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Thread: Circle calculation
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26th Sep 2018, 11:03 AM #31Senior Member
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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
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26th Sep 2018, 06:24 PM #32Golden Member
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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
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26th Sep 2018, 06:41 PM #33Gear expert in training
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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
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26th Sep 2018, 06:45 PM #34Golden Member
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26th Sep 2018, 07:24 PM #35Pink 10EE owner
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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.
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26th Sep 2018, 07:41 PM #36Senior Member
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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
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26th Sep 2018, 08:10 PM #37Golden Member
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27th Sep 2018, 08:06 PM #38Senior Member
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Whoops. 2.27
yeah, fat fingers.
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27th Sep 2018, 08:09 PM #39Gear expert in training
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28th Sep 2018, 06:05 PM #40Senior Member
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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.
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29th Sep 2018, 09:01 PM #41Gear expert in training
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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
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1st Oct 2018, 08:49 PM #42Senior Member
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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
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1st Oct 2018, 10:01 PM #43Golden Member
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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
Originally Posted by lamestllama
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.
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1st Oct 2018, 10:55 PM #44Senior Member
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So what you're saying is ?
Diameter 2.jpg
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1st Oct 2018, 10:58 PM #45Golden Member
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