General aviation force feedback yoke for Cessna sim

[IanH1960]

Morning OP,

All the vibrations effects can be adjusted or disabled completely. Run the BFF_CL_Setup application, open your config file and go to the Vibration 1 & Vibration 2 tabs - tick the checkboxes for the vibrations you wish to disable. You can also alter the magnitude and frequency gains to adjust the vibration effect strengths.

Ian

[Cessna172sim]

Sounds too good to be true.

[IanH1960]

Hi OP,

...all the force effects are adjustable. I suggest you start with the "Bonanza" config file in the software zip package - and then tune this to suit your requirements once you get your system commissioned and operating.

There's a bit of work required to do this as there isn't a single set of settings that will suit all cockpit builds and aircraft. I hope to write a step-by-step "tuning" guide soon to help builders through it.

Ian

... it's not as complicated as I'm making it sound!

[Cessna172sim]

Maybe there could also be C172 flight dynamics as a default since (I think) it is most commonly used GA flight model. How about it?

[IanH1960]

Maybe there could also be C172 flight dynamics as a default since (I think) it is most commonly used GA flight model. How about it?

... once you get your configuration settings sorted I could include it as the default C172.

Ian

[Tom_G_2010]

. . . As far as I remember, when there's no any springs or cables connected to the Y-column and all parts are clean/greased, the friction (aileron + elevator) is not so big.

(If) the FFB motor power is anything like the GA Frasca trainers I have been test flying you have to use huge amount of opposite force against the yoke if the plane is not trimmed correctly. With this kind of force the yoke friction is nothing. Of course I cannot tell what's the case with this DIY unit.

As far as I can remember from real life PPL flying, in C172 the most powerful control surface forces affects the elevator. This should be the one that should be working as real as possible. Then comes the ailerons and at the end the rudders.

Over the weekend I made some very basic force measurements on the controls in my fuselage. I have the full Y bar with the 3 pulley pairs linking the aileron movement between the yokes, the ~18" push rod and cam are still there from the elevators, and both rudder bars. With no return springs the force needed to overcome mechanical friction and move the Y for elevator control will be less than 2 pound. For the rudder pedal about 1 pound. However, it took just under 10 pounds of force to rotate the yoke shaft at the chain linkage for aileron movement.

So that 9 to 10 pounds is what has my attention. I suspect with a more thorough cleaning and additional lubrication I could reduce that some, but it does seem like a lot of mechanical friction to me. Rolling the yokes by hand feels relatively smooth so there's ample mechanical advantage at the yoke handle to overcome the friction. However back at the shaft were I measured it a rotational force of 9 to 10 pounds was needed.

I may have this all wrong (I have a background in Electrical Engineering, not a Mechanical), but if I read the FFB spec correctly the system was designed to create about 9 LBF of aileron feedback force. So, does that mean, in my case, that most of it's normal output would be consumed just to overcome the mechanical resistance and rotate the yokes before providing any appreciable feedback? If so, can it handle producing 18 LBF to deliver a net 9LBF of feedback to the controls? If not, I may need to rework that linkage to reduce mechanical friction.

[IanH1960]

Hi Tom,

I'm not properly familiar with the details of your mechanism. Can I assume that the "chain linkage" force you measured for the Aileron friction is effectively the chain tension needed to induce rotational movement of the aileron?

If so then you would expect it to be higher than the hand force required at the control wheel. In rotational systems it is torque we are primarily interested in. Torque off course is force x distance - so the 2lbs control wheel force x the larger control wheel diameter should (all things being equal) = the 10lbs chain tension force x the chain sprocket pitch circle radius. If you need 2 lbs at the wheel but 10 pounds at the chain it suggests your wheel radius is about 5 x the PCR of the chain sprocket.

The aileron figure I gave is the force output at the wheel - ie in a low friction system the FFB aileron drive will produce a force of approx 10lbs at the wheel (when it is held single handed). So you would compare your measured 2lbs at the wheel to the 10lbs spec of the FFB system.

In general however it is will be wise to try and remove as much friction as possible from the mechanism so as to not loose any low level subtle force cues and not to inhibit low level vibration effects from coming through to the pilot's hands.

Incidently the aileron force level is based on a speed reduction ratio of 3:1 between the motor shaft and the yoke tube - this is a convenient ratio for DIY belt drives so suits the DIY yoke. If this gearing ratio is increased then the FFB force level goes up in direct proportion - significantly greater force levels could be obtained by designing-in different gearing in the transmission....

Ian

[Tom_G_2010]

Hi Tom,

I'm not properly familiar with the details of your mechanism. Can I assume that the "chain linkage" force you measured for the Aileron friction is effectively the chain tension needed to induce rotational movement of the aileron?

If so then you would expect it to be higher than the hand force required at the control wheel. In rotational systems it is torque we are primarily interested in. Torque off course is force x distance - so the 2lbs control wheel force x the larger control wheel diameter should (all things being equal) = the 10lbs chain tension force x the chain sprocket pitch circle radius. If you need 2 lbs at the wheel but 10 pounds at the chain it suggests your wheel radius is about 5 x the PCR of the chain sprocket.

The aileron figure I gave is the force output at the wheel - ie in a low friction system the FFB aileron drive will produce a force of approx 10lbs at the wheel (when it is held single handed). So you would compare your measured 2lbs at the wheel to the 10lbs spec of the FFB system.

In general however it is will be wise to try and remove as much friction as possible from the mechanism so as to not loose any low level subtle force cues and not to inhibit low level vibration effects from coming through to the pilot's hands.

Incidently the aileron force level is based on a speed reduction ratio of 3:1 between the motor shaft and the yoke tube - this is a convenient ratio for DIY belt drives so suits the DIY yoke. If this gearing ratio is increased then the FFB force level goes up in direct proportion - significantly greater force levels could be obtained by designing-in different gearing in the transmission....

Ian

Ian,

Thanks for the additional info and clarification. Your assumption about my measurement at the chain linkage is correct. I'll be working on reducing the mechanical friction and proceed from there.

Again, Thanks!
Tom G.

[Cessna172sim]

Some pictures of the project so far: (all wood parts were cnc-machine cutted)
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[Cessna172sim]

Still one more new image from the first 'loose-test' fitting:
6.jpg