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Math Help - Integration of a cycloid.

  1. #1
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    Integration of a cycloid.

    Howdy yall, this is my first post, looking for some help with a question I have on a project that I am working on at work.

    So what I have here are two equations that give the (x,y) coordinates to plot a Cycloid having a radius of r, and rolls along a time/position of t.
    I am trying to build a cam path that operates as smoothly as possible and my limited understanding of calculus has left me puzzled as to how I can integrate these…twice.
    x=r(t-sin(t))
    y=r(1-cos(t))
    The idea that I am using a Cycloid form in the acceleration portion of my cam path, so that at the start and finish there is zero acceleration that leads right into my constant velocity sections of the cam hopefully making the machine operate very smoothly at high speed.
    The trick is I will know what my maximum allowable acceleration is once follower design is done based on the force that the strength of the follower can handle. So when I know this… how do I work this cycloid acceleration plot down to a displacement plot so that I can actually design the cam path to be cut?
    My first thought was to integrate them both separately… but I don’t know enough if this will work or not. The next thing was looking on Wikipedia they have
    x=r*acos(1-(y/r))-(y*(2*r-y))^(1/2)
    I plotted that out… and to tell ya the honest truth… I can’t figure out how to make this thing equal to y and get my ti-89 do the work.

    Anything would be great! Thanks!!!

    (quick link to wikipedia cycloid Cycloid - Wikipedia, the free encyclopedia)
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  2. #2
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    Re: Integration of a cycloid.

    Try looking at 'integration parametric equations' on Google.
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  3. #3
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    Re: Integration of a cycloid.

    Wow. amazing what knowing the right word does. THANKS! I will have to sort thru this and see if I can wrap my brain around it. (my calc skills are rusty)
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  4. #4
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    Re: Integration of a cycloid.

    Integration of a cycloid.-cycloidimage.jpg

    Alright... would anyone care to take a look at this thing...

    The blue line is a cycloid of r=1 (consistent with the equations above)
    the red line is my velocity (area) plot.... I like it... very smooth increase in velocity starting and ending with constant velocities... just like I wanted...
    The green line... Well to be right honest with ya... I'm not sure it's what I want. This should be my cam follower position plot. It actually looks a lot like what I was expecting... with the exception of its start not being at the origin. Anyone care to take a stab at explaining that?
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  5. #5
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    Re: Integration of a cycloid.

    ...Still working on it... The green line is 1.875 @ 0.
    It just seems meaningless.
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  6. #6
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    Re: Integration of a cycloid.

    Okay... so with my parametric cycloid, I figured out that my first integral is the "area" under the cycloid... is the "arc length" the 2nd integral?
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  7. #7
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    Re: Integration of a cycloid.

    No, but do integrate again to get the position function. Get the first integral as a function of t (instead of a definite integral), then integrate again with respect to x, using the chain rule again as with the first time round.



    I'll do a pic for the integration in a few minutes...

    Just in case a picture helps or is of interest for the chain rule process...





    ... where (key in spoiler) ...

    Spoiler:


    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (x or t), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).


    Full size


    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; March 29th 2012 at 12:05 PM.
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  8. #8
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    Re: Integration of a cycloid.

    I love you!

    I get it. I neglected to the ∫ y*(dx/dt)*dt to the second integral as well. Never heard of a parametric before yesterday!

    THANK YOU SO MUCH!
    Last edited by Heloman; March 29th 2012 at 11:25 AM.
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  9. #9
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    Re: Integration of a cycloid.

    Okay... your graph looks like what I want. My graph is doing something funny at the top... any ideas? Integration of a cycloid.-cycloid-fixed-i2.jpg
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  10. #10
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    Cool Re: Integration of a cycloid.

    Okay.... so I put my calculator in parametric mode... and it worked. Great... how to get excell to do parametric.

    EDIT: Got it!!!!!
    Last edited by Heloman; March 29th 2012 at 12:15 PM.
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  11. #11
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    Re: Integration of a cycloid.

    Oh hey! look at that shnazzyness you put up there. I will be drooling over that for a while!
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  12. #12
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    Re: Integration of a cycloid.

    Anyone happen to know what it means for the Cycloid to be "Dual Frequency"?

    Well thankfully now at least I can go between the cam Displacement, velocity, and acceleration curves... now I just have to stick my nose in my cam design book I just got and see if I can figure out how to use my new tool you all helped make!

    Way helpful here... I might have to go give my 2 DOF diff eq a spin in the other diff eq board in a few days!
    Last edited by Heloman; March 29th 2012 at 08:32 PM.
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  13. #13
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    Re: Integration of a cycloid.

    So now that I have had some time to play with this tool... would someone kindly check me on this thought...
    Imagine that the cycloid is representative of accleration. and its units were inches/sec^2. then the red plot in this threads post #9 would be inches/sec, thus the green plot would merely be inches.

    I am having a hard time beliving that the magnitude of these are all relating to eachother properly. As in if I have an acceleration has a peak value of inch/sec^2 (gravity) the plot for the displacement is MASSIVE.

    I do hope I am missing something.
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  14. #14
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    Re: Integration of a cycloid.

    Quote Originally Posted by Heloman View Post
    a peak value of inch/sec^2 (gravity)
    So scale the y values by 4.9 ...? (Not 9.8!)

    And how does that compare with, say, displacement due to constant gravity, y = 4.9 x^2 ...?

    Btw time is x not t, yes? There is room for confusion on this point, since you mentioned a 'time/poisition of t' to begin with... but this looks like you meant a cycling-wheel-angle of t, because from then on it's about a cycloid-shaped acceleration curve, which would make it logical to integrate twice with respect to x. If that were the problem, anyway, I don't see it causing your current consternation. Whereas the doubling might.
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  15. #15
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    Re: Integration of a cycloid.

    a peak value of 386 inch/sec^2 .... oops... forgot to type the real number for some reason. Its not what I'm going for... but I thought its a value to play with for now.

    Anyway... Going back to my first post... and who knows maybe I REALLY don't understand what I'm doing... but the idea is that I am using a cycloid equation merely because its has the right mathematical properties for what I am trying to do. Basically an equation that has high enough order to be integrated twice and give a continuous value... and be differentiated once and at least have a continuous solution.

    I'm working under the thought that the diameter of the circle driving the cycloid (thus the peak value of the cycloid) represents the peak acceleration of a cam follower. It starts and stops at 0 acceleration. I care about the acceleration because F=m*a, and by knowing the mass of my follower I can figure out how much force will be applied to my system to make sure it is strong enough to handle dynamic force at the speed I need to run it at. Then, assuming that all of these factors tie together (which I did and am now having a problem believing)... I was hoping that I would be able to put in a max acceleration value, and I would be able to figure out how fast my follower will be going at the end of the acceleration (The red plot in posts #7 & #9). I actually have a specific velocity in mind... and figured if I rearranged the equation for the red plot to solve for acceleration for a given velocity I could massage the whole system into giving me the appropriate cam displacement profile (the green plot) to actually cut the cam profile out of a piece of metal.

    I have been able to solve the thing out graphically (basically solving the system of the 3 equations with descriptive geometry) ... so I know I at least have what I need... I was just thinking that the numbers would make a little more sense... but It doesn't look like I have the understanding to know what is really going on. I'm probably missing something simple.
    Last edited by Heloman; April 21st 2012 at 11:11 AM.
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