How to calculate the "lift-force" of a electro magnet?
Patrik Johansson
lift. Or rather to know the formula for designing a magnet to lift a
certain weight. Is the "push-force" equal the "lift-force"??
For example:
If I would like the magnet to be able to lift 10kg using 12V. How
would that formula look like??
/Patrik
cncguy
"Patrik Johansson" <pat...@patadata.net> wrote in message
news:6b5955c6.02022...@posting.google.com...
Frank Bemelman
Patrik Johansson <pat...@patadata.net> schreef in berichtnieuws
6b5955c6.02022...@posting.google.com...
10 Kg, that magnet will have the size of a bucket, if you want one
with a plunger and - let's say 5 cm of pulling distance. Current at
12V about 65 amps.
Well, these are just my gut feelings about it. Any formula that
would give you smaller solenoid, I would not trust until proven ;)
Perhaps you can search the web for manufacturers of electromagnets
and solenoids. Browsing through their catalogs may give you an
idea of what is possible and what not.
--
Thanks,
Frank Bemelman
(remove 'x' & .invalid when sending email)
Jim Thompson
pat...@patadata.net (Patrik Johansson),
In Newsgroup: sci.electronics.design,
Article: <6b5955c6.02022...@posting.google.com>,
Entitled: "How to calculate the "lift-force" of a electro magnet?",
Wrote the following:
I'm trying to remember, something to do with Hamilton's principle,
energy versus position of the load... calculate/measure inductance
versus slug/load position and go from there (sorry for the
vagueness... it's been 30 years since I worried about that :)
...Jim Thompson
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cncguy
of time or the rate at which work is done, or energy expended. Work is
measured in joules (J). One joule is the work done when an object is moved a
distance of one meter against an opposing force of one newton. One newton is
the force required to cause a mass of one kilogram to accelerate at a rate
of one meter per second squared. Time is measured in seconds (s), power is
therefore measured in joules per second. The electrical unit of
measurement for power is the watt (W). One watt is the rate of doing work
when one joule of work is done in one second.
In other words power (hp) = [(distance (ft) / Time (s) ) * weight (lb)] /
550
10kg = 22.046 pounds
So to raise 22.046 pounds one foot in one second would be
.040 horsepower
Since 1 horsepower = 746 watts = 550 foot-pounds/second
That would be 29.84 watts or 22.046 ft-pounds/second
Expressing this in equation form:
P = W/t
P = EI = E (E/R) = E²/R
P= (IR) I = I²R
where P = power, in watts
W = work, in joules
t = time, in seconds
I = current in amps
R= resistance in ohms
E= power source in volts
"Patrik Johansson" <pat...@patadata.net> wrote in message
news:6b5955c6.02022...@posting.google.com...
Bob
(mu0)(Ae)(NI)^2/(2(Lg)^2) where in MKS units, (mu0) = 4*pi*10^(-7), Ae =
effective core area in meters squared, N = number of turns of wire, I =
current in amps, Lg = length of the air gap in meters. F is in newtons.
This formula neglects the length of the magnetic core assuming that it
has a high mu, greater than 2000 or so and that the core is not
physically long. It also neglects fringing and other second order
effects. Note, there is always an air gap even when the pole pieces
(touch) the work piece. Even if they are highly polished, there is a few
micro meters of gap. Also note that if we are talking about a single
pole magnet like wire wound on a nail the gap is very long consisting of
the distance from the work to the other end of the nail. It's easy to
see that because of the squared gap term, the gap has to be short for
the magnet to be strong. That's why magnets are usually some form of "U"
shape. Hope this helps.
Bob
Adrian M Jansen
Where B = magnetic field in tesla
A = area over which field is active in sq m
u0 = permittivity of free space ( 4 pi * 10e-7)
F is in newtons
The hard part is to calculate a good value of B from the core and coil
dimensions and current, and since its B^2 which is important, errors get
badly magnified.
Good luck.
--
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Regards,
Adrian Jansen aqja...@oqzemail.com.au
Design Engineer J & K Micro Systems
Microcomputer solutions for industrial control
Note reply address is invalid, remove q's from address above to reply.
Jim Thompson wrote in message ...
John Wasser
In article <3C7560D0...@yahoo.com>, Bob <nsmon...@yahoo.com>
wrote:
> Patrik Johansson <pat...@patadata.net> wrote:
> > I would like to calculate the force that a electro magnet is able to
> > lift. Or rather to know the formula for designing a magnet to lift a
> > certain weight. Is the "push-force" equal the "lift-force"??
> >
> > For example:
> > If I would like the magnet to be able to lift 10kg using 12V. How
> > would that formula look like??
>
> The simple formula for force in an electromagnet is:
> F = (mu0)(Ae)(NI)^2/(2(Lg)^2)
> where in MKS units,
> (mu0) = 4*pi*10^(-7),
> Ae = effective core area in meters squared,
> N = number of turns of wire,
> I = current in amps,
> Lg = length of the air gap in meters.
> F is in newtons.
> This formula neglects the length of the magnetic core assuming that it
> has a high mu, greater than 2000 or so and that the core is not
> physically long. It also neglects fringing and other second order
> effects. Note, there is always an air gap even when the pole pieces
> (touch) the work piece. Even if they are highly polished, there is a few
> micro meters of gap. Also note that if we are talking about a single
> pole magnet like wire wound on a nail the gap is very long consisting of
> the distance from the work to the other end of the nail. It's easy to
> see that because of the squared gap term, the gap has to be short for
> the magnet to be strong. That's why magnets are usually some form of "U"
> shape. Hope this helps.
If you want to hold 10Kg against gravity (9.8m/s^2) I think you need a
force of 98N (98Kgm/s^2):
98 = 4 * pi * 1E-7 * Ae * (N*I)^2 / (2*Lg)^2
You need to pick a core cross-sectional area and a gap size. I would
start with a U-shaped bar about 3cm in diameter (about 22 cm^2 or
2.2E-3 m^2 area). I would guess that a gap size of 1mm (1E-3 m) would
be achievable.
98 = 4 * pi * 1E-7 * 2.2E-3 * (N*I)^2 / (2*1E-3)^2
98 = 2.8E-9 * (N*I)^2 / 4E-6
3.9E-4 = 2.8E-9 * (N*I)^2
3.9E-4 = 2.8E-9 * (N*I)^2
139E3 = (N*I)^2
373 = N * I
(Somebody please check my math!)
In theory you could put one turn of very heavy wire around the core and
shove about 400 amps through it but the battery wouldn't last long. In
theory you could use about 400 turns of the heaviest wire that will fit
on the core and use a resistor to limit the current to about one amp
but that would be a waste of energy.
A more efficient solution is to find a guage of wire where the current
in the wire is minimized (thin wire, lots of turns), the wire is heavy
enough to carry the current (possibly thicker wire) and the total
resistance of the wire just matches the necessary current and voltage
given the number of turns required (multiple simultaneous equations to
solve).
Bob
I calculate 470 amp turns for NI. I think there are a couple of errors
in the math. The area of 3cm dia is 7.07e-4 M^2 and the 2 in the
denominator is not squared. Ninty eight newtons is right. With 3000
turns of wire, 157ma would be required. On six volts this is 38ohms.
Assuming a mean wire path of 20 cm, the total length would be 600
Meters, 1968 feet. Number 22 or 23 AWG wire will do this. The winding
area would be about 2.4 in^2, 15.4cm^2.
Bob
georgethom...@gmail.com
Thanks
whit3rd
> Hi frank, could you explain how you would work this out? I have been tasked with finding a power rating and material size for lifting 1.8 tonnes using an iron electromagnet.
and not magnetic. If it's steel, with a flat surface. If it's scrap iron, with maybe
10 lbs of iron within the first centimeter distance from the magnet's poles.
If this is for a crane operation, be aware that magnetic lifting probably is NOT
safe for a populated worksite because it's not a calculable problem situation.
Jasen Betts
What shape is your iron in?
Is the elecromagnet lifting the iron or is it starting in contact with
the iron and a mechanism is lifting the electromagnet?
BTW this Heinekin advert is totally fake.
https://www.youtube.com/watch?v=6LU6nmwbbTM
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Spehro Pefhany
years old, and unfortunately Frank has not been seen in these here
parts for quite some time.
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Best regards,
Spehro Pefhany
