Wrapping juggling balls

Question

'Twas the night before Christmas and throughout the net,
Not a question was posed, at least---not yet.
When what to my horror I suddenly realized:
One present was not wrapped, a present most prized.
For a juggler of the future, three balls all the same.
But how best to foil-wrap, within a tight frame?

Q1. What is the smallest square that can wrap three unit-radius balls, without cutting the square?

To wrap means to completely cover their convex hull. I mention "foil" above because one may want to crinkle the wrapping over sphere caps, analogous to how Mozartkugeln are wrapped.¹

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Likely easier is this question, which may only require rough bounds:

Q2. Which of the two configurations shown above is easier to wrap, easier in the sense that a smaller square suffices?

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¹ The square of diagonal $2 \pi$ is the smallest square that wraps a unit-radius sphere. Demaine, Erik D., Martin L. Demaine, John Iacono, and Stefan Langerman. "Wrapping spheres with flat paper." Computational Geometry 42, no. 8 (2009): 748-757. Journal link.

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Added. I thought I would compute the surface areas of the convex hulls of the two configurations. For the linear configuration,

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I compute $$4 \cdot 2 \pi + 4 \pi = 12 \pi \approx 37.7 \;.$$ For the triangular configuration,

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I compute $4 \pi$ for the three $\frac{1}{3}$ spheres, three times $2 \cdot \pi$ for the cylinder pieces, and two flat $\sqrt{3}$ triangles: $$4 \pi + 6 \pi + 2 \sqrt{3} = 10 \pi + 2 \sqrt{3} \approx 34.9 \;.$$ Of course, this does not address which is easier to wrap with a square.

Answer 1

The situation becomes easier, if we look at the orthogonal projection of the balls onto the euclidean plane and accounting for the need to wrap up to the equators of the balls by convoluting the images of the balls with disks of radius $\pi/2$; the analogous construction is done with a projection of the balls onto the plane $z=2$ and covering the convoluted images with one of the parts obtained by cutting the square sheet of paper along a diagonal.

for the linear arrangement of the balls we thus get as a lower bound for side length of the square $4*\sqrt{2}+2*\pi/2\approx 8.80$

for the depicted equilateral configuration we have as the side length $\sqrt{2}$ times the height of the bounding right triangle depicted here:
the height is $\pi/2+\sqrt{3}/2+\pi\sqrt{2}/2$ and the sidelength of the covering square is $(\pi/2+\sqrt{3}/2)*\sqrt{2}+\pi\approx 6.59$

From the image suggests that the bounding triangle could be made smaller if the balls were placed at the corners of a right triangle with equal catheti:

in this case we get for the sidelength $\pi/2+1+1+\pi/2 +\pi\sqrt{2}/2=\pi(1+\sqrt{2})/2+2\approx 5.79$

so the answer seems to be that the juggling balls in the depicted right-angular triangle configuration yields the optimum.