To me, the statement about rotation about two axes simultaneously
doesn't really make sense.
To begin with, as clarified by the answers and comments
to your previous question,
the way mathematicians use the word "rotation", it's only about the
initial and final state, not about the actual motion that takes place
in between. When we say things like "rotation by an angle of $\phi$
about the $z$ axis" it is easy to be fooled into thinking that this
must take place by really spinning something continuously around the $z$ axis.
But really, all it means is the transformation from an initial to a final
state that's described by the rotation matrix
$$
\begin{pmatrix}
\cos\phi & -\sin\phi & 0 \\
\sin\phi & \cos\phi & 0 \\
0 & 0 & 1
\end{pmatrix},
$$
regardless of what happened along the way.
If you really want to describe the process of spinning around the
$z$ axis, you need to use a time-dependent rotation matrix
(or a time-dependent quaternion, or however you represent your
rotations). For example, we can say that at time $t$ the body
is rotated (compared to the initial position at time zero) by the rotation
$$
\begin{pmatrix}
\cos t & -\sin t & 0 \\
\sin t & \cos t & 0 \\
0 & 0 & 1
\end{pmatrix}.
$$
Any type of spinning or tumbling motion of a rigid body
(with the center of mass fixed, for simplicity)
can be described by a time-dependent rotation matrix,
but in general the time-dependence will of course be much more complicated
than this simple "spinning around a single axis" case.
The type of motion actually performed by a rigid body in empty space
is described by something called
Euler's equations.
They are expressed in terms of the angular velocity vector $\omega$,
which is related to the time derivative of the time-dependent
rotation matrix, so even if one manages to solve the Euler equations
for $\omega$ there is still some integration left to do in order to
really get the rotation.
Anyway, Euler's equations simplify when no torques act on the body,
because the right-hand side becomes zero, but they are still not
completely trivial. In that case there are some simple solutions
which just consist of pure spinning around a single axis (namely one
of the principal axes of inertia of the body), but most solutions are
more complicated, like the tumbling motion of your asteroid, or the
book in that video when the astronaut tries to spin it around the
"middle" axis. (Spinning about that axis is possible in principle, but
unstable, so that the slightest deviation will make the rigid body
tumble erratically instead.) I have no idea what those people mean who
say that these tumbling solutions are described by "rotating around
two axes simultaneously". The most concrete geometric description of
the tumbling is given by Poinsot's ellipsoid, which
is what that "polhode" stuff refers to. (The quotation is from a
footnote in the section about this in Goldstein's book Classical
Mechanics.) In short, for any rigid body, one can define an abstract
"inertia ellipsoid" which is imagined to be attached to the body, and
the body tumbles in just the same way that this ellipsoid would tumble
if it were rolling on a certain abstract "invariable plane" which is
imagined to be fixed in space (in just the right way for this whole
thing to work, of course).
I suggest you try to find some mechanics textbook that you like,
and read about rigid body dynamics there; it's a bit too involved to try to
explain in detail here, I think.
I always think of this sort of stuff in terms of forcing the the result to have the units you want and/or the only units that can possibly make sense.
In the first case, you are given $\frac{rotations}{minutes}$ and $feet=\frac{feet}{1}$ and you want to end up with velocity, which has the general form $\frac{distance}{time}$. In this case, the pertinent units for distance and time are feet and minutes, respectively. The key, though, is the following: one rotation equals $2\pi$ times the radius (in feet). That is, $$\frac{rotations}{minutes}=2\pi\cdot radius\cdot\frac{feet}{minutes}.$$
But remember that in the end we just want an expression for velocity; that is, $\frac{feet}{minutes}=$ something. That's simple, though. Just divide both sides of the above equation by $2\pi\cdot radius$.
The second and third ones are very similar, but angular velocity has units $\frac{rotations}{minutes}$.
The last thing you asked is a specific instance of this sort of dimensional analysis, except that you have to convert everything to miles in the end. Alternatively, you can think of it as follows: if the velocity is 50 feet/second and the angular speed is 100 revolutions/second, that means in one second, the apparatus is making 100 revolutions - and furthermore, that these 100 revolutions = 50 feet. From this you get that one revolution is 1/2 a foot. That means that the circumference of the circle drawn out by the motion of the apparatus is 1/2 a foot, meaning that the radius of the circle is $\frac{1}{2}\cdot\frac{1}{2\pi}$ feet. To turn this into miles, just divide by 5280. Hope that helps.
Best Answer
From Pythagoras $\sqrt{115^2+200^2}\approx 231$ Similarly, $\frac {200}{\cos 30^\circ}\approx 231$