# How to calculate the angular velocity ratio of a cardan shaft?

Recently this question has been bugging me, and while it might seem simple, I was not able to find a satisfactory answer.

Suppose we have a cardan coupling like in the following picture, where the ingoing and outgoing shafts both are provided with a U-shaped end. The U-shaped ends point to one another, are rotated an angle of 90 degrees with respect to each other, and are connected with a cruciform intermediate peace that allows the shafts to revolve about their own axis. In the image below, the blue ingoing and the green outgoing shafts make an angle of 30 degrees in the plane of the paper (screen). The angular velocity of the ingoing shaft is ωin and the angular velocity of the outgoing shaft is ωuit. See the image below. How do I calculate the ratio of the angular velocities ωuit / ωin given that the U-shaped end of the ingoing shaft lies in the plane of the screen (left), and perpendicular to the plane of the screen (right)? Is there a systematic way to do this?

left: ωuit / ωin = 1/cos(30)

right: ωuit / ωin = cos(30)

• That whole second paragraph could be "Suppose we have a U-Joint...". – JPhi1618 Jun 2 '16 at 13:29

I'm not convinced it's as simple as that formula. I think you've got three solid bodies involved in the conversation.

If it were me, I'd design the system in SolidWorks, then put the pieces together with centerlines and contact surfaces mating. Part of the issue here is ID/OD and axial clearance / slop. Its not much in real life, but enough to be a factor in input vs output curve, particularly when non-zero torque loads are involved. I'd spin the input shaft 2.5 degrees of arc, and measure the output rotation. Add 2.5 more degrees of twist and measure output again. Repeat so you have a complete map (I would think 90 degrees of arc total should suffice...) I doubt the input to output is a pure linear relationship. Or you could do it with 3D angular relationships, and a really good appreciation for Descriptive Geometry. Although two of the parts involved are the same design, you've still got to manage three bodies in the analysis.

How good are you at calculating steradians? Me, not so much.

I've got a good grasp of Descriptive Geometry, but I think this is a hard problem to do in the purely theoretical.

Another approach is to use matrix calculations with a 3d rotational operator. I think you'll find more on the transformations matrices in the field of robotic arm design.

Edit / Update: Check out this reference. It has a nice output graph showing the speed variations. It shows formulas similiar to yours but again without explanation on how those formulas were identified. What is pretty bizzare is the peak off standard doesn't always occur at 45 degrees... it gets offset a degree or more, depending on how far the u-joint is 'bent'. That makes me suspect this data is obtained thru experimentation rather than calculation but what do I know?

Update #2: Here's a pretty detailed reference, from Dario Governatori University of Rome Tor Vergata. Look at slide #13, Kinematics. The author creates a relationship for each half of the u-joint cross, based on the angles of the input/output. He takes dot product = 0 as they are perpendicular to each other. Substitute angle inputs, and then differentiate the relationship of the input and output angles with respect to time. That makes sense but certainly is not intuitive. I think the overview illustrations here are excellent. I will say, I found the "Equation of Motion" description in wikipedia much more complete and easier to understand (even if the wiki illustrations are a bit weak.) Using both references together makes things much clearer.

So, er, yeah. Scratch the SolidWorks stuff from my answer. What the heck was I thinking?

• Thanks for the answer. However, I did not state there is a linear relationship. And also, maybe there is no formula, but there must be some systematic way to derive the answer, since this question originated from an old exam, and those are the answers provided (by them). – BlueRine S Jun 1 '16 at 7:13
• @BlueRineS See update #2 above. – zipzit Jun 1 '16 at 16:13