How many ordinals can we cram into $\mathbb{R}_+$, respecting order? I've been pondering the following question.

How can we measure the amount of "space" above an element $p$ in a partially ordered set $P$?

One way would be to try to cram the elements of increasingly large ordinals into the space above $p$, respecting order. The least ordinal whose elements "don't fit" is then a measure of the amount of space above $p$ in $P$.
This suggests the following definition. (Recall that an order monomorphism preserves strict order relationships).
Definition. Given a partially ordered set $P$ and an element $p \in P$, the cramming ordinal of $p$ is the least ordinal $\alpha$ such that there is no order monomorphism $f : \alpha \rightarrow P$ satisfying $f(0)=p$.
This begs the following question.

What is the cramming ordinal of the element $0 \in \mathbb{R}$?

(Actually, I presume all elements of $\mathbb{R}$ have the same cramming ordinal in $\mathbb{R}$, but lets just consider $0$ for concreteness).
Now observe that $\omega$ is a lower bound for the cramming ordinal of $0$, because $f$ is witnessed by the identity function.
Observe also that given a monomorphism $f : \alpha \rightarrow \mathbb{R}$ satisfying $f(0)=0$, it holds that $\tan^{-1} \circ f$ is also a monomorphism. However, the range of $\tan^{-1} \circ f$ is bounded above in $\mathbb{R}$. So we can copy the idea of $\tan^{-1} \circ f$ repeatedly, in fact $\omega$ many times. So if $\alpha$ is a lower bound for the cramming ordinal of $0$, then $\alpha \omega$ is a lower bound, too.
Thus since $\omega$ is a lower bound for the cramming ordinal of $0$, so too are $\omega^2$, $\omega^3$ etc.
My question is, what is the cramming ordinal of $0 \in \mathbb{R}$?
 A: Here's a more elementary argument that any countable ordinal can be embedded into $\mathbb Q$. The trick is to show something stronger: any countable total order can embedded into $\mathbb Q$:
Let $(\mathbb N,{\sqsubset})$ be an arbitrary countable total order. (We can assume wlog that the underlying set is $\mathbb N$ -- that's what it means to be countable!) We defined an order monomorphism $f: (\mathbb N,{\sqsubset})\to(\mathbb Q,{<})$ by ordinary long induction on $\mathbb N$:


*

*Arbitrarily set $f(0)=42$.

*For $n>0$ suppose we have already decided what $f(m)$ is for $m<n$.
If $n\sqsubset m$ for all $m<n$, then let $f(m)=\min_{m<n}f(m)-1$.
If $m\sqsubset n$ for all $m<n$, then let $f(m)=\max_{m<n}f(m)+1$.
Otherwise let $f(m)=\dfrac{\max_{m<n, m\,\sqsubset\, n}f(m) + \min_{m<n, n\,\sqsubset\, m}f(m)}{2}$
In fact essentially this construction shows that any countable total order can be embedded into any dense subset of $\mathbb R$.
Once we've mapped something into $\mathbb R$, we can remap it into $\mathbb R_+$, by composing with, say,  $x\mapsto e^x$.
A: First of all, the technical term is "cofinality of the cone generated by $p$", not "cramming ordinal of $p$".
Now, for the case of $\Bbb R$, note that the order is homogeneous, so the actual point we choose to "cram" above is irrelevant, but this also means that $0$ is as good as any.
More to the point, since the rationals are dense in $\Bbb R_+$, we can embed any countable into $\Bbb R_+$. This is proved by transfinite induction. 


*

*Clearly it's true for $0$; 

*If you can embed $\alpha$ using $f$, then without loss of generality you can embed it into $[0,1)$, then extend $f$ by setting $f(\alpha)=1$. This is clearly order preserving.

*If $\beta$ is a limit ordinal and we can embed every smaller ordinal than $\beta$, write it as the union of intervals $[\alpha_i,\alpha_{i+1})$ each with order type strictly less than $\beta$ (e.g. take a strictly increasing sequence of ordinals, $\alpha_i$ with limit $\beta$). 
By the assumption we can embed every ordinal smaller than $\beta$, so we can embed the ordinal isomorphic to $[\alpha_i,\alpha_{i+1})$ into the interval $[i,i+1)$ in the rationals. Clearly we have embedded $\beta$ into the rationals by gluing the maps. $\square$
Furthermore if we do embed $\omega_1$, then we can cut $\Bbb R$ into $\aleph_1$ disjoint open intervals, this is impossible because there is an injection from every family of disjoint open intervals into $\Bbb Q$.
Therefore the cramming ordinal is $\omega_1$.
