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Old 21st Apr 2022, 12:11
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Originally Posted by JimEli View Post
Prouty, R., Helicopter Performance Stability, and Control, 1986.

On page 306, Prouty estimates the total equivalent flat plate area of an example UH-60 sized rotorcraft to be 19.3 ft-sq (zero angle of attack). The contribution of the horizontal stabilator to the total drag is estimated to be 0.2 ft-sq. or roughly 1% of the total area at 115 kts. So the contribution of the horizontal stabilizer is really insignificant.
That 1% drag is at zero AoA. A forward CG will induce a noze down attitude that increases the iduced drag so the drag will be higher than this for a forward CG and closer to this value at aft CG.

I did a calculation of the difference in main rotor lift needed on a 10.6T NH90. Between max forward CG and max aft CG there is a reduced need for lift from the main rotor of about 327kg due to the reduced need for “negative” lift from the stabilizer.
At 3000’ +15C and at MCP the TAS increase is about 4kt between 10.5t and 10t gross weight, so for 327kg we gain about 2.6kt from the reduced load on the main rotor only.
There is no data to find the extra drag that max forward CG cost due to the induced drag from the stabilizer, but probably more than zero as lift seldom comes for free
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