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Number of blades
5

Number of blades

Number of blades

(OP)
What is the effect of different number of blades on various helicopter rotors. I am interested in the hovering power efficiency - static thrust etc.? Articles/reports/books. This information is very difficult to find in books or on the Internet.

RE: Number of blades

Your referring to the "Figure of Merit" ratio which is the measure of a rotors efficiency, as well as "Hover Power Req'd", and Power correction factors, etc etc etc.

The best way to answer this question is to email me at moygr1@home.com (that's a one not an L) and state exactly what you need to know.


This gets deep pretty quick.

Figure of Merit Ratio:

.707 * (Ct^ 3/2) / Cq <-- Torque Coefficent

or .707 * Sqrt(CT^3)/ Cq

same thing


now your Cq is for the main rotor and is calculated by:

Q / (pPiR^2)(omega R)^2 <-- again density and Pi

of which to find your torque you take:

5252 * SHP / RPM
which is SHP = Shaft horsepower

so if i had a 2,5544 shp at the main bevel gear for the main rotor mast, at 258 rpm the torque would be 51,990.728 ft/lbs of torque. Some times you need to estimate this if you dont have exact figures for the transmissions. An easy way to "guess" is to take total engine power, take 25% of that, and figure torque at that rating. usually you lose 1/4 of total power due to transmission losses and accessory drive losses. etc.

what id do for quick estimates is this: Say the UH-60 Blackhawk, which is what these numbers are for.

Take 1650 Eng. SHP *2 = 3300 hp
then .77 * 3300 = 2,541shp

5252 * 2541 / 258 = 51,726 not bad when you say its only an esitmate, how far is this off by?
.005 one half of one percent!
  

now omega is angular velocity in Radians per sec. and R is the rotor radius.


Hover induced Power req'd.

HPi = (T/550) * sqrt(w/(2pB^2)) <--p = air density for sealevel use .002378 for 4000 feet use .0021109 and for say 14000 feet use .0015455 ( have a sheet i can fax you if you desire.

T= thrust
w = disc loading lbs/ft^2
B = Tip loss factor

w = thrust / (Pi*R^2) <-----disc area


B = 1 - sqrt (Ct/2R) <---Thrust Coefficient/rotor diameter


while your writing:

the Ct =  T/(piR^2)* p * (Vt^2)

(piR^2) <-- Disc area

p <-- Air Density in Slugs per cubic foot. note: the pi above is the mathematical "pi" 3.141,,,,, not the density.

Vt^2 <-- tip speed squared which is 2(pi)Rn

where R = radius in feet
where n = revolutions per second

so to get revolutions per second. take the RPM and * by 60

if you have a fax, i can send out some stuff for you.


Joe
Efax 419-818-1506

RE: Number of blades

There is another Tip loss factor, one based off of geometery.

B = 1 - Ctip <-- Ctip is the Tip chord (width) in feet.
        ----
         2R  <-- R is the rotor Radius




a common number in the industry is to have a tip loss factor of .95 to .97 dont expect 1.00 or unity, you wont get it.



Also, Thrust of a rotor, eh,


T = (pAv) * v^2

rho or air density, p times the Disc Area, times the induced velocity of the rotor, get that quantity, then take the induced velocity, square it, and then multiply to two results together, this is good for approximations.


This is where you will have to experiment a bit to find a unity of sorts. Unless you feel like doing some calculus.


v = sqrt(T/(2pA))


RE: Number of blades

The simple way to put it what do you want in a rotor system. the more blades you have ,can reduce the blade eliment loading but your autorotation will be over less distance (with the hughes 300 look at your heels thats where you go a Bell 206 look at the lower part of the windshild and thats where you go,and so on.) then you have the parts count, the more blades you have the more grips,bearings,links,dampers,pieces tec.so its more than a numbers game but a mission or goal set to accomplish.how much storage space? a multi bladed a/c needs more room unless its a folding system then you get more complex and the last thing the world needs is a more complex helo to maintane, so what ever you do keep it simple. more is some times not better just diffrent.

RE: Number of blades

I asked the helicopter test pilot which do he
prefer. He said that the difference is huge in terms
of vibration levels. He prefers tri baldes compared
to two even for light helicopter. The pilots feel
vibration down to their spine, and the helicopter is
their working space, so we should hear their voice, also.


RE: Number of blades

Here's are three reasonable equations that show the effect of the number of blades on thrust.  Equations from Stepniewski "Rotary Wings Aerodynamics".

Eqn 1) T = 4 * pi * R^2 * density * integral(vi^2 * x, dx, x=R0 to Rtip)

Eqn 2) vi = Omega * { - a * s / 16 + sqrt [ (a * s / 16)^2 + a * s * x * theta / 8]}

Eqn 3) s = Nb * R * c / (pi * R^2)


T - rotor thrust (lb)
pi - 3.14...
density - 0.0023769 lb-sec^2/ft^4 (slug/ft^3)
R - rotor radius (ft)
R0 - blade cutout radius (ft)
Rtip - rotor radius reduced for tip effects (ft) ~ 0.98R
vi - induced velocity (ft/sec)
x - differential rotor radius (ft)
Omega - rotor speed (radian/sec)
a - lift curve slope (5.2 / radian)
s - solidity; ratio of blade area to disk area
theta - collective (radians)
Nb - number of blades
c - chord (ft)

The solidity is proportional to the number of blade.  The induced velocity is almost proportional to solidity.  Induced velocity is proportional to solidity if you replace collective (theta) by collective divided by solitity.  And thrust is proportional to induced velocity squared.  From this, you can see that keeping everything the same, that thrust is approximatly proportional to Nb^2.

RE: Number of blades

2
One reason from a pilot's point of view is that more blades equates to smaller rotor diameter for the same thrust (lift). When landing clearings are a big problem as in the northern provinces, smaller blade radius is preferred.

Having said that, one disadvantage is that the rotary momentum, or angular inertia of the unpowered rotor system during autorotation is less in a rotor disk of smaller radius. Hence the incredible difference betwen the D model MD-500 and the Bell 206B in autorotational bottom end recovery to a soft touchdown.

But the main reason for more blades is this: at high speed, high altitude, hot weather, and heavy loads, a helicopter needs to maintain a higher peak angle of attack on the retreating blades, in order to balance lift. At some point, the angle becomes too high, the retreating blade stalls, and the pilot will experience two or three things. First, a sudden lurch to the left. Secondly, a sudden lifting of the nose, as gyroscopic precession acts 90 deg later to force the rotor disk upwards at the front, automatically correcting for the retreating blade stall, thus un-stalling it, and thirdly, a possible slight moistening of the undershorts.

So add more blades and you can go faster and carry more.

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