Geometry question

Discussion in 'AutoCAD' started by Chip Harper, Aug 4, 2004.

  1. Chip Harper

    Chip Harper Guest

    How do you obtain a radius given only two points???

    Had an engineer call wanting the raidus placement of some columns and I was
    totally lost. We have a regtangular shaped building. At both ends there is a
    circular beam (basicly an arc) overhead that is supported by two columns
    placed at either side of the entry way. Our discussion was concerning the
    foundation sheet but I had to go to the roof sheet, offset the beam by half
    to place an arc dead center over the columns and get the radius of that.
    This guy was obtaining a radius from the center point of the two columns
    somehow.... can anyone explain or post an example of how to do this?
     
    Chip Harper, Aug 4, 2004
    #1
  2. Chip Harper

    Paul Turvill Guest

    There are literally an infinite number of arcs that can pass through two
    points; all will have radii greater than half the distance between the
    points, but the value of the radius is, of course, indeterminate.

    .... unless, of course there is some other constraint, such as the arc being
    a semi-circle (180 deg.), or subtending some other known angle.
    ___
     
    Paul Turvill, Aug 4, 2004
    #2
  3. Chip Harper

    Walt Engle Guest

    How about posting the dwg - really not sure, but think I know what you're
    talking about. Need to look at the dwg.
     
    Walt Engle, Aug 4, 2004
    #3
  4. Chip Harper

    Walt Engle Guest

    Tried three time to unzip, but each time get message says "No files
    found".
     
    Walt Engle, Aug 5, 2004
    #4
  5. Chip Harper

    Walt Engle Guest

    Well.....I am having difficulty understanding what you have posted, i.e.,

    "At both ends there is a circular beam (basicly an arc) overhead that is supported by two columns
    placed at either side of the entry way. Our discussion was concerning the foundation sheet but I
    had to go to the roof sheet, offset the beam by half to place an arc dead center over the columns
    and get the radius of that. This guy was obtaining a radius from the center point of the two
    columns
    somehow"

    Some of the confusion is due to me because there is a difference between beams and columns. You say
    "circular beam" - is that the arc over the entrance?
    You say "two columns placed at either side of the entry way" - are those the partial arcs?
     
    Walt Engle, Aug 5, 2004
    #5
  6. Chip Harper

    Chip Harper Guest

    This is the foundation drawing (plan view), the beam is shown on the roof
    framing sheet.
    The (2) C2/C4 columns (grid line A2 and A3) support a circular shaped beam
    overhead. The beam starts at the half circle shaped bump out (green- layer
    is S-COL-CONC-BELOW) and you see two yellow dots [vert. rebar], runs over
    the two columns and back to the wall again. Given the center points of the
    two 4x4 TS columns, can you obtain a radius value?
     
    Chip Harper, Aug 5, 2004
    #6
  7. Chip Harper

    Walt Engle Guest

    Let me look at it again now that I have a better guide.
     
    Walt Engle, Aug 5, 2004
    #7
  8. Chip Harper

    Walt Engle Guest

    Let me know if what I sent is right.
     
    Walt Engle, Aug 5, 2004
    #8
  9. Chip Harper

    Chip Harper Guest

    Excellent, center point for the start and end are obvious enough, but how
    did you obtain the third point for the arc?
     
    Chip Harper, Aug 5, 2004
    #9
  10. Looks like a 3P circle drawn through 3 of the centers... and then trimmed.
     
    Tracy W. Lincoln, Aug 5, 2004
    #10
  11. Chip Harper

    Stephen R Guest

    Two points define a line in space. If you want to define an explicit
    radius, you must have three minimum.

    If you only have two points, then a design assumption must be made. I think
    in your case, you would offset an existing arc (or radius) the distance from
    the existing radius center line to the new radius center line. Thus ending
    with a new radius with a radius larger by the offset amount (this assumes
    the two points are not on the same arc center line of the existing arc.

    Stephen R.
     
    Stephen R, Aug 5, 2004
    #11
  12. Chip Harper

    Chip Harper Guest

    Awww...light comes on....Excellent. Thanks Tracy and Walt.
     
    Chip Harper, Aug 5, 2004
    #12
  13. Chip Harper

    Walt Engle Guest

    No, not a three point circle - a three point ARC, starting with center of
    either half-circle, second point mid-point of line (you drew) through 4x4TS
    and third point center of other half-circle.
     
    Walt Engle, Aug 5, 2004
    #13
  14. Chip Harper

    Joe Butler Guest

    I don't know if this will help.

    1.) draw a line (chord=C) between the two points.
    2.) draw a line (middle ordinate=M) form the mid point of the chord,
    perpendicular to the chord to the arc.
    3.) radius=(C squared/8M) + (M/2)
    JoeB
     
    Joe Butler, Aug 6, 2004
    #14
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