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Hinged three-bladed rotor like C30 avoids the drag vibrations but unfortunately not "ground resonance". Hinged two-bladed rotor restore drag vibrations, now transformed into cyclic differences in individual centrifugal forces, and keeps the serious downside of ground resonance
2. Engine power for a ultralight

Yes, undoubtedly
3. Engine power for a ultralight

Words and definitions changes depending on the messages. For example when you say that in vertical descent the speed forward is zero. Since "forward" speed refers to movement, then "forward" is below in a vertical descent and forward speed is the speed of descent.
4. Engine power for a ultralight

The aerodynamic definition of "lift" is the force perpendicular to the direction of the speed. So, in vertical descent, the rotor has zero lift, just a vertical drag, and L/D = 0
5. Electric Pre-Rotator

If 200 rpm of the electric spinner seems sufficient to you, why do not they seem sufficient to you with the mechanical spinner, since then the thrust of the propeller and the gripp of the wheels discharged by the rotor, do not yet create problem ? 200 rpm instead of 300rpm requires only 44% of...
6. Engine power for a ultralight

Yes, but my words were "To cancel the impact speed ..." During this phase of slowing down, the state is no longer "steady" and the helicopter can use part of the kinetic energy of a heavy rotor, while the gyro cannot: A heavy rotor don't help him
7. Engine power for a ultralight

During a gliding descent, aircrafts are not like a brick which regularly increases its speed of fall during the descent. To say that the descent is stabilized means that the potential energy is dissipated as it goes. To cancel the impact speed of a helicopter when descending vertically, it is...
8. Engine power for a ultralight

The minimum mass for the helicopter concerns the mass of the blades, not that of the load. It is simply a matter of giving the pilot enough time to reduce the collective pitch before the Rrpm decreases below that the stall of the blades. The rotor of a gyroplane does not have this concern...
9. Engine power for a ultralight

Certainly. And today, a 12 Hp engine weighs about 8 kg and it allows to takeoff in less than 300ft , rise at 600 ft/mn and flight 50 mph in cruise.
10. Engine power for a ultralight

The rotor adds about 5 kW of profile drag, which requires a more powerful and heavier motor, which requires a more solid structure, and more fuel. This additional weight now increases the induced drag which ... etc, etc. It's just a "snowball effect" PS Thank you for making me discover this...
11. Engine power for a ultralight

A FW can fly with much less power. My JCD03 of 175 lbs and span 23 feets was powered by a 12 hp SOLO engine and a propeller 28 inchs rotating at 5800 rpm. But my calculations always gave more than 35 hp for a gyro.

Entering the characteristics of MTOsport (392 kg) in my spreadsheet, I obtain a maximum L/D of 3.7 at 29 m/s At this moment, Parasitic drag = 360N Rotor profile drag = 426 N Induced drag = 250 N

I also can't post an Excel file, while I did in the past A possible formula for profile power when forward speed is nul is: P = ¼ Cd.ρ.c.ω^3. R^4 with Cd = 0.011 So, when ρ = 1.225 kg/m3, c= 0.18 m R= 3.5 m and ω = 36.6 rd/s, then Profile power = 4400 w, let's say 4800 w at 20 m/s forward

This is only true for an FW, but false for a gyro because the profile power of a rotor keep almost constant . In my opinion, relying on the cD min of 8H12 measured in a wind tunnel with low turbulence, H. Dudda underestimates the drag of the rotors . Thus, to justify the power absorbed in...

Just change Cd*S from 1 (with no fairing) to 0.35 (about clean) at 50 mph allows 25% power less in cruise. Are you sure that everything is done to avoid the separations of the flow behind the pod?

Compressibility phenomena are not worrying on slow blade tips of our gyros unlike fast helicopters. We better worry flow separations on the rear parts of the fuselage. Much more to gain with less research. Unfortunately, we look only the cosmetic of nose .
17. Engine power for a ultralight

Xavier's calculation is correct, because the finesse he evaluates corresponds to that of the nose up angle of 18 degrees. But he neglects add the weak parasitic drag of the airframe at this time. On the other hand, his speed of 18 m/s is suitable for lifting the weight of 430 lbs at this angle...
18. Engine power for a ultralight

No simple formulas, Xavier. Just my many simulations with the best ratio power/weight of engine, the best ratio load/ weight of airframe and the best ratio weight / diameter of rotor. I have never obtained a result lower than 35 hp to takeoff with a nose-up rotor of 18 degrees. This angle...
19. Engine power for a ultralight

It depends on the diameter of the rotor and the diameter of the propeller. With a 23 'x 7 "rotor and a 63" diameter propeller, more than 35 hp are required for a total weight of 430 lbs
20. Lift balancing on the Gyrhino

Wyne, Because of the lift, the blades always remain above the plane of the hub and browses a coning. With the forward speed, the browsed cone is tilted backwards relative to the plane of the hub. It is due to the blade rises from 6 o clock. until 12 o clock when it is on the advancing side, then...