Posted on May 17, 2011 at 12:48arctan(Ax/sqrt(Ay^2 + Az^2)) gives the angle of the vector with the YZ plane, calling the X direction ''positive''. it fails when the vector is null or parallel to X; use atan2(Ax, sqrt(Ay^2 + Az^2)) to avoid that problem.
a few basics:
lets call any vector which modulus is 1 a ''versor''. a versor indicates a direction but not a modulus.
vector acceleration measurements taken on earth with static sensors should be versors when scaled by (1/g), but they won't be because of errors.
normalization obtains a versor from a non-null vector, with the same direction and sense. to normalize v into versor N(v):
-calculate the modulus m(v)=sqrt(vx^2 + vy^2 + vz^2)
-scale each coordinate of v by (1/m(v))
so N(v) = (1/m(v)) * v for v != 0
the dot product of two vectors is a scalar (simple number) defined like this:
v1 . v2 = v1x*v2x + v1y*v2y + v1z*v2z
so: m(v) = sqrt(v . v)
and it's commutative: a . b = b . a
it's a property of the dot product that the product of two versors is the cosine of the angle between them. so if v1 and v2 are versors, the angle in radians between them is:
acos(v1 . v2)
now on to your question:
let v be a tri-axial vector measurement of acceleration using a static sensor. check that m(v) is close to 1g. (if not, the sensor is not static and tilt info is not reliable, so you abort.) normalize it into versor vn:
vn = N(v)
lets call x the versor whose components are (1, 0, 0). x is a versor that has the direction of the positive X axis.
so the angle between v and the X axis (0 to pi) is:
ax = acos(vn . x)
note that since x = (1, 0, 0), vn . x = vnx, where vnx is simply the first component of versor vn. so:
ax = acos(vnx)
also, if you prefer the angle between v and the YZ plane (-pi/2 to pi/2) calling the X direction ''positive'' (like in your equation), then that angle is:
ayz = asin(vn . x) = asin(vnx)
note that with the formula a = asin(N(v) . d) you can calculate the angle between a vector v and the plane perpendicular to any versor d, calling the direction of the versor ''positive''. (d must be a VERSOR!) (if d is one of the x,y,z axis versors, then the calculation can be simplified a bit, as shown.)
so...
calibrate your sensor by taking a measure of the ''up'' direction: leave it flat and still and measure vector u. m(u) should be around 1g, otherwise discard u. then normalize u:
un = N(u)
now you're ready to measure tilt. measure vector v. discard it if m(v) is not around 1g. then normalize v:
vn = N(v)
then the tilt angle (0 to pi) relative to the calibrated ''up'' direction is simply:
acos(un . vn)
where 0 means leveled and any deviation increases the angle.
but you probably want more of a joystick behavior. you can establish x and y directions by calibrating on two orthogonal sides. (xn . yn should be close to zero if it's true that x and y they are nearly orthogonal. the same goes for xn . un and yn .un, but you don't really need to measure u for this calculation.)
in this case, the angles against planes YZ and XZ (-pi/2 to pi/2) are:
tiltx = ayz = asin(vn . xn)
tilty = axz = asin(vn . yn)
which is probably what you want.
pheew! I thought this post was going to be much shorter!
🙂
