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Thread: Toric Generators

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    Master OptiBoarder lensgrinder's Avatar
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    Toric Generators

    I was wondering if anyone could help me to determine the math behind how a toric generator produces its cross curve. I have attached a diagram of a generator wheel as if the person were looking at it straight on as opposed to from the top. As you can see, as the angle between the base and cross increase the radius of the right side of the wheel decreases, which causes the cross curve to increase. Another analogy is if you hold a coffee cup with the mouth facing you you see a complete circle, if you rotate the cup to the left it creates an ellipse. Mathematically why does this happen?
    Attached Thumbnails Attached Thumbnails Gen Example.jpg  

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    I can't help with the math. I am REALLY lacking in that area. But, have you thought about sending an e-mail to Loh, Coburn, Schneider,etc? I would imagine that there would be some tech at one of those companies just chomping at the bit to explain exactly what you inquire about. Might be worth a shot.

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    ATO Member HarryChiling's Avatar
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    Last edited by HarryChiling; 02-21-2007 at 04:47 AM.
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    Thanks Harry. I have attached more diagrams of what I am talking about. I am talking about generator wheels, the old style. Like the Coburn 108, 113, 2113 or Optek 863.

    I do understand what you are saying, but what I am having a hard time trying to wrap my mind around is the radius of the wheel does not change, but when the wheel is rotated it creates an ellipse and how do you find what the curve of the side of the ellipse will be?

    The diagram I attached shows how the base is created(top view), it moves side to side to create a certain radius, this I understand.

    To create the cross curve(side view) the wheel is rotated away from you as the curve increases(creates a steeper ellipse) and toward you as the curve decreases(creates a flatter ellipse).

    Thanks for any help
    Attached Thumbnails Attached Thumbnails Gen Base Cross.jpg  

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    ATO Member HarryChiling's Avatar
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    Last edited by HarryChiling; 02-21-2007 at 04:58 AM.
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    Thanks Harry, I will play with it a little more. You know I have that book, but it is at my office and I had just thought of the question on my way home the other day. Here is another question, do you know the radius of a diamond wheel edge and the diameter? I am thinking around 5mm for the radius and about 100mm diameter. I will check when I get in on Mon, but use these numbers.

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    ATO Member HarryChiling's Avatar
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    Last edited by HarryChiling; 02-21-2007 at 04:48 AM.
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    Picture and formulas attached
    Attached Thumbnails Attached Thumbnails Topffräser_a.jpg  

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    Thanks Rafael. Where is the picture from?

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    Thumbs up lensgrinder

    1/8th inch radius on a 3.5 inch wheel

  11. #11
    ATO Member HarryChiling's Avatar
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    Last edited by HarryChiling; 02-21-2007 at 04:58 AM.
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    Hallo Lensgrinder, the pictures are from my old boss, where i have learned.
    We must describe all positions and then he take a look over, everytime. So with the years there....but all things in german

    Here some more, perhaps you can need them too:



    Steps on generating:


    The flatter curve is resulting from the distance of the generator tool to the
    centre of rotation. The generator tool is tourn around the centre of rotation.
    The steeper curve is resulting from tilt of the generator tool. If there is no tilt
    the surface will be spherical.

    MB = Centre of the basis curve
    MC = Centre of the cross curve

    rB = Radius of the basis curve
    rC = Radius of the cross curve

    Attached Thumbnails Attached Thumbnails 1Step.JPG   2Step.JPG   3Step.JPG   4Step.JPG  

  13. #13
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    Thanks Harry and Rafael.
    Those are cool diagrams Rafael, I could use them, thanks again.

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