


Joined Jun 2003
752 Posts

servo torque for a 6 to 8 foot rotor?
How do you determine how much servo torque is needed for a given rotor size for roll tilt?




Joined Nov 2004
4,756 Posts

Quote:
Here is a diagram showing the forces involved. Here is also a table showing the torque required for a 1 lb gyro, with 5 degrees of rotor tilt and a 3 lb gyro with 8 degrees of tilt. The chart is for various coning angles and flapping offsets. Clearly the least load on the servo is small flapping offsets Small flapping offsets create unstable rotors with very little control authority. The design optimization is toward the largest flapping offset that the servo can take. Hope this helps. mickey reason for edit : stupidity reason for additional edit : additional stupidity 



Joined Nov 2004
4,756 Posts

Quote:
Previous post was assuming a DC head. If you are using cyclic pitch and the blades pivot at the aerodynamic center the similar calculation is always 0 torque. 



Joined Jun 2003
752 Posts

Wow, thats pretty complicated to figure! Can you do an example of my gyro?
It weights 6.6 lbs, coning angle unkown? I have 15o upstops but it doesnt cone up much. it just has a slight upward coning angle. flapping offsets are 1.5 inches from the rotor shaft. Would this be a simple way to determine servo tourqe needed? Simply hang the model by the rotor shaft and see if the servo is strong enough to tilt the body under the rotor? If it can swing the body to the side it should be strong enough right? 


Joined Nov 2004
4,756 Posts

Quote:
Lets say each blade is 3 feet by 3 inches. The rotor will have to turn around 400 rpm to fly level. If each blade were to weigh 3 ounces, the centrifigal force on each blade is 18 pounds. This gives a coning angle of 7 degrees. Your flapping offsets are 1.5 inches. When you run the formula you get 121 ounce inches. Suppose you really start flying fast and the RPM gets up around 600. Then the servo torque required is: 212 ounce inches. If the blades weigh 4 ounces each these numbers change to 162 and 327 ounce inches respectively. If the blades weight 2 ounces each: 80 and 121 ounce inches repectively. I'd plan on something > 100 ounce inches no matter what. Does this help? 



Joined Nov 2004
4,756 Posts

The 3 ounce blade weight is a pretty good value.
Rotor blades are supposed to weigh 20  30 lbs / cubic foot. Lighter than that and the bending forces are too high. Heavier than that the centrifugal forces are too high. A 3foot by 3 inch blades is .75 square feet. If the airfoil is 9% thick then they are .27 inches thick, or about .135 inches (.135/12 feet) thick on average. This is .75 * .135/12 = .0084 cubic feet. * 20 lb/cubic foot = .168 pounds = 2.7 ounces. at 30 lb/cubic foot = 4.00 ounces. So the blades should weigh between 2.7 and 4 ounces. For best controllability the coning angle is supposed to be less than 9 degrees. If you put this in the calculator. The 3 ounce blade generates between 8 and 11 degrees of coning and 111 ounce inches of torque on the servo. The 4 ounce blade generates between 6 and 8 degrees of coning and 148 ounce inches of torque on the servo. I'm still estimating that you need > 100 ounce inch servos....(or two 50's on each control) Keep in mind that this does not include the drag load on the elevator servo caused by the head being higher than the pivot point. That is on top of the load just to move the head around. Probably looking at Hitec HS645MG's (133 ounce inches) or HS945MG's (153 ounce inches). You might even consider the HS5745MG (250 ounce inches) for elevator. You are bumping into a real issue with DC heads, servo loading. The HS5745MG uses 4 AMPS! when it is stalled, 1 amp running with no load, so be careful about your radio battery. Two of these guys running is 2 amps. This could take out a standard 500 mah RX battery in 15 minutes. mickey 
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