Topic: Dividing an arc into equal segments

[Maverick®]

  Alright I know I should remember this from high school math but my brain is in lockdown.  (See attached)  I am drawing a deck.  Customer wants a curved deck but doesn't want to pay for the labor of having one so.......... I need to divide the arc into even segments.  (Blue lines)  Someone please show me how much I have forgotten from Math class.

[CAB]

Here is a lisp solution, also one by Joe Burk but I can't put my finger on it at the moment.

http://www.theswamp.org/index.php?topic=12154.msg150674#msg150674

[Dommy2Hotty]

Wouldn't the "DIVIDE" command work?

[Maverick®]

whoa!  Sorry guys.  I am looking for a mathematical solution.  I don't use Acad.

I really should have known to put that in the question.

[Dommy2Hotty]

whoa!  Sorry guys.  I am looking for a mathematical solution.  I don't use Acad.

I really should have known to put that in the question.

:LOL:  Yeah, that threw me off there...

[Atook]

whoa!  Sorry guys.  I am looking for a mathematical solution.  I don't use Acad.

I really should have known to put that in the question.

I'd imagine you just find the length of the arc and take it from there.

Length of arc: http://www.themathpage.com/aTrig/arc-length.htm

You could also divide theta into units, and draw in the incremental angles.

[Maverick®]

  I probably made it harder than it is but......  I found the radius of the arc using this. Drew lines from the endpoints of the arc to the center point.  Found the angle and divided by 5. Drew lines from center point adding those angles and connected the dots.  Turned out perfect to the 32nd but probably could have been done easier.

  Oh well...... *skips off to the coffee maker*

[paulmcz]

  I probably made it harder than it is but......  I found the radius of the arc using this. Drew lines from the endpoints of the arc to the center point.  Found the angle and divided by 5. Drew lines from center point adding those angles and connected the dots.  Turned out perfect to the 32nd but probably could have been done easier.

  Oh well...... *skips off to the coffee maker*

What is the radius on the arc then?
As I understand the problem, the height of an arc connecting all 6 vertexes is actually larger than 2’. The measurement of 2’ in your sketch is taken from the flat surface of the middle segment. Something is not right here?!?!

[solo]

Hi Maverick,

I came up with an equation that will give you the length of each straight line segment.  You just have to plug in the length and width of the overall arc and the desired number of segments.  The only problem is every time i try to include an image it never shows up on the forum.  Could you, or anyone, please let me know how to attach an image or a file.

Thanks,
Solo

[solo]

Hi Maverick,

Here is the formula i mentioned.  I combined equations for the arc radius and included angle into this formula.  If you would prefer to see several individual steps leading up to the final chord length calculation i could post those as well.

Hope this helps,
Solo

<edit: image type changed>

[Maverick®]

Thank you Solo! That is what I needed.  I will be able to use that in the future.

[Maverick®]

What is the radius on the arc then?
As I understand the problem, the height of an arc connecting all 6 vertexes is actually larger than 2’. The measurement of 2’ in your sketch is taken from the flat surface of the middle segment. Something is not right here?!?!

   That is true Paulmcz.  I wasn't looking for the vertices to hit on the arc.  Just to have equal segments using the endpoints and center point of an arc.  4 of the vertices are actually outside the arc as you say. 

  Thanks for the help everyone.  A lot of work for a deck. But I've learned something!