This is a brief summation of what was learned to date, with some corrections.
Restate the Basics:
- 24-inch tire = 0,61 m diameter (d), traveling at
Velocity (v) = 30 mph / 48,3 km/h linear speed, using
Power (P) = 2 hp
Windings and Effective Length:
- tire spins at 7 rps
Angular Velocity (Ãâ€°) = 7 rps * 2Ãâ‚¬ = 44 rads/s
Current (I) = 44 Amps
Torque (Ãâ€ž) = P/Ãâ€° -> Ãâ€ž = 1491.4/44 = 33.9 Nm
- F = IL x B
Ãâ€ž = 2rF = 2rBIL
B = 0.5 T
- Wire Length (L) = Ãâ€ž / 2rBI -> 33.9 / (2 * 0.3048 * 0.5 * 44) = 2.53 m
However, as noted - our winding is circular
, and the effective length is Â½ the circumference, we therefore correct as follows:
- L = d; actual Lw = Circumference (C) / 2
Circumference (C) = 2rÃâ‚¬ = dÃâ‚¬;
1/2C = dÃâ‚¬/2; Lw = (2.53 m * Ãâ‚¬) / 2 = 3.974 m
The winding will likely not be circular; it could be trapezoidal, rectangular, or oval. Therefore the actual effective length of Â½ the winding will need to factor in the shape and adjust for losses where corners do not contribute.
Rectangle with full-radius ends (racetrack shape). The most effective part of the coil is the straight length that is tangent
to the radius-arm. Therefore a relationship between the effective length and the overall length should be determined in advance.
The quickest method is through CAD: Develop the desired shape, measure the effective length, and contrast with measurement the parameter; this provides a rough correction ratio. If the effective length is not entirely tangential then more rigorous calculations are in order.
Correct Efficiency and Voltage:
Let us presume then that the effective length of our winding is 3.974 m however it is divided up.
- R = 4.1328 ohms/km for 11 AWG -> (4.1328 * 3.974) / 1000 = 0.01642 ohms
P = I^2 * R = V^2 / R
P = 44^2 * 0.01642 = 31.8 watts
Calculate the Efficiency:
- Pe = (Pi Ã¢â‚¬â€œ Pr) / Pi -> (1491.4 Ã¢â‚¬â€œ 31.8 ) / 1491.4 = 97.9%
Calculate the Volts:
Halbach and Simple Magnet Arrays:
- 1491.4w / 97.9% = 1524w; 1524 - 1491.4 = 32.5w (the amount of energy required to overcome the inefficiency, in theory).
Reverse to get to R:
Given P = I * V, solve for V:
- V = P / I -> 1524 / 44 = 34.6V
This conversation is about Axial Flux machines
, having a Rotor-Stator-Rotor
arrangement. A Simple
magnet array is constructed as a series of magnets in a circular arrangement with alternating Up-Down/North-South faces. Assemblage would require a backing plate to constrain the potential flux leakage. This is considered the less-expensive solution.
array consists of at least twice as many magnets which are configured by alternating the orientation N-L-S-R and so on that effectively doubles the flux on one side through redirection, thus saving the potential weight of the backing plate. However this assembly is more difficult to achieve. In addition, if a shape other than arc is used then high-inductance/low-reluctance filler material would enhance the flux field rather than retard if the air gap remained.
In a Simple array we want a small air gap between the magnets; in a Halbach array we want the magnets to touch.
A coreless stator does not have iron or ferrous material, thus we eliminate cogging
, which increases the efficiency of the motor. However our torque value drops because the coreless design cannot focus and concentrate the flux in the same manner than an iron stator. Therefore the closer the two rotor faces are to each other the greater the flux density in the air gap between; the goal is to have high flux density
Windings (Teeth) to Poles (Magnets) Ratio:
Though not exclusive, the more optimum ratios of Stator Teeth (Windings) to Magnet Poles are close to but not exactly 1:1. There is a direct relationship between the increase in the amount of torque per rotation and the amount of magnetic poles. Increasing the count of pole-pairs also directly increases the frequency of the signal pulse, given as
- ÃŽâ€ Magnet Pole pairs (p) = ÃŽâ€ Signal Frequency (f)
If we had 10 pole-pairs in a motor, each pair would provide 1/10th of the total toque, and by the same token require 1/10th of the amount of current per pair.
We can reduce Current (I) further by increasing the number of phases per cycle. There is a direct relationship to the number of phases to the number of winding teeth, and the ratio for the amount of current per pulse is specified as the square root of the number of phases; therefore the current pulse of a 3-phase motor is given as:
- I1p = I3p / Ã¢Ë†Å¡3 Ã¢â€°Ë† I/1.732
Thus 44 A / 1.732 = 25.4 A per phase.
If we had a 3-phase motor with 10 pole-pairs (p) and 21 teeth, calculate the amount of current per phase.
- 21 Teeth / 3-Phase = 7 windings per phase, wound as:
Tire rotation is 7 rps, therefore 7 * 10 p = 70 Hz
Itooth = Itotal / 10 / Ã¢Ë†Å¡3 => 44 / 10 / Ã¢Ë†Å¡3 = 2.54 A
=> 2.54 A @ 70 Hz Ã¢â‚¬â€œ at least theoretically
If there are no questions I should like very much to move onward and recalculate the Force
as we move it originally from the Wheel circumference towards the smaller Hub working diameter.