geometry - Math Parrallelogram Level Generation -

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i developing mobile runner. therefore need platforms allow transition 1 height (2d).

img

given: triangle afb has rectangle alpha, triangle cde has rectangle @ delta, line bc = ef, line ed = ab , height of parralelogram ed(=ab)
looking for: af = cd (any of both)

i can't find solution.

tip: have formula case when can give me parralel line long sides of parrallelogram able work out rest. couldn't find parrall line though.

i'll try re-interpretation "planar mechanism" possibly still not right in input description.

input: solid fixed triangle @ bottom short perpendicular sides length a , c. top triangle same, rotated 180°, not fixed. triangles connected bars of length b rotationally flexible , fixed giving angle phi.

  • bottom triangle: a=(0,0), f=(c,0), b=(0,a)
  • lower bar: f=(c,0), e=(c,0)+b*(cos(phi), sin(phi))
  • upper bar: b=(0,a), c=(0,a)+b*(cos(phi), sin(phi))
  • upper triangle: e=(c,0)+b*(cos(phi), sin(phi)), d=(c,a)+b*(cos(phi), sin(phi)), c=(0,a)+b*(cos(phi), sin(phi))

if not angle given, total height @ segment cd, 

h=a+b*sin(phi),

then horizontal component can calculated via trigonometric identity

b*cos(phi) = sqrt( b^2 - (h-a)^2 )

set bsin=b*sin(phi)=h-a, bcos=sqrt(b*b-bsin*bsin) then

c=(  bcos, a+bsin) d=(c+bcos, a+bsin) e=(c+bcos,   bsin)

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