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Thread: Locating A Corner on a Mill

  1. #1
    Supporting Member rgsparber's Avatar
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    Locating A Corner on a Mill

    This work was inspired by a very clever tool presented by Marv Klotz on
    Homemadetools.net (http://www.homemadetools.net/forum/one-tool-two-uses-
    27184 ).

    Given a block held at an arbitrary angle in a mill vise, how do you locate the top corner? Marv's tool is elegant. Of course, I'm always looking for other ways to do the same thing. Here is a method that uses no special tools. It assumes the faces of the block are flat but not necessarily perpendicular.

    If you are interested, please see

    http://rick.sparber.org/LACOM.pdf


    Your comments are welcome. All of us are smarter than any one of us.

    Thanks,

    Rick

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    Rick

  2. The Following 4 Users Say Thank You to rgsparber For This Useful Post:

    Jon (Aug 8, 2017), kngtek (Aug 9, 2017), mklotz (Aug 8, 2017), Paul Jones (Aug 10, 2017)

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    Supporting Member mklotz's Avatar
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    Well, disliking the term "geek", I prefer to describe myself as a math enthusiast, On that basis, let me make a stab at deriving Rick's equations...

    The (presumed straight) line running from x3 to x2 has the equation...

    x32 = [(x2 - x3)/z1] * z + x3

    and the (presumed straight) line running from x2 to x1 has the equation...

    x21 = [x1/z1] * z

    When these two lines cross at (x0,z0) it must be true that x32 = x21, so

    [(x2 - x3)/z1] * z0 + x3 = [x1/z1] * z0

    which leads, after a bit of algebraic gymnastics (an exercise for the student) to

    z0 = x3 * z1 / (x1 + x3 - x2)

    plugging this result into the equation for x21 above, we have

    x0 = (x1 *x3) / (x1 + x3 - x2)

    consistent with Rick's result.

    Rick's equation for z0 follows immediately.

    Well done, Rick. There aren't many folks out there who appreciate how helpful math can be in the shop; and I'm not just talking about learning the basic trig relationships.

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    Last edited by mklotz; Aug 8, 2017 at 04:47 PM.
    ---
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    Supporting Member rgsparber's Avatar
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    Marv,

    Thanks for checking my work :-)

    Here is another challenge - when using the edge finder, do you have to account for its radius. Given that we touchdown on both sides of the corner, I would not be surprised it this radius doesn't matter.

    Rick
    Rick

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    over my head

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    Supporting Member mklotz's Avatar
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    Quote Originally Posted by rgsparber View Post
    Marv,

    Thanks for checking my work :-)

    Here is another challenge - when using the edge finder, do you have to account for its radius. Given that we touchdown on both sides of the corner, I would not be surprised it this radius doesn't matter.
    Should be easy to test. Simply add a constant 'r' representing the radius of the probe to each of the 'x' measurements and see if the equations reduce to the result you published. You may have to introduce an 'x0' measurement to replace the zero you assumed. [In general, it's good practice to never set a zero before deriving relationships; assign a variable and let it go to zero after the algebra is done.]
    Last edited by mklotz; Aug 9, 2017 at 09:00 AM.
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    Supporting Member mklotz's Avatar
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    OK, Rick, you made me do it. Here's the proof of your conjecture that the diameter of the edge finder doesn't matter.

    You'll have to refer to the scan of my handwritten work. I didn't have the patience to convert all that math to typewritten text.

    Instead of your "X=0 & Z=0", I labeled that point (x0,z0). With that notation I wrote the equations for the left line (xL) and right line (xR) in your diagram. By setting xL = xR, we arrive at equation (1) for "z", [By setting x0 = 0 and z0 = 0 this equation reduces to yours as it should.)

    Now, if we want to include the effect of the edge finder, we need to make the replacements shown on the sheet. Imagine the x measurements being made with a DRO. When you touch off on x0, the DRO will read x0 - r, whereas when you touch off on x3 it will read x3+r.

    Below the "replace", you see what equation (1) becomes when we make these replacements. When the equation containing the replacements is simplified, all the "r"s disappear and the result is equation (2) which, you will note, is identical to equation (1).

    I think that proves your conjecture very nicely.


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    Last edited by mklotz; Aug 9, 2017 at 11:36 AM.
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    I agree with Marv and Rick. Algebra and Trig are indispensable solutions for converting 'abstract' figures into factual articles. The problem many face stem once again, from removal of vocational electives from public schools. Theory and rote memorization never make the point of utility better than usage in real situations, reinforced by visual results. Math was my biggest adversary in classes, but a breeze in shop! With 45 years machining, and 26 navigating my attention is better than predictions indicated. My solutions tend along with Rick, using different means. I respect Marv's ability to illustrate sophisticated calculations, I'm restricted to smaller bites.
    I prefer trig overall for various solutions. I probably use algebraic techniques unknowingly, by reasoning out digits to work with over letters.
    But tell algebra this; quit asking us to find your X; she's long gone and not coming back!
    Sincerely,
    Toolmaker51
    ...we'll learn more by wandering than searching...

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    I salute you guys who still posses your math skills. I lost a large portion of mine years ago when I had a severe blood flow discrepancy to my brain that I didn't even realize was happening until one day I woke up and thought I was having a heart attack. It turned out to be 90% Arterial blockage to my LAD the artery that supplies the Heart caused by scar tissue build up pressing on the outside of the artery which was most likely caused by a missed fragment of Shrapnel dating circa 1971 or 2 . A stint solved the problem but I had already spent nearly a year without a decent new coherent design thought. That was about the time I decided it best that I retire and put my shingle in a drawer
    Toolmaker 51 Never underestimate the power of "n" when looking for "X" or you will wind up having to cosign the tangent.
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    Depending on the height I have my mill table cranked to I have always found it easy to locate the corner with either my knee, hip or elbow with very little difficulty

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    Jon (Aug 10, 2017)

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    Quote Originally Posted by 12bolts View Post
    Depending on the height I have my mill table cranked to I have always found it easy to locate the corner with either my knee, hip or elbow with very little difficulty
    Pretty easy to find all of the movement handles as well, especially in the sark
    Never try to tell me it can't be done
    When I have to paint I use KBS products

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