# Calculating Gyration forces on a propellor shaft

I am a Co-op student trying to calculate the forces exerted on a prop shaft as the plane goes into a turn or incline/decline.

this is my method of finding the moment exerted on the propeller for a plane in a turn

If the propellor is rotating about a horizontal axis (X) and the plane is in a turn, rotating about a vertical axis (Y) then there is a moment exerted on the propeller about a axis perpendicular to both the Y and X axis (Z axis) which is calculated by:

Mz = Iprop*$$\varpi$$prop*$$\varpi$$plane

where $$\varpi$$plane is calculated based on the radius of the turn the plane is making and the airspeed of the plane then $$\varpi$$plane = speed of plane / radius of turn * unit conversion to rad/s

Is that the extent of the forces?

Sorry if this isn't very concise

Maybe if i write out my calculations it will be easier to understand.

The prop rotates at 1000RPM so that's 9.551 Rad/s

the plane has a minimum turning radius of 1.5 Miles and an airspeed of 374 dividing the speed by the radius and converting gives ~ .069 rad/s.

to calculate the moment of inertia of the prop I assumed the propeller blades were triangles 2.166m high with a .3048 wide base and a thickness of .0508m and all four blades roughly weigh 500lbs total. This gave me a density of 3436.3.

I then calculated the moment of inertia using a triple integral of r^2 dV where r^2 = x^2 + y^2 and dV = dz*dy*dx with bounds Z: 0 to 0.0508, Y -(.3048/(2*2.1336)x to .0714x, and x 0 to 2.1336 and ten multiplying it by the density. This yielded a moment of inertia per blade of 129.3.

I then took all these values and subbed them in to M=I*(rotational speed of plane)*(rotational speed of prop) to get M=225 N/M or 15.417 lb/ft which seems very small.

Does this seem reasonable? are my calculations correct?