This is the Power Required for Prerotation

Dennis, iv been tinkern with this same stuff.
Im not go'n to argue your numbers, coz im not that smart, but in the lift columb, for 200rrpm you have 50.
50 wut?

And iv found if you set the AOA at bout -3.5 degrees ona 8H12, youll have 0 lift.
Theres still go'n to be more drag than a symetrical section, but no lift.
 
Dennis, iv been tinkern with this same stuff.
Im not go'n to argue your numbers, coz im not that smart, but in the lift columb, for 200rrpm you have 50.
50 wut?.

That is how much lift the rotorsystem will produce at 200 RPM's.


And iv found if you set the AOA at bout -3.5 degrees ona 8H12, youll have 0 lift.
Theres still go'n to be more drag than a symetrical section, but no lift.

Same thing I found out back around 1988 when I made the jump-takeoff Air Command. The problem is at that degrees the blades produces the same amount of drag as they do at zero when the lift is creating the drag. So there is no real benefit to pitching the blades that far down, unless your propeller is producing so much thrust you can't hold back with that lift load off the wheels during braking. I used a variable pitch propeller.
 
The problem is at that degrees the blades produces the same amount of drag as they do at zero when the lift is creating the drag. So there is no real benefit to pitching the blades that far down, unless your propeller is producing so much thrust you can't hold back with that lift load off the wheels during braking.

That's not quite accurate. Although the parasitic drag remains about the same, the induced drag resulting from producing lift is eleminated.
 
Thanks Dennis!!!
 
How is the lift so low? even at 350 rpm the lift is only about 150 pounds. What rpm do the blades have to spin at to get enough lift to fly? If my gyro flies at 320 rpm, isn't my lift going to be the gross weight of my gyro with me in it? like 500 pounds?

It seems like the HP required numbers on the right would be for about 50 rpms less than what is shown on the left. 150 rpm to 300 instead of 200-350.
 
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Dennis I understood your numbers, and solved some unanswered guestion for me and I also understand the lift coefficient due to the fact the blades are not yet under load, the lift factor will increase greatly once the aircraft is moving and the blades inter the collious effect. Is this correct thinking.

Dan
 
the lift factor will increase greatly once the aircraft is moving and the blades inter the collious effect. Is this correct thinking
The lift facter will increase greatly wen you increase the AOA.
 
Whoever did the rotor calculations that Dennis posted got the trends correct but the numbers are overly optimistic.

The NACA 8H12 was developed in the 1940s in the hope of improving helicopter performance (the “H” in the designation stands for helicopter) but in real world testing, offered no improvement over the NACA 0012. The 0012 is easier to fabricate and provides higher maximum lift.

The wind tunnel data, developed in a wind tunnel of essentially zero turbulence and using a polished 24” chord mahogany model does look good.8H12.JPGHowever, on a helicopter that creates its own turbulence (constantly varying angle of attack, centrifugal force acting on the boundary layer, etc.), the low drag “bucket” of the 8H12 vanishes (the dipper shaped region of L/D plot).

I measured the inflight lift/drag ratio of several different rotors a number of years ago*. At a steady 50 mph on the same gyro, a SkyWheels rotor had substantially better L/D ratio than most of its contemporaries; ~7:1 whereas the others were in the range of ~5:1.
The only rotor with better L/D ratio than SkyWheels was a DW rotor at ~8:1. The DW rotor uses the 0012 thickness distribution with cambered and reflexed meanline.

Bensen was the first to use an 8H12 and others followed the leader. The Bensen version of an 8H12 suffered from centrifugal air pumping through the gaps in the upper skin segments.

The 8H12 is easy to hand start but offers no performance advantage over an 0012.

Here is the calculated profile drag of a 7” chord x 23’ diameter rotor at zero lift using a realistic drag coefficient:

100 rpm----------0.17 hp
200 rpm----------1.35 hp
300 rpm----------4.6 hp
400 rpm----------10.92 hp
500 rpm----------21.32 hp

*Inflight L/D ratio of a rotor can be measured by aligning a straightedge with the rotor’s tip plane. The L/D ratio is then 1/sin of the straightedge angle. My tools were a straight board for a straightedge, a post for the board to pivot on, a set of drawstrings to align the board and a Smart Level.
 
Thank you Chuck!!
 
How is the lift so low? even at 350 rpm the lift is only about 150 pounds. What rpm do the blades have to spin at to get enough lift to fly? If my gyro flies at 320 rpm, isn't my lift going to be the gross weight of my gyro with me in it? like 500 pounds?

It seems like the HP required numbers on the right would be for about 50 rpms less than what is shown on the left. 150 rpm to 300 instead of 200-350.

John,

During the time of prerotation, the rotorblades are only being powered by an on-board mechanical device to the point where you can make a running take-off, or to the point of performing a jump take-off if you have variable pitch control.

If this were a helicopter, then much more on-board power and pitch would be required for hover.

Look at the lift (rotor thrust) figures. They increase exponentially with RPM. A helicopter rotor this diameter would need to spin around 550 RPM with around 4 degrees pitch, requiring around 40hp to hover.

Furthermore, when a gyroplane begins it's take-off run, the rotors are now in autorotation with a greater external source of power.
 
Whoever did the rotor calculations that Dennis posted got the trends correct but the numbers are overly optimistic..


Chuck, you always have to lead with an insult. Are you that insecure? I'm sorry your jealousy makes you the way you are. I did the calculations myself, as I said. I know that it kills you to think so, but after all, I did manage to build over 1700 rotorcraft all of my own design.... and counting.


Here is the calculated profile drag of a 7” chord x 23’ diameter rotor at zero lift using a realistic drag coefficient:

100 rpm----------0.17 hp
200 rpm----------1.35 hp
300 rpm----------4.6 hp
400 rpm----------10.92 hp
500 rpm----------21.32 hp


You need to read what I wrote. As I said, I did these calculation for my own project, not an existing rotorblade that is on the market now.

To try and make your gab, you failed to introduce a few facts that would support your figures as a comparison to mine.

First, the airfoils are not the same or the same size.

Next, I used only 2 degrees pitch.

Finally, your comparison is with rotorblades that have flat sawed-off tips, which create a great deal of drag at the absolute worse place you want to have resistance... at the end of a fulcrum. This type of tip belongs to the dinosaurs, and I don't know why anyone would still use it.

My rotorblade tips are designed to dramatically reduce the tip vortices to reduce this drag, making for a more efficient flying rotorsystem that requires less power for prerotation, and that is why my figures are as they are.

I did a lot of work making these calculations, and out of courtesy I shared them with the group. Yet you have to find a way to muddy the water with your insults, as usual.
 
Dennis my friend you are getting way to sensitive.
Although you are often confronted here regarding your past we appreciate your information and all of your new products.
I assure you Chuck has no reason to be jealous!!
So far I've never found a mistake Chuck has made and we learn every time he speaks.
 
OK Dennis, you did those calculations.

How about posting the equation you used to arrive at profile drag?

How does your tip vortex reduction scheme fit into the profile drag equation?
 
Dennis I understood your numbers, and solved some unanswered guestion for me and I also understand the lift coefficient due to the fact the blades are not yet under load, the lift factor will increase greatly once the aircraft is moving and the blades inter the collious effect. Is this correct thinking.
Dan

Dan,

I'm sure you mean Coriolis Effect. Yes, the lift will increase with airspeed, but Coriolis effect has nothing to do with the lift, unless I'm not understanding what you are asking.

Coriolis effect is a result of the rotorblade Center of Mass moving inward and outward from the center of rotation while the advancing and retreating blade rises and lowers, causing the blades to lead and lag. Or on a semi-ridged system to speed up and slow down equally in forward flight during each revolution. Yes, some lift differential does occur during this, but only enough to equalize the advancing blades lift to the retreating blades lift.
 
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