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enter image description here

Please excuse my poorly drawn doodle here, I'm almost inept at drawing.

I'm attempting to compute i2, j2, x2, y2.

Knowns: x1, y1, xk, yk, i1, j1, the arc is circular

Constraints:

  • resulting arc is circular
  • cartesian co-ordinate system
  • intermediate calculations can be in any co-ord system that makes sense
  • problem may be arbitrarily rotated, and so can't assume y2=j2 or i2=xk, etc.

    x2 and y2 are not known, but the arc length from x1,y1 to x2,y2 is an arbitrary but fixed, known distance, say 100 units.

I suppose where I'm getting most confused is that there's an arc involved. If this were triangles even I could solve this problem. I just don't know my properties of arcs well enough to make valid assumptions.

Is it a valid approach to manipulate the arc length formula to supply x2,y2 and then use similar triangles to solve back for i2,j2?

Can we assume that the line formed by x2,y2-i2,j2 is always parallel to x1,y1 to i1,j1 regardless of rotation because of the constraints (ie since the initial and resulting arcs are circular, do these lines not form a right angle wtih the line (i1,j1), (xk,yk) ) ? Cannot make this assumption, counterexample let the angle = 180 degrees.

This is not a homework problem, it's something I need to understand for a project I'm involved with.

Any help is greatly appreciated.

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    Next time you may want to use [GeoGebra](http://geogebra.org/) for drawing such a picture :)2011-07-12
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    @Theo Buehler Looking into it now... but it's only useful if it can help me derive formulas.2011-07-12
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    @Theo Buehler It's not explicitly stated, and I don't believe we can make that assumption... I think the conditions all hold if the angle were 180 degrees (for example)...2011-07-12
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    I'm confused by $(i_1,j_1)$ and $(i_2,j_2)$—how are they defined? If they are the centers of the circles containing the given arcs, then shortening the arc won't change the center. If they're something else (especially given the remarks about rotating the picture), I'm not clear on how they're defined.2011-07-12
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    (i1, j1) is the center point of the original, given circular arc (call it a1). (i2, j2) will be the center point of a non-circular arc that traces the same path as a1 if you traverse "k" units along a1 to point (x2,y2)2011-07-12
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    I'm seeing a problem with the way i2,j2 is defined. If we can simplify the problem down to finding x2, y2 I think that will suffice.2011-07-12
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    I'm sorry but I'm still confused. Let me make a guess: Are you trying to divide a circular arc from $(x_1,y_1)$ to $(x_k,y_k)$ with center $(i_1,j_1)$ into $(k-1)$ (or maybe $k$) equal pieces and want to know the coordinates $(x_2,y_2)$ (and maybe $(x_3,y_3)$, etc later on)? @Isaac: ping!2011-07-12
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    @Stephen: If you start with a circular arc (some contiguous piece of a circle), and remove a piece, the result is still a circular arc, still part of the same whole circle as the original arc, so it has the same center. Now, if I understand your simplification, you know the endpoints and center of the original arc, you know the length of the piece removed (and from which end it's removed), and you want to know the new endpoint of the arc?2011-07-12
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    @Isaac @Theo What is actually happening is that a machine is tracing a circular arc with center i,j and start/end points as described. After k distance along the arc, it has to stop and perform a mechanical compensation. The machine can ONLY work in circular arcs. To resume tracing the original arc, it needs a new center point (i2,j2), starting from where it stopped, ending at the original end points that describes the same arc. Isaac's comment earlier, asking about i2, j2 made me realize that this is in fact impossible because the curve was uniquely described by the original parameters.2011-07-12
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    The part about rotation that seems to have thrown confusion into the mix, was that the start and end points could be in any quadrant, and would not necessarily describe a quarter circle or some other convenient configuration.2011-07-12
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    @Stephen: So $(i_2,j_2)=(i_1,j_1)$, since the remainder of the arc has the same center as the original arc.2011-07-12
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    Yes, of course you're correct ... a great light bulb has dimly flickered on. Somewhere in the description of the problem I was under the impression that i and j (as the inputs to the computer functions) had to change in a particular was and this kept me from seeing what you were saying. THANKS SO MUCH for everyone's patience.2011-07-12

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I'm going to make some assumptions based on the comments. We have a circular arc, which is a part of a circle with center $(i,j)$, and the endpoints of the arc are $(x_1,y_1)$ and $(x_k,y_k)$, and we want to remove a piece of length $\ell$ from the $(x_1,y_1)$ end of the arc and find the new endpoint of the arc $(x_2,y_2)$.

From the given endpoints and center, we can find the radius $r$ of the arc (find the distance between either endpoint and the center). The arc we're removing has length $\ell$, so it has measure $\theta=\frac{\ell}{r}$ (if $\theta$ is the measure of an arc, the length of the arc is $r\theta$).

We now want to rotate $(x_1,y_1)$ by $\theta$ along the given arc, about $(i,j)$. I'm going to assume this is a clockwise rotation (which may or may not be correct). The result of this rotation is $$\begin{align}(x_2,y_2)&=(i+(x_1-i)\cos\theta+(y-i)\sin\theta,j+(y-j)\cos\theta+(i-x)\sin\theta)\\&=(i+(x_1-i)\cos\frac{\ell}{r}+(y-i)\sin\frac{\ell}{r},j+(y-j)\cos\frac{\ell}{r}+(i-x)\sin\frac{\ell}{r})\end{align}.$$

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    Thanks! I was in the process of writing up a substantially less attractive version.2011-07-12
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    @Isaac For a counter-clockwise rotation do I change the x2 formula to i+(x1−i)cosθ-(y−i)sinθ or do I simply subsitute -θ for θ as I've seen suggested elsewhere?2011-07-12
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    @Stephen: For a counterclockwise rotation, you could just change $\theta$ to $-\theta$, or change the $y-i$ in front of $\sin$ in the $x$ term to $i-y$ and the $i-x$ in front of $\sin$ in the $y$ term to $x-i$. ($\cos(-\theta)=\cos\theta$, so the cosine terms are unaffected, but $\sin(-\theta)=-\sin\theta$, so the sign of the sine terms changes, which is equivalent to swapping the order of the subtraction in their coefficients.)2011-07-12
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    Thank you for the follow up :)2011-07-12