Thanks for posting the formula. Let me see if I understand it
correctly. For the hypothetical small helicopter:

Blades are 10' long and the cg is half the length (no tip weights). R
would be 5' plus the distance from the center of the shaft to the
blade mounts. I'll call R as being 5.5' in this example.

I'll say the blades are 25 lbs each and the RPM is 600 (450 MPH tip
speed).

FORCE = (W * R * RPM^2) / 2900 = (25 * 5.5 * 600^2) / 2900
= 17069 lbs.

This is assuming that the units are supposed to be lbs and feet.

The result is less than I thought, but still pretty impressive.

By adding 2 lb tip weights, the CG force increases to about 23,500
lbs. That's a pretty big jump, but the biggest jumps obviously come
with the RPM.

Did I do it right?

Dennis.


(Hennie) wrote:

The centrifugal force generated by a rotor blade
is (W*r*rpm^2)/2900 W is the blade weight. r is the distance from
the hub center to the CG of the blade

Dennis Hawkins
n4mwd AT amsat DOT org (humans know what to do)

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