How does an aft center of gravity affect the minimum control speed on ground (Vmcg)? Does it reduce all V-speeds?
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1$\begingroup$ By changing the load on the nose wheel. More forward cg position requires less rudder for lateral control. $\endgroup$– Peter KämpfJan 22, 2017 at 9:02
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$\begingroup$ This document from Airbus says that VMCg depends mainly on engine thrust and pressure altitude. No mention of CG. $\endgroup$– ryan1618Jan 22, 2017 at 15:50
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$\begingroup$ @RyanBurnette: The document also says that steering is not used in certification. But it continues "steering would be helpful in controlling the aircraft". And steering clearly works better with more load on the nose wheel. $\endgroup$– Peter KämpfJan 22, 2017 at 17:35
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1 Answer
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$V_{mcg}$ is the minimum speed on the ground at which control is possible using rudder alone upon failure of critical engine. From FAR Part § 25.149:
VMCG is the minimum control speed on the ground, and is the calibrated airspeed during the takeoff run at which, when the critical engine is suddenly made inoperative, it is possible to maintain control of the airplane using the rudder control alone (without the use of nosewheel steering), as limited by 150 pounds of force, and using the lateral control to the extent of keeping the wings level to enable the takeoff to be safely continued.
As the center of gravity moves forward, the rudder arm increases, improving rudder effectiveness. As a result a forward cg results in a lower $V_{mcg}$. This is the reason the tests for $V_{mcg}$ is carried out at the rear most possible center of gravity. From Advisory Circular 23-8C: Flight Test Guide for Certification of Part 23 Airplanes:
For rudder limited airplanes with constant aft c.g. limits, the critical loading for VMC testing is most aft c.g. and minimum weight. Aft c.g. provides the shortest moment arm relative to the rudder thus the least restoring moments with regard to maintaining directional control.
as,
There are variable factors affecting the minimum control speed. Because of this, VMC should represent the highest minimum airspeed normally expected in service.
answered Jan 22, 2017 at 9:49
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1$\begingroup$ Wrong, on the ground the airplane does not pivot around the cg. $\endgroup$ Jan 22, 2017 at 11:10
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$\begingroup$ @PeterKämpf, and neither does it in the air, right? Because a torque is caused by pair of forces and the arm between them is what determines the torque and CoG only matters as action point of inertial forces (including gravity). But it this case all the forces are aerodynamic. $\endgroup$ Jan 22, 2017 at 21:14