Shaft Coupling

My first thought: No. This is based on taking the situation with the driving shaft at 0°, 90°, 180° and 270°. Clearly, in each of these cases, the driven shaft will be at the same angle.

My second thought: Are you talking about angular or radial misalignment, or a mixture of both? If both, it may need another pint or two of deep thought.

_{[I assume that there was never any suggestion of the average velocities differing ].}

"Clearly, in each of these cases, the driven shaft will be at the same angle." ...

After the first pint of reconsideration, I'll stand by 0° and 180°. Intermediate angles are under review.

May need to model it in the cool clear light of dawn.

The average angular velocity over 1 (or more) revolution(s) will be the same as it will be back to the same position at the end of each rev'. But I suspect there will be some variation over the course or 1 revolution as the pin slides in the slot.
Del

Agreed, for radial misalignment. The pin travels at constant speed. The output rpm is lowest when the point of contact is furthest from the shaft of the driven part, and vice versa.

For angular misalignment, no variation.

Cheers........Codey

Del, I do agree, GA fro me.

the average speed, over one turn, will be equal.

the angle synchronization will not always be equal.

also my compliments and GA for the exact explination further down this threat.

I did't read that far already, when I gave the GA

Hello,

The PIN in the sketch given, is meant only to transmitting the rotational power from the driver shaft-A to the Driven shaft-B to fecilitate the off-set of te two shaft axis.

So,in my view the velocity ratio of the two shafts is always constant.

Thanks.

There will not be any difference in velocity. At any angle in the rotation, the driving pin on shaft B will always make contact at the same spot on the follower on shaft A. This is a strange looking coupling. It's more of an eccentric drive.

Just MHO, as per complete revolution, no difference in RPM.

Fractional, with too much misalignment increase and resp. fractional decrease in rotation speed will occur.

usable as accellerator /decellerator for certain seeves. (The Maltezer cross is a dividing coupling and more odd).

You do know that there are couplers you can use that will connect two shafts and allow some misalignment, right? A universal joint, for example.

Same RPM. I would prefer to see the pin and hole repeated under the shafts. That would still work and would eliminate the eccentric load on the shafts.

I'd be concerned about vibration from the radially unbalanced load with this design. A counterweight would help. If there is any lateral misalignment, you would not have a constant angular velocity at the output, and torque would likewise vary over the course of a rotation. If the misalignment equals or exceeds the radial distance of the pin from the shaft, the mechanism will crash, stall, or just rock back and forth instead of turning. If there is significant misalignment, lubrication of the pin becomes an important consideration. Angular misalignment will tend to bend the pin back and forth when it is out of the plane of the misalignment angle unless the slot sides are rounded. However, rounding the slot sides will decrease the contact surface, with a corresponding increase in contact pressure, thereby compounding the lubrication issue. If instead you make the slot wider to permit angular misalignment, you'll have a lot of hammering at out-of-plane rotational angles.

This is looking at extreme-case situations. The mechanism may tolerate small misalignments, but it wouldn't take much more to drastically shorten its time between failures.

A drive shaft with universal joints on both ends and a sliding splined coupler may cost quite a bit more, but it would avoid a lot of these problems. (That's what motor vehicles use to accommodate suspension movement.) Do you have a space limitation that will not accommodate a U-joint drive shaft?

I thought that too, but I couldn't remember the name.

Play with one here.

GA from me.

Oldham coupling is much better and gives a constant speed to both shafts without periodic oscillations. Provided the misallignement is small!

No velocity difference, but it would be advisable to incorporate a better coupling if given the choice to do so.

What are the shaft sizes, speed and load?

Awsome man.. i was measuring velocity of shaft B and I was beating my head against the wall .. but this explains it.. -economist

Del, Please supply the equations so that this matter can be further elucidated.

The equations are obtained by a projection of the triangle O1O2P on the x and Y axis:

R1 sin (φ)=R2 sin(θ) → θ = asin (sin(φ)*R1/R2)

R2 cos(θ)= e+R1 cos(φ) R2 = R1 *(1+(e/R1)^2+2*(e/R1)*cos(φ))^0.5

V1= ω1*R1 V2=ω2*R2 = V1*cos(φ-θ) = ω1*R1*cos(φ-θ) → i= ω1/ω2=(R2/R1)/cos(φ-θ)

For a relative eccentricity of 0.2 the instantaneous ratio is in the following picture.

In the last comment before this one a word was missed : average

As one sees the variation is as mentioned in one of the first comments proportional to e/R1.

Hope now you have all input in the form you expected.

Any by a relative eccentricity of 0.2, you mean what?

e/R1

Dear Mr.Nickname,

Pl. provide the equation to me by mail to sd610@rediffmail.com

Thanks,

DHAYANANDHAN.S

Which other equations do you think about ? For the coupling all is above and you can make the effort to copy them.

The speeds should be the same, but the vibration would be horrible!

If the shafts are misaligned and one is driven at a constant velocity, the velocity of the other will not be constant. The shaft closer to the point of contact will turn faster.

You should not leave this statement half the way even after hearing the nice explanation from Nick Name. We will put this way:

If shat A rotates 'n' times, shaft B will also rotate 'n' times as well.

If shaft A rotates 'one' revolution, shaft B will also rote one revolution, but:

When shaft A rotes at constant speed and makes one rotation, shaft B will cover the first half (nearly) at a faster speed (or slower speed) than shaft A and rest half (nearly) shaft B will cover at a slower speed (or faster speed) than shaft A. Speed here refers angular velocity.

Del is right and my opinion is that it will work like a universal with more angle than it should have becaude of what will become side slap.