# How does this Bell 206 cyclic control linkage work?

I'm studying the Bell 206 Maintenance manual to understand how the pilot's control inputs get converted to swash plate movements.

This diagram seems to explain the cyclic well, but I can not figure out how the lateral stick movement is transformed.

Here is an example of how the crank mixer is operated by the cyclic (Y bar 9)

The forward-aft movement of 12 (the stick) causes member 10 to rock back and forth, moving 9 back and forth, which operates the mixing cranks at the end (connected to 7 and 8).

However, it looks like the stick can't mover laterally at all, since 10 can't move laterally (?) I can't figure out even why 10 has a provision for motion in the vertical plane in the first place.

Perhaps I'm misunderstanding how 12 moves? I'm assuming it pivots around the base (15).

A more detailed view of the linkage:

Thanks!

• Does this answer your question? How do a helicopter's cyclic and collective transfer commands to the swash plate? Commented Feb 5, 2022 at 2:10
• I have read those answers thoroughly. I understand the principle. I wish to know how this particular design works. There is something about the lateral motion of the stick here I do not understand, since the linkage 10 looks like it will lock any lateral motion. Commented Feb 5, 2022 at 2:17
• You should provide a link to your source, both for attribution and so people can see any related material. Different views sometimes clarify operation. Commented Feb 5, 2022 at 2:23
• @Pilothead done, thanks. Commented Feb 5, 2022 at 12:41
• That's good, and your second illustration shows the spherical bearings at 5 and 16. Commented Feb 5, 2022 at 12:44