Bicycles Asked by DB77 on June 6, 2021
Possible to engineer a new front bike wheel hub that would spin more freely, and for a longer duration, due to the use of magnets ?
The mechanical advantage is (not motor) and uses magnetic forces only. The objective is less friction, and more rotational force, to therefore enable the wheel to spin more freely and for longer duration.
Using magnets instead of traditional ball bearings offers no practical benefits.
Using them in any other way is drawing inspiration from perpetual motion nonsense.
Answered by whatsisname on June 6, 2021
Most maglev bearings are active and require a power source. Passive ones are hard to get right (expensive, fragile, limited in the range of cross-forces they can deal with). They can achieve low friction in ideal circumstances, but you're not going to get rotational forces without some form of power.
You'd need to support the wheel side-to-side somehow as well. Using repulsive magnetic forces there would lead to very strange handling, and anything mechanical would bring the friction back. Bearing friction is a very small component of drag on bikes, and scales linearly with speed so there's no real benefit.
People have tried magnets in hubs, but as part of a freewheel mechanism, not as bearings
Answered by Chris H on June 6, 2021
Bicycles are powered by humans, sometimes with assistance from a lithium ion battery. Humans are awesome, even with relatively little athletic training. However, our power to weight ratios are much poorer than gasoline powered vehicles, or even all-electric vehicles. Any solution to reduce friction has to not add too much weight and aerodynamic drag that it offsets the gains.
Also, at least from a glance at Wikipedia, present magnetic levitation technology appears to require electrical current as input. I would assume that no known ferromagnetic materials have sufficient field strength to levitate a shaft from the rest of the hub. The power input would presumably need to come from a generator, which induces drag and isn't perfectly efficient.
I'm not sure what range of friction losses are possible from bearings. I think I've heard that at a rider input of 250W, there are around 20W of total friction losses in the drivetrain (i.e. not including the hubs) on a performance road bicycle. Losses from the hub bearings would most likely be lower than this.
Answered by Weiwen Ng on June 6, 2021
You cannot use permanent magnets alone (it's a physic impossibility, a consequence of Earnshaw's Theorem). You need to lawyer your way out of it, which you can do either by removing "permanent" (i.e. employ active electromagnets also; this is what is commonly done with magnetic suspensions for lamps etc.) or by removing "alone".
it has been pointed out to me that Earnshaw's Theorem only holds with ferro- and paramagnetic sources. The addition of diamagnetic materials does allow stable configurations with permanent magnets alone. Werner Braunbeck's trick, though, only holds with diamagnetic materials, and those have a strength that's whole orders of magnitude below high-flux rare earth ferromagnets. So, unless the wheel is large enough to support a half ton of diamagnetic bismuth shield, I'm afraid passive levitation isn't a thing for bikes.
Earnshaw's Theorem actually applies to a whole class of forces, so you can use no combination of those to wiggle out of "alone". No electrostatic fields, no gravitational fields. You're pretty much left with superconducting pinch-effects and, again, mechanical bearings (you can replace a mechanical bearings assembly with permanent magnets as long as you keep at least one bearing).
You could have a spokeless, hubless wheel where the weight is borne by magnets between an inner rim and the tyre, with mechanical bearings to keep the two rims aligned (the bearings would only be subject to lateral forces). This allows the magnets to spread out the weight and thus exert a minimum force.
And at this point you need to keep grit out of the space between the magnetic poles. You need a seal, and that's going to need to be mechanical, and problems with friction will resurface. Also, the magnets will be heavy. A toroidal N41 magnet, circularly magnetized (north pole in the hole, south on the periphery), 4cm wide, 2cm thick and 5cm of inner radius, weighs about six kilograms; you need another wider segment to stay above, and that will probably weigh two to three kilos; this will get you a repulsion of around 200 kg, i.e., unless you thrust 2000 newton of force downwards, the magnets won't touch. One advantage is that it will double as suspension. A big disadvantage is that anything magnetic will be attracted to the hub with great force (you can harm yourself pretty severely with those kind of babies).
Answered by LSerni on June 6, 2021
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