Calculus challenge for a cam design 4

[gravesity]

Hello any math wizards out there,

I'm designing a machine that includes a large, heavy arm which pivots on a shaft. When the unit is in operation, we hold the arm up with an electromagnet; when the magnet is shut off, gravity swings the arm down through 35 degrees until it hits a stop.

To lift the arm up again we want to use an electric actuator with a roller bearing, rather than a pin, so the actuator can retract out of the way and the arm can fall freely when it needs to.

The actuators doesn't like to take side loads, so I want to contour the contact surface on the arm so that the roller is always pushing normal to it. I can't seem to figure out a proper equation for it.

Essentially I want to control the slope of the contact surface. I can get that in terms of the swing angle, very easy. Presumably I take the integral from there, and that's where I'm struggling. The slope is a cartesian quality, but describing angles in terms of swing angle is a polar situation.

See the attached PDF of my final attempts before giving up. The vertical dotted line is the path of the circular roller. The shape of the contour is a straight segment (about 6 inches long) followed by a short curve (the contact area) which I have approximated here with a circular chord. I've shown it here in three positions; 0 degrees, 35 degrees, and somewhere in the middle. Ignore the graphical error here where the ending position (at alpha = 0) is a tad beyond horizontal; in reality it will have to end at horizontal so the roller isn't traveling back along the contour in the reverse direction (the 3.500 dimension will have to grow a little).

With the contour at alpha = 0, I imagine an x-y plot of the equation we seek. x varies from Xo onward, and y from 0 onward.

At first I thought it was a simple problem of defining the slope as a function of x, and then integrating. Slope as a function of beta is easy, but slope as a function of x is baffling.

Now my mind starts to lose hold of the problem. This could be a parametric equation with x and y as functions of alpha or beta (beta = -alpha), or perhaps polar?

I'm sure I can come up with a contour, freely-sketched if necessary, that gets me within a percentage of the desired slope (since in theory the actuator can handle some very minimal side load) but I’d be more comfortable with an explicit solution, since one must exist.

Is anyone interested in tackling this? I'd be much indebted to you.

Thanks,
Simon

 

[btrueblood]

" actuators doesn't like to take side loads"

Right, no bending moments. So, put a rollear bearing on the end of the actuator rod, and a mating notch in the arm. On the opposite end of the actuator put a similar roller bearing. On actuation, the roller homes (rolls "downhill") on the notch to bed into a mating radius in the arm, and lift starts from there on through the 35 degree rotation. Constrain the actuator to limit how far it will rotate as it is retracted, so it will always find the ramp portion of the notch on the next cycle.

What you are trying to do in your cam is not really possible, I think - you want a zero-angle contact patch...but as soon as the arm rotates a few degrees, that contact patch is now sloped towards the outer edge of the arm, and the roller will want to roll that way. If the "cam" is cut to force the roller to roll inwards, then that curvature will tend to give a steeper slope at smaller angles, i.e. the roller will want to run to the innner end of the cam for all angles, so upon first contact the roller immediately runs to the inside end of the pocket. You have at best a neutrally stable, and more likely an instable situation, balancing a ball atop another ball is the picture.

 

[gravesity]

btrueblood, extremely astute point about balancing a ball on a ball. To build the machine like I have described it, the actuator would have to make first contact beyond the pivot axis plane, and then the always-normal contoured surface as I described it could be concave instead of convex.

But you are right-- if the actuator is allowed to pivot, and the bearing sits in a notch, it doesn't matter what the contact angle is, and the actuator is still free to pull away, as long as we can get it to fall back against its home stops. I'll explore that and see if our geometry constraints will allow that. Otherwise, I'll try for the concave version I described above.

In which case I would still need a math wizard.

Thanks for the insight, and for getting me out of tunnel vision mode.

-Simon

 

[rb1957]

i think your sketch was the way btb replied ... line contact. but concave or convex, you'll still have line contact, no?

designing the concave surface also implies that you've got a radius notch at the two edns of the travel (as opposed to the tangets drawn) ... but this could be details understood and omitted for clarity !?

if the actuator is pin ended (like btb suggests) then it'll be axially loaded even if it is inclined.

question (asked quite possibly in complete ignorance) ... if your actuator is balancing on a line contact, what's stopping it from sliding sideways (say maybe 1lb side force applied against a zero stiffness) ?

 

[zekeman]

Here are 2 equations that satisfy your requirements.

A=R*cos(@-w)

tan(w)=R/{dR/d@)

where

R=R(@) is the curve you seek in polar form
A= horizontal distance from pivot point to roller centerline



@= angle pivot point to any point on curve

w= angle that point R(@) rotates until it reaches the x position, A.

Unfortunately I couldn't get a closed form solution by
eliminating w in the set of equations, but you should be able to do it numerically with good accuracy.

 

[gravesity]

Yes, that sketch was simplified for clarity.

If I mounted the actuator rigidly, then contact area would be chosen so that the contour is concave, rather than convex, so there would be bias against deflecting left and right. That is, of course, if everything is aligned perfectly, which would be needlessly difficult to achieve.

So at this point I'm pursuing BTB's suggestion (pinned actuator and an open notch on the arm), as long as it ends up that we have enough room in there.

 

[gravesity]

Zekeman, thanks for the equation. I'll chug through it and see how it fits.

-Simon

 

[zekeman]

correction

The second equation should be

tan(@-w)=R/{dR/d@)

 

[gravesity]

Zekeman, will you please upload a sketch to illustrate your terms? I'm not quite getting it.

As I understand you: R is a function of @, a polar equation with its origin at the axis of rotation. Are you suggesting that w is a constant that can be eliminated between the two equations?

Also, forgive me; it's been eons since my last math class: I don't know what to do with your {dR/d@) term.

Thanks.

 

[zekeman]

Since the correction, I think I have a closed form solution

@=sqrt(x^2-1)+ invtan(1/Sqrt(x^2-1)]+constant

x=R/A

get the constant from initial condition

R/A=secant( 35) =1.22

35=.7*180/Pi+55+constant
constant=-62 deg

Using x as the independent variable, you should get @, point by point.

BTW, if I did it correctly,this solution assumes a zero slope of the tangent at the contact with the roller (condition for no lateral force on actuator) and the curve is concave upwards if I am not mistaken.

 

[zekeman]

Disregard my previous post, since there is yet another error in the second term of the 2 equations I presented earlier.

My new set is

1) A=R*cos(@-w)

2) tan(w)+[tan@+R/{dR/d@]/[1-tan@*R/dR/d@]=0

Note that w is the angle swept from a point on the R(@) curve until it reaches the vertical x=A line.I will attempt to upload a graphical description subsequently.

Now a closed form solution appears impossible but can be done numerically after eliminating the terms in w. I did that and get
A^2/R^2=[cos^2(@)+sin(2@)*Z+sin^2(@)*Z^2]/[(1+Z^2)]
where
Z=[tan@+R/{dR/d@]/[1-tan@*R/dR/d@]

Now the trick is to solve for dR/d@ from this mess and then you can integrate numerically for an answer.

You have a difficult problem here , but your idea has merit over any others I have seen. See if you can do something with this and if you have problems, maybe I can help further.

 

[gravesity]

Zekeman, you've outdone yourself on this one. I'll have to return to this on Monday, so it may be a while before I get a chance to validate it. Thank you for your dedication. I hope it's been fun for you.

Before you go further, I have to warn you that we just decided the actuator bearing would land in a notch (tip of the hat to btrueblood) so, alas, we don't need the special contour anymore! But I would like to see the equations through, on my own time, so I can understand the problem for future applications if they arise. And to satisfy my curiosity, or I'll continue having calculus dreams every night.

 

[zekeman]

Trying to upload the cam drawing.Will post details later.

 

[zekeman]

Trying again

 

[zekeman]

I have finally settled on the solution which I have been dancing around for a while.

It is the closed form solution I presented on 5/11 except that the constant of integration is -130.11 degrees, not -62.

The corrected form is:

@=sqrt(x^2-1)+ invtan(1/Sqrt(x^2-1)]+constant

x=R/A

get the constant from initial condition

R/A=secant( -35) =1.22

-35=.7*180/Pi+55+constant
constant=-130.11 degrees
I showed a few points on the cam curve in my just previous post. You can get as many points you need by entering values of R from 7.32 down to 6.025
I think it looks ok and should be a viable solution.

I like your concept and from your sketch it looks like you got pretty close to the required contour.

The Btrueblood solution is also very nice, and it may objectively be the better , but I would hope your company didn't take it because you couldn't solve the problem. Very few designers or engineers can handle this type of specialized problem, but fortunately there are plenty of math and cam consultants out there that do this on a daily basis and engineering departments do use them.I would hate to see a design compromised for lack of the math. (this is not a solicitation since I have been retired a while (after having done 100's of cams-- but none like this) but I like an occasional challenge as this forum sometimes offers)

Finally, I would be happy to discuss the solution further to alleviate your calculus dreams.

 

[unclesyd]

You might want to check out doing away with your cam with Non-Circular gears from Cunningham Gear. If you proceed as posted you might want to to replace one of the suggested bearings with a C-Flex bearing. They are self centering.

http://www.cunningham-ind.com/index.html

http://www.c-flex.com/

 

[gravesity]

Hi Zekeman,

We did choose the notch-and-constrained-actuator approach instead of the cam, because it requires far less precise alignment when we are building up the machine. But I agree with you about not ruling a good design out for lack of math. Newton didn't just work it all out for his personal use.

Thanks again for your volunteer efforts on this one. I'll post again once I have a chance model it up and try it out.