Manifolddesigner
Automotive
- Apr 29, 2009
- 63
Hello,
My company has been building a breakaway device for years now with a pair of magnets firmly mounted to the device. The mating piece is a piece of mild steel. They have the magnets oriented in opposite poles; they (I believe erroneously) call it a Halbach array. There's a steel backing piece to transfer the magnetic flux from one magnet to the other.
During my playing w/ magnets, when all magnets are oriented parallel, the force to pry off the steel puck is about the same as when they are oriented opposites. Now, I must note that it is MUCH easier to get the bloody magnets to orient NSNSNS than NNNNNN. Even stuck to a thick pc of steel, they want to flip flop.
Since the permeability of the N52 magnets is about the same as air, the magnetic circuit that I draw has the same amount of resistors and the same amount of motors regardless of orientation.
My question,
If I had a mild steel rod say 1.25” OD w/ a 1” ID counterbore 3/8” deep. Say the end is ¼” thick. (the whole thing kinda looks like a socket w/o the hole in the middle)
Then I stick a single cylinder magnet (say 3/8”dia, 3/8” tall) in the base of this. Same height as the counterbore. When I stuck this to a steel plate, the circuit that I draw shows much less resistance…wouldn’t this mean higher force than simply sticking the magnet to it?
This implies that we could get the same amount of force using less $$$ N52.
I understand that if the ID of the steel counterbore were reduced to near the same size as the dia of the magnet, then it would ‘short-circuit’ and all the flux could easily travel from the front of the magnet to the rear and it wouldn’t need to stick to the mating pc of steel.
Am I hallucinating here? Some of the guys I work w/ have been doing magnets at MIT and I don’t want to step on any toes, but it seems like it would work.
I suppose the obvious solution is to go play on the lathe … In the time it took to write this…
Jason
“Major consumer electronics manufacturer” Mechanical Engineer.
My company has been building a breakaway device for years now with a pair of magnets firmly mounted to the device. The mating piece is a piece of mild steel. They have the magnets oriented in opposite poles; they (I believe erroneously) call it a Halbach array. There's a steel backing piece to transfer the magnetic flux from one magnet to the other.
During my playing w/ magnets, when all magnets are oriented parallel, the force to pry off the steel puck is about the same as when they are oriented opposites. Now, I must note that it is MUCH easier to get the bloody magnets to orient NSNSNS than NNNNNN. Even stuck to a thick pc of steel, they want to flip flop.
Since the permeability of the N52 magnets is about the same as air, the magnetic circuit that I draw has the same amount of resistors and the same amount of motors regardless of orientation.
My question,
If I had a mild steel rod say 1.25” OD w/ a 1” ID counterbore 3/8” deep. Say the end is ¼” thick. (the whole thing kinda looks like a socket w/o the hole in the middle)
Then I stick a single cylinder magnet (say 3/8”dia, 3/8” tall) in the base of this. Same height as the counterbore. When I stuck this to a steel plate, the circuit that I draw shows much less resistance…wouldn’t this mean higher force than simply sticking the magnet to it?
This implies that we could get the same amount of force using less $$$ N52.
I understand that if the ID of the steel counterbore were reduced to near the same size as the dia of the magnet, then it would ‘short-circuit’ and all the flux could easily travel from the front of the magnet to the rear and it wouldn’t need to stick to the mating pc of steel.
Am I hallucinating here? Some of the guys I work w/ have been doing magnets at MIT and I don’t want to step on any toes, but it seems like it would work.
I suppose the obvious solution is to go play on the lathe … In the time it took to write this…
Jason
“Major consumer electronics manufacturer” Mechanical Engineer.
