# Engine power for a ultralight

#### bryancobb

##### Junior Member
1) In my opinion the airflow directed downward by the driven region in the vertical descent is what slows the descent.

2) Significant how?

3) In my opinion the horsepower consumed by the rotor goes down as the indicated airspeed increases.
1) You are sooo correct Vance, according to everthing I have always been taught.
2)
3) That is true up to the ideal autorotation speed (best L/D). Above that speed induced drag goes up significantly and to keep RPM up, the rotor uses
more HP. The "HorsepoweSeconds Needed Curve looks like an upside down "U" if airspeed is the x axis and power required is the y axis.

#### bryancobb

##### Junior Member
In vertical autorotation, and from the reference plane of the descending rotor disk, there's no downward airflow

As JC writes in a recent post:

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

1) If you add forward speed, the power involved will be even bigger... But remember that we're talking about vertical autorotation....
1) NO we are not. We are talking about steady state autorotation at the best autoration forward speed, where best L/D occurs and the shallowest descent angle occurs. This is the emergency procedure airspeed to be used if an engine failure happens. A vertical zero-airspeed descent is definitely
NOT the speed you would choose if your engine failed. If your engine fails, you should try to get to the best speed pretty quickly without abrupt stick movements. That will give you the most options for a dead stick landing safely in a suitable spot and give the most saved-up energy for that final flare.

#### XXavier

##### Member
1) NO we are not. We are talking about steady state autorotation at the best autoration forward speed, where best L/D occurs and the shallowest descent angle occurs. This is the emergency procedure airspeed to be used if an engine failure happens. A vertical zero-airspeed descent is definitely
NOT the speed you would choose if your engine failed. If your engine fails, you should try to get to the best speed pretty quickly without abrupt stick movements. That will give you the most options for a dead stick landing safely in a suitable spot and give the most saved-up energy for that final flare.

I was talking about vertical autorotation, and Vance too. See his messages 50, 52, 54...

#### XXavier

##### Member
1) You are sooo correct Vance, according to everthing I have always been taught.
2)
3) That is true up to the ideal autorotation speed (best L/D). Above that speed induced drag goes up significantly and to keep RPM up, the rotor uses
more HP. The "HorsepoweSeconds Needed Curve looks like an upside down "U" if airspeed is the x axis and power required is the y axis.

Theres no L/D to speak of, because the gyro is in purely vertical descent. The mentioned autorotation is vertical, as is clear in Vance's words. Vance writes: 1) In my opinion the airflow directed downward by the driven region in the vertical descent is what slows the descent.

#### Vance

##### Gyroplane CFI
Theres no L/D to speak of, because the gyro is in purely vertical descent. The mentioned autorotation is vertical, as is clear in Vance's words. Vance writes: 1) In my opinion the airflow directed downward by the driven region in the vertical descent is what slows the descent.
The definition I use for Lift to drag ratio is the amount of lift generated by an airfoil moving through the air divided by the drag produced.

In my opinion as long as an airfoil is moving through the air with a positive angle of attack lift is produced and there is a lift to drag ratio.

I have not seen my gyroplane rotor stop during a vertical descent so in my opinion the airfoil is still producing lift and has an L/D ratio even in a vertical descent in a gyroplane.

#### Jean Claude

##### Junior Member
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

#### Vance

##### Gyroplane CFI
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
The driven region of my rotor blades keep going around in a vertical descent.

If the driven region of the rotor blade didn’t have drag they would not need a driving region to keep them going around.

In my opinion if the airfoil didn’t produce lift perpendicular to the direction of the speed I would descend much faster than 1,400 to 1,600 feet per minute in a vertical descent in The Predator.

#### WaspAir

##### Supreme Allied Gyro CFI
Keep in mind...during climb, energy from the burned gasoline becomes prop thrust which pushes the gyro upward through the air as it climbs, with the rotor tilted aft for the climb. The energy in the additional airflow, put there by gasoline spins the rotor even though you are going uphill.

In an unpowered, steady-state, autorotational descent, the only energy you can use to spin the rotor is the energy of falling. If you stopped the rotor, you'd fall like a brick. When airflow from falling spins the rotor, it acts like a parachute. There is definitely a low weight that cannot autorotate, in a helicopter or a gyro.
The discussion has moved on since I last checked, but I think a reply to this earlier post is important. My assertion is that a gyro too light to autorotate effectively in a power-off descent will not be able to autorotate effectively in a power-on climb to get into that situation, so the problem is not real. Gyros autorotate at all times; helicopters do not, and the comparison doesn't work.

Responding to your statements, the prop does not push the gyro uphill in a climb. The prop pushes the aircraft forward creating airspeed. The rotor then produces the lift that enables the climb, in response to that airspeed and and the g-load. The rotor is tilted aft while climbing, but also while cruising, and descending to land, even with the throttle chopped, because it takes its airflow from underneath the disc. Gyros do not climb as a result of downward prop thrust any more than a Cessna hangs on its prop to climb. The rotor/wing does the lifting. Do the trig for a typical small deck angle and you will see that the vertical component of prop thrust is a tiny fraction of aircraft weight.

G-load doesn't change when the engine dies. Blade profile, rotor diameter, collective pitch, and every other rotor characteristic remain the same. As with any glider, potential energy from height is converted to airspeed and flight continues.

If a gyroplane is too light for the rotor design to maintain adequate rpm in autorotation in descent, it will likewise not be able to maintain autorotation with that same rotor and weight in cruise or climb. You just can't get into the situation that you're worried about.

The helicopter comparison is invalid, because a helicopter is not in autorotation during climb and cruise, and its rotor characteristics (if unwisely chosen) could permit powered-rotor flight at a given load but not permit effective autorotation with that same rotor in an emergency.

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#### WaspAir

##### Supreme Allied Gyro CFI
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
I think you are using the wrong "speed" for that definition. We loosely talk about the lift of the "rotor" for simplicity, as if it were a fixed wing, but total rotor lift is the sum of the lift from the individual blades.
The speed/direction used to define lift is the airspeed seen by the airfoil, not the velocity of the airframe. The rotorblade airfoil is spinning quickly as well as descending. If it made zero lift in this conditon, it would be stalled, and would produce no autorotative force to keep it spinning. That's obviously not what happens.

P.S. Maybe we could be clearer in our terms. In a vertical descent, the glide ratio is zero. But there is clearly both lift and drag being produced, so can one not talk about L/D in that sense?

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#### Aviator168

##### Newbie
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
No. As far as the blades are concerned, the relative wind direction is not up and the blades are generating lift. Part of the lift componment is driving the rotor to spin. Just like a FW glider, part of the lift of the wing is pulling the glider forward.

Maybe we could be clearer in our terms. In a vertical descent, the glide ratio is zero.
To the blades of the a gyro, there is no vertical descent, the blades are always gliding. Just like the wings of a FW glider.

#### WaspAir

##### Supreme Allied Gyro CFI
In the sailplane world, we use the term glide ratio to refer to the airframe as a whole, which is what I was trying to get across -- there could be more clarity if we use one term for the whole ship, and another for the airfoils that produce that behavior, since in rotorcraft the two values are not at all the same.

#### Aviator168

##### Newbie
In the sailplane world, we use the term glide ratio to refer to the airframe as a whole, which is what I was trying to get across -- there could be more clarity if we use one term for the whole ship, and another for the airfoils that produce that behavior, since in rotorcraft the two values are not at all the same.
When analysing aero dynamic for rotorcraft, you can't use one term for the whole ship because not all the subcomponment is moving in unison.. When the whole rotorcraft is in vertical descent, there is a lot more happening on the blade than just generating lift. From blade tip to root, the blade from generating lift to stalling.

#### WaspAir

##### Supreme Allied Gyro CFI
When analysing aero dynamic for rotorcraft, you can't use one term for the whole ship because not all the subcomponment is moving in unison.. When the whole rotorcraft is in vertical descent, there is a lot more happening on the blade than just generating lift. From blade tip to root, the blade from generating lift to stalling.
I think you are now trying to tell me what I have already previously said first in post #69 above. If you don't like my proposed terms to separate whole craft behavior from blade behavior, don't use them. It is, however, universally accepted to describe whole ship flight behavior accurately with terms such as vertical descent, glide ratio, and indicated airspeed, without simultaneously attempting to characterize the flow at any given blade station accurately by those same measures. My initial point was that the glide ratio in a vertical descent is zero but lift is not, hence L/D and glide ratio are different measures. Do you disagree with that?

#### Aviator168

##### Newbie
I just like to look at things in a different way. Like, I wouldn't look at vertical descent of a gyro with a glide ratio of zero.

#### WaspAir

##### Supreme Allied Gyro CFI
Well, it seems pretty natural to me to see it as zero feet forward for each foot of altitude lost. My sailplane goes 43 feet forward for each foot lost at best L/D speed (glide ratio 43:1, a descent angle from horizontal of about 1.3 degrees), a VASI approach is about 19 feet forward for each foot down (19:1, or 3 degrees) my Bell in autorotation might optimally go 4 feet forward for each foot down (4:1, a descent angle of about 14 degrees), the Space Shuttle with split rudder open would go about 1 foot foward for each foot lost (1:1, a 45 degree descent angle), and a vertically descending gyro is going zero forward for each foot of descent (0:1, a 90 degree angle). (It"s the cotangent of the descent angle.)

If that doesn't seem natural to you, we can just move on.

#### Aviator168

##### Newbie
If that doesn't seem natural to you, we can just move on.
I agree and I can see where you are coming from.

#### thomasant

##### Member
Common sense tells me that a vertical descent is exactly as it implies. No forward motion for any amount of vertical motion depending on the ROD. Obviously that is a 0 glide ratio.

#### thomasant

##### Member
430lb I indicated is total weight including fuel and payload. Empty weight is about 200lb. I am just asking the power required for such a gyroplane.
If you go by Martin Hollman's book of "Modern Gyroplane Design" he advocates 9 lb/hp. That would give you a requirement of 47.78 hp for adequate performance. Empty weight of 200 lbs is very optimistic.

P.S. I paid \$200 for that book 10 years ago when I first thought about getting into gyroplanes.

#### XXavier

##### Member
Common sense tells me that a vertical descent is exactly as it implies. No forward motion for any amount of vertical motion depending on the ROD. Obviously that is a 0 glide ratio.

The discussion on the L/D needs some re-focusing, I believe.

If we are talking about the L/D of a given aircraft, it's obvious that it's zero for that craft descending vertically, since L is defined, by convention, as the component of the aerodynamic force perpendicular to the relative wind, clearly zero for a stone (or a gyro) falling in a vertical trajectory. If we speak about the 'glide ratio', it's also obvious that it's a ratio between velocities, (or distances, for a given interval of time) and not forces, but always numerically identical to the L/D.

But if we are talking about lift and drag of the blades in vertical autorotation, the L/D ratio is obviously not zero in most (if not all) elementary sections of the blade...

Anyway, when L/D was first invoked in this thread, it was related to the aircraft as a whole... It's clearly zero in vertical descent, in autorotation or not...

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#### Vance

##### Gyroplane CFI
The discussion on the L/D needs some re-focusing, I believe.

If we are talking about the L/D of a given aircraft, it's obvious that it's zero for that craft descending vertically, since L is defined, by convention, as the component of the aerodynamic force perpendicular to the relative wind, clearly zero for a stone (or a gyro) falling in a vertical trajectory. If we speak about the 'glide ratio', it's also obvious that it's a ratio between velocities, (or distances, for a given interval of time) and not forces, but always numerically identical to the L/D.

But if we are talking about lift and drag of the blades in vertical autorotation, the L/D ratio is obviously not zero in most (if not all) elementary sections of the blade...

Anyway, when L/D was first invoked in this thread, it was related to the aircraft as a whole... It's clearly zero in vertical descent, in autorotation or not...
I agree with J.R. and Antony.

The definition I use for Lift to drag ratio (L/D) is the lift of an airfoil divided by the drag of the airfoil.

In my opinion it does not go to zero because a gyroplane is not moving forward (a vertical descent) because part of the airfoil is still providing lift and the airfoil still has drag as it provides lift.

The definition I use for Glide Ratio is the distance the aircraft flies forward divided by the sink.

There is much more involved in Glide Ratio than the L/D ratio of the airfoil.

In my opinion Glide Ratio is what goes to zero in a vertical descent.