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Cam -follower displacement as a time function

  1. Apr 18, 2005 #1
    There is a cam with offset of 10mm. A follower rests on it vertically. When the cam rotates at rpm N. I want to describe the displacement of the follower vertically with respect to time.
    I can be shown that the displacement is 10*sin(wt) where w=2*3.14*N*t/60. But i want to derive it mathematically. Any diagrams will also help.
  2. jcsd
  3. Apr 18, 2005 #2


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    Okay, your cam is an offset circle and you are modeling it by taking its "average height" as displacement 0. That means the maximum displacement will be 10 and minimum -10. Parametric equations for a circle of radius R are x= R cos s, y= R sin s where s is the angle, in radians, around the circle. Since the "displacement can be measured in either x or y direction, you could use either 10 cos s or 10 sin s for your model. Apparently you have chosen 10 sin s. Since s is in radians, one complete rotation corresponds to 2 pi (2*3.14 approximately) radian. You want to write s= wt so that one complete rotation: s= 2 pi corresponds to t= 1/N minutes (N revolution per minute corresponds to 1/N minutes per revolution). That is, when s= 2pi, you want
    s= 2pi= w(1/N) so w= 2pi N. That's NOT what you had because I to t to measured in minutes (when I said "t= 1/N minutes") and you are NOT! It's a really good idea to specify what units your variables are in when you state the problem! Because I know that t seconds corresponds to 1 minute, I can see that, if we measure t in seconds, as you are doing, we have w= 2 pi Nt/60.
  4. Apr 18, 2005 #3

    I'm new here and this is my first post.

    The following is the derivation of the equation for displacement you've got:

    Amplitude = 10mm

    for an oscillation we can say it follows a sinusodinal path:

    Displacement = ASin(wt)

    where w = Angular Frequency
    t = Time
    A = Amplitude

    we calculate W as follows:

    1. The cam rotates N times per minute, so rotates N/60 times per second.
    2. This is already an angular frequency, but angular frequency must be in Radians/second so we multiply N by 2Pi to convert it to Radians, giving

    So we substitute these values into the origional equation giving:

    D = 10Sin[(2Pi(N)/60)t]

    I've included a diagram showing the displacement of the cam and how this relates to amplitude:

    http://mobyhost.no-ip.com/cam.jpg [Broken]

    This is the simplest way I can think of deriving this formula, I think its correct.
    Last edited by a moderator: May 2, 2017
  5. Apr 18, 2005 #4


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    Yes but is the ampletue of that offset circle cam actually sinusoidal? I think not, just look at the displacement at various angles of rotation (taking the maxium displacement as zero degrees) :

    0 degrees : R+10 (max)

    180 degrees : R-10 (min)

    So far so good, but in that case we must have a displacement of R at 90 degrees however this is not the case.

    90 degrees : sqrt(r^2 - 10^2).
  6. Apr 18, 2005 #5
    EDIT: Yarr... I'm confused... scratch that...
    Last edited: Apr 18, 2005
  7. Apr 18, 2005 #6
    yes, it is sinusodinal in the vertical displacement with respect to time.

    I've drawn a little diagram to illustrate this:

    http://mobyhost.no-ip.com/sin.jpg [Broken]

    Hope it helps

    Edit: Assuming you take zero displacement to be when the cam is at 90 degrees rotated. (which for this question is correct as your plotting the movement of the cam follower - see diagram)

    Last edited by a moderator: May 2, 2017
  8. Apr 18, 2005 #7


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    Sorry I must be looking at it wrong because I cant see it. It still seems to me that the vertical displacement at 90 degrees will be sqrt(r^2 - 10^2). Can anyone see what I'm doing wrong. Maybe I'll have to draw a diagram to show how I'm interpreting the problem but I don't have time right now.
    Last edited: Apr 18, 2005
  9. Apr 18, 2005 #8

    what you do is you are looking at the vertical displacement of something riding up and down on top of the cam.

    If you take the displacement to be zero when it is rotated 90Deg then it will show a sinusodinal relationship about this point as it rotates.
  10. Apr 18, 2005 #9


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    Ok, here is a really rough sketch of how I'm picturing it (sorry I don't have any vector graphics package on this PC right now so it's a very rough freehand.)

    Can you see what I mean? The top circle shows the cam at maxium vertical displacement (zero degrees) and the bottom circle shows the cam at the 90 degree point. Im assuming that the cam follower is a piont that rides directly vertically above the point of rotation.

    Attached Files:

    • cam.gif
      File size:
      2.8 KB
    Last edited: Apr 18, 2005
  11. Apr 18, 2005 #10
    EDIT: Everything I wrote below is wrong

    Thinking about what you said though, I think that the amplitude of a follower for an eccentric circular cam is not 10, it is 10 - (r^2 - 10^2). But it is still sinusodinal.

    For what I was talking about (with an amplitude of 10) would be a cam that looks like this ( I assume this is what is being referred to in the question):

    http://mobyhost.no-ip.com/cam2.jpg [Broken] - Sorry I drew the cams incorrectly in my other diagram.

    If not, then specify what sort of cam it is ?

    Because right now, I'm getting super confused

    Uart - yea I see what you mean but this just changes the amplitude, not the sinusodinal motion

    EDIT: Ok, thinking about that pear cam, I think thats wrong becasue that would give a truncated sine wave of just bumps.
    Last edited by a moderator: May 2, 2017
  12. Apr 18, 2005 #11


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    Ok now I see how you can get a sinusoid out of the offset circle cam. The follower has to be a line (horizontal line) rather than a point, that works ok.
  13. Apr 18, 2005 #12
    your right, that is crazy. I'm totally confused now.

    I'm no cam expert, only a lowly physics student. Any of those mechanical engineer types around ???

    Edit: Ok I stand by my origional derivation

    ....I've gotta stop posting on this topic ! I'm changing my mind every 3 seconds :-S
    Last edited: Apr 18, 2005
  14. Apr 18, 2005 #13
    If you take the follower to be the highest point on the outside of the circle, then your equation is sinosoidal.

    If you take the follower to be the point on the outside of the circle directly above the point of rotation, your equation is somewhat different.

    I get


    Which is probably wrong due to some careless mistake... I didn't check it.
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