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Reading through the manual I see that spins are not allowed below 10000 msl in this turboprop aircraft, the t6b, the reason that is written in the manual is as follows, quote:

"Spins below 10k msl are prohibited due to high stress on the propeller which occur during the spin maneuver with the propeller RPM below 80%. To prevent high stresses from occurring, the power management unit (PMU) artificially maintains propeller RPM at 80% with the PCL at IDLE when the aircraft is above 10000 ft pressure altitude."

A bit of background on the maneuver, it is executed with PCL idle airspeed 150kts and 30 degrees nose up. So I understand that at low altitudes the air is much denser (below 10k) and as the aircraft comes down with PCL at idle apparently at or above or less? (I dont know) 80% rpm the air would push the blades more than the rpm that is giving out by the engine? Like turning more than is supposed to, damaging the engine (its designed to move the air and not the prop being moved by the air)? I am missing I guess a/some concepts with torque RPM and propeller RPM? If someone please would help me understand why or how below 10k could damage the engine/prop during a spin I would be very grateful. the engine used is a Pratt & Whitney Canada PT6A-68 turboprop, 1,100 shp (820 kW). there is only one lever to control the power (PCL), it is a variable pitch propeller but its automatic, we only control the PCL for torque power the PMU and the Propeller interface unit (PIU) control the speed (Np) and blade angle

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  • $\begingroup$ Just an editorial comment: The quote in the second paragraph answers the question. Perhaps consider rephrasing to something like "what causes propeller overstress in a spin below 10K'?" The rest is good background as to why you are asking... $\endgroup$ Commented Oct 8, 2022 at 15:31

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Here's my take. The propeller is under huge centripetal forces when at operating speeds, and this force is essential for keeping the blades from bending too far forward (thrust) or against the direction of rotation (drag). This is why most helicopters have some sort of a hinge mechanism, they let the blades free and allow them to position themselves in a cone for the thrust and centripetal force to balance each other out. Reducing prop RPM and placing the aircraft in a spin does 2 things. -It reduces centripetal force which is necessary for placing the force vector on the prop blade axis -Airspeed drops and angle of incidence between the prop disk and incoming airflow is varied wildly. This increases thrust (props generate the most thrust at zero airspeed) and places loads on the blades and thus the propeller shaft under loads that vary in magnitude and direction.

These loads can easily propagate through the reduction gear and the free turbine shaft.

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  • $\begingroup$ Thank you for the comment, do you think you could draw the angle of incidence you are talking about in relation to the prop disk and incoming ariflow. I am trying to picture what you said in my mind. Thank you for your time Efe! $\endgroup$ Commented Nov 1, 2022 at 21:49

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