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In this question it is asked, "how do you find the length of a belt between two pulleys" Given two circles, find the length of a pulley belt that connects the two.

this question assumes that the pulley is on the outer edge, and asks only about the length.

use case: i have a 3D printer CAD simulation where the pulleys may be of any arbitrary size, and the belt may at times cross over from outer edge to inner edge, and back. my first attempt to code this up (in python) was not entirely correct:

enter image description here

you can see in the case of the (very) large bearing the belt layout is incorrect (leftmost part of belt in picture), also the pulley which doubles-back on itself - swapping from outer-to-inner and then inner-to-outer in the process - the "belt" is overlapping inside the pulley (lowest one in the picture)

so there are two cases where i need to know the length of the belt, and, unlike the other question posted i also need to know the angle that the belt segment makes between the two bearings. we also assume a starting angle in each case from the first to the second bearing

  • given a starting angle and two pulleys of arbitrary size, the belt wraps around the outer edge of the first and then continues in a straight line until it contacts the second bearing on the OUTER edge.

  • given a starting angle and two pulleys of arbitrary size, the belt wraps around the outer edge of the first and then continues in a straight line until it contacts the second bearing on the INNER edge.

enter image description here

if there is a simpler way to express this, please feel free to express then answer the simpler question :) if anyone would like to see the actual python code it is here, as GPLv3+ licensed source.

all help greatly appreciated.

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    $\begingroup$ ok so let's now provide the link to the images, which were previously prevented and prohibited. image 1: hands.com/~lkcl/foldable3dsandwich200/belt_example.png $\endgroup$ – lkcl Aug 8 '17 at 5:55
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    $\begingroup$ and now image 2: hands.com/~lkcl/foldable3dsandwich200/belt_diagram.png $\endgroup$ – lkcl Aug 8 '17 at 5:57
  • $\begingroup$ I edited your question to insert pictures and link. What is the "starting angle"? $\endgroup$ – Intelligenti pauca Aug 8 '17 at 8:13
  • $\begingroup$ thank you - stackexchange should be... more trusting :) you could see where the images should have been. the "starting angle" is the angle at which the belt is already on one pulley when the next one is added. so, in diagram 2 it's the red (or green) bottom end of the belt. that comes in "straight" from a previous pulley, at a certain angle "starting_angle". called so because to calculate the length of the arc you need to know that "starting" angle. i belieeeeve i have all the information needed, will reply to your other answer. $\endgroup$ – lkcl Aug 10 '17 at 9:32
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In the case of an external common tangent, a trapezoid is formed and elementary geometry gives: $$ L=\sqrt{d^2-(r_B-r_A)^2}, \quad \varphi_B=\pi-\varphi_A=\arccos{r_B-r_A\over d}, $$ where $d$ is the distance between circle centers (see diagram).

enter image description here

In the case of an internal common tangent, in a similar way one gets: $$ L=\sqrt{d^2-(r_B+r_A)^2}, \quad \varphi_B=\varphi_A=\arccos{r_B+r_A\over d}. $$ That is all you need to know. enter image description here

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  • $\begingroup$ absolutely superb! thank you! i will come back with a link to where it's implemented, with some links to confirm it. really appreciated, aretino $\endgroup$ – lkcl Aug 10 '17 at 9:26
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    $\begingroup$ okay. so. this is fascinating. so, Ra and Rb are parallel (have to be), and L is at right-angles to both. thus, yes, ha! PhiA and PhiB would be related by pi... then you draw another line parallel to L but going from A to meet Rb and now the distance from there to B is (Rb-Ra)... that makes a triangle where you know L, you know (Rb-Ra) so yes! awesome! $\endgroup$ – lkcl Aug 10 '17 at 9:38
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    $\begingroup$ now, because you know both PhiA and PhiB, the piece of belt that lays across from A starting at an angle named "starting angle", the arc simply goes from "starting angle" to PhiA and then follows the line marked L. and when you add the next bearing, the "new" PhiA and the "old" PhiB can be used to work out the next arc. superb! i'm really looking forward to seeing if this works. $\endgroup$ – lkcl Aug 10 '17 at 9:41
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    $\begingroup$ ok so i marked it up to illustrate, from the original diagram you did aretino: extended the parallel line (length L) from A, marked the triangle that makes in yellow, marked up (Ra-Rb) to the right... it looks so obvious, i wonder how i missed it! :) hands.com/~lkcl/foldable3dsandwich200/2rbQ2.jpg $\endgroup$ – lkcl Aug 10 '17 at 13:22

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