Blade manufacturing story

[cavuman1]

I hate to admit that the last time I opened a Physics textbook was in college - one half-century ago! I have about five hours right seat time in a Bell 206 LR and was always fascinated at the exquisite design as well as inestimable strength of helicopter blades. Too lazy to look up the necessary equations to calculate load distributions, I found the following on Copters.com: "The rotating blades of a helicopter produce very high centrifugal loads on the rotor head and blade attachment assemblies. As a matter of interest, centrifugal loads may be from 6 to 12 tons at the blade root of two to four passenger helicopters. Larger helicopters may develop up to 40 tons of centrifugal load on each blade root. In rotary-wing aircraft, centrifugal force is the dominant force affecting the rotor system. All other forces act to modify this force."

No wonder these magnificent machines are known as "frantic palm trees"! I am astounded that they hold together as well as their fixed-wing cousins, which, as we all know, are simply "groups of rivets flying in close formation"!

- Ed

For the Geeks among us:
If you know the velocity of the object, simply use the following formula:

Code:

F = mv²/r

where:

• Code:

  F

  is the force (expressed in newtons);
• Code:

  m

  is the mass of the object;
• Code:

  v

  is the velocity; and
• Code:

  r

  is the radius.

If you know only the angular velocity

Code:

ω

, you can recalculate it to normal velocity by simply multiplying it by the circumference of the circular path. Use the following equation:

Code:

v = ω2πr

in case your ω is in Hz (1/s).

Or the formula:

Code:

v = ωr

for ω in rad/s

Time to grab a beer....