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Thread: Fixing the length

  1. #1

    Fixing the length

    Is it possible to fix the length of a line segment so you can move it around (including rotation) and not change the measure?

  2. #2
    Make a segment whose length is dependent on a parameter

    Number > New Parameter (Shift+⌘+P Mac, Shift+Ctrl+P Windows)

    Enter the number of cm you want to have the segment be

    Choose cm for the unit

    Make a segment using the parameter as the length

  3. #3
    Hi Steve,

    I'll elaborate a little on Kcoder's suggestion. After you have created a length parameter, construct a point. Then select the point and the length parameter and choose Construct | Circle by Center + Radius. That will give you a circle with a radius equal to your length parameter. Use the Segment tool to construct a radius of the circle, and then hide the circle. The radius segment can be rotated or translated, and the length will remain constant unless you change the length parameter.

    For more tactile control, construct a reference segment instead of the parameter. Then construct a point, select the point and the reference segment, and then follow the steps above to construct a circle with that radius. If you want to vary the length of your radius, you can adjust the length of the reference segment.

    Happy sketching,


  4. #4
    I am a newbie and I want to make a parallelogram where I have already specified two adjacent sides by their paths being concentric circles. Now I want to specify the adjacent sides' respective opposite sides as being the lengths of their respective adjacent sides, but cannot yet find a reference of how to symbolically specify the dynamic length of an object to be used in setting the length of another object...



  5. #5
    Perhaps I wasn't clear enough, so let me try again. One vertex of the parallelogram is the center of the two concentric circles, and the two circles are the paths (loci) of two vertices of the parallelogram. I want to specify the third point by hanging two lines off the two vertices, specifying their lengths as being the same as those of their opposite sides, and then joining
    the two free ends of the two lines to make the fourth vertex of the parallelogram. Only I do not know how to set the lengths of the two lines to those of the existing two lines, ie radii of the circle.

  6. #6
    Hi Shiraz,
    The simplest solution to your question is by means of a transformation. Let's call the common center point B and the points on the two concentric circles A and C. These are three of the four vertices; let D be the fourth. Side CD of the parallelogram must be parallel to side BA and the same length as BA. You can satisfy both properties if you translate point C by the vector from B to A. (You could instead translate point A by the vector from B to C.)

    If you prefer a construction that doesn't involve a transformation, select segment AB and vertex C and choose Construct | Parallel Line. Then select segment BC and vertex A, and again choose Construct | Parallel Line. The intersection of the two parallel lines is vertex D.

    By the way, I highly the Sketchpad tutorials, the Sketchpad Tips, and the Reference Center. You can get to any of these by choosing Help | Using Sketchpad. Once you get there, for instance, you can click Sketchpad Tips, then Transform, and then Translating to get to the Sketchpad Tip on using the Translate command.


  7. #7
    Hi Scott.

    I haven't progressed enough to do a transformation, so Construct|Parallel Line it was, and worked.


  8. #8
    Thank you, Elizabeth. I'm new to this group but had Steve's question. I have a further problem. I'm investigating linkages, i.e. chains of line segments of fixed length joined at their ends. By the circle construction you describe I can do so, except when I want to make a loop, i.e. to anchor the free end of my chain to a reference point. That is to say, I want to identify the free end of the chain and the reference point as one and the same. Is there a construction which allows me to do this?

    Many thanks, P.S.

  9. #9
    Hi P.S.,.
    You can come pretty close to the construction you're looking for, by considering the constraints carefully. By the time you're ready to construct the last segment (and its endpoint, the free end of the chain), it may not be possible to place that last segment so that its end lies on the reference point. In fact, the only way that the free end can lie on the reference point is if the other endpoint of the last segment lies on a circle of the required radius centered on the reference point. So that next-to-last-point must constructed be at the intersection of two circles. To make this clearer, let's take an example, of a chain of 5 segments, labeled a, b, c, d, and e, and with starting points A, B, C, D, and E. (Thus the first segment a runs from point A to point B, and the last segment e runs from point E to point A, completing the chain.) To construct this last segment, e, point E must be exactly the required distance from the desired destination A. So once point D has been constructed, along with a circle with radius equal to the required length of segment d, you can also construct a circle about A with radius equal to the required length of segment e. To close your chain, point E must lie at one of the two intersections of these circles.
    Once the construction is finished, you can drag point A freely, and you can drag each of the points B, C, or D around their respective circles. But you cannot drag point E freely, since it must lie on the intersection of the two circles of which segments d and e are the radii. Further, depending on how you drag the earlier points, you can place D far enough away from A that the remaining two segments aren't long enough to reach back to A. In this case the two circles don't intersect, point E doesn't exist, and the chain cannot be closed.
    I know this doesn't feel like a completely satisfactory solution to your question, but at least it helped me to think about the problem of what has to happen in the chain so that the last segment can be made to have the required length!
    Best regards, Scott

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