Coordinates with Radius?

[fenomeno]

I'm using AutoCad 2000 and I'm trying to make arcs with radius' that have surveyor coordinates. I can do straight lines, but don't know how to make the radius.

Here's an example:

Arc of curve having a radius of 1,940.00' a central angle of 11 degrees 37' 06" and a chord that bears south 09 degrees 23' 43" West 392.72'


I know that [south 09 degrees 23' 43" West 392.72'] is written as @392.72'<s09d23'w,13" but, how do I write [radius of 1,940.00' a central angle of 11 degrees 37' 06"]???????????

 

[CarlAK]

Usually a curve that is part of a boundary or road design is tangent to segments entering and leaving the curve. If not, it should be identified as "non-tangent". You can use this property to help draw the arc. Offset the segment leading in to the arc, by the radius. The end of the offset line will be the arc's radius point, by definition of tangency.

You can draw the chord as defined, and the endpoints of the chord give you 2 points on the arc (beginning & end). You could draw 1940' circles centered on each end of the chord, they will intersect at the arc's radius point.

You don't need the central angle to draw the arc, but it's helpful for arc calculations. For example, the central angle, in radians, equals the arc length divided by the radius (delta=L/R). So for your arc, when drawn, should be 393.39'

Hope this explanation helps!

 

[fenomeno]

Hi


Yes, this helps alot!

Thank you very much CarlAK.

It seems so easy now, after the explanation.


Have a nice day,

il fenomeno