Joining Polylines, without changing lengths, unknown angels,.

[jonestim324357181]

Hi, I'm having a bit of trouble getting my polygon closed without changing the dimensions of the polygon. The drawing attached or at http://img13.imageshack.us/img13/7494/drawingz.png you can see I've got a polygon. However I do not know the radius of the angles A, B and C. I know that all the lines have to be the dimensions marked. Can Autocad automatically bend the angle say, at A without changing the dimensions in order to complete the polygon? I tried this http://forums.augi.com/showthread.php?106618-Joining-polylines but found it wasn't what I wanted.

[Ed Jobe]

Tim, Draw the 2060 and 2900 circles with their center at the endpoints. Then draw a 4000 circle using TTR (tangent, tangent, radius). Then draw the lines by using the intersections of the circles.

[Renslo]

This is close, but you lost the '4000' distance;
instead, after you've drawn the 4000 circle, connect only the 2900 line to intersection 1,
then draw another R4000 circle with the midpoint at intersection 1,
where the 4000 and the 2060 meet is intersection 2.

[Richard McDonald]

Tim, Draw the 2060 and 2900 circles with their center at the endpoints. Then draw a 4000 circle using TTR (tangent, tangent, radius). Then draw the lines by using the intersections of the circles.


Ed the Chord of the circle r=4000 wont give you a Line length of 4000.

This task is a little like the question how long is a length of string you need one more defined angle to solve.

Sum of the interior angles of a polygon = (n-2)*180 where n=the number of sides.

(5-2)*180 = 540 minus the known leaves 450 degrees. the division of the angles could be anything within the constrains of the lengths given.

option 1 Option 2
A=135%%d 146%%d
B=90%%D 78%%d
C=94%%d 110%%d
D=131%%c 116%%d *note: These are not the extents just two integers chosen at random.

To get this empiracly you could strick line "A" creating angle A then create 2 circles one r=2060 to create angle "D" and the other from the end of line "A" 4000 long.
The intersection of the circles is the end point of line "B" and Line "C".

[jaberwok]

As Richard says - there are too many unknowns for this to be solved by geometry.
One more known angle would do it.

[arshiel88]

Take advantage of the new features of AutoCAD. You wouldn't consider it cheating, do you? :D

Put Aligned Dimensional Constraints in all members then put Coincident Constraint on all endpoints. (Parametric menu)
After the last Coincident constraint, all members will join to form a polygon. Finally, put an Angular Dimensional Constraint on the 3500 and 2320 member. Then, behold :D you will see the angels clearly.

Similar to my solution on this thread. (http://forums.augi.com/showthread.php?145822-Join-the-dots-a-bit-of-Autocad-%28any-version%29-fun)

see attached drawing.

Edit: You can move the endpoints to see other possibilities. You may also add a Fix constraint on the lines forming the right (90°) angle so the polygon will not rotate as you move the points.

[nextvkin]

This is one solution - see attached. The centre of the blue circle (4000mm radius) can be anywhere on either of the magenta or red circle as long as it intersects both, thus resulting in a few solutions.

[LaserGuy]

Delete any two Angular Dimensional Constraints, then apply a Coincident contraint between the endpoints.

89064

[Ed Jobe]

Ed the Chord of the circle r=4000 wont give you a Line length of 4000.

This task is a little like the question how long is a length of string you need one more defined angle to solve.

Sum of the interior angles of a polygon = (n-2)*180 where n=the number of sides.

(5-2)*180 = 540 minus the known leaves 450 degrees. the division of the angles could be anything within the constrains of the lengths given.

option 1 Option 2
A=135%%d 146%%d
B=90%%D 78%%d
C=94%%d 110%%d
D=131%%c 116%%d *note: These are not the extents just two integers chosen at random.

To get this empiracly you could strick line "A" creating angle A then create 2 circles one r=2060 to create angle "D" and the other from the end of line "A" 4000 long.
The intersection of the circles is the end point of line "B" and Line "C".
Correct, I didn't really think it through. It was the first thing that came to mind. But I did succeed in getting some responses for the OP. ;-)

Thanks

[jonestim324357181]

Correct, I didn't really think it through. It was the first thing that came to mind. But I did succeed in getting some responses for the OP. ;-)

Thanks

Hi,

Thanks for the replies :D (Hi there LaserGuy)

I realized this won't be necessarily correct, as there are multiple pentagon shapes that have the same amount of sides (5) with different dimensions, but different angles (other than the 90 degree angle). Using the intersections on these circles will work out one specific shape aligned to the midpoint and endpoints.

This became rather apparent to me when I used the parametric constraints option that JDMather mentioned here: http://forums.autodesk.com/t5/AutoCAD-2013/Joining-Polylines-without-changing-dimensions/m-p/3778374 (and later posts in this thread mentioned)

For some reason when I posted this question it didn't occur to me that there is actually many answers.

[jonestim324357181]

This is one solution - see attached. The centre of the blue circle (4000mm radius) can be anywhere on either of the magenta or red circle as long as it intersects both, thus resulting in a few solutions.

The name of that PDF made me smile, yes I realized spelling mistake this morning when I checked the the thread. I'm quite sure the angels are quite angular.

[John_C]

Hi, I'm having a bit of trouble getting my polygon closed without changing the dimensions of the polygon ...

Dredging in the deep dark corners of my memory, what you are attempting to do is exactly the same as error correction for a closed loop traverse in surveying.

Scroll down to the page marked 2/14 for an explanation of one method ... http://www.scribd.com/doc/53112628/Surveying-Traverse-Calculation

hth