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== Pre-Design Research == | == Pre-Design Research == | ||
=== Helium === | === Helium === | ||
The whole shark will have a certain weight on which a gravitational force is applied. The total amount of gravitational force is given in equation X.X.<br><br> | The shark will float through the air by using helium. To determine how much helium is needed to make the shark neutrally buoyant [http://en.wikipedia.org/wiki/Buoyancy Archimedes’ principle] is be used. This principle states that any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object. The upward lift is defined in equation X.X. <br> | ||
<math>F_z = M \cdot g = (M_{shark} + M_{helium} ) \cdot g = (M_{shark} + \rho_{helium} \cdot V) \cdot g </math>< | <br> | ||
To make the shark neutrally buoyant the force from the lift has to be equal to the force applied by gravity, which is displayed in equation X.X. If the upward lift equals the gravitational force, no net force in vertical direction will be present. <br><br> | <center><math> F_a = \rho_{air} \cdot g \cdot V \qquad (1)</math> </center><br> | ||
<math> F_a = F_z </math> < | The whole shark will have a certain weight on which a gravitational force is applied. The total amount of gravitational force is given in equation X.X.<br> | ||
Substituting the terms in equation X.X. with the terms from equation X.X. and X.X., a relation between the mass of the shark and the amount of helium needed can be derived. This relation is given in equation X.X.<br><br> | <br> | ||
<math> \rho_{air} \cdot g \cdot V = (M_{shark} + \rho_{helium} \cdot V) \cdot g </math><br><br> | <center><math>F_z = M \cdot g = (M_{shark} + M_{helium} ) \cdot g = (M_{shark} + \rho_{helium} \cdot V) \cdot g \qquad (2)</math></center><br> | ||
<math> M_{shark} = (\rho_{air} - \rho_{helium} ) \cdot V </math><br><br> | To make the shark neutrally buoyant the force from the lift has to be equal to the force applied by gravity, which is displayed in equation X.X. If the upward lift equals the gravitational force, no net force in vertical direction will be present. <br> | ||
<math> \rho_{air} = \dfrac{P}{(R \cdot T)} </math>< | <br> | ||
<center> <math> F_a = F_z \qquad (3)</math></center> <br> | |||
Substituting the terms in equation X.X. with the terms from equation X.X. and X.X., a relation between the mass of the shark and the amount of helium needed can be derived. This relation is given in equation X.X.<br> | |||
<br> | |||
<center><math> \rho_{air} \cdot g \cdot V = (M_{shark} + \rho_{helium} \cdot V) \cdot g \qquad (4)</math><br> | |||
<br> | |||
<math> M_{shark} = (\rho_{air} - \rho_{helium} ) \cdot V \qquad (5)</math><br> | |||
<br> | |||
<math> \rho_{air} = \dfrac{P}{(R \cdot T)} \qquad (6)</math></center><br> | |||
If the assumption is made that the temperature is equal to 293K and the pressure is 1 bar, the density of air is equal to 1,204 kg per cubic meter. At the same conditions the density of helium is 0,1787 kg per cubic meter. Using these values for the densities of air and helium, 1 cubic meter of helium can make 1 kg neutrally buoyant. | If the assumption is made that the temperature is equal to 293K and the pressure is 1 bar, the density of air is equal to 1,204 kg per cubic meter. At the same conditions the density of helium is 0,1787 kg per cubic meter. Using these values for the densities of air and helium, 1 cubic meter of helium can make 1 kg neutrally buoyant. | ||
Using relation from equation X.X, the amount of helium needed to make the shark neutrally buoyant can be determined. | Using relation from equation X.X, the amount of helium needed to make the shark neutrally buoyant can be determined. | ||
Revision as of 17:21, 7 July 2011
Project by Twan Haazen
Supervisor Rob Janssen
This wiki will contain information about my Bachelor Final Project.
Introduction
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Pre-Design Research
Helium
The shark will float through the air by using helium. To determine how much helium is needed to make the shark neutrally buoyant Archimedes’ principle is be used. This principle states that any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object. The upward lift is defined in equation X.X.
The whole shark will have a certain weight on which a gravitational force is applied. The total amount of gravitational force is given in equation X.X.
To make the shark neutrally buoyant the force from the lift has to be equal to the force applied by gravity, which is displayed in equation X.X. If the upward lift equals the gravitational force, no net force in vertical direction will be present.
Substituting the terms in equation X.X. with the terms from equation X.X. and X.X., a relation between the mass of the shark and the amount of helium needed can be derived. This relation is given in equation X.X.
[math]\displaystyle{ M_{shark} = (\rho_{air} - \rho_{helium} ) \cdot V \qquad (5) }[/math]
If the assumption is made that the temperature is equal to 293K and the pressure is 1 bar, the density of air is equal to 1,204 kg per cubic meter. At the same conditions the density of helium is 0,1787 kg per cubic meter. Using these values for the densities of air and helium, 1 cubic meter of helium can make 1 kg neutrally buoyant. Using relation from equation X.X, the amount of helium needed to make the shark neutrally buoyant can be determined.
Design
Tale Section Design
The first thing that was designed was the tale section of the shark. This part is responsible for the propulsion of the fish. According to the movement of a shark, this part will translate in horizontal direction. To make this translating movement a flexible rod will have to be used. The start of it will be connected to the main part of the shark and at the end a large part of foam will be attached to provide the proper amount of area to ensure the propulsion. Surrounding the flexible rod some in between part will be attached to form the shape of the shark tale. A thin line of fish wire will go through these parts and will be connected to end of the tale. If pulled on one of these ends, the flexible rod will deform in such a way that it will bend in the direction of the side where the force is applied. If afterwards the force is applied on the other side of the tale, the flexible rod will bend the other way. With this bending back and forth, the desired translation of the tale is acquired. A small but strong servomotor will provide the proper amount of force to bend this rod. Below a sketch of this design is shown from the top. Also a 3d CAD-design is shown.
image
Prototypes
Tail Section Prototype
After the designing phase, a prototype has been build. The flexible rod is a part of flexible plastic and the in between parts are made out of MDF. Small holes have been drilled into these parts to allow the fish wire to run through it. When applying force into these wires the tale of the shark bends as required. This is shown in the figures below.