Best margin above a take-off obstacle

Than you for your explanation Jean C.
The assumption you are doing about getting airborne at 58 km/h seems reasonable. However my point is what if I don´t get airborne until being very closed to 75 km/h. Then the induced drag during the take off roll would be weaker and would use less than 50 meters to reach 75 km/h.

This is a variable that depends very much of the take technic used.
There are no discussions about this question. The safest Take off technique is to keep a moderate nose up attitude in the take off roll. In my experience, the lower the attitude the better the take off performance. To get airborne at 58 km/h implies a moderate nose up attitude...

PS: Totally agree about the head wind point. And if you are flying at high density altitude Vx will be faster that what is stated in the Manual... But this is another question, our machines flying manuals are very poor, consequently you need to solve the lack of information with knowledge.
 
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Completely agree Ferran and it’s something that isn’t discussed enough in training. We have an absurd situation where we might test our student at the end of his training with knowledge of items in the POH but as far as it relates to take off performance literally have no idea how that data was achievedand therefore can’t have trained the pilot to achieve it....
 
However my point is what if I don´t get airborne until being very closed to 75 km/h. Then the induced drag during the take off roll would be weaker and would use less than 50 meters to reach 75 km/h.
Ferran,
The drag of the rotor during the takeoff roll is not parasitic: It mainly translates the energy absorbed to increase the kinetic energy of rotation of the rotor: Rotation energy = Drag * distance
Therefore, not taking the air before 75 km / h will give indeed less rotor drag, but will increase the distance of run. So, the 50 meters gained for reach 75 km/h after takeoff are precedently lossed during the rolling.
 
Jean Claude,. I've never said nothing about parasitic drag. I've talked about induced drag... And the induced drag for a flying gyrocopter is much bigger at 58 km/h than at 75. But if the gyro is rolling below 70 km/h is still much lower.

What you say about the energy transfer to the rotor would be correct if the efficiency in this transfer was the same at 58 km/h than at higher speeds. However the efficiency is better at faster speeds.
Ferràn
 
I misspoke, Ferràn. I meant that the induced drag you are talking about is a useful drag (not parasitic) even when you are rolling, because it is only the lift of the rotor which produces the increase of the rpm.
Thus, rolling faster, the energy required (induced drag * distance) will remain the same to reach the same rpm.
 
Ok, thank you, Jean. I agree that the energy required will be the same. However not all induced drag is transformed in rotor rpm... The efficiency depends on rotor blades angle of attack, that is to say forward air speed.

It would be interesting to know how interact airspeed, rotor rpm antirotative lift and autorotative lift in the take off roll, in order to find what is the best combination to get the best performance.
 
Ferràn,
I simulated with good accuracy, the standard takeoff of an ELA07 (100 hp) pre-launched at 260 rpm, with the stick fully drawn, ie hub at + 18 °, then balanced on the main gear,
Assumed conditions: Mass 450 kg, Initial propeller thrust 2000 N, Parasite drag : 0.7 m2, Rolling coef. on grass 4%, Ground effect for Z / R = 0.6 True rotor inertia 170 kg m2 or imaginary 120 kg m2 , Aerod. pitch setting 2.8 degrees:

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With the rotor inertia of 120 kg m2, instead 170 kg m2, the distance is shorter of 40 meters and the take-off speed little lower.
This shows:
Difference energy of the prop. thrust = 1900 N (average)* 40 meters = 76 kJ
Difference in rolling energy = 2250 N (average) * 4% * 40 meters = 3.6 kJ
Difference on forward speed energy = 0.5 * 450 Kg * (22.2^2 - 20.8^2) = 13.5 kJ

Thus, energy of forward speed gived to the rotor during the take-off run is 76 - 3.6 - 13.5 = 58.9 kJ
while rotor energy difference = 0.5 * (170-120) * (36.6)^2 = 33.5 kJ

It suggests an average "pneumatic efficience" of 0.57
 
Phil,
I did not seek the optimum. I entered the X and Z positions of the hub and main wheels relative to G and that of the thrust line of the propeller. When the sum of moments around the main whells becomes positive, I enter line by line (every 0.2 s) the angle of the hub to maintain it zero. To be honest, it's a tedious juggling between the rotor spreadsheet (which gives the lift, drag and extra torque as a function of the angle and speed forward) and that of the gyro (which calculates the rpm reached, the forward acceleration, the speed reached and the moment on the main wheels). Automatic correction of the propeller thrust and parasitic drag with the speed forward, correction of the rolling drag with the lightening of the wheels on the ground.
The rotor spreadsheet also takes into account the ground effect according to the forward speed (not negligible)

The results agree very well with the take-off distances and forward speed of the Cierva C30 precisely measured by the Aeronautical Research Committee, under various conditions of pre-launch and load (Report n°1859 )
 
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Thank you very much, Jean C. It is a great job. I don´t know how you are calculating everything, however the results seem to be realistic.
Ferràn
 
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