I've always put paper or plastic or other material shims in and I've never hit problems but this has always been on small production runs. Getting the gap dimension is fairly simple too. The basic formula is: -
\$\mu_e = \dfrac{\mu_i}{1+\dfrac{G\cdot\mu_i}{l_e}}\$
Where G is gap and \$l_e\$ is effective length of the core. \$\mu_i\$ is initial permeability (before gapping) and \$\mu_e\$ is effective permeability (after gapping).
I know that you want to use N49 material but you can get perfectly good clues about gapping N49 if you look at the table for N41 material: -
So, for an effective length of 70mm the permeability of N41 (1890) with a 1mm gap becomes 67.5 i.e. pretty close to the number quoted of 70. In fact it's pretty much the gap that defines the permeability now. For instance, if you had a material with a permeability of (say) 1000 and did the math with the 1mm gap, the new permeability comes out at 65.4.
Don't forget that a 1mm gap on the centre limb of the core translates to a 0.5mm gap all round.
So, for N49 material the new value of permeability is pretty much going to be about 65 for a 1mm gap. How does this affect saturation? Firstly you need more turns because the inductance will have dropped by the proportion 1000:65. Inductance is proportional to (turns)\$^2\$ so now, to restore the inductance, you need 3.922 times as many turns as you previously had.
This makes the current in the primary identical to as before but, ampere turns have increased by 3.922 and therefore the H field is 3.922 times bigger BUT, and this is the important thing, B = \$\mu H\$ and, because \$\mu\$ has lowered by 15.38 times (3.922\$^2\$), B has effectively lowered by 3.922 and the risk of saturation is much smaller.
Regarding the sanding down of ferrite - it's very easy to do but a little tricky to measure how far you have sanded down. I've done this once and didn't have any problems other than it being a bit fiddly but you can take off ferrite with reasonably fine grade sandpaper quite easily by hand.
If you are building a radio receiver probably it could be better to wind them separated on two pieces of insulating tubes in order to find the best position on the ferrite rod and the best distance between them. This because, depending from the circuit, the coupling coefficient and the inductance can be independently tuned. The 23 uH probably will load the 470 uh tuning coil and the loading is varied by distance. If you need some ferrite rods, i bought many about 30 years ago for another use.
Best Answer
You can find some calculators to allow you to approximate the inductance of coil wrapped on an open rod. If I plug your numbers into this one, I find 15 turns should be sufficient to yield 40uH with your stated dimensions (it won't let me spread the turns out as much as yours).
There is a link in the calculator to the equations they've implemented and I suggest you study the equations rather than trusting this (or any other) online calculator.
The inductance is much less than it would be with a closed magnetic path because the permeability of air is 1/2000 that of your rod.
A rather thorough treatment of how the effective Al varies with parameters can be had in this document.