#### MaD Dog

##### Flapping Member

- Joined
- May 10, 2004

- Messages
- 97

- Location
- Trinity, N.C.

- Aircraft
- PropCopter...Shoffman Co-Ax

- Total Flight Time
- 400 hrs on Gyro

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- Thread starter MaD Dog
- Start date

- Joined
- May 10, 2004

- Messages
- 97

- Location
- Trinity, N.C.

- Aircraft
- PropCopter...Shoffman Co-Ax

- Total Flight Time
- 400 hrs on Gyro

- Joined
- Oct 30, 2003

- Messages
- 1,346

- Location
- Deep River, CT

- Aircraft
- none currently

- Total Flight Time
- 80 gyro; 65 helicopter(R22)

Al hit the nail on the head.

What are you calculating?

What are you calculating?

lift = Cz * 1/2 *R * V2 * S

cz depends on airfoil, i guess its the 0.002378

R radius

V wind speed

S surface (main disc - central disc)

but this equation (i gave to dean first then, deleted it) is not accurrate as it is used to calculate the lift for WINDMILLS, assuming a wind normal to the disc.

like christian said in other post, another method should be more accurate

Victor, your and Al's formulas are derived from momentum theory; Force = Mass x Acceleration.

The factor; 0.002378 is air density, slugs/ft³ (weight of air, lb. per ft³ divided by acceleration of gravity)

Momentum theory assumes uniform flow through the rotor disc and ignores the influence of ground effect.

Power required is down wash velocity x gross weight divided by 550. The answer will be optimistic by 10% or so and to that must be added rotor blade profile drag power.

The factor; 0.002378 is air density, slugs/ft³ (weight of air, lb. per ft³ divided by acceleration of gravity)

Momentum theory assumes uniform flow through the rotor disc and ignores the influence of ground effect.

Power required is down wash velocity x gross weight divided by 550. The answer will be optimistic by 10% or so and to that must be added rotor blade profile drag power.

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ok for air density (i m dumb considering a Cx of 0.002 )

why 550 ? can you explain or give a link ?

i have various formula and the only method i consider is : perform them all and design a little above of all.. (hope my english is understandable)

thanks

James Watt discovered that Welsh mine ponies could move coal carts at a rate of 550 ft/lb/sec over the course of a working day. A drawbar pull of 100 lb., moving at the rate of 5.5 fps (3.75 mph) = 1 HP.

Therefore, one Imperial HP = 550ft/lb/sec.

Velocity (fps) x force (lb.) divided by 550 = HP.

*That would be 76 kg/meters/sec but you round off to 75 kg/m/sec.

During the prime of the Citroen 2 CV, I saw horses pulling sugar beet wagons in the south of France that would equal 3 Welsh mine ponies.

The 2 CV, by the way, probably was 15-20 hp. The 2 CV came from the tax collector's formula.

Therefore, one Imperial HP = 550ft/lb/sec.

Velocity (fps) x force (lb.) divided by 550 = HP.

*That would be 76 kg/meters/sec but you round off to 75 kg/m/sec.

During the prime of the Citroen 2 CV, I saw horses pulling sugar beet wagons in the south of France that would equal 3 Welsh mine ponies.

The 2 CV, by the way, probably was 15-20 hp. The 2 CV came from the tax collector's formula.

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LOL okay C beaty, i m in Metric system

thanks

thanks

The ~10% figure results from slipstream rotation and non uniform velocity over the rotor disc.

I agree that blade profile power, fuselage negative lift and transmission losses can substantially increase hovering power. Profile power alone can amount to as much as 40% of induced power, depending on blade loading and resultant tip speed.

Although, I don't visit PPrune as often, I usually seek out your posts there.

For those of you who don't know of him, Nick is the S-92 Program Manager at Sikorsky Aircraft.

For decades, Western aerodynamic texts have mathematically shown that the tail rotor wastes 8 - 10% of the power. Recently, Kamov has stated in an article entitled Aerodynamic Features of Coaxial Configuration Helicopter that the tail rotor wastes 10-12% of total power. In addition, the latest aerodynamic text (year 2000), by Leishman (University of Maryland) supports Kamov's position.

If a craft's gross weight to empty weight is 2:1, then the above might result in a difference in payload of up to 24%.

Dave J.

Welcome Nick, looking forward to your input in the forum.

hello dave, christian, nick i v not be introcuded but pleased any way .. did i miss something ? coax or not coax? hoax ?

cheers

cheers

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