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Length change of diagonal in deformable triangle

  1. Dec 28, 2013 #1
    Hi

    I have a problem understanding a figure in my lecture notes. The figure is the following one

    Y9loP.jpg

    It shows the deformation of a triangular element from time [itex]t[/itex] to time [itex]t+dt[/itex]: So at t it is a isosceles triangle and at t+dt it is deformed. According to my lecture notes (page 16, eq. 28), the increase in the diagonal AC is given by

    [tex]
    \delta(AC) = \frac{a+d}{\sqrt{2}} + \frac{b+c}{\sqrt{2}}
    [/tex]

    It is not clear to me why that is the case. I would just have said it should be
    [tex]
    \sqrt{(a+d)^2 + (b+c)^2}
    [/tex]
    but this seems not to be the case. Does anyone see how one arrives at the first expression? I'd be very happy to get some help, I am pretty stuck.

    Best,
    Niles.
     
    Last edited: Dec 28, 2013
  2. jcsd
  3. Dec 28, 2013 #2

    Nugatory

    User Avatar

    Staff: Mentor

    That's the correct calculation for the distance that the vertex C moves. However, the vertex C is not moving directly along the original AC line, so that's not the amount by which the distance AC changes.

    If I am five km to the north of you, and I move one km, will the distance between us change by exactly one km? Only if I move due north or due south.
     
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