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Kinematic Problem

  1. Jan 17, 2008 #1
    Hey,
    Would anyone be able to solve this Kinematic Problem?
    At least approximately.

    The mechanism in question is a cylinder, piston and a slider. The cylinder is fixed on a pivot in the middle of its body, obviously from the outside.

    I have found some approximate solution, but I am not sure if it is good enough.

    Thanks in advance,
    Atif
     

    Attached Files:

  2. jcsd
  3. Jan 17, 2008 #2

    Mech_Engineer

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    This looks like homework to me, why don't you post your attempt at a solution for it first and we can point out any errors?
     
  4. Jan 17, 2008 #3
    I came here to ask for help, if you’re not going to help, don’t reply. I don't want this to discussion to burst into an argument about nothing. I have a problem and I am asking this community to help me solve it.

    If you have nothing of value to post, simply do not post.

    Whoever can solve this, is a genius.

    It is for my father.

    He talked about using some program called Mathematician 5? He said that anyone that knew how to use the program, could solve this problem in a relatively short amount of time...

    Please help, It would be greatly appreciated. I know there are some exceptional minds on this forum, so I am asking those people for help. Even if it looks like homework... it is something that we need closure on and fast...


    Thank you.

    p.s. Sorry If i sound hostile Mech_Engineer, I am just stressed.
     
  5. Jan 17, 2008 #4
    I don't understand what is being asked here. All I see is a picture, what are you trying to solve for? I see a couple of equations, what have you worked out so far? The forum has a pretty strict policy conserning showing work before requesting help.
     
    Last edited: Jan 17, 2008
  6. Jan 17, 2008 #5

    FredGarvin

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    First off, can the attitude. Simply posting a question and then waiting for people to solve your problem for you is not how we operate here.

    Next. Formulate a proper question. What are you looking for? From your picture, all I can tell is that you are trying to figure out the path the face of the piston will follow in space.
     
  7. Jan 17, 2008 #6

    stewartcs

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    Amen.

    It seems of recent there is more and more of this here.

    CS
     
  8. Jan 17, 2008 #7
    Sorry,

    I didn't know the forum has a policy on this type of thing..
    I have to talk to my dad about what he wanted to find out. He didn't tell me anything. I don't even know what kinematics is. He just told me to post this and ask if anyone can solve this... I figured if he said nothing, it must be self-explanatory. It obivously isn't.
     
  9. Jan 18, 2008 #8

    FredGarvin

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    It does appear that he is looking for the position of the piston with the changing values of the dependent variables. However, it is nice to have it stated explicitly.

    BTW, kinematics is the process of describing an object's motion without worrying about forces, i.e. positions, velocities and accelerations but no forces.
     
  10. Jan 18, 2008 #9
    I think you are right Fred; he is trying to find something like that out?
    He told me that the equation is at the top... which he thinks its obivous what needs to be found out. I think I drew the graph badly.. haha... I'm a graphic designer... not a engineer so I haven't got a clue about this type of thing.

    Again, sorry for the trouble, and thank you for seeing past my Jackass-ness.
     
  11. Jan 18, 2008 #10
    Ok, I think C,H denote the y,x position of the pivot's center, constatn. Find the constants for cylinder width, and radius of rotation around the pivot which will give the dependant variable for the stroke. You should be able to work something out from there.
     
  12. Jan 18, 2008 #11
    Wow, thanks for the help mate!
    I will tell my dad right away.

    Thanks!
     
  13. Jan 25, 2008 #12
    try this:

    Define the angle the cylinder makes with the ground to be theda.
    tan(theda) = (y-H)/(C-x)

    also
    tan(theda) = y/sqrt(L^2 - y^2)

    eliminate theda
    (y-H)/(C-x) = y/sqrt(L^2 - y^2)

    since H, C, L are constants, you have a function of y with respect to x.
     
  14. Jan 25, 2008 #13
    thanks Mike,
    you are indeed the man! :)

    thanks again!
     
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