.
 

Vance

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Actually, the g load is decreased in a straight path descent, just like a climb. But I agree with everything else in the post.
I don’t know what would cause a decrease in g load in an un-accelerated descent.

I do not notice a decrease in rotor rpm when I am in a stabilized descent to land.
 

ferranrosello

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It’s just what math’s say about it. Usually if you descend with power the airspeed increases and the rotor rpm follows and increases too. If you reduce power and hold the air speed there is no increase in rotor rpm.
 

J ro

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The accident aircraft, I believe.


1656880-large.jpg
Any idea why this Jro crashed?
 

Aaron R

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I don’t know what would cause a decrease in g load in an un-accelerated descent.
For unaccelerated flight (steady climb, descent, or level), the forces resolve. So for level flight, lift = weight. For either climb or descent, lift is decreased because the weight vector partially opposes either thrust or drag. Lift is only produced equal to the weight component opposing lift (weight*cos(gamma)). Load factor (g load) is lift/weight. This diagram may help. The aircraft type doesn't matter. It depicts a climb, but if you draw out the descent, the lift reduction works out the same.
 

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WaspAir

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For a steady state climb in a gyro, the effect is negligible because the angles are typically small. Assume a 1000 fpm climb at 60 mph, and the flight path is only about 11 degrees above horizontal. The cosine of that is 0.98. I have yet to meet a pilot who can distinguish 0.98 g from 1.0 g. Cockpit g-meters cannot be read to that precision.

It is important to remember that pitch attitude and climb angle are not the same. Just pointing the nose up does not mean you will climb at the incline of the nose (extreme slow flight is a good example, with high pitch, high power, and struggling to stay level). Gyros don't climb like rockets or fighter jets (or even like helicopters). Likewise, few gyro pilots dive at high power settings, descending more commonly with throttle back and little thrust.

Brief zooms with high pitch and high climb rate are not "steady state". Practically speaking, I suggest as a rule of thumb, you should able to hold the same pitch, power, airspeed, and rate of climb (or descent) for a full minute to consider it steady and avoid being misled by transient conditions.
 

Aaron R

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the effect is negligible because the angles are typically small
True, I was nitpicking, just to make the point that the effect is the same for climb and descent. The fact is negligible in most conditions. But in extreme cases it may come into play. For example, when the ELA demo guy does his dive, I'll bet the spiraling maneuver is to avoid the unloading that would occur in a straight path dive. I have no desire to do that, but I think it doesn't hurt to understand the effect.
 

Jean Claude

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I don’t know what would cause a decrease in g load in an un-accelerated descent.
I do not notice a decrease in rotor rpm when I am in a stabilized descent to land.
In a stabilized descent for landing, the propeller does not push, so the rotor thrust is lightly reduced.
With a usual angle of attack of 10 degrees, the difference of rotor thrust is so small that the effect on the Rrpm is almost unnoticeable (√(COS 10°) = 0.99)

Usually if you descend with power the airspeed increases and the rotor rpm follows
In flight the forward speed are no effect on the rotor thrust, thus the Rrpm should not change.
It is only due to the extension of the stalled area on the retreating blade, that a slight increase of rrpm occurs on its own, by compensation.
For example if 353 rpm at 50 mph, then 360 rpm at 100 mph
 
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ferranrosello

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In a stabilized descent for landing, the propeller does not push, so the rotor thrust is lightly reduced.
With a usual angle of attack of 10 degrees, the difference of rotor thrust is so small that the effect on the Rrpm is almost unnoticeable (√(COS 10°) = 0.99)


In flight the forward speed are no effect on the rotor thrust, thus the Rrpm should not change.
It is only due to the extension of the stalled area on the retreating blade, that a slight increase of rrpm occurs on its own, by compensation.
For example if 353 rpm at 50 mph, then 360 rpm at 100 mph
Yes!!! and a reduction of the Angle of Attack in the advancing blade. The effect is noticeable!!!
 

OuterMarker

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I completely agree with Doug. There have been so many papers written on the science of rotorcraft design and rotorcraft flight. The math programs are out there to check the safety of the gyroplane of interest. Word of mouth or slick marketing can place a pilot in a dangerous position where dollars are king. And in some cases, no matter how experienced a pilot is, it's just not enough to have a successful outcome on that last flight. When I applied for my repairman's certificate through the FAA, the examiner, after looking through my documentation on my build, he said he had a one question oral test. The question? "What is the definition of "experimental?" His answer was this: "every flight you make in your experimental aircraft is just that, an experiment which hopefully always ends successfully. If not, then we, the FAA come out to investigate." That chat sobered me up. I was a builder, mechanic, and chief pilot. I went back to check my math...
 

Jean Claude

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Yes!!! and a reduction of the Angle of Attack in the advancing blade. The effect is noticeable!!!
With an imaginary blade without stalling, the Rrpm would not increase at all, despite the forward speed changes.
 

TyroGyro

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Any idea why this Jro crashed?
The accident aircraft was originally misidentifed apparently as a J-RO.

It seems it was a Magni M24.


I have updated the thread title accordingly.

Pilot-Instructor Marc Picard, aged 66. RIP
 
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Tyger

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It seems to me that most of the prior discussion on this thread has little to do with this particular accident. The pilot was certainly very experienced. Does anyone have any inkling of what actually went wrong?
 

TyroGyro

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It seems the BEA originally identified it as 05-OQ (a DTA J-RO), but now they are saying it was 05-TD (a Magni M24)...


although several sites are still noting it as 05-OQ. Confusing...
 
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