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  1. #31
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    Quote Originally Posted by lamestllama View Post
    The circle centres should be 2mm apart in the direction of the pitch and 1.0666666666666.... (ie D - 1.2mm) apart in the other and as stated above the circles should be 2.2666666666666666.... in diameter (where the dots indicate the last numeral is recurring).
    Hi lamestllama...... I believe you are correct here. Maths has never been my best subject so I'm having trouble understanding your explanation of the formula (post #24). Are you able to interpret it graphically ?
    I have redrawn the triangle (base 4mm, Height 1.2mm) to show the true representation of the thread dimensions. the triangle points are at the tangents horizontal to the circle centres.

    plastic thread.jpg

  2. #32
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    Phaser, the triangle I am concerned with is the one in the attached sketch. Because the pitch is 4 mm then the base of this triangle is 2mm. Because they both have equal diameter then the hypotenuse is D (the diameter). Now as there is 1.2mm of overlap between the circles in the other direction then the length of that side of the triangle is D - 1.2mm. You can see this by considering the outside dimension of the circles shown on the far right of the sketch it is 2D minus whatever overlap (in this case 1.2mm). The rest of my post was just putting it in pythagoras's theorem and solving for D.
    IMG_20180926_164619.jpg

  3. #33
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    Ok, I'm with it now; you draw your pitch/depth triangle and you just draw a 3-point circle using the midpoints of the angled lines and the apex as your 3 points.

    To try and put the whole diameter-depth thing to images:

    Imagine you just have 2 lines of circles of equal diameter whose tangents touch each other. This is effectively a thread with zero depth and the distance between centres = the diameter
    As you increase the depth of the thread, you decrease the distance between centres by the same amount: therefore centre distance = diameter - thread depth

    Clipboard02.jpg

  4. #34
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    Quote Originally Posted by elanjacobs View Post
    O

    As you increase the depth of the thread, you decrease the distance between centres by the same amount: therefore centre distance = diameter - thread depth
    That's exactly how I thought of it initially.

  5. #35
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    It would be pretty easy in Fusion 360. Just draw it as a coil. Yes I know it is not what you want to hear, but might be easier in different software.
    Gold, the colour of choice for the discerning person.

  6. #36
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    Sorry Elan, didnt see the question. I drew it in the following method:

    - draw 4 circles in a rough zigzag, tangental to each other. Add the constraint that all circles are equal diameter
    - vertically align the two left circles, same with the right circles so that the zigzag is symmetrical. I also aligned one row of circles with the origin (just so it didnt complain about undefined values)
    - dimension the vertical distance between the first two circles at 2mm.
    - Add a single point on the right hand side of the left circle, add a point on the left hand side of one of the right circles. Set the distance between those two points as 1.2 on the horizontal plane.

    Drawing is now fully defined. Grab the measure tool and ask it what the diameter of the circles are. 2.7mm

  7. #37
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    Quote Originally Posted by scottyd View Post
    Drawing is now fully defined. Grab the measure tool and ask it what the diameter of the circles are. 2.7mm
    A typo on your part? or garbage in garbage out? This answer is provably wrong.

  8. #38
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    Whoops. 2.27

    yeah, fat fingers.

  9. #39
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    Quote Originally Posted by scottyd View Post
    Sorry Elan, didnt see the question. I drew it in the following method:

    - draw 4 circles in a rough zigzag, tangental to each other. Add the constraint that all circles are equal diameter
    - vertically align the two left circles, same with the right circles so that the zigzag is symmetrical. I also aligned one row of circles with the origin (just so it didnt complain about undefined values)
    - dimension the vertical distance between the first two circles at 2mm.
    - Add a single point on the right hand side of the left circle, add a point on the left hand side of one of the right circles. Set the distance between those two points as 1.2 on the horizontal plane.
    Thanks. Anyone know how to get AutoCAD to work to constraints like that?

  10. #40
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    Sorry mate, its been that long since Ive used autocad. I jumped ship at solidedge 17 and never looked back, not even for a second. The learning curve was super easy and in no time at all, you could be very quickly productive.

  11. #41
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    No problem. I was using autocad for a few years at the last place I worked and they do free 3 year licences for students so I've stuck with it. My new place uses Solidworks, but the student version isn't free. Might have to see if TAFE has an educational licence for us

  12. #42
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    Quote Originally Posted by lamestllama View Post
    Phaser, the triangle I am concerned with is the one in the attached sketch. Because the pitch is 4 mm then the base of this triangle is 2mm. Because they both have equal diameter then the hypotenuse is D (the diameter). Now as there is 1.2mm of overlap between the circles in the other direction then the length of that side of the triangle is D - 1.2mm. You can see this by considering the outside dimension of the circles shown on the far right of the sketch it is 2D minus whatever overlap (in this case 1.2mm). The rest of my post was just putting it in pythagoras's theorem and solving for D.
    IMG_20180926_164619.jpg
    I still don't quite understand your theory here but it seems to work. Will it work for other lengths and heights ?
    That means we've got two formulas now.
    On another forum I go to, I posed the same question which had them stumped until one of the guys came up with this and it works perfect for all heights and lengths :
    The formula for calculating a circle radius from its chord but substituting diameter for radius.

    For a circle chord
    Radius.jpg
    For the wave calc
    Diameter.jpg

  13. #43
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    Quote Originally Posted by phaser View Post
    I still don't quite understand your theory here but it seems to work. Will it work for other lengths and heights ?
    That means we've got two formulas now.
    On another forum I go to, I posed the same question which had them stumped until one of the guys came up with this and it works perfect for all
    You don't have two formula. It is the same formula but with the H and W you specifically asked about. I put in H=1.2mm and W/2=2mm as these were the values you asked about.

    You might recall I said
    Quote Originally Posted by lamestllama
    There is a right angled triangle with hypotenuse D, base 2mm (half the thread pitch) and height is D-1.2. Using pythagoras you get
    D^2 = (D-1.2)^2 + 2^2



    If for 1.2mm substitute in H and for 2mm substitute in (W/2) as 2mm was half of the pitch then simplify as follows.

    D^2 = (D-H)^2 + (W/2)^2
    D^2 = D^2 -2HD + H^2 + (W/2)^2
    2HD = H^2 + (W/2)^2
    D = H/2 + W^2/8H

    ie the same formula as you got from the other forum but thats what you would expect when there is only one correct answer.

  14. #44
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    Quote Originally Posted by lamestllama View Post
    You don't have two formula. It is the same formula but with the H and W you specifically asked about. I put in H=1.2mm and W/2=2mm as these were the values you asked about.

    You might recall I said



    [/FONT]If for 1.2mm substitute in H and for 2mm substitute in (W/2) as 2mm was half of the pitch then simplify as follows.
    [/FONT][/COLOR]
    D^2 = (D-H)^2 + (W/2)^2
    D^2 = D^2 -2HD + H^2 + (W/2)^2
    2HD = H^2 + (W/2)^2
    D = H/2 + W^2/8H

    ie the same formula as you got from the other forum but thats what you would expect when there is only one correct answer.
    So what you're saying is ?
    Diameter 2.jpg

  15. #45
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    Quote Originally Posted by phaser View Post
    So what you're saying is ?
    Diameter 2.jpg
    Exactly!

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