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  1. #1
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    Help with angled slot!!!

    Hi guys. I am new to cincinatti controllers and also hadn't been using G code programming for a brave while and today i've got oil manifold to make for next week with slot located at an angle. Thinking here what steps to take? Guys if you may i need solution please. Controller is acramatic 1200,which is so different from what i used to that i in frustration. Plus last time i calculated something like that was 2004 and even then i didn't do it often,using CAD. Now,i don't have CAD and i stuc. Would you help me please?
    Here i tried to represent all known dimensions (angle i calculated myself) Datums are in higher left corner of the billet. Attaching a picture with hope
    Attached Thumbnails Attached Thumbnails slot angled drawing.JPG  

  2. #2
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    This is pretty easy to calculate using a little trigonometry. All you have to do is work out four points, the end points of the arcs. You already know where the arc centers are, and you have the radius and angle, so this is pretty easy.

    Read up on Sine, Cosine and Tangent here.
    Sine, Cosine, Tangent

    Also read about the Pythagorean theorum here.
    Pythagorean theorem - Wikipedia, the free encyclopedia

    Hope this helps.
    Frederic

  3. #3
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    Tnx Frederic. Not that i expected that "back to school" answer but nevertheless. I know that i have to calculate tangent points but that's where i stuck. I am not sure how to do and how to apply trigonometry in this case. How do i find end of the arcs? Honestly it's been such a long time since i did that i can barely bring myself to think about it. And i've got to programm it on the controller which i never used before. All this pressure just makes me nervous.

  4. #4
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    angle slot

    two solutions to your prob no math required.
    start point and end point if given use G1.
    if not find a single point, write your program in a straight line on X or Y then use G51.1 and rotate to the desired angle (rotate is Rand the angle).Remember to divide by 60 for partial degrees. If it works (it did for me) let them think you are a genious!
    Then ask for a raise!.Good Luck!

  5. #5
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    Dmitriy,

    Here's a longer answer. Remember, often people who post here are pusting during their free time at work, so you will often just get some pointers from somebody who has to get back to doing what they're paid for. So you won't always get step by step instructions. That was the case with me earlier today.

    Also, many people don't want to do the math for you, because if they mess up, they have just scrapped your part and possibly damaged your machine.

    It's the evening now, so I have more time.

    I've also had time to pour myself some scotch, so I suggest that you double check my calculations. :cheers:

    Angle (purple) is known to be 36.9

    Length of hypotenuse (red) is known to be 14mm. It's the radius of the arcs.

    Length of the Opposite (green) is unknown.

    Length of the Adjacent (yellow) is unknown.

    Apply the Sine Function:
    sin(θ) = Opposite / Hypotenuse

    Plug in the numbers we know:
    sin(36.9) = Opposite / 14mm

    Use Windows Calculator in Scientific mode, or a scientific calculator. Type in 36.9 and hit the sin button. You should get 0.60042022532588404976462454989735.

    0.60042022532588404976462454989735 = Opposite / 14mm

    0.60042022532588404976462454989735 * 14mm = Opposite

    8.4058831545623766967047436985629 = Opposite.

    So now we know two sides of the right triangle. Apply the Pythagorean theorem. (I don't know why they call it a theory. It's a fact.)

    8.4058831545623766967047436985629^2 + Adjacent^2 = 14^2

    70.658871608155533318381412284272 + Adjacent ^2 = 196

    Adjacent^2 = 125.34112839184446668161858771573

    Adjacent = 11.19558521881926754161335652704

    So, the point we're looking at is 11.1956mm up, and 8.4059 over from the center point of the bottom arc.
    Some quick addition gives the coordinates. X = 21.5941 Y = -38.8044

    Since both arcs are the same, and they are 180 degree arcs, the other three points can be found by adding or subtracting 11.1956 and 8.4059 from the center point coordinates as needed.
    Attached Thumbnails Attached Thumbnails slot angled drawing.JPG  

  6. #6
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    Huh?

    I'm not sure what you're trying to do but it's easy to find the points in CAD with no calculating. Just add a line with an endpoint at one of the points you're trying to find and look at its properties. I did the drawing in about a minute from your drawing. No problem finding any of the coordinates.
    Attached Files Attached Files
    Gary Shepherd
    www.16tracks.com

  7. #7
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    Thanks a lot guys for answers!!! @ DroopyPawn: i don't have an access to CAD at the moment,otherwise it wouldn't be a problem
    @ TXRED Thanks a mllion mate,i'll have something to rely on It's hard for the first time when u on ur own and under the pressure. You gave me something to hold on to!!
    @ Valimarsbro: Wow! That's techinque i didn't know about. Will definitely give it a go!!! Thanks man.

  8. #8
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    Excellent TXFred. It's good to see folks take the time to thoroughly answer a question. I know I appreciate that when I ask something.

    OT - By the way, a theorem is a mathematical statement that has been rigorously proven. It is not a theory or proposition ... anymore.

    Cheers,
    Rich

  9. #9
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    You can download plenty of free CAD programs. If you're a student, AutoCAD is free. You can download the student version (which is identical to the Pro version) for free and it has a 36 month license.

    Personally, I prefer eMachineShop for simple drawings like this. It's free.
    Gary Shepherd
    www.16tracks.com

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