# Thread: Integration of a cycloid.

1. ## Re: Integration of a cycloid.

Are you sure need acceleration to be a cycloid? If so, then hopefully you can at least stretch it vertically, other wise,

Originally Posted by Heloman
a peak value of 386 inch/sec^2
is going to take 193 pi seconds to reach. Apart from that, though, I don't see the problem with the numbers. E.g. what is your 'specific velocity in mind'?

Anyway, why not use something simpler, like...

?

2. ## Re: Integration of a cycloid.

The velocity I am trying to get to is 390 inches per second... I figured that if I dropped that into the equation for the velocity plot it would give me back an acceleration of an appropriate magnitude.

As far as why I am using a cycloid... Short answer is I am trying to figure out how it was done before. I have a cam design book that I picked up and it compared your sin profile, with a cycloid profile, and a "modified trapazoid" profile. of the 3 it shows that the cycloid has the least amount of required driving torque of the three. I do get kind of annoyed when end of the day the position plot to make the profile between the three different profiles... is almost negligible as far a manufacturing tolerances. That being said, at the high speed that this machine is running at... we need to make it as efficient as possible and that points to the cycloid.

The other reason for the cycloid is another little portion of the information that the machine came with... said that it's a "dual frequency" cycloid to keep vibration down. I was hoping that if I learned about the cycloid I would eventually come across what they meant by "dual frequency"... so far I'm still lost.

What we are working on is one acceleration corner of a cam profile going from a dwell to a constant velocity section that is lifting a follower. Then I have just had to use the same profile mirrored and reversed to bring the follower from constant velocity to dwell again at the top of the follower lift. As far as I can tell to maintain the same constant velocity coming out of acceleration then going into de-acceleration to go to zero... both of the corners have to be of identical profile otherwise they wont match up.

Anyway... the point of this all is to try to understand a machine that was designed and built by very smart people in the 1950's...of which the engineering records no longer exist and I am sadly no where near as smart as those old guys who managed to get this all done with 3 significant digits on a slide rule.

3. ## Re: Integration of a cycloid.

Originally Posted by Heloman
The velocity I am trying to get to is 390 inches per second...
$390 = 3 \pi r^2$

$\Rightarrow r = \sqrt{\frac{130}{\pi}}$

So,

$\displaystyle{x = \sqrt{\frac{130}{\pi}}\ (t - \sin t)}$

... throughout, and we have,

acceleration:

$\displaystyle{y = \sqrt{\frac{130}{\pi}}\ (1 - \cos t)}$

velocity:

$\displaystyle{y = \frac{130}{\pi}\ (\frac{3}{2}t - 2 \sin t + \frac{1}{2}\sin t \cos t)}$

displacement:

$\displaystyle{y = (\frac{130}{\pi})^{3/2}\ (\frac{3}{4}t^2 - \frac{3}{2} t \sin t + \frac{1}{2} \cos t + \frac{5}{4} \sin^2 t + \frac{1}{6} \cos^3 t - \frac{2}{3})}$

Do we get feasible numbers?

4. ## Re: Integration of a cycloid.

Okay Tom, I had to disappear for the weekend sorry about that. I'm not following the 390=3*pi*r^2...

5. ## Re: Integration of a cycloid.

$3 \pi r^2$ is the value of the first integral, i.e.,

$r^2 (\tfrac{3}{2} t - 2 \sin t + \tfrac{1}{2} \sin t \cos t)$

at $t = 2 \pi$, i.e. the area under one cycloid. (Cycloid - Wikipedia, the free encyclopedia.) That will be the velocity reached when acceleration returns to zero at the end of the cycle. E.g. 390.

I omitted r from the integrations for simplicity, but the displacement function comes out with r^3:

$\displaystyle{y = r^3 \ (\tfrac{3}{4}t^2 - \tfrac{3}{2} t \sin t + \tfrac{1}{2} \cos t + \tfrac{5}{4} \sin^2 t + \tfrac{1}{6} \cos^3 t - \tfrac{2}{3})}$

So an actual cycloid acceleration curve reaching a final velocity of 390 gives these numbers...

(You can't see the blue cycloid at this scale.) Don't know if these are remotely realistic, or the kind of thing that you were expressing doubts about. (Compare with constant acceleration though, as I said?)

Originally Posted by Heloman
I have a cam design book that I picked up and it compared your sin profile, with a cycloid profile, and a "modified trapazoid" profile. of the 3 it shows that the cycloid has the least amount of required driving torque of the three.
I've been browsing myself, to try and understand whether you are allowed to scale the cycloid in one direction (?), and I found this book, don't know if it's the one you've got... it compares a 'double harmonic', 'cycloidal' (not a cycloid as such) and a 'polynomial', and chooses the third...

Cam Design and Manufacturing Handbook - Robert L. Norton - Google Books

Explain some more about what you're doing and the information you're relying on?

6. ## Re: Integration of a cycloid.

Wow. I cant even begin to say how rusty my mechanical engineering brain is. My displacement plot was r^2 not r^3... so at least I got that figured out... but yeah... its basically a definite integral at 2pi for each of the equations. So at least I know I'm not insane. but yes that is what I was expressing doubts about. I cant make a 7900 inch... 200 yard tall cam.

But yeah... I tried to scale it to fit in my overall cam displacement and period. My thought was at least I have the "Ideal" geometery If I play with the acceleration, I can eventually get the plot curves to line up within the displacement and period I need... and it would be good enough.

As you can see... Its not an exact solution... but it's pretty close to the original.

That book will be here tomorrow!

7. ## Re: Integration of a cycloid.

... been reading some of that book online... I have some studying to do.

8. ## Re: Integration of a cycloid.

Okay... Finally back to this

I need to know the slope of my position plot at 2π.

The whole parametric thing is throwing me. at first I was thinking (dy/dt)/(dx/dt)=dy/dx thus I tried...

(r*((3/2)*2π-2*sin2π+(1/2)*sin2π*cos2π))/(r*(2π-sin2π)) and it's not 1.5.

Thoughts...

9. ## Re: Integration of a cycloid.

Originally Posted by Heloman
at first I was thinking (dy/dt)/(dx/dt)=dy/dx thus I tried...

(r*((3/2)*2π-2*sin2π+(1/2)*sin2π*cos2π))/(r*(2π-sin2π))
I haven't looked at that properly - please enlarge/clarify, although...

Originally Posted by Heloman
I need to know the slope of my position plot at 2π.
... well, that would just be the height of your velocity curve at 2π.

10. ## Re: Integration of a cycloid.

...yup that's it.

On to the next part....

11. ## Re: Integration of a cycloid.

Okay...
So what I am trying to do now that I have learned how to manipulate these functions... is make it fit in a rectangle.

Base length of the rectangle is X2
Height length of the rectangle is Y2

Then Looking at my cycloid's 3rd integral (green line in post #20) The end of it is defined as:

X1=2πr
Y1=3π^2*r^3

The slope (m)... at the end of that line... like you said is the velocity value which at 2π = 3πr

What I need is a line that goes from the end of the green line, at the same slope as the green line to the far corner (X2,Y2) of my rectangle.

So what I tried to do is...

Point slope:
(Y2-Y1)=m*(X2-X1)

If I plug in the functions from above... with my rectangle dimensions as inputs I need to figure out how to solve for r.

(Y2-(3π^2r^3))=(3πr)*((X2)-(2πr)) solve for r.

Well If I plug in my values for
Y2=5.245
X2=6.675

and go on wolfram alpha or my calculator...

((5.245)-(3π^2r^3))=(3πr)*((6.675)-(2πr)) solve for r.

it gives me appx. .0907773

It's not the right r to solve the problem.

Graphically I have been able to solve the problem with r=.35335

What am I missing!?

/Thanks ahead of time if it's probably something really simple. I just dont have anyone on my level to bounce thoughts off of.

12. ## Re: Integration of a cycloid.

Originally Posted by Heloman
The slope (m)... at the end of that line... like you said is the velocity value which at 2π = 3πr
You mean 3πr^2 ...?

Originally Posted by Heloman
and go on wolfram alpha or my calculator...

((5.245)-(3π^2r^3))=(3πr)*((6.675)-(2πr)) solve for r.

it gives me appx. .0907773

It's not the right r to solve the problem.

Graphically I have been able to solve the problem with r=.35335
Yeah, that could account for it, then...

((5.245)-(3&pi;^2 * r^3))=(3&pi;r^2)*((6.675)-(2&pi;r)) - Wolfram|Alpha

... since r = 0.31266 looks pretty close.

And apart from that I could relate your equation to the drawing, but I still have only the foggiest notion what a cam actually is. Is it an engine part that you're making on a lathe (which is the cam), or is it an actual cam-shaft, or what?! Please enlighten.

13. ## Re: Integration of a cycloid.

Yeah... that looks right.

But...

Here is the graphic representation of what I am trying to do. That r value gets me pretty close but not quite.

Does the concept of what I'm trying to do at least check out mathematically? I need to flip and invert the cam corner to the opposite side of the rectangle so that they feed into each other.
At this point I wouldn't be surprised if my graphics program or my equations in it are not doing something right. I will have to double check.

As for what it is, It's a cam drive for a sorting machine that has to run at pretty high speed. If we figure it out I might have to invite you over to check it out.

14. ## Re: Integration of a cycloid.

Technically the graphical version above... as far as I can tell will work.

If you look at the top of the rectangle there is the equation (red sigma) driven dimension that is the size X2, then there is the grey dimension below it... then the angle off to the right... that shows how well the line off the cycloid plots would line up with each other.

At this point (I actually drew the above diagram a few months back) I could make the thing and I know it would work... but it would be nice to have the whole thing in excel also so as to make the power requirements equations easier... and make me feel better that I actually could learn how to do the math involved.

15. ## Re: Integration of a cycloid.

Er, well really I would just like to get the faintest understanding of, say,

Originally Posted by Heloman
I need to flip and invert the cam corner to the opposite side of the rectangle so that they feed into each other.
Where does the rectangle come from. Please dumb it down a lot! Really simple, like what is it that is accelerating and in what direction. And when I say simple I don't mean this guy's idea of simple. What is a cam drive, and in what way do you adjust it, or manufacture it?

But yes, like I say I could follow your reasoning from the drawing in post 26. Just not what the drawing represents.

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