1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Length change of diagonal in deformable triangle

  1. Dec 28, 2013 #1

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


    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

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

    It is not clear to me why that is the case. I would just have said it should be
    \sqrt{(a+d)^2 + (b+c)^2}
    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.

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


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook