[Cam Design] - Force and torque curves

[magicfrog]

TL;DR Summary

[CAM DESIGN] - Force and torque curves

Hi everyone, I hope this is the right place to post this thread.

I am having a little difficulty in carrying out a study concerning a cam spring follower mechanism.
The mechanism consists of a cam, formed by different arcs of circumference (I attach a sketch - the edges are connected with R1), and a circular follower (Ø14). The eccentricity between cam and follower is zero, so the centre of rotation of the cam and centre of rotation of the follower are aligned.

What I would like to obtain are the force (spring) and torque variation curves as a function of the angle of rotation of the cam (an example of a 45° angle is shown in the figure) using excel.

My approach is to consider geometry. As the angle of rotation of the cam changes, the point of tangency between cam and follower changes, and therefore the tangent line and normal at each point change. With these changes I get different strokes and different torque at each point.

I just can't convert this reasoning into excel.

Can any of you give me some hints? Or, is there a faster or easier method to achieve my goal?

I am open for further explanations.

Thank you to anyone who can help me.

 

[Baluncore]

Welcome to PF.

Can any of you give me some hints? Or, is there a faster or easier method to achieve my goal?

Simplify the rotation by fixing the cam in space, centred on the origin. Then find the path taken by the centre of the circular follower as it orbits the cam.

For each angular step about the origin, the follower centre will be the maximum radius in that direction from all the possible cam contacts. Track the follower centre point and the contact point on the cam for each step.

To find lift, determine the radius. To find torque, determine the slope at the cam contact.

 

[magicfrog]

I am not sure if I have understood correctly.

What I was trying to do was to fix the centre of rotation of the cam, vary the angle of rotation of the cam and obtain the position of the rotated points using the rotation matrix. At each step of the rotation angle, derive the tangency point between follower and cam (tangency point between two circumferences). By noting that point I can derive the slope and then say spring force and torque.

Am I missing something?
How can I set this up in Excel? With Matlab it would be instant, I think.

 

[Baluncore]

What I was trying to do was to fix the centre of rotation of the cam, vary the angle of rotation of the cam and obtain the position of the rotated points using the rotation matrix.

I am suggesting that you do not rotate the cam. Instead, you geometrically roll the follower around the fixed cam. For each step of the rotation, the follower centre, and the point of contact, are then fixed in Cartesian space.

The shape of the cam can be defined by simple arcs, then the point of contact for each angular step can be computed as it progressively moves along the sequence of piece-wise arcs.

It is following the contact curve that is needed, and determining when the contact passes from one segment to another. I believe the path can be followed numerically, by the osculation of the follower circle, with the many circular arcs of the cam.

 

[magicfrog]

I thought about what you suggested, but I still have a doubt.

By keeping the cam fixed and rotating the follower around, I have an angular and a radial variation of the centre of the follower.

I can't see how I can manage the radial (orange) variation as the angle changes.

Conversely, if I rotate the cam and keep the follower fixed, I also have the possibility of rotating all the centres of the arcs of the cam profile (by means of the rotation matrix) and therefore, by using the equation of two tangent circumferences, I can obtain the point of tangency at each step.

Am I missing any steps?

 

[magicfrog]

I will add an answer.

I followed the advice you gave me. In doing so, I had an epiphany which I post here with the following image.

Using trigonometry, I was able to write the distance x(ϑ).

The variation of x(ϑ) respect to x(ϑ=0) is the variation in stroke of the spring pushing my follower and, consequently, using Pythagoras, I obtained the distance in order to calculate the torque.

This was possible because I knew the geometry of the cam, in particular the position of the centres of the circumference arcs that describe it.

But thinking more generally, having a cam given not by circumference arcs but managed with splines, how could it be managed? What approach would be correct to use?

Thank you!

 

[Baluncore]

But thinking more generally, having a cam given not by circumference arcs but managed with splines, how could it be managed? What approach would be correct to use?

A spline or a Bézier, will be defined as a parametric curve. For each parametric step along that curve, you can numerically evaluate lines of points, that are one follower radius outside the cam surface.

There will possibly be segments of the cam surface, where those lines of follower centre points cross. In that situation you must select points from the line with the greatest distance from the cam centre.

 

[Lnewqban]

Welcome, @magicfrog !

In practical applications, the cam, when lifting, has also to overcome the friction force on the slider-guide directing the follower wheel linear movement.

 

[magicfrog]

Welcome, @magicfrog !

In practical applications, the cam, when lifting, has also to overcome the friction force on the slider-guide directing the follower wheel linear movement.

Thank you for your suggestion.

But in this approximation we can consider it negligible for the purpose of calculating the acting force and torque, correct?

Especially in the case of considering components acting on each other with pure rolling.

 

[Lnewqban]

But in this approximation we can consider it negligible for the purpose of calculating the acting force and torque, correct?

You could consider it negligible, but its not in a real calculation.
Valves’ guides and seals in internal combustion engines get worn out from that component of the force you have represented, which is perpendicular to the line joining both axes.

That resistive force increases for abrupt changes of the cam profile, like the three shown in your diagram.

Just imagine the extra force a car has to apply to have one of its tires come out a pothole which radius matches the radius of the tire.
Many accidents of street scooters happen due to the very small diameter of their tires.

 

[magicfrog]

You could consider it negligible, but its not in a real calculation.
Valves’ guides and seals in internal combustion engines get worn out from that component of the force you have represented, which is perpendicular to the line joining both axes.

That resistive force increases for abrupt changes of the cam profile, like the three shown in your diagram.

Just imagine the extra force a car has to apply to have one of its tires come out a pothole which radius matches the radius of the tire.
Many accidents of street scooters happen due to the very small diameter of their tires.

Yes I understood the concept, which is why I wanted to emphasise that I consider it negligible in the first approximation.

It must also be said that the next step is to take it into account when calculating the torque acting on the cam. This does not only include the force normal to the surface point by point, but also the tangent force (which I am neglecting at this first stage). This will certainly make an incremental contribution to the torque.

In order to minimise this increase, will I only have to act on the surface roughness and the coefficient of friction of the connected materials?

 

[Baluncore]

But in this approximation we can consider it negligible for the purpose of calculating the acting force and torque, correct?

Especially in the case of considering components acting on each other with pure rolling.

That is correct.

The coefficient of dynamic friction, µ, for rolling bearings, as quoted by NSK are:
µ value. Bearing Type:
0.0010 Cylindrical Roller Bearings.
0.0010 Self-Aligning Ball Bearings.
0.0011 Thrust Ball Bearings.
0.0013 Deep Groove Ball Bearings.
0.0015 Angular Contact Ball Bearings.
0.0015 Needle Roller Bearings with Cages.
0.0022 Tapered Roller Bearings.
0.0025 Full Complement Needle Roller Bearings.
0.0028 Spherical Roller Bearings.
0.0028 Spherical Thrust Roller Bearings.

 

[magicfrog]

I share my steps and also the curves obtained with excel through this process.

In particular, based on the angle from 0° to 90°, I divided the spline into three curves described by circumferences (known centres and radii) that I considered interesting for the study.

I have obteined from quota ‘x’ from trigonometry (present in some images in previous posts). Noting ‘x’, I derived the distance perpendicular to the axis of application of the force and then derived the force exerted by the spring and torque as the angle varies.

​

Given this cam curve, recovered for the purpose of the exercise, the two points of ‘discontinuity’ correspond to the change of curvature.

I cannot say if they are acceptable or not as behaviour in reality.

 

[Baluncore]

I cannot say if they are acceptable or not as behaviour in reality.

An instant change in the path of the follower, represents an instant change in velocity, so an infinite acceleration. That will generate a noisy collision, or a bounce, where the follower contacts the cam at two points, while switching between trajectories.

 

[magicfrog]

An instant change in the path of the follower, represents an instant change in velocity, so an infinite acceleration. That will generate a noisy collision, or a bounce, where the follower contacts the cam at two points, while switching between trajectories.

Would this be true even if the two curves have a tangency constraint? Or does it depend on the curvature alone?

 

[Baluncore]

Would this be true even if the two curves have a tangency constraint? Or does it depend on the curvature alone?

If the curves meet without a step change in direction, there will be no problem. That is, assuming that the follower radius is smaller than the radii of all concave curves that define the cam profile.

 

[magicfrog]

If the curves meet without a step change in direction, there will be no problem. That is, assuming that the follower radius is smaller than the radii of all concave curves that define the cam profile.

Then I would say that this is not the case since the follower diameter is slightly smaller than the master cam diameter. And certainly those small radii do not help to have a ‘continuous’ curve.

What if instead you wanted to draw the cam profile curve known as the torque curve with respect to the angle? Would it be sufficient to apply an inverse procedure?

Eventually, I could impose control points with desired torque values. For example, starting torque, ending torque and some points in between. From there, can I derive the cam point that returns this desired values to me? And finally connect everything with parametric curves?

I guess that's what you do during design.

 

[Baluncore]

And certainly those small radii do not help to have a ‘continuous’ curve.

You must consider what happens on convex and concave parts of the profile. Small radii are OK on convex curves, but not on concave curves.

 

[Lnewqban]

Would this be true even if the two curves have a tangency constraint? Or does it depend on the curvature alone?

Angular velocity is the main limiting factor.

When spinning fast enough, real cams can make the follower float (unable to fully follow the contour or profile of the cam), because the spring force and inertia of the slidding assembly.

That problem occurs in addition to sudden accelerations and spiking forces and associate noise, vibration and wear, mentioned in post #14 above.

Please, see:
https://en.m.wikipedia.org/wiki/Valve_float

In the attached picture, please note how dramatic changes of profile are used in mechanisms intended for radially positioning shafts.