At possible questions point - I got a little confused. If the blade near the centre is stalled, I had then assumed that drive would come further out. But looking at the diagram, it looks something like stall/near stall/lift/ lift/ drag/ drag. I always kind of understood the part of autorotate since the tomahawk flight sim on the zx spectrum (don't laugh) - as many times I ran out of fuel, or lost engines. It was common then to cut all collective, and then flare on landing. I understood how to do that. Your video is showing me the science behind how I did that. I guess for most people they learn the theory and then practice. I'm assuming that in hot/high conditions, there is a devil in the detail. Autorotation of getting speed into the blade in thin air easier, but I'm assuming you will have to burn up energy faster/harder when you flair because you'll need perhaps more collective/AoA. I have always liked flight sims, I think I understand flight models from doing, and not from theory - which is probably its own disaster :) Also, Mr Kumar asks a good Q below, because in the tomahawk sim, you certainly had to angle forward and ideally keep above 60kts for maintaining rotor speed. I am assuming that forward tilt provides additional energy in terms of rotor speed. Somewhere in my head I have the idea that above a certain speed, even at low pitch/AoA, you then get some lift from the rotor.. Many thanks for the excellent vids!
Yes the blade stalls at the root as the relative rotation speed is very slow coupled with a high rate of decent flow. As you go towards the centre of the blade you start to balance this out by increasing that relative rotational speed (it increases as the blade travels further in the same time as the tiny route the blade root takes). But as you get closer to the tip the blade rotational speed is so high in comparison to the rate of decent flow you end up with a shallow relative airflow and therefore a small angle of attack and a lift vector tilted rearward (due to that shallow relative airflow) which essentially creates drag.
The drag vector component always seems arbitrary not only in this video but in every drawing including the FAA materials. How is the drag vector component length determined and how can its accuracy be determined? Can it be calculated? Is it irrelevant as long as a drag vector is shown/included?
Why and how does the inter blade move slower than the outer blade??? It is one solid attachment from one end to the other. Does not make physical sense.
You are now discovering the difference between angular velocity and linear velocity. They have the same angular velocity. The entire blade rotates 360 degrees in the same amount of time, but since the tip of the blade is farther from the center, it must move faster to cover a greater distance in the same amount of time as the root. The root moves around the center in a small circle while the tip moves in a much bigger circle, but it takes the same amount of time. Angular velocity is degrees over time, or rotations over time (RPM). Linear velocity is distance over time. Time is constant but the distance covered is much higher for the tip than the root due to a higher radius from the center. This results in a much higher linear velocity for the tip.
A little confused...why looking at points A, B, C, D do we assume that the lift vector has the same magnitude at each? Using the lift proportional to 0.5*p*V^2, V is increasing going toward the tip yielding (all things being equal like lift coefficient at each station), higher lift. What am I missing?
Sorry, but I think this is rubbish. Forget all your diagrams, take a model airplane propeller, hold it against an airflow, it will spin with a negative incidence. Try spinning it the opposite way round, it will slow down and spin the opposite direction. Rotors of gyrocopters have a negative 1.5deg incidence. Think of the rotors as glider wings, instead of going forward they spin in a circle. And as we know gliders wings point downwards, otherwise they would stall. Basically your diagrams are incorrect. Firstly it would be more clear if you show the centre of forces going through the centre of the spar. The force on the foil should be shown at right angles to the plane of the foil. You will then see that in all the cases the drag is always backwards
This explanation is not rubbish. It is very much correct. The lift force should not be drawn at right angles from the plane of the foil, but at right angles from the relative wind, like he has done. Together with the drag it will create the resultant force. He is always drawing the drag backwards in his diagrams, however, with the angle of the lift pointing forwards enough (90 degrees of the relative wind) the resultant force is pointing forward, dragging the rotor forward driving the rotordisk. This is a correct explanation and I would advise you to watch it again and maybe find some other explanation which will show you very much the same thing.
The rotation of the rotor is one part of the relative wind, the other is the air coming from below due to the descending profile during a autorotation. This relative wind is at an angle causing the lift vector to point forward. Due to it being a rotor the outward part of the rotor is experiencing a much higher horizontal velocity the the part closer to the rotorhub. Due to this difference in speed at the ends of the rotor the resultant aerodynamic force (combination of drag and lift) point to the rear, decelerating the blade. However, closer to the rotorhub the horizontal airflow velocity is less, however the vertical velocity remains the same. This increases the angle of the relative wind causing the lift vector to point more forward. The resultant force will therefor still be pointing forward, even though the drag is still pointing to the rear, accelerating the blade. Setting the correct pitch on the rotorblade will cause a equilibrium between the driving and driven part of the rotor. You are trading altitude for rotorspeed.
I finally got my head round this. What I know understand is that the aerodynamic neutral plane of an aerofoil is perhaps 3 deg down from the geometric plane. Therefore a positive incidence to the aerodynamic neutral plane is in fact a negative 1deg. I define the aerodynamic neutral plane of an aerofoil is when the foil goes neither up nor down in an airflow.
![HIWWHEE | OHM [Official MV]](http://i.ytimg.com/vi/OJJLtTLqLag/mqdefault.jpg)
Super explanation of the process. Thanks.
At possible questions point - I got a little confused.
If the blade near the centre is stalled, I had then assumed that drive would come further out. But looking at the diagram, it looks something like stall/near stall/lift/ lift/ drag/ drag.
I always kind of understood the part of autorotate since the tomahawk flight sim on the zx spectrum (don't laugh) - as many times I ran out of fuel, or lost engines. It was common then to cut all collective, and then flare on landing. I understood how to do that. Your video is showing me the science behind how I did that. I guess for most people they learn the theory and then practice.
I'm assuming that in hot/high conditions, there is a devil in the detail. Autorotation of getting speed into the blade in thin air easier, but I'm assuming you will have to burn up energy faster/harder when you flair because you'll need perhaps more collective/AoA.
I have always liked flight sims, I think I understand flight models from doing, and not from theory - which is probably its own disaster :)
Also, Mr Kumar asks a good Q below, because in the tomahawk sim, you certainly had to angle forward and ideally keep above 60kts for maintaining rotor speed. I am assuming that forward tilt provides additional energy in terms of rotor speed. Somewhere in my head I have the idea that above a certain speed, even at low pitch/AoA, you then get some lift from the rotor..
Many thanks for the excellent vids!
Yes the blade stalls at the root as the relative rotation speed is very slow coupled with a high rate of decent flow. As you go towards the centre of the blade you start to balance this out by increasing that relative rotational speed (it increases as the blade travels further in the same time as the tiny route the blade root takes). But as you get closer to the tip the blade rotational speed is so high in comparison to the rate of decent flow you end up with a shallow relative airflow and therefore a small angle of attack and a lift vector tilted rearward (due to that shallow relative airflow) which essentially creates drag.
The drag vector component always seems arbitrary not only in this video but in every drawing including the FAA materials. How is the drag vector component length determined and how can its accuracy be determined? Can it be calculated? Is it irrelevant as long as a drag vector is shown/included?
Why does driven section move inboard when we raise the collective level ?
Very aptly explained.
I would really appreciate if you can touch upon autorotation with forward speed.
Apologies it took so long to upload auto in fwd flight. Hope it’s still useful.
Why and how does the inter blade move slower than the outer blade??? It is one solid attachment from one end to the other.
Does not make physical sense.
They’re probably connected through a planetary gear of a different ratio or something.
You are now discovering the difference between angular velocity and linear velocity. They have the same angular velocity. The entire blade rotates 360 degrees in the same amount of time, but since the tip of the blade is farther from the center, it must move faster to cover a greater distance in the same amount of time as the root. The root moves around the center in a small circle while the tip moves in a much bigger circle, but it takes the same amount of time. Angular velocity is degrees over time, or rotations over time (RPM). Linear velocity is distance over time. Time is constant but the distance covered is much higher for the tip than the root due to a higher radius from the center. This results in a much higher linear velocity for the tip.
A little confused...why looking at points A, B, C, D do we assume that the lift vector has the same magnitude at each? Using the lift proportional to 0.5*p*V^2, V is increasing going toward the tip yielding (all things being equal like lift coefficient at each station), higher lift. What am I missing?
Amazing explanation, can you also explain auto rotation in forward flight
Apologies it took so long but I finally got time to upload the fwd flight one
Sorry, but I think this is rubbish.
Forget all your diagrams, take a model airplane propeller, hold it against an airflow, it will spin with a negative incidence. Try spinning it the opposite way round, it will slow down and spin the opposite direction.
Rotors of gyrocopters have a negative 1.5deg incidence.
Think of the rotors as glider wings, instead of going forward they spin in a circle. And as we know gliders wings point downwards, otherwise they would stall.
Basically your diagrams are incorrect. Firstly it would be more clear if you show the centre of forces going through the centre of the spar. The force on the foil should be shown at right angles to the plane of the foil. You will then see that in all the cases the drag is always backwards
This explanation is not rubbish. It is very much correct. The lift force should not be drawn at right angles from the plane of the foil, but at right angles from the relative wind, like he has done. Together with the drag it will create the resultant force. He is always drawing the drag backwards in his diagrams, however, with the angle of the lift pointing forwards enough (90 degrees of the relative wind) the resultant force is pointing forward, dragging the rotor forward driving the rotordisk. This is a correct explanation and I would advise you to watch it again and maybe find some other explanation which will show you very much the same thing.
The rotation of the rotor is one part of the relative wind, the other is the air coming from below due to the descending profile during a autorotation. This relative wind is at an angle causing the lift vector to point forward. Due to it being a rotor the outward part of the rotor is experiencing a much higher horizontal velocity the the part closer to the rotorhub. Due to this difference in speed at the ends of the rotor the resultant aerodynamic force (combination of drag and lift) point to the rear, decelerating the blade. However, closer to the rotorhub the horizontal airflow velocity is less, however the vertical velocity remains the same. This increases the angle of the relative wind causing the lift vector to point more forward. The resultant force will therefor still be pointing forward, even though the drag is still pointing to the rear, accelerating the blade. Setting the correct pitch on the rotorblade will cause a equilibrium between the driving and driven part of the rotor. You are trading altitude for rotorspeed.
I finally got my head round this.
What I know understand is that the aerodynamic neutral plane of an aerofoil is perhaps 3 deg down from the geometric plane. Therefore a positive incidence to the aerodynamic neutral plane is in fact a negative 1deg.
I define the aerodynamic neutral plane of an aerofoil is when the foil goes neither up nor down in an airflow.