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B Drawing an Archimedean Spiral

  1. Aug 11, 2017 #1
    I have a concern about a method for drawing an Archimedean spiral.

    If a spiral is drawn by rotating a string around a rod, it would only be an approximation of an Archimedean spiral. Starting from the edge of the circle, each time the string goes around once, the change in distance from the center would be ##\sqrt {r^2+(2πrx)^2}## (using the Pythagorean Theorem) and the change in the angle would ##2πx-\arctan(\frac{2πrx}{r})=2πx-\arctan(2πx)## (x being the number of rotations around the rod). When dividing the two and plotting the graph, it doesn't show a perfectly horizontal line, which would be expected if it were truly Archimedean: https://www.desmos.com/calculator/v5c9rsbhd3

    Do you think this approximation could be a problem? How would the accuracy of this method compare to plotting points and then connecting them free-handed? Thanks in advance.
     
  2. jcsd
  3. Aug 11, 2017 #2

    haruspex

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    It depends on the application.
    The error is worst closest to the rod. At the extreme, the tangent to the curve goes through the centre of the rod.
    I can visualise a better mechanism. The axis of the rod could be mounted on an axle which is itself on an arm length r (radius of rod) from a fixed point. Some gearing mechanism would ensure the rod is irrotational as its axis rotates about the fixed point. The string would be always tangential to the rod at the fixed point.
     
  4. Aug 11, 2017 #3

    mfb

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    Is there any application for high precision Archimedean spirals drawn manually?
    It is not a problem if you just want to make a nice pattern on paper. If you want to produce parts that have to fit together, you'll use machines anyway, they will just follow an x-y-path given by a computer, they don't have that problem.
     
  5. Aug 11, 2017 #4
    Do you think drafting might require it (especially for architecture or something similar)? I also could see it being useful in school, maybe in a geometry class. I'm not entirely sure, though.
     
  6. Aug 11, 2017 #5
    After starting this thread, I've thought of a way of drawing a true Archimedean spiral (I can't say I'm sure it will work, though).
    Disc A is placed flat on the paper. Disc B is in contact with this one in this orientation. In this video, Disc A would be the wood disc and disc B would be one of the pink discs (ignore the differential):

    Disc B is connected to a shaft and is allowed to slide across this shaft to change the gear ratio. This shaft is connected to a gear, and the gear is meshed with a rack. The edge of the rack would hold a pencil. Disc B, the shaft, the gear, and the rack, are allowed to rotate along the axis of rotation which goes through the center of Disc A. As it rotates, it causes Disc B to spin, causing the gear to spin, and causing the pencil to move outward or inward.
    I'm not sure if I got the idea across clearly, but I think it's an interesting idea.
     
  7. Aug 11, 2017 #6

    haruspex

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    You could simplify that a bit by using a threaded shaft, like a worm gear, and a corresponding thread in the hole of disk B.
     
  8. Aug 11, 2017 #7
    And the threaded shaft would be unable to rotate and instead be forced to move straight through while disc B rotates?

    I think that may cause the system to get stuck and/or disc B to slip, especially if the threads were densely packed. But I may just not be imagining it correctly.
     
  9. Aug 11, 2017 #8

    haruspex

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    No, the shaft would be fixed to a vertical axis, unable to to rotate on its own axis and unable to migrate across the vertical axis. As it rotates about the vertical axis, the disk rolls on the ground, and so rotates on the shaft. That drives the disk towards/away from the vertical axis.
     
  10. Aug 11, 2017 #9
    I should have clarified that when I wrote it "would be unable to rotate", I meant rotating around its own axis. I'm assuming that disc B and the shaft are both rotating together around the vertical axis (causing the pencil to rotate around this axis as well).

    This is quickly drifting away from math, so if we want to keep this going, I guess it should be moved.
     
  11. Aug 11, 2017 #10

    haruspex

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    It was the shaft "moving straight through" that confused me. The disk moves along the shaft. Apart from rotating about the vertical axis the shaft does not go anywhere.
     
  12. Aug 11, 2017 #11
    I'm imagining disc B rotating around the vertical axis at a fixed distance from the center while the shaft (which is connected to the pencil) moves inward or outward while rotating around the vertical axis. This keeps the gearing ratio the same while drawing your spiral.

    It seems like you're imagining that disc B is the one moving inward or outward.
     
  13. Aug 11, 2017 #12

    haruspex

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    I was, with the pencil mounted adjacent to it. But it could work either way.
     
  14. Aug 11, 2017 #13
    Wouldn't moving disc B toward or away from the center change the gear ratio, making the spacing of the spiral uneven? (You would also need to prevent the pencil from rotating around the axis of disc B). If that weren't the case, moving the disc instead might be easier than moving the entire shaft.
     
  15. Aug 11, 2017 #14

    haruspex

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    Ha! This approach cannot work. It will produce a geometric spiral.
    As the shaft rotates dθ about the vertical axis the point of contact of the wheel with table moves rdθ (near enough). That will drive the wheel further away by αrdθ for some constant α. dr/dθ=αr.
     
  16. Aug 12, 2017 #15
    The camera quality is terrible, but this a rough orthographic drawing from top, front, and side as I would imagine it to look like. It uses a rack and pinion, the pencil is shown, and disc B is replaced by a ball (which I think would be more effective).
    WIN_20170812_19_19_33_Pro.jpg
     
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