TITLE: Covering a unit ball with balls half the radius
QUESTION [32 upvotes]: This is a direct (and obvious) generalization of the recent MO question, "Covering disks with smaller disks":

How many balls of radius $\frac{1}{2}$ are needed to cover completely a ball of radius 1?

The answer may be in the paper "Covering a Ball with Smaller Equal Balls in $\mathbb{R}^n$,"
by Jean-Louis Verger-Gaugry, which I cannot immediately access (Springer link here).
If anyone knows the answer for $\mathbb{R}^3$, I'd be curious to learn of it.
And it would be especially interesting if there were a proof as satisfying as  Noam Elkies's for seven half-disks covering one in $\mathbb{R}^2$.  Thanks!
Update. [29May2012] A series of contributions by Will Jagy, Gerhard Paseman, Karl Fabian, and zy, have reduced the upper bound from $56$ to $33$ to $22$, with a lower bound of $16$.
Update. [5Aug2012] Ed Wynn settled the question with a careful analysis: $21$ balls are needed, and suffice!

REPLY [24 votes]: Here is (I'm fairly sure) an optimal solution, building on the ideas in other answers.
We know that we can generate maximum 30° caps on the sphere, by placing the half-balls at radius sqrt(3)/2.  We deduce from http://neilsloane.com/coverings/index.html that we will need 20 half-balls to cover the outer surface, plus a 21st to cover the centre.  So, 21 is a lower bound.
To get a 21-ball solution, scale the 20 coordinates of http://neilsloane.com/coverings/dim3/cover.3.20.txt to that radius, and add one at the origin.  Note that the balls at sqrt(3)/2 give caps on the central half-ball with the same angle as the outer caps, so the cover works on the inside too. 
Here are the coordinates of 21 half-balls to cover the ball: {{+0.060111,-0.479945,-0.718359},{-0.559060,+0.562759,+0.347496},{-0.164091,-0.848680,-0.053063},{-0.509173,-0.375273,-0.591535},{-0.140963,+0.208121,+0.828743},{+0.690970,+0.512659,-0.098697},{+0.377263,-0.700146,+0.342736},{-0.250648,-0.504068,+0.658097},{-0.743210,+0.007958,+0.444494},{+0.550076,+0.074555,-0.664724},{+0.521328,-0.588831,-0.362624},{+0.088527,+0.791576,+0.339956},{-0.158465,+0.221391,-0.822116},{-0.718586,-0.473046,+0.099309},{-0.339530,+0.726217,-0.327609},{+0.439846,-0.237215,+0.707294},{+0.853710,-0.140538,+0.037800},{+0.536789,+0.335659,+0.590924},{+0.243253,+0.709202,-0.433429},{-0.778146,+0.197643,-0.324694},{+0.000000,+0.000000,+0.000000}}

There is some "fat" in the surface covers (using 30° caps when 29.6..° is required), and each non-central sphere bulges convexly out of the cone that it's responsible for, so it's easy to believe that this is indeed a cover.  Also I've checked it thoroughly, though I'll be grateful if someone checks it with less makeshift methods.  At the positions stated, the radius of the balls can be as low as 0.49812.  By moving the original coordinates further inwards, to radius 0.8595, the radius of the balls can be as low as 0.49439.

Ed Wynn, 4 August 2012.

[Graphic added by J.O'Rourke:]