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

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!  :-)

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...

Cheers,


Howard Winter
St.Albans, England


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