Robert's HW - Revision history

Silent protagonist at 23:51, 1 November 2010

2010-11-01T23:51:37Z

← Older revision		Revision as of 16:51, 1 November 2010
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	<math> \varphi(t)= \mathcal{L}^{-1} \{ \Phi(s) \} = \varphi_0 cos \left( - \dfrac{ (D) (GM) }{I} t \right)u(t) </math>		<math> \varphi(t)= \mathcal{L}^{-1} \{ \Phi(s) \} = \varphi_0 cos \left( - \sqrt{\dfrac{ (D) (GM) }{I}} t \right)u(t) </math>

	This implies that, like a pendulum, a boat has a natural period of oscilation determined only by it's own physical properties. This period is important when considering more dynamic problems such as the motion of a boat in a storm.		This implies that, like a pendulum, a boat has a natural period of oscilation determined only by it's own physical properties. This period is important when considering more dynamic problems such as the motion of a boat in a storm.

Silent protagonist at 23:49, 1 November 2010

2010-11-01T23:49:37Z

← Older revision		Revision as of 16:49, 1 November 2010
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	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.).		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.).

	[[Image:~~Listing Boat~~\|thumb\|~~left~~\|~~220px~~\|]]		[[Image:Righting_arm.png\|thumb\|right\|500px\|]]






	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:		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:

Silent protagonist at 23:42, 1 November 2010

2010-11-01T23:42:46Z

← Older revision		Revision as of 16:42, 1 November 2010
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	<math> \varphi(t)= \mathcal{L}^{-1} \{ \Phi(s) \} = \varphi_0 cos \left( - \dfrac{ (D) (GM) }{I} t \right)u(t) </math>		<math> \varphi(t)= \mathcal{L}^{-1} \{ \Phi(s) \} = \varphi_0 cos \left( - \dfrac{ (D) (GM) }{I} t \right)u(t) </math>

			This implies that, like a pendulum, a boat has a natural period of oscilation determined only by it's own physical properties. This period is important when considering more dynamic problems such as the motion of a boat in a storm.

Silent protagonist at 21:45, 1 November 2010

2010-11-01T21:45:22Z

← Older revision		Revision as of 14:45, 1 November 2010
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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>

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

Silent protagonist at 21:37, 1 November 2010

2010-11-01T21:37:24Z

← Older revision		Revision as of 14:37, 1 November 2010
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	This is a simple ODE that may be solved using Laplace transforms.		This is a simple ODE that may be solved using Laplace transforms.

			<math>\displaystyle\mathcal{L} \left\{ \ddot{\varphi} - \dfrac{ (D) (GM) }{I} \varphi \right\} = \left ( s^2 \Phi(s) - s\varphi_0 - \dot{\varphi_0} \right ) - \left( \dfrac{ (D) (GM) }{I} \Phi(s) \right) </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>

			Converting back to time domain:


			<math> \varphi= \mathcal{L}^{-1} \{ \Phi(s) \}

Silent protagonist: Created page with '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…'

2010-11-01T17:48:06Z

Created page with '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…'

New page

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.).

[[Image:Listing Boat|thumb|left|220px|]]

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:

<math>\sum M_G = (D) (GM) \varphi = I \ddot{\varphi}</math>

Rearranging:

<math> \ddot{\varphi} - \dfrac{ (D) (GM) }{I} \varphi = 0</math>

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