other

We calculate the main rotor thrust. If we look at the surface (area F), swept the screw as it rotates, like an impenetrable plane, we see that this pressure is applied on top of plane pi, and the bottom pressure r2 and p-2 larger pixels.

From the second law of mechanics we know that the mass is accelerated only when acted upon by a force kakaya-. Moreover, this force equals mass times acceleration, and is directed toward the acceleration (in this case downwards).

What is this power? On the one hand it is obvious that this force is the result of exposure to air screws. On the other hand, is it? the strength of the third law of mechanics must comply with the equal in magnitude and opposite impact on the air screw. The latter is none other than the thrust screw.

However, if we look at the dynamometer, measure the actual thrust of the screw, we determine that our count more inaccurate. In fact, the thrust will be less, as we believe the perfect job screws and do not take into account the energy losses due to friction and twisting stream of air for the screw.

In fact, the air particles are suitable for the screw, having not only an inductive speed in the axial direction perpendicular to the plane of rotation but also the rate of twist. Therefore, when calculating rates of induction, inductive and dropping them i2 counted as twisting the air during rotation of the rotor.

The formula traction lift coefficient cy similar to the coefficient of traction; flight speed corresponds to the peripheral speed of the ends of the propeller blades having a radius r and angular velocity, the wing area 5 area corresponds to drive the sweeping screw lg2. The coefficient is determined from the curve of the purging screw at various angles of attack.

The magnitude of the dimensionless ratio of thrust for a particular already created screws operating in this mode, you can calculate by dividing the rotor thrust T, expressed in kilograms, by the product of the other parameters of the screw, which also has the dimension kg traction.

We found that if the lift of the aircraft is created by dropping down the air wing, the lift force is created by a helicopter dropping down the air rotor.

When the helicopter has a forward speed, then, of course, the amount of air ejected downwards increases.

For this reason, the investment of the same power rotor helicopter, having a translational velocity develops more thrust than the screw hanging helicopter.

Conversely, to generate the same thrust on the helicopter rotor having a forward speed, it is necessary to transmit less power than the screw hanging helicopter.

Reduction of power requirement with increasing velocity occurs only up to a certain value of the velocity at which the increase in the air resistance of the helicopter motion, not only consumes power gain, but even the latest demands increase.

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