I am currently in the midst of designing a 2 Speed Dog Shifting Gearbox. My question is regarding the elipse cutout for the dog gear to slide. As seen here, 254 uses a center distance of .49", which makes sense. It seems to me because they are using a 1/2" Stroke Piston. One question that arises is, “Why do they make it .49” instead of .5"?
But my real question is with regards to 973’s output shaft. Instead of using a center distance of somewhere around .5", they have a center distance of .39. Why would they use that center distance, if the stroke is 1/2 an inch?
Wouldn’t that cause some sort of jamming?
Thanks,
Adam
DISCLAIMER: I do not have much experience with Pneumatics, so I’ll leave it up to the CD community for answers.
We use a 1/2" stroke piston because they are easier to get a hold of and do not require stroke limiting spacers. We have used ~.375" spacing before, but did not like having to account for the spacer and did not want to order a shorter stroke custom cylinder.
I am not sure why you see a .49" spacing. It should actually be larger than 1/2" to account for tolerence stack up and to give the dog some room to disengage before engaging in the other dogged gear.
Alright, so I understand that if we do not want to use a stroke limiting spacer, the center distance should be >1/2". How much greater? And would it be possible to simply keep the spacing at 1/2"? The width of the teeth on the dog is .45", so wouldn’t that allow the dog to fully disengage between strokes?
As Travis touched on, we mechanically limit the motion of the shifting action by limiting how far the external shift block can move (the block threaded onto the cylinder that interfaces to the small internal shaft via bearings).
I have a personal obsession with trying to make things smaller, and doing this allows us to shave the maximum amount of length off the gearbox by reducing the space between the gears (where the dog is), and minimizing how much the cylinder sticks out in the rear.
I have very little experience with pneumatics, so this may be a fairly trivial question. But, from what I gather, when the piston is extended (or retracted) and hits a gear before it completely extends or retracts, the piston will remain locked into that position.
Adam. If the center distance is shorter than the stroke length, will a problem arise of the system jamming or the free spinning dog gears being pushed out of place?
What engineering thought should go into the dimension for those center distances?
DO NOT SHEAR THAT PIN! they suck to get back in, or get one out (especially if it snapped inside). unless you have a tool to do it (we dont), and i have no idea if there even is a tool.
Since I’m just entering into Mechanical Engineering as an Undergraduate next year at UCLA, I am unfamiliar with the engineering principles and guiding formulas to test shear stress applied at certain conditions, and the shear stress that a screw (I’m thinking #4-40) will withstand before succumbing to failure.
Can anybody please enlighten me, or is this not worth calculating?
screws don’t like shear, also, 4-40 is rather small, the amount of shear force the cylinder would apply would be equal to the force the cylinder applies (as in all of the load applied to the screw by the cylinder are shear forces) so that would be (one half of the bore) squared times pi times the system gauge pressure
there may be some additional forces on the screw from the transmission itself, but the shaft should take most of this so it should be negligible.
NOTE: I believe in 2010 AM tried using screws in their shifters and found that they broke a lot more often then pins did.