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Reverse circular curves to connect two non parallel lines

Posted on 2013-01-04
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Last Modified: 2013-02-12
If I have two parallel lines  like this

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                                                      --------------------------

I can connect the two with two circular curves using a compass, straight-edge and a pencil and without any measurement or calculation with the steps
draw perpendiculars from the end points towards the other line
connect the two end points of the given line
draw perpendiculars to the connecting line at the quarter points
the intersection of the perpendicular to end point and perpendicular to the quarter point will be the centre of the curve.

Can we have a similar construction if the lines are not parallel?
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Question by:Saqib Husain, Syed
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by:d-glitch
ID: 38743907
Still thinking..
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by:TommySzalapski
ID: 38743963
I'm not fully understanding your algorithm, but why not draw line segments that are parallel that touch the two lines and just follow the same pattern?

If you want the new parallel lines to be the same angle off of the original lines (which we'll call A and B) then:
Draw a line parallel to B through the near endpoint of A. Bisect the angle that it makes with segment B. Call that segment C.

Now draw a line parallel to C through the near endpoint of B.

This will give you two line segments that are parallel that touch the near endpoints of both of the original lines. Then just follow your original steps.
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by:TommySzalapski
ID: 38743971
To follow up, I'm not sure why it would need to be so complicated.
If you want a curve that touches the points, just connect the points, bisect that new segment, and draw a half circle.

Or if you want it to be smaller than a half circle:
Connect the two lines
Draw the perpendicular bisector of that segment
Pick any point on that line and draw a circle through one of the original points. It will connect the line segments.
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by:BigRat
ID: 38743979
I'm afraid I find your description rather vague.

Two parallel lines of a given length (same length?) AB and CD, let us say AB has a higher y coord than CD.

Then

   "draw perpendiculars from the end points towards the other line"

I presume draw a perpendicular from A and B towards the level of CD and from C and D upwards to the level of AB?

   "connect the two end points of the GIVEN line"

What line was GIVEN?


   "draw perpendiculars to the connecting line at the quarter points"

What now is the connecting line? And whose quater points?

Clearly the construction is giving a center for the circle and its radius. I presume so that one circle passes through AC and another BD? But any circle can pass through two points - there are an infinite number of such circles. Your algorithm may select just one of these which is only valid if the lines are parallel. To prove that we must be a little more detailed about the question.
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by:ozo
ID: 38744333
Call the end point of one line A, and the end point of the other line A'
Draw the perpendiculars a through A and a' through A' as before and find their intersection B
Draw the line b through B, which bisects the angle between a and a'
Find the perpendicular bisector c of the line segment between A and A'
Find the point C at the intersection between b and c
Find the perpendicular bisector d of AC and the perpendicular bisector d' of A'C
The centres of the circles will be the intersection of d and a and the intersection of d' and a'
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by:aburr
ID: 38744755
Question which I think you are asking and which I will answer.
Given two distinct line segments connect their ends with circles using only the geometers tools
of straight edge and compass.

draw line connecting the two closest ends. Bisect the new line (a standard procedure which I can detail if necessary). Using the compass draw a circle which will go through both of the closest ends.
Repeat the process with the ends farthest apart.
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by:Saqib Husain, Syed
ID: 38744933
Ozo, I think you have understood the problem correctly but I have not succeeded in implementing it. Please see this image
Reverse-curve-not-working.bmp
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by:TommySzalapski
ID: 38744934
Oh. One more possible interpretation.

You want to connect the two ends of the line segments with concentric curves.

In that case, draw both connecting lines and the perpendicular bisectors of both connecting lines.
Where the perpendicular bisectors intersect will be the center of both curves.

This will work for line segments that are parallel or not parallel unless your bisectors happen to be parallel which could happen in either case.
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by:TommySzalapski
ID: 38744943
Oh, ssaqibh posted.

Any curve drawn on the perpendicular bisector of the connecting line that passes through one point with connect them. I think you are making it harder than it needs to be.
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by:ozo
ID: 38747232
Hmm, would you accept a method for compass, straight-edge, pencil, and string?
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by:Saqib Husain, Syed
ID: 38747269
Looks like it is not possible otherwise. In that case....yes
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by:ozo
ID: 38747293
Starting with parallel lines, the circles can simultaneously satisfy equal radii, equal arc angles, and tangency.
With non-parallel lines, you may have to give up at least one of those constraints.
Do you have a preference for which to give up?
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by:Saqib Husain, Syed
ID: 38747379
I am looking for equal radii and tangency.
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ozo earned 500 total points
ID: 38750801
I hope this description is clear enough,

With two strings and two pencils, take point A, A' and lines a, a' as before,
pick a point B far down a, in the direction of A', and a point B' far down a' in the direction of A.

Anchor the ends of string with lengh AB+A'B' to B and B'
One pencil p slides along a, the othter p' slides along a', and the string goes from B, along a around p, across to p', then along a' to B'

Anchor the ends of another string with length 2AB+BA' to B and A'
The string goes from B, along a to p, around p, back along a to B, across to A', along a' to p', around p', back along a' to A'

Sliding p along a and p' along a', pull both strings tight.

The pencils should be marking the centres of the circles.
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by:Saqib Husain, Syed
ID: 38822696
Ozo, I am very sorry i have not been able to try this out. Please bear with me.
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by:Saqib Husain, Syed
ID: 38881400
Ozo

I am still stryuggling to get this to work. It may be due to the knotting or the variation in string tightness at various stages.

Anyways I have verified it with calculations and this is very much accurate so I think it is time to close the question.

Thankyou so much for the help and the patience.

Saqib
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