Magnetic Levitation Experiments with the Electrodynamic Wheel

Transcription

1 Magnetic Levitation Experiments with the Electrodynamic Wheel Society of Physics Students, Northern Virginia Community College, Annandale, Virginia Vincent Cordrey, Angel Gutarra-Leon, Nathan Gaul, Walerian Majewski Supported by a Sigma-Pi-Sigma Undergraduate Research Grant from the Society of Physics Students and by grants from the NVCC Educational Foundation

2 General Experimental Setup Force gauges to measure lift and propulsion Conductive plate Electrodynamic wheel

3 Exp. : High-Density, Externally Driven Electromagnetic Wheel Exp. : Low-Density SelfDriven Electromagnetic Wheel

4 The Conductors

5 Halbach Field Cancellation Fields above the array (shown in orange) are in opposite directions and cancel. Fields below the array (shown in green) are in the same direction and augment each other. Drawing adapted from original drawing by J.C. Mallinson (97). 5

6 FEMM calculation of the magnetic fields produced by the circular Halbach Array. Note, the field in the center is virtually completely cancelled. 6

7 Theory Induced voltage ε and current I in each circuit of inductance L and resistance R from variable magnetic flux of amplitude Φ resulting from relative motion of plate w/r to magnet, with velocity v - are related by the circuit equation: ε = LdI/dt + RI = ωφ cosωt, where ω=(π/λ)v, λ is the space period of the magnet. Solving explicitly for I and using the general magnetic force formula F = IxB, B being the field of the magnet, we can find components of the force acting on magnet. Original calculation was done for linear motion, here we compare it with circular motion of large radius. 7

8 = F " = B R o w kl ωl +( R e &$'( ωl )$ Lift= F ( = B o w kl./ = 7. = $9:9; 5 8 < +( R ωl )$ e &$'( 9. 8 Theory λ =wavelength of the Halbach array; k= π/ λ R = resistance in each closed circuit v = Ω r, r is the radius of the wheel; ω = πv/λ = (πr/λ) = NΩ L = circuit inductance (self-inductance + inductive coupling) B = peak field strength of the Halbach array w = width of the inductor Ω= angular velocity of wheel y =distance between the lower surface of the Halbach and the geometric center of the inductors. 8

9 Solving for current I and then using the force between the circuit and magnet s field given by F = IxB, one obtains for the ratio of the force components: F(lift)/F(drag) = (L/R) ω, for large ω Y = m X ω=(π/λ)v λ is the space period of the magnet For our wheel, Nd magnets produces full North to South oscillations per revolution ω=ω 9

10 Lift and Aluminum Large Wheel Aluminum Disc y =.8x +.56 R² = y =.8x +.79 R² =.9979 Small Wheel Aluminum Disc () () Large Wheel Aluminum Rectangle y =.95x +.57 R² = Small Wheel Aluminum Rectangle y =.9x +.5 R² = () () Large Wheel Aluminum Split Guideway. y =.9x R² = (Traction) () Small Wheel Aluminum Split Guideway. y =.99x -.89 R² = ()

11 Lift and Copper Large Wheel Copper Rectangle y =.5x +.9 R² = () y =.5x R² =.9995 Small Wheel Copper Rectangle () Large Wheel Copper Split Guideway. y =.87x +.66 R² = () Small Wheel Copper Split Guideway. y =.6x R² = ()

12 Lift and (Lift / ) Ratio at 5 ec Large Wheel Copper Rectangle Large Wheel Aluminum Disc Large Wheel Aluminum Rectangle Small Wheel Copper Rectangle Small Wheel Aluminum Disc Small Wheel Aluminum Rectangle Large Wheel Copper Split Guidway Large Wheel Aluminum Split Guideway Small Wheel Copper Split Guidway Small Wheel Aluminum Split Guideway Lift ()

13 Higher RPM and Lift-Off Small Wheel Aluminum Disc Small Wheel Copper Rectangle Radians / Sec Radians / Sec

14 Analysis We compared the efficiency of the lift to drag ratio on the metal plate on when both the wheels spinning with 5 radians per second and we saw that the aluminum disc was more efficient at producing lift than the aluminum rectangular plate. We saw that the Aluminum Split and the Copper split metal plates were both inefficient in producing lift. Explanation: the split has broken (interfered with) the eddy currents responsible for the lift. The larger wheel is more efficient than the smaller wheel at the same angular velocity ω.

15 Conclusions Our Experiment, generally, supported theoretical prediction The larger the inductors (larger circular disc vs. rectangular plate), the slower a speed was needed to reach the point where lift overtook drag. The closer the wheel of magnets was to being a straight line (the larger the radius), the higher the lift to drag ratio at the compared angular velocity ω of 5 radians / sec. 5

16 Applications Maglev Cargo and/or Passenger Vehicles Non-contact gear coupling Non-contact conveyor belts based on EDW Liquid metal pumps Projectile launchers 6

17 References J.Bird, T.A. Lipo, University of Wisconsin-Madison, College of Engineering, Wisconsin Power Electronics Research Center, research report 5-9 An Electrodynamic Wheel with a Split-Guideway Capable of Simultaneously Creating Suspension, Thrust and Guidance Forces, 5. K. Halbach, Journal of Applied Physics, vol. 67, 9 Applications of Permanent Magnets in Accelerators and Electron Storage Rings, 985. J.C. Mallinson, IEEE Transactions on Magnetics, Vol. Mag- 9, No., One-sided Fluxes - A Magnetic Curiosity?, December 97. R.F. Post, D.D. Ryutov, UCRL-JC-859 preprint, The Inductrack Approach to Magnetic Levitation,. 7

18 Acknowledgments We would like to thank Dr. Walerian Majewski for his guidance, the Society of Physics Students, the NVCC Educational Foundation and the Virginia Community College System for their generous financial support. 8