Hello,
Can someone please help me out? I am trying to determine the longest sections of straight pipe I can run between two curves with the least amount of intersections. The caveat is that the intersection points must always be on the centerline between the two boundaries/curves. Lines (pipes) must be within the two boundary curves or can be tangent at most. Every intersection of pipe saved saves the client money.
Can someone suggest a workflow to help solve this? I am trying to find a AutoCAD tool or math equation which would yield the 100% best scenario every time to have the least amount of intersections and least amount of pipe. I know that mathematically there should only be one correct answer/scenario for any given curve and the equation/tool to find that out is my goal right now. There are also limits on single lengths of pipe that can be used. Thank you in advance.
Draw a LINE from anywhere on the center-line curve, PERpendicular to the inboard bounding curve. ROTATE that Line 90° about its inboard end in either direction. EXTEND it both ways to the center-line curve.
If the result is not too long, you can use MIRROR, with the mirror line being from an ENDpoint to the CENter of the curve, to get adjacent runs.
If it's too long, you can SCALE it down to the maximum length using the Reference-length option. Then select it, grab the grip at one end, hit Enter until the Rotate option comes up, pick the Copy option within that, and rotate the copy 90° in the direction that puts the copy aiming outboard, i.e. crossing the center-line curve. MOVE the source Line from the common ENDpoint to the INTersection of the rotated copy with the center-line curve. MIRROR as above for adjacent runs.
Either way, you will probably need to ROTATE the collected runs about the CENter of the curves, to get the ends to land in a workable place.
Hi,
you might solve this using parametrics.
- alfred -
Assuming the "curves" in your drawing are arcs, the length of the straight line in your drawing is:
L = 2 * Sqrt( rc^2 - ri^2)
where rc = radius of center line arc, ri = radius of inside curve
This ignores the diameter of the pipe. To include the pipe diameter use:
L = 2 * sqrt(rc^2 - (ri + diaPipe/2)^2)
where diaPipe is pipe diameter
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