Robert's HW: Difference between revisions

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Solving for <math>\Phi</math>:
Solving for <math>\Phi</math>:


<math> \Phi(s)= \dfrac{ s\varphi_0 + \dot{\varphi_0} }{s^2- \dfrac{ (D) (GM) }{I} } = \varphi_0 \left( \dfrac {s}{s^2- \dfrac{ (D) (GM) }{I}} \right) + \dfrac{\dot{\varphi_0}}{s^2- \dfrac{ (D) (GM) }{I}} </math>
<math> \Phi(s)= \dfrac{ s\varphi_0 + \dot{\varphi_0} }{s^2- \dfrac{ (D) (GM) }{I} } = \varphi_0 \left( \dfrac {s}{s^2- \dfrac{ (D) (GM) }{I}} \right) + \dot{\varphi_0} \left( \dfrac{1}{s^2- \dfrac{ (D) (GM) }{I}} \right) </math>


Assuming no initial angular velocity (<math> \dot{\varphi_0}=0 </math>) and converting back to time domain:
Converting back to time domain:




<math> \varphi= \mathcal{L}^{-1} \{ \Phi(s) \}
<math> \varphi(t)= \mathcal{L}^{-1} \{ \Phi(s) \} = \varphi_0 cos \left( - \dfrac{ (D) (GM) }{I} t \right)u(t) </math>

Revision as of 13:45, 1 November 2010

I decided to model the natural response of a boat given a small initial list on flat water.

Assume a boat of arbitrary geometry, with a given displacement(weight) D, mass moment of inertia I, and metacentric height GM (The metacenter is a theoretical point in a boat through which the buoyant force always passes for small angles of list.).




As the boat lists at angle φ, the centroid of the displaced volume of water shifts in the same direction, causing the buoyant force to be offset, resulting in a moment acting to right the boat. This righting moment is equal to the displacement of the boat times the righting arm GZ, where:

GZ = GM*sin φ

Which at small angles of φ can be approximated:

GZ = GM*φ

Thus, summing moments about the center of gravity of the boat:

Rearranging:

This is a simple ODE that may be solved using Laplace transforms.


Solving for :

Assuming no initial angular velocity () and converting back to time domain: