On 3/23/07, Jon S. Berndt <[EMAIL PROTECTED]> wrote:
> > > My question:
> > > If I put the aircraft (Cub or 172) into a constant bank and
> > > pitch angle turn ("trimmed"), I expect phidot and thetadot to
> > > be near zero, and psidot non-zero... Yet, I see thetadot near
> > > zero, and phidot and psidot non-zero (as a matter of fact:
> > > phidot is almost as "large" as psidot). Surely this can't be
> > > right? According to the FlightGear (and MathsWorks / Matlab)
> > > documentation, this is not p,q,r but Euler/Gimbal angular
> > > rates in rads/sec... not body rates...
> >
> > Sorry, but I don't have ANY time to respond to this, but keep in mind that
> > you are looking at BODY-AXIS rates and accelerations. If you are
> > in a turn,
> > the pitch and roll are going to show values.
> >
> > Bill
>
> p,q,r are body rates. In a *steady* turn, I would expect body rates (p,q,r)
> to be zero. Phi, theta, psi (Eluer angles) are measured with respect to the
> local frame. In a steady turn, the euler *rates* would be non-zero.
>

Hi,

I'm sorry but its sounding a little anomalous to me.

>From the kinematic relationships;

p = phidot - psidot sin(theta)

q = thetadot cos(phi) + psidot cos(theta) sin(phi)

r = psidot cos(theta) cos(phi) - thetadot sin(phi)

now for a steady turn, theta can be approximated close to zero and for
the thetadot is also close to zero and negligible.  So we are left
with roughly;

p = phidot

q = psidot sin(phi)

r  = psidot cos(phi)

If we p q and r were to be measured in the body axis, I guess they
would be zero.  But in the inertial frame;

p = phidot = 0     since its in a level turn the bank angle would be constant

q = psidot sin(phi) = non zero since phi has a value and in the
inertial frame the heading is changing

r =  psidot cos(phi) = non zero again

Therefore q and r should have values and so should psidot, but
thetadot and phidot should be zero or close to zero.

Am I missing something?  Please do let me know.

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