# Vertical Rate of Climb

During climb the helicopter center of gravity moves along the flight trajectory with the velocity V The climbing velocity vector can be repre­sented as the vector sum

where is the horizontal component of the velocity;

V is the vertical rate of climb.

У

The horizontal component of the velocity equals V ^ cos 0, but since the climb angle is usually small and does not exceed 10°-12°, then ^ V^.

This means that the same power is expended on horizontal displacement of the helicopter during climb as is expended in horizontal flight at the same speed.

Analyzing (32) from this viewpoint, we can say that the excess power AN = GVy is expended on vertical displacement of the helicopter. Hence, knowing the helicopter weight and the excess power, we find the vertical rate of climb

The excess power can be found from the power required and available curves for helicopter horizontal flight (see Figure 63a). The maximal excess power corresponds to the economical speed (for the Mi-1, = 80 km/hr or

about 23 m/sec). Therefore, climbing should be performed at the economical speed. Moreover, from the power required and available curves for the Mi-1, we can conclude that vertical climb is impossible when using rated engine power. This means that the static ceiling when using rated power is equal to zero, i. e., helicopter hovering is possible only in the air cushion influence zone.

When using takeoff power vertical climb is possible, but the rate of climb will be lower than when climbing along an inclined trajectory. Conse­quently, this once again confirms that climbing should be performed along an inclined trajectory. Vertical climbing is performed only when it is necessary to clear surrounding obstacles. It must be kept in mind that takeoff power can be used only for a brief period, not to exceed five minutes.

Increase of the flight altitude involves change of the power required and the power available, and, therefore, change of the vertical rate of climb.