Speeding through the primes until they look uniform Imagine you are traveling out the $x$-axis such that your velocity at $x$ is $v(x)$.
If $v(x)=1$, then you pass the primes at increasingly longer intervals, on average (of course there are close primes, e.g., twin primes).
Is there some function $v(x)$ so that the event of passing a prime becomes uniformly random,
in that the expected time between passing primes becomes a constant?
For example, $v(x) = x / \log^2 x$ doesn't quite work, in that the delay between
passing primes still increases with $x$.

My closest interpretation to your question is if there is a function $v(t)$ such that
$$\pi\left( \int_0^T v(t)dt\right)\sim aT$$
for some "prime-hitting" frequency $a$. The answer is yes. We'll look at a basic example. Denote the position function as $F(T) = \int_0^T v(t)dt$, then, using PNT (Wikipedia / MathWorld), transform the equation into a more tractable form
$$ \frac{F(T)}{\ln F(T)} \sim aT.$$
We will do one better and ensure equality of the above expression. (This is not the same as equality in the original asymptotic formula with $\pi(\cdot)$, but still entails our original desired form as a logical consequence.) Take the negative reciprocal of both sides,
$$ \frac{1}{F(T)} \ln \frac{1}{F(T)} = -\frac{1}{aT},$$
and then use the Lambert W function (Wikipedia / MathWorld) to simplify,
$$ \ln\frac{1}{F(T)} = W\left(-\frac{1}{aT}\right)$$
$$ F(T) = \exp\left( - W\left(-\frac{1}{aT}\right)\right)$$
which gives:
$$ v(t) = \frac{d}{dt} \exp\left( - W\left(-\frac{1}{at}\right)\right).$$
Note you can use a derivative formula for $W(\cdot)$ found in the linked articles alongside the chain rule if you so desire. And you can use even better asymptotes of the prime counting function with the same inverse-function reasoning in order to obtain velocities with more uniform prime-hitting.
EDIT: There might be problems in realizing this empirically due to the domain of the $W$ function and branch cuts. I'm having problems figuring how how to get a working graphic off of Alpha, or if my formula needs to be augmented to address some technicality I'm not seeing.
