Creating an arc with only radius, delta, and length?

[sherloc987]

Hello! I’ve got an old boundary survey I’m practicing with and I’ve discovered that the surveyor only gave radius, delta, and length measurements for the curves. Here is a screenshot if someone could point me in the right direction.

https://m.imgur.com/u84bINQ

[chriscowgill7373 · Accepted solution]

From the end of the line (doesnt matter which direction you are coming from, make a LINE with a right angle 1925 units. draw a CIRCLE with the center at the end of the line and radius 1925, trim the circle with EDGEMODE on using the line, make the length of the remaining arc 1474.26 using the LENGTHEN command.

This post has some code:

https://forums.autodesk.com/t5/visual-lisp-autolisp-and-general/draw-arc-by-length-radius-and-delta/...

And there is a tool in Civil 3D that can do it:

https://knowledge.autodesk.com/support/civil-3d/learn-explore/caas/CloudHelp/cloudhelp/2019/ENU/Civi...

[Kent1Cooper]

You may be able to use ArcLength.lsp, >here<.

[rkmcswain]

In addition to the valuable input from @chriscowgill7373 and @Kent1Cooper - you might take a look at this: Entering Survey Data using AutoCAD

[WashingtonG]

Giver the situation you presented in your attachment and assuming tangents to the arc, there are two easy methods you could use.

1. There are 2 parallel arcs  150' apart with the same angle, etc.....  Since these are parallel, then the end points of the tangent lines to each arc will allow you to create a line through the ends points which will be a perpendicular line passing through the center of the circle. Do this for each pair of arc ends and where they cross is the center of the circle.

2. The ends of the each arc are the same distance from the center of the arc's circle .  Drawing a circle with radius 1925' centered on each of the arc ends will create an intersection at the center point of the arc you need.

3.  I have another method as well, but it would require a pic. to explain which I would have to create.