On Fri, Jun 30, 2006 at 04:24:13PM +0100, Howard Winter wrote:
> Peter,
> 
> On Mon, 26 Jun 2006 09:23:18 -0400, Peter Todd wrote:
> 
> >...
> > Nah, what I'm really worried about for that wireframe project
> > is I have to figure out the math to take a x,y,z reading from an
> > acellerometer and convert it into a two part roll (whatever is the
> > proper terminology for that) to rotate my wireframe model.
> > 
> > Don't have the clue how I'm going to do it. Don't even know if it's an
> > easy problem or a hard one. At least with an education that touched it
> > I'd have a vague memory of where to start, even if it took me another 4
> > days to relearn all the details I'd forgotten!
> 
> Isn't it fairly straightforward Trigonometry?  Solve the vector for x & y then for this and z?
> 
> Where:
> x is the magnitude of the x-axis movement (have to resolve the previous speed and the current acceleration 
> into a distance moved in a fixed, small, time period)
> y is the magnitude of the y-axis movement
> z is the magnitude of the z-axis movement

I think those definitions are incorrect. x, y, and z are acellerations
along their respective axises. However I think to try to integrate them
into a distance moved isn't possible.

Essentially, take the device and orient it with z perpendicular to
gravity. Now rotate it around the x axis. The y value will look like a
sign-wave as it's value is essentially something based on sin()

So how are you going to know if that is meaning accelleration, or
rotation, without a gyroscopic component to directly measure rotation?

> By Pythagoras:
> 
> A = SQRT(x^2 + y^2)  this is the magnitude of the resultant of x and y.  
> The angle between this resultant and the x axis is arcTan(y / x)
> 
> B = SQRT(A^2 + z^2)  this is the magnitude of the final resultant.  
> The angle between this and the x-y plane is arcTan(z / A)
> 
> So you now know how far to move (B), and the direction to move (the two angles).  You'll have to work out the 
> signs to get the overall direction right, but a quick test should soon solve that - probably quicker than 
> thinking it through!  :-)

As above, if I'm not mistaken your approach may work, but I think only
if the orientation of the device is fixed. My problem is the opposite.


My intuitive thinking on it is essentially that if I were trying to
calculate small angular movements, say +-10degrees on x and y, it'd be
some simple trig to convert that into "rotate about x, then rotate about
y" angular co-ordinates.

But given that I have three coordinates, x, y, z, my intuition tells me
that turning that into two values will be tough. Especially given that
one of the two value's, y, has a meaning that changes depending on what
the other is because of that "rotate about x, *then* rotate about y"
meaning.

Oh well, I'm sure reading up on more angular coordinate math will help.

> Of course, it's over thirty since I did this stuff, but it flashed back to me.  Hopefully correctly, but you 
> never know!  :-)  As for the matrix calculations, you're on your own - I haven't done those since I was 11...

Damn smug and lazy child prodigys! :)

-- 
pete@petertodd.ca http://www.petertodd.ca
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