Length change of diagonal in deformable triangle

  • Thread starter Niles
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  • #1
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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:

Answers and Replies

  • #2
Nugatory
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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.
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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