Lecture # 10: Hydrostatic Skeletons

1. Lecture # 10: Hydrostatic Skeletons

2. 1 cell cellular sheet cellular bilayer one way gut ecto- derm endo- derm mouth anus bilayered canister Body Plan Evolution cephalization mesoderm

5. 1 cell cellular sheet cellular bilayer one way gut ecto- derm endo- derm mouth anus bilayered canister Body Plan Evolution cephalization mesoderm

8. ectoderm coelom endoderm gut mesoderm coelom

9. pseudocoelom ectoderm mesoderm endoderm gut

20. r P d P= internal pressure r = radius d= thickness slice in half How does stress in a worm depend on geometry? Consider a hollow spherical animal…. P P what is stress in wall? Define tension, T, as force/length then T = s x d = r p / 2 T = ½ r p s = force/area = (p r2 p) / (2 p r d) = (r p) / (2 d) LaPlace’s Law: Tension in wall of sphere is proportional to radius and pressure. disk area ~ p r2 rim area ~ 2 p r d

21. 1) longitudinal slice 2) slice in half Consider a cylindrical animal…. Equivalent to spherical case, Thus longitudinal tension, TLis same as in sphere of equal radius: TL = ½ r p 3) cap with hemisphere

22. d sc 1) transverse wedge slice area =2rd rim area =2 d d r d Circumferential or ‘hoop stress’ is twice than longitudinal stress. TC = 2 x TL Consider a cylindrical animal…. sc = force/area = (2 r d p) / (2 d d) = r p / d Again, TC = sc x d TC = r p

23. Implications of LaPlace’s Law: • Small worm withstand greater pressure than large worms. • Large worms should have thicker walls. • Square cross sections should be rare. P P P P Pierre-Simon Laplace 1749-1827 tension is infinite

24. Solve for dV/dq: V = D3 sin2q cos q Maximum volume at q = 54.73o 4 p L Consider a helical worm: L Volume = p r2 L Solve for volume in terms of q (helical angle): D = L cos q r = D sin q /(2 p )

25. muscle action Permissible Morpho-space V = d3 sin2q cos q 4 p elliptical profile circular section

28. Ontogenetic scaling of burrowing forces in the earthworm Lumbricus terrestris • Kim Quillin • J Exp Biol 203, 2757-2770 (2000)