Hi guys. I'd like to discuss the equation for magnetic field strength. It looks to me like it predicts overunity:
H = Amps * Turns / Length
Note, there's nothing in there which allows you to determine power. You've only got amps, not volts.
The 2 things that ultimately determine how much power it takes to create a given field are:
- Size of the coil
- Resistivity of the conductor material
So, if we increase the physical size of a coil - we get more field for the same input power. There's a practical limit to this where each turn takes so much wire that the law of diminishing returns applies, but there's still scope for a vast range of field strengths for a given input.
If we reduce the temperature of the coil, things get really interesting... The resistivity drops with temperature and there's a point around -230C, even before superconductivity, where the resistance of the coil would be negligible - meaning you can get your desired magnetic field for much less input - even just a few microwatts of power.
Now, assuming that we can turn a varying magnetic field into useful work - we have, at some point on the scale - overunity... The question then becomes - where?
From what I can tell - OU is just outside the physical range we commonly use. A standard sized solenoid is underunity. Make it 10-20 times bigger, attach it to a crank, and my calcs say it's well OU.
Here's some examples - of standard solenoids with a mobile iron core. All these coils use 48 watts of input power:
Length CoreDiam CoilDiam Hfield OutputPower
100mm 10mm 20mm 48,7561 4w
100mm 20mm 30mm 37,756 23w
100mm 20mm 50mm 55.358 34w
100mm 40mm 60mm 37,801 93w
100mm 100mm 200mm 48,815 755w
100mm 100mm 300mm 59,792 925w
100mm 100mm 3000mm 81,751 1,265w
1000mm 500mm 1000mm 15,436 59,752w (80Hp!)
The output power values are determined from:
- Max possible force (from book: http://archive.org/details/solenoidselectr01undegoog)
F = AMPTURNS * AREA / LENGTH
- average pull throughout power stroke
PULL = MAXPULL * sin(0.77 * PI * (CoilLength / InsertedDistance))
- Throw = Coil Length.
- Crank size is half throw.
- Duty cycle is 50%.
Done.
Hi Tim,
Maybe there is a typo in the last raw of CoilDiam column: is it not 5000 mm instead of the 500? You can use the Modify feature for 12 hours from the time of posting, after that you cannot edit.
Thanks :). Can you delete the superfluous posts too?
Hi Tim,
My questions:
1) When considering the coil current needed for the AmperTurns in the calculations, have you thought of the coil's self inductance which will greatly modify the coil impedance (think of a series R-L circuit)? This means that you need to use higher input voltage for insuring the needed current within a certain ON time for the coil and this involves eventually a higher than calculated input power. Did you include this in the calculations shown or you think it is not an issue? (Because I assume you will switch ON and OFF the coil current with 50% duty?)
2) I have not seen any loss calculation occuring in the shorted coil and its effect on input power? How many turns are involved approximately in the shorted coil with respect to the main input coil? Have you considered this question?
Greetings, Gyula
Hi Gyula. Did you mean to post this under the 'Magnet Piston Engine' thread?
You're right, and I've not included inductance & loss calcs because:
- I'm not sure how to do that
- the shorted coil could be replaced with a PM. I have no idea how many turns it would need at this stage...
Instead I've just multiplied the required ampturns by 10 - on the basis that if i plan to give it 10x more than it should require - then that should cover most eventualities. So where 800AT should saturate the core, I've gone for 8000AT... Note, It still goes OU if I increase the assumed power in by 100x, or even 1000x - just at a bigger size...