The key to the question is "Normal"
Today I went for a normal walk
A "normal" in geometry is not something "common" or "ordinary". A geometrical "normal" is always perpendicular to something else.
Let us say that this something is...
...a line that extends from a fixed point, to the walker.
If the walker then walks along the "normal" of this thing, they will...
...always walk in a circle. The point is the center of the circle. The line in question is the radius of the circle. Walking along the normal of a radius of a circle means you walk along the circle itself.
If you walk this way, one foot will move further than the other.
Now let us do the Maths:
Let us assume that the distance between the walker's feet are, well, one foot, 30 cm, or 0.3 meters. Let us assume they went $x$ number of laps around the circle. So the distances walked is $(r + 0) \cdot x \cdot 2\pi = 3000$ for the inner foot, and $(r + 0.3) \cdot x \cdot 2\pi = 3100$ for the outer.
Divide one by the other and we get:
$\frac{(r + 0) \cdot x \cdot 2\pi}{(r + 0.3) \cdot x \cdot 2\pi} = \frac{3000}{3100} \Rightarrow$
$x\cdot 2\pi$ cannot be zero so we can cancel that factor out
$\frac{r}{r + 0.3} = \frac{30}{31} \Rightarrow$
$31\cdot r = 30\cdot r + 9$
$r = 9$
Hence...
...they walked along a circle that had a radius of 9 meters, with the inner foot directly on the circle. I leave it to the reader to figure out how many laps it was.