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Thread: simple machine

  1. #1

    simple machine

    Hi Everyone,

    I am new to the forum, so please bear with me as I adjust.
    I have attached ( I think) a PDF image of something I have been working on and would really appreciate some input concerning the resulting force generated by the cam when the roller is pushed under it with a 250 lb force. To make it easier, I have made the assumption that there is nearly no friction as the rollers and surfaces are all hardened and ground, and grease lubricated. The cam (wedge) angle is 5 degrees. I have come up with a trig formula of R=cot 5 x 250 lbs = 19,779 lbs. Am I headed in the right direction? Any help or suggestions will be greatly appreciated.
    Regards, Jim
    Attached Files Attached Files

  2. #2
    Principle Engineer
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    Formula looks ok but answer is wrong.
    And how do you take out the 250 lb sideload?

  3. #3
    Thanks for the reply, The rest of the system is not shown due to file size. The rollers inside the cam groove are attached to an air cylinder (2"bore) and the cam is attached to a 4 post die set, the cylinder rod is captured in the lower half of the diesset frame, the cam moves the upper half pf the frame up and down 1/4". The cylinder body is shown but the rod is omitted. Please elaborate on the answer to the equation.

  4. #4
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    Quote Originally Posted by machineengineering View Post
    Please elaborate on the answer to the equation.
    Run it on your calculator again. You either entered an incorrect angle for the Cot or an incorrect force.

    I get 19,672.251688

  5. #5
    Thanks PinkertonD, I must have pushed the wrong button, I got what you got! I am in denial about needing glasses, this pushed me along. I am having doubts about the formula I used now, would you be so kind as to give the drawing another look and let me know your opinion please.

  6. #6
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    Quote Originally Posted by machineengineering View Post
    I am having doubts about the formula I used now, would you be so kind as to give the drawing another look and let me know your opinion please.
    Well, you haven't stated what you were trying to achieve, I merely used the formula to arrive at a number. The Zeke is usually pretty spot on with his stuff and I just stepped in to elaborate on his behalf.

    If you give us more information we can give better answers. I am in denial about being a mind reader.

    Until we know your "track" it is really difficult to decide if it is the right track for you.

  7. #7
    Sorry I know its vague. The system is actually used to print cylindrical containers at high speed (around 2000 per minute). The cylinder is being used to drive a roller under a 5 degree wedge (cam) and has a 2" bore at 80 psi = 250 lbf. The cam is being used to open and close the 4 post die set that supports the printing mandrel. I am interested in how much up force (opening force) is generated by the cam angle and cylinder force. There is a video of our system on our website www.machine-engineering.com which should provide a much better description than is permitted here, Please check it out on the products page by clicking on the ME 250 TS image. The cylinder is supported by hardened rollers and the rollers in the cam are also quite hard Rc 60. This product is headed for market and I am just double checking everything first. Thanks for your interest and advise. Regards, Jim

  8. #8
    Technical Fellow jboggs's Avatar
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    Double check your formula. Unless my old eyes (and memory) are failing me, I think your direct upward force is: R = 250 lbs / cot 5 = 2,857 lbs.
    I am a very visual person. Draw the force vector diagram to scale in AutoCAD or Solidworks and measure the length of the resulting vector. That is ALWAYS a good back check on trig formulas.

  9. #9
    Thanks J, I think you are right, according to my free body diagram I get the same thing. Anyone else agree?

  10. #10
    Lead Engineer RWOLFEJR's Avatar
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    I think your resulting force is half that... 1,427.5 lbs.
    ??

  11. #11
    Lead Engineer RWOLFEJR's Avatar
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    Oops... I hit the go button before some explanation. I used a wedge formula. It seemed to better represent the mechanics you are dealing with. Then I checked it on a toggle joint formula which really isn't the same motion but results in the same force by its nature, (just changing the input direction via pins) and got the same number.

  12. #12
    Thanks for your input. Please post the formula or link to the program/software. I would really like to give it a try. Did you have a look at our web site video?

    Regards, Jim

  13. #13
    Lead Engineer RWOLFEJR's Avatar
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    The formulas are in Machinery's handbook...
    I'm sure you could probably get them in the google world, or the world wide web, but to me it's easier to flip the pages in a book. I know where most sections are in the book from frequent reference to it. The mechanics section is near the front.

    I could be wrong about my formula selection and if so I'm anxious to hear from folks so I can get it straight in my head... and avoid possible errors down the road.

  14. #14
    Lead Engineer RWOLFEJR's Avatar
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    Just checked out your video on the website. Slick machine...
    Noticed the use of a Ringfedder coupling... I'm a big fan of those things. One of the best inventions since sliced bread.

  15. #15
    Thanks for your comments, we have been working on this for a long time. We have obviously done physical testing and it's hard to argue with success, but I want to have the documentation correct to go with the package. I will check out the handbook. As to the ringfeder, you are correct, we use them in several areas, two on the same shaft to align the spindle disc, we can pull it around to within .0002/.0001" True Position. I will be setting up a strain guage fixture to try and gather actual real world data that would include friction, losses due to deflection etc. so wish me luck.

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