alwynhartman

So heres the skinny,
Lets suppose for a moment that there exists two complete airframes with identical mtow and empty weight

Which of the two ships will exhibit better overall performance if one is a tandem rotor with specifically zero % overlapped rotors and the other is of conventional config?

minigundiplomat

I would guess that the tandem would have a [roughly] 5% power advantage based on not having that power diverted to anti-torque [tail rotor].

SASless

Why the concern about over lapping rotors?

Look back through the designs of Tandem rotor helicopters....almost all have over lapped blades.

Also....why not consider "side by side" rotor systems....like the Germans and Russians have tried.

alwynhartman

Well SAS, I thought it would be interesting to see how performance would compare if only one variable was changed, in this case, an extra rotor. To introduce overlap (proper practice or not) would not be the point of this exercise and would thus yield results not answering the original question.
However, it is indeed a good idea to have some overlap due to practical considerations as the ship will become super long if no overlap was present.

For side by side rotors the total flat plate drag area of the airframe would be much higher if all things are kept equal due to the outriggers on which the rotors ride. The machine itself would also become even bigger because it now also has a tail to park somewhere on the flight deck.

alwynhartman

So, I seem to have answered the question myself. Please feel free to critique.
Consider the following, design parameters:
MTOW 700 kg
Rotor Dia 7 m
Chord 0.18 m
Number of blades 3
Tip speed 192 m/s
Zero Lift Drag co. 0.008
To avoid super long calculations, lets employ basic aerodynamic design formulas.

First, what is the induced power for the above mentioned rotor?
P=(K(T)^1.5)/sqrt(2pA)
Assuming the industry standard starting value of 1.15 for K and standard air density of 1.225 kg/m3 for p, we can find that:
P= (1.15 * (700*9.81)^1.15)/sqrt(2*1.225*38.48)
= 67394 Watt = 67.39 kW
Plus the profile power calculated from the solidity ratio and zero lift drag co.
Solidity ratio = total blade area / disk area = 0.0491
Profile power = 0.125*p*A*(Vtip)^3*sol ratio*zero lift drag co
= (0.125*1.225*38.48*(192)^3*0.0491*0.008)/1000
= 16.39 kW
Now the total power becomes 83.78 kW for a conventional helicopter. Plus 10% for the anti torque rotor the total power = 83.78 * 1.1 = 92.15 kW

For the case of the tandem layout, induced power for each rotor will be calculated assuming a 50:50 sharing of thrust. Profile power remains the same but must be multiplied by 2 due to two rotors.
So, induced power per rotor = Pi = 23.83 kW
Adding profile power brings us to 23.83*2 + 16.39*2 = 80.44 kW

The above is still missing transmission losses and various other sources of parasitic drag but it does give a good view of what to expect. Thank you for the answers though