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Calculating required disc loading


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#1 voyagerB

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Posted 21 May 2017 - 05:14

Can anyone tell me the disc loading figure needed to lift say a 2200 lbs (1000 kg) off the ground, assuming that there must me some sort of 'equilibrium' between prop diameter and amount of hp.

 

For instance, the Robinson R22 uses a 130 hp (397 Nm / 5.24 liter) engine that drives a 7.67 m. (25 ft) diameter prop to lift a maximum of 620 kg (1366 lbs). The R22 has a disc loading of 2.61 lb/ft² (or 13.7 kg/m²).

 

The Robinson R44 four-seater (max. takeoff weight 998 kg) has a disc load of 10.5 kg per sq.m (2.15 lb/ft²). But then with a prop diameter of 10.1 meter, using a 245 hp engine (almost twice the hp of that of the R22).

 

A typical RC heli may weigh 800 grams (0.8Kg) and have a rotor diameter of 70cm (0.7m), giving a disk loading of: 0.8/(pi x (0.7 x 0.7) is 2.1 kg per sq.m. or 0.43 lbs per sq.ft.

 

Anyone who has info regarding contra-rotating props and how their disc load compares to normal props.

 

220px-Contra-rotating_propellers.gif


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#2 r22butters

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Posted 21 May 2017 - 11:02

Damn!

Is this the extra credit question at the end of a test, or is teacher punishing you for calling a rotor "props"? :o
The only dream I have left is to live long enough to see the pilot shortage. Its been about fifteen years since they first told me it was coming, so,...

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#3 voyagerB

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Posted 21 May 2017 - 11:58

Damn!

Is this the extra credit question at the end of a test, or is teacher punishing you for calling a rotor "props"? :o

 

As a Dutchman I am not too familiar with the right English terms. I use rotor and prop for the same thing that spins. Wrong? I know the difference between FPP and VPP...



#4 r22butters

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Posted 21 May 2017 - 13:05

...I know the difference between FPP and VPP...


Ah' everyone knows that, but what about APP?
The only dream I have left is to live long enough to see the pilot shortage. Its been about fifteen years since they first told me it was coming, so,...

Aaaaaaaany day now! :D

#5 voyagerB

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Posted 21 May 2017 - 13:17

Ah' everyone knows that, but what about APP?

 

The one you use to hail an Uber flying taxi?



#6 r22butters

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Posted 21 May 2017 - 15:47

The one you use to hail an Uber flying taxi?


Close, they're the dudes that gas me up before I fly. :D
The only dream I have left is to live long enough to see the pilot shortage. Its been about fifteen years since they first told me it was coming, so,...

Aaaaaaaany day now! :D

#7 iChris

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Posted 22 May 2017 - 02:07

Can anyone tell me the disc loading figure needed to lift say a 2200 lbs (1000 kg) off the ground, assuming that there must me some sort of 'equilibrium' between prop diameter and amount of hp.

 

Anyone who has info regarding contra-rotating props and how their disc load compares to normal props.

 

Power required by the rotor to produce thrust equal to weight can be determined from the momentum theory based on Newton’s for every action there is an equal and opposite reaction. For a helicopter in hovering flight, the action is the development of a rotor thrust equal to the gross weight. The reaction is in the acceleration of a mass of air from above the rotor to a velocity below the rotor. In other words, we need to move a mass of air equal to the gross weight each second.

 

T= (m/sec) (∆v)   or    E/sec = (force) (velocity) = T v ft. lbs./sec

 

Since 550 ft. lbs./sec is equivalent to one horsepower, the ideal power is:

 

hp = Tv/550    since v = √DL/2p   we end with the following approximation for the power required to hover (sea level) taking into account a Figure of Merit (F.M.) of 0.75.

 

Losses that occur from the engine, transmission, tail rotor, generator, hydraulic system, etc. must be made up for by the engine and are additive to the power required by the rotor.

 

CLICK BELOW TO ENLARGE

 

6mf7ggN.jpg


Edited by iChris, 25 August 2017 - 20:02.

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Regards,

Chris

#8 voyagerB

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Posted 22 May 2017 - 06:42

I noticed that if you do: disk load x disk area for the Robinson R22 and R44 and you compensate for max. takeoff weight, you end up with approx. the same figure.

 

R22 (TO weight 620 kg) disk load x disk area: 1364.2

R44 (TO weight: 998 kg) disk load x disk area: 2198.6

 

998: 620 = 1.6

1.6 x 1364.2 is 2196

 

Then you can start calculating how much oomph and hp you need to take a certain weight off the ground. Like I mentioned in my previous mail the step from propelling the R22 to the R44 is huge (130 hp / 320 Nm versus 245 hp / 688 Nm). Sorry for the metrics. And sorry for being a newbie to the stuff, learning as I study.



#9 adam32

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Posted 22 May 2017 - 10:10

Study for what???

#10 voyagerB

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Posted 22 May 2017 - 11:29

Study for what???

study = reading



#11 r22butters

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Posted 22 May 2017 - 11:41

Not really getting the point in making such calculations, I feel I must ask. Does it matter that your figures are off a bit?

Max (allowable) takeoff power in the R22 is 131 hp

Max Gross Weight is 1370 lbs (622 kg)

Max Gross Weight R44 II is 2500 lbs (1134 kg)

Plus, what book are you reading?

Edited by r22butters, 22 May 2017 - 11:59.

The only dream I have left is to live long enough to see the pilot shortage. Its been about fifteen years since they first told me it was coming, so,...

Aaaaaaaany day now! :D

#12 voyagerB

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Posted 23 May 2017 - 01:20

What is the percentage lost to powering the tail rotor? I read different figures: 10 to 30%...



#13 iChris

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Posted 23 May 2017 - 03:55

What is the percentage lost to powering the tail rotor? I read different figures: 10 to 30%...

 

The main rotor absorbs most of the helicopters power, but there are other losses as well. The engine and transmission absorb 3% to 5% of the total power with turbine engines, or 5% to 9% with reciprocating engines. The turbine engine has larger transmission losses since its high rotational speed requires more reduction, whereas the piston engine has significant losses for cooling.

 

The tail rotor absorbs about 7% to 9% of the total helicopter power in a hover and as much as 10% to 15% during max sideward flight to hold heading.  You’ll have to see what flight regime they were testing to get that 30% number on the tail rotor.  

 

Additional loss of about 2% due to aerodynamic interference rotor-fuselage and rotor-rotor. The tail rotor and aerodynamic interference power losses are much smaller for the helicopter in forward flight. Because of the fuselage download produced by the rotor wake, the rotor thrust is 4% to 7% larger than the gross weight, producing a corresponding increase in required power.

 

What you need to gain from this is the relationships between the dynamics, such as, how the Disk Load affects the performance and power required, not always an exact number, but an approximation. I can’t count the number of flight test that I have done, were after all the data had been compiled, the Chief Engineer (name prefixed with "Dr.") comes back and says, we don’t really understand why the data doesn’t match our calculations, we’ll need to do additional testing to understand what the dynamics are and what’s missing.

 

In your search for exactness, don’t feel bad if it sometimes alludes you. Sometimes it’s not achievable under the circumstances.

 

 

It is the mark of an educated mind to rest satisfied with the degree of precision which the nature of the subject admits and not to seek exactness where only an approximation is possible.

 

Aristotle, Philosopher


Edited by iChris, 23 May 2017 - 04:09.

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Chris

#14 voyagerB

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Posted 22 August 2017 - 07:30

Chris, you wrote: "Since 550 ft. lbs./sec is equivalent to one horsepower"...

 

Does that mean: 1 hp to lift one lb 550 ft up in the air per second?

 

I am still in the subject of how to calculate the amount of hp or kW needed to lift a certain weight based upon the aircraft's disk loading (lb per ft2). I came across this:

http://www.heli-chai...namics_101.html

 

Where an "empiracly defined formula to calculate the thrust loading (after McCormick)" was introduced. TL [lb/hp]= 8.6859 * PL.

Thrust loading is defined as lb per hp = 8.6859 x disk loading (power / disk area)... 

 

It specifies two examples: 

25 hp with the use of a 12 ft diameter rotor yields 347 lb of lift.

300 hp with the use of a 6 ft diameter rotor yields 1,251 lb of lift. 

 

So, I am still trying to figure out what minimum power is required to lift 900 lb with a 10 ft rotor that has a disk loading of around 11.5 (lb per ft2)... Left aside the power loss on the tail rotor.

 

I got lost... But perhaps I can say that if a 300 hp with a 6 ft diameter rotor generates 1251 lb of lift... the same amount of hp with the use of a 10 ft diameter rotor can lift 2.77 as much: 3470 lb. Meaning that in order to lift 900 lb approx. 78 hp is required.

 

Correct me if I am wrong....



#15 AS350 pilot

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Posted 22 August 2017 - 12:27

Sometimes (most times) I just like to boil things down to their simplest. As pilots we pick things up and we put things down. That's really it!



#16 iChris

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Posted 25 August 2017 - 23:54

Chris, you wrote: "Since 550 ft. lbs./sec is equivalent to one horsepower"...

 

Does that mean: 1 hp to lift one lb 550 ft up in the air per second?

 

It specifies two examples: 

25 hp with the use of a 12 ft diameter rotor yields 347 lb of lift.

300 hp with the use of a 6 ft diameter rotor yields 1,251 lb of lift. 

 

So, I am still trying to figure out what minimum power is required to lift 900 lb with a 10 ft rotor that has a disk loading of around 11.5 (lb per ft2)... Left aside the power loss on the tail rotor.

 

I got lost... But perhaps I can say that if a 300 hp with a 6 ft diameter rotor generates 1251 lb of lift... the same amount of hp with the use of a 10 ft diameter rotor can lift 2.77 as much: 3470 lb. Meaning that in order to lift 900 lb approx. 78 hp is required.

 

Correct me if I am wrong....

 

Power defined as the "rate of doing work" or the "rate of using energy". 1HP being the work needed to raise 550 pounds a distance of one foot in one second or the rate of work needed to move 33,000 pounds a distance of one foot in one minute.

 

I6MCU4c.png

 

The equation below (taken from post #7 above) is a good first approximation for the “rotor power” required at sea level.

 

Under perfect conditions the actually power and induced power would be equal, Figure of Merit (F.M. = 1.0); however, under real world conditions F.M. tops out at 0.75 - 0.82, State-of-the-art rotors may have figures of merit approaching 0.82.

 

If you plug those numbers in from their two examples, you’ll end up around the same HP. Remember we’re talking about the power required by the rotor. You’ll want to take into account other losses that would increase the overall power required.

 

kocy3kX.png


Edited by iChris, 26 August 2017 - 00:23.

Regards,

Chris




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