Spaghettification inside a black hole? Taylor and Wheeler in "Exploring Black Holes" calculate that the spaghettification time, measured from feeling a 1g tidal difference head-to-toe to disintegration at the singularity, is a constant, a little less than one second.  For small black holes (3 solar mass) this happens well outside the event horizon. But for large black holes it is inside.  But how is one "spaghettified" inside the black hole.  The singularity is time-like relative to both your head and feet - so it is not a different distance away and so how does the tidal force arise?
 A: Taylor & Wheeler's spaghettification time is valid for the case of "raindrops", a particular motion where the astronaut fell from rest far away from the black hole (as Brent Meeker commented).
As for inside the horizon, maybe it is unhelpful for you to focus on the description of Schwarzschild $t$ and $r$-coordinates swapping roles (as John Rennie commented). Understand that any astronaut anywhere measures 3 dimensions of space and 1 dimension of time, in their local vicinity (the technical term is orthonormal frame or tetrad). The spaghettification is calculated relative to the astronaut's own space and time.
Update: the details involve writing down some vectors. I'll work in Schwarzschild[-Droste] coordinates. The $r$-coordinate vector in this case is $(0,1,0,0)$, which indeed is timelike inside the horizon. Now the raindrop 4-velocity is
$$\Big(\frac{1}{1-2M/r},-\sqrt{2M/r},0,0\Big)$$
The radial direction for the raindrop is not $(0,1,0,0)$. Instead we seek a spatial vector orthogonal to the 4-velocity. This is:
$$\Big(-\frac{\sqrt{2M/r}}{1-2M/r},1,0,0\Big)$$
which is indeed spatial, so Taylor & Wheeler's calculation is well-grounded.
