> -----Original Message-----
> From: piclist-bounces@mit.edu [mailto:piclist-bounces@mit.edu]On Behalf
> Of James Newtons Massmind
> Sent: Thursday, February 17, 2005 1:27 PM
> To: 'Microcontroller discussion list - Public.'
> Subject: RE: [EE] Low cost encoder. Is it a compass?
>
>
> Yep, that fried my brain... But I'm very interested in this...
>
> Does "unit radius vector" (Rhat) mean the distance from the magnet to the
> hall effect sensor?
>
> Xhat is then the value we want to solve for? No, it must be X we are
> interested in. So I don't understand what "unit vector of deflection from
> the axis of the magnetic field" means. Help?
>
> Once I understand the math a bit better I can start the hell of trying to
> solve for X and convert it into code.
>
> The need for the slope could be removed by having another sensor at 90' to
> the first?
>
> Aside from being slightly less than human with regard to math (RAH says
> "people who  don't understand at lease calculus are at best
> sub-human") I'm
> also pretty clueless on analog electronics.
>
> Does anyone have a schematic for connecting a hall effect sensor
> to a PIC in
> such a way that the reading is available as an analog value? All
> I can find
> are go/no-go sort of interfaces where the sensor is being used to count
> pulses.
>
>
> ---
> James.
>
>
>
> > -----Original Message-----
> > From: piclist-bounces@mit.edu
> > [mailto:piclist-bounces@mit.edu] On Behalf Of Mike Hord
> > Sent: 2005 Feb 17, Thu 07:34
> > To: Microcontroller discussion list - Public.
> > Subject: Re: [EE] Low cost encoder. Is it a compass?
> >
> > Okay, math heavier.  This is going to suck, doing in plain
> > text, but here goes.
> >
> > The arbitrary field strength from a magnet dipole is
> >
> > H~ = (m/4*pi*(R' )^3)(Rhat*2cosX + Xhat*sinX)
> >
> > where H~ is the magnetic field strength vector at the sensor,
> > m is the magnetic moment of the dipole, R' is the magnitude
> > of the distance from the dipole, Rhat is the unit radius
> > vector and Xhat is the unit vector of deflection from the
> > axis of the magnetic field.  This is in spherical
> > coordinates; the Yhat vector plays no role since it
> > represents rotation about the magnetic field axis.
> > Classically, X should be theta and Y should be phi, but
> > including those in a plain text e-mail is just asking for trouble.
> >
> > Anyway, in this case, we have a magnet and sensor, between
> > which the distance isn't changing, we can write off the first
> > portion of the RHS, (m/4*pi*(R' )^3), as a constant k.  That leaves
> >
> > H~ = k(Rhat*2cosX + Xhat*sinX)
> >
> > The single sensor in this case gives us only magnitude, not
> > direction.  So we're really interested in |H~|.  Nothing in
> > the magnitude operation is going to alter or remove those two
> > sinusoidal functions, so the resulting measurement will
> > clearly be sinusoidal.
> >
> > The current position along a sinusoid can be fixed if we know
> > magnitude and slope, so the sensor must be both reading the
> > current value and comparing it to prior values to establish
> > the current position.
> >
> > It has to have some help from the user to figure out which
> > direction it's turning, though, because without that info,
> > the sensor won't know if it's gone, say, 10 degrees clockwise
> > versus 350 degrees anti-clockwise.
> >
> > I hope this is clear enough.  It's been a while since I did
> > too much of this sort of thing.
> >
> > Mike H.
> > --
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