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As a former helicopter pilot, I have some familiarity with translationalbrian whatcott <inet@intellisys.net>
lift. The reason a helicopter experiences it has to do with lift on an
airfoil being proportional to the square of the velocity of air over the
wing. In a helicopter, this velocity varies as a function of r, the distance
from the hub (axis of rotation), the rotational speed of the rotor blade, and
the speed of the helicopter through the air. Consider a section of the
airfoil some distance r from the hub such that, in a hover, the velocity of
air over this section is 50 m/s. In this hover, the velocity will be 50 m/s
through the entire rotation of the blade. Two opposite points on the blade
will have an amount of lift equal to some constants times (50 m/s) squared,
or a total of 2 times k times 50 squared, or 5000 k.
Once the helicopter starts to move, however, the relative velocity of air
over the rotor will change as the blade rotates. If the helicopter is moving
forward at, say 10 m/s, (about the speed pilots notice the phenomenon) then
on one side, the advancing rotor blade will see air approaching at 50 + 10 =
60 m/s; on the other side, the retreating blade will see air approaching at
50 - 10 = 40 m/s. Now the total lift from these two points is 60 squared
times k plus 40 squared times k, or 3600 k + 1600 k = 5200 k, an increase of
200 k units of lift. Summing this type of contribution over the length of the
blade and the whole rotational cycle gives the total translational lift.
Well, it's not quite that simple, but I hope that gives some sense of the
phenomenon.
Bob Yeend