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  1. #46
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    Quote Originally Posted by lamestllama View Post
    Exactly!
    Ok well I still don't get it, prob because I don't have a maths degree.
    To me, your formula appears to calculate from that hypotenuse (D), but I don't know what that is until after I've drawn it out. It's like it's a formula to prove that D is correct.
    Can we try this again ?
    I have a thread wave 15 long and 3.7 high..... Where do I start with that ?

  2. #47
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    Quote Originally Posted by phaser View Post
    Ok well I still don't get it, prob because I don't have a maths degree.
    To me, your formula appears to calculate from that hypotenuse (D), but I don't know what that is until after I've drawn it out. It's like it's a formula to prove that D is correct.
    Can we try this again ?
    I have a thread wave 15 long and 3.7 high..... Where do I start with that ?
    It's probably easier to stick to this one as all the algebra shuffling has been done and it gives you a straight answer: D = H/2 + W^2/8H

    So:

    D = (3.7/2) + (15^2)/8 x 3.7
    = 1.85 + 225/29.6
    = 1.85 + 7.6 (ok, 7.601351351)
    = 9.45(135135)

  3. #48
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    Quote Originally Posted by elanjacobs View Post
    It's probably easier to stick to this one as all the algebra shuffling has been done and it gives you a straight answer: D = H/2 + W^2/8H

    So:

    D = (3.7/2) + (15^2)/8 x 3.7
    = 1.85 + 225/29.6
    = 1.85 + 7.6 (ok, 7.601351351)
    = 9.45(135135)
    Ok thanks for that. it's the formula I put on here and is easy to use.
    Your breakdown of the equation is great too.

  4. #49
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    Quote Originally Posted by phaser View Post
    Ok thanks for that. it's the formula I put on here and is easy to use.<br>
    Your breakdown of the equation is great too.
    Oops, didn't see your post where you had that one.

    EDIT: Now that I go back and look at the whole thing, calculating from a chord is definitely the most logical visually. The chord length is simply half the the hypotenuse on a triangle where the other sides are the thread depth and half the pitch. Don't know why I didn't see that before
    EDIT EDIT: No...that doesn't work...you have to have the segment height or the angle as well at the chord length. Too late for math, time for bed....

  5. #50
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    Quote Originally Posted by elanjacobs View Post
    Oops, didn't see your post where you had that one.

    EDIT: Now that I go back and look at the whole thing, calculating from a chord is definitely the most logical visually. The chord length is simply half the the hypotenuse on a triangle where the other sides are the thread depth and half the pitch. Don't know why I didn't see that before
    EDIT EDIT: No...that doesn't work...you have to have the segment height or the angle as well at the chord length. Too late for math, time for bed....

  6. #51
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    Ok, now I'm with it. Chord length is half the pitch, chord height is half the depth. Plug it in to a circle calculator, job done.

    My previous ramblings is what happens when I try to do math at midnight...

  7. #52
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    Quote Originally Posted by phaser View Post
    Ok well I still don't get it, prob because I don't have a maths degree.
    To me, your formula appears to calculate from that hypotenuse (D), but I don't know what that is until after I've drawn it out. It's like it's a formula to prove that D is correct.
    It is a formula to show the relationship that D must have with the rest of the geometry if D doesn't have that relationship with H and W then then the arcs will not meet tangentially. There is only one D that will work for a given H or W. So yes it is in a sense a formula to show D is correct but since it can manipulated be to give D in terms of W and H it is a formula that can be used to calculated D given you know W and H. The logic as to how we derive the formula is as follows

    1. The distance between the centres of the adjacent arcs must be the unknown diameter D you seek because the arcs meet tangentially and they both have the same radius. So we know this distance is D but we don't yet know how to calculate D

    2. We know this unknown distance D between the centres of these arcs represents the hypotenuse of a right angle triangle with sides of length W/2 and D - H (we covered this before)

    3. we know that Pythagoras's theorem applies to right angled triangles so we can write D^2 = (D - H)^2 + (W/2)^2 even though we have no idea of what D H or W are yet because we know they must have this relationship by pythagoras

    4. Now all is left to do is use algebra to rearrange this so that we have D on one side by itself so that it is easy to use in the way that you want which is to calculate D given H and W

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

    We can expand (D - H)^2 to (D - H)(D -H) which multiplied out gives D^2 - 2HD + H^2 so we can now write

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

    Now subtract D^2 from each side of the equation and add 2HD to each side of the equation to get

    2HD = H^2 + (W/2)^2

    Now divide both sides by 2H

    D = (H^2) / 2H + ((W/2)^2) / 2H

    and simplifying the fractions (H^2) / 2 H becomes (H*H) / 2H which is H/2 and ((W/2)^2) / 2H becomes ((W^2)/4) / 2H which is W^2/ 8H so now we can write

    D = H/2 + W^2/ 8H



    Up until 15 years ago I was writing third party applications in C and C++ for AutoCAD and other AutoDesk related products so this stuff comes kind of naturally. Can you tell me which step above makes no sense to you?

  8. #53
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    Quote Originally Posted by elanjacobs View Post
    Ok, now I'm with it. Chord length is half the pitch, chord height is half the depth. Plug it in to a circle calculator, job done.

    My previous ramblings is what happens when I try to do math at midnight...
    You could trivially construct the arcs you want geometrically just plain old draughtsmanship. A compass and a ruler would suffice but the CAD program of your choice will allow you to enquire as to the precise radius of the arcs you have drawn.

    You have enough information to place three points that lie on the arc you need. ie The the end points of the arc and the mid point of the arc are points you can just measure out using W and H/2.

    Now you can find the centre of the arc by constructing two lines that bisect and are normal to lines drawn between the points you placed that represent the midpoint and each endpoint of the arc. The point where these lines intersect is the centre of the arc and the radius is from that point to any of the three points you originally placed.

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