Calculate total angular momentum of object rotating about 2 axis (e.g. Earth) Consider the Earth. It rotates about its own axis (going through the poles) with some angular velocity $\vec\omega$, and around the sun, with some angular velocity $\vec\Omega$.
In every textbook/web page I've seen so far, I have seen the angular momentum due to orbiting the sun being calculated separately from the angular momentum due to the Earth's rotation about its own axis.
Great. But how do I get the complete angular momentum of the earth?
I am aware of the following answer: Angular momentum of rotating and revolving body (earth), but I do not think it answers the question. The answer uses an angular velocity $\vec {\boldsymbol{\omega}}$ -- but how would you even obtain that velocity if the object is rotating about 2 axis? Euler's rotation theorem won't apply, because one of the axis is not on the object.
Therefore, I will restate the question: Given an orbiting angular velocity $\vec\Omega$ and angular velocity about the earth's axis $\vec\omega$, how would I find the total angular momentum of the Earth (or an object exhibiting a similar rotation description with 1 rotation axis on the body, the other one off)?
 A: Surprisingly, the rule for adding up two angular velocities does not depend on whether the "axis of these angular velocities" go through the object or not, and whether they intersect or not.
Angular velocity of a body does not depend on your choice of inertial frame of reference. Suppose we have some arrow attached to the body; at moment $t_0$ this arrow pointed to a distant star $A$; at moment $t_1$ this arrow pointed to another distant star $B$ - well, if it is true, than it is true in all inertial frames of reference. And how fast the orientation of body changes - it does not depend on frame of reference (as long as the frame of reference is inertial).
Now let's measure the total angular velocity of Earth. It is possible first to measure it in the frame of reference attached to the Sun and rotating in such a way, that the Earth's velocity is zero. Let's say the angular velocity of Earth in this frame of reference is $\vec\omega$. The angular velocity of the frame of reference is $\vec\Omega$, so the total angular velocity of Earth is $\vec\omega + \vec\Omega$. It is a vector directing towards Polar star, it's magnitude is approximately $1/86164sec$ - where 86164 is number of seconds in sidereal day, that is the period of rotation of Earth relative to distant stars.
Now to second part of your question: "In every textbook/web page I've seen so far, I have seen the angular momentum due to orbiting the sun being calculated separately from the angular momentum due to the Earth's rotation about its own axis."
This time the frame of reference is attached to the Sun and it is inertial. "Fair" way to calculate total angular momentum of the Earth in this frame of reference is to split the Earth into many small parts, calculate momentum of each part and sum up the results. Easier way would be to calculate the momentum around Earth's center of mass, than calculate the momentum of Earth as if all it's mass is located in it's center of mass and add up these two vectors. Total result would be the same - it's a simple mathematical theorem.
Note, that the momentum due to the Earth's rotation around it's axis is much smaller then the momentum due to rotation of Earth around the Sun. More importantly, not only the total Erath's momentum (that is sum of these two vectors) is constant in time, each one of these components is constant itself! (we ignore influence of the Moon and other planets). So, if you want to calculate the details of how velocity of Earth depends on distance to Sun (Keppler's laws) - you can safely ignore the "rotation around own axis" part of the Earth's angular momentum.
A: First, consider that Earth's spin is at an angle to the orbital axis.

Here $$\begin{array}{r|c|c|c}\\ \text{Quantity} & \text{Symbol} & \text{Value} & \text{Units} \\ \hline
\text{orbital distance} & R & 1 & \text{AU} \\
 & & 1.496\cdot 10^{11} & \text{m} \\
\text{orbital speed} & \Omega & 1 & \text{rev/year} \\
& & 1.991\cdot 10^{-7} & \text{rad/s} \\
\text{spin} & \omega & 1 & \text{rev/day} \\
 & & 7.2921\cdot 10^{-5} & \text{rad/s}  \\
\text{axial tilt} & \theta & 23.4 & \deg \\
 & & 0.4084 & \text{rad}
\end{array}$$
The combined rotation (given the title about the negative x-axis from above) is
$$ \vec{w} = \pmatrix{0 \\ 0 \\ 1.991 \cdot 10^{-7}} + \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & -\sin \theta & \cos \theta \end{bmatrix} \pmatrix{0\\0\\7.2921 \cdot 10^{-5}
} = \pmatrix{0 \\ 2.8961\cdot 10^{-5} \\ 6.7123\cdot 10^{-5} }\; \text{[rad/s]} $$
which can be translated into
$$ \vec{w} = \pmatrix{0 \\ 5.9735 \\ 13.845 } \; \text{[deg/hr]}$$
What is interesting is that you can calculate the instant center of rotation of the earth relative to the earth $(c_y,c_z)$ ($c_z$ shown negative below). This is the point which the earth is actually spinning about.

To find the point calculate the orbital speed (positive x-axis is out the page)
$$ \vec{v} = \vec{\Omega} \times \pmatrix{0\\-R\\0} = \pmatrix{ 2.9785\cdot 10^{4} \\ 0 \\0} \;\text{[m/s]}$$
and then the center of rotation
$$ \pmatrix{ 0 \\ c_y \\ -c_z} = \frac{ \vec{w} \times \vec{v}}{ \| \vec{w} \|^2} = \pmatrix{0 \\ 3.7410\cdot 10^{8} \\ -1.6141\cdot 10^{8} }\;\text{[m]} $$
which is interesting considering in lunar distance units (1 LD = 384402000
m)
$$ \pmatrix{ 0 \\ c_y \\ -c_z} = \pmatrix{ 0 \\ 0.9732 \\ -0.4199 }\;\text{[LD]} $$
which is almost one LD towards the sun always, and one-half LD under the earth in the summer solstice, and one-half LD over the earth in the winter solstice.
Now that the kinematics of the earth are established, we can talk about dynamics.
The earth is rotating with $\vec{w}$ and so its angular momentum at the center of the earth is $$\vec{L}_E = \mathrm{I}_E\, \vec{w}$$ where ${\rm I}_E$ is the mass moment of inertia of the earth.
But since the earth is translating also, it has linear momentum $$ \vec{p} = m_E \vec{v}$$.
To calculate the angular momentum of the earth about the sun, then we combine both quantities with the following rule
$$ \vec{L}_S = \vec{L}_E + \pmatrix{0\\-R\\0} \times \vec{p} $$
If you do the calculation you will find the majority of angular momentum along the z-axis, with a small component along the y-axis.
What is interesting is that you can find the location in space where the axis of percussion of the earth passes through. In a similar fashion to above, this point is
$$ \pmatrix{0\\h_y\\h_z} = \frac{ \vec{p} \times \vec{L}_E}{ \| \vec{p} \|^2} $$
The significance of this point in space is that if you were to apply an equal and opposite momentum $\vec{p}$ to the earth through the center of percussion, the earth would not only stop orbiting but also stop spinning. You can remove all the kinetic energy of the earth with one impulse through this point. It would stop the earth on its tracks.
