Installing an aerial hoop in the house

I bought my stepdaughter a one-metre aerial hoop for Christmas and then had to work out how to hang it. My preference was for her to use it outside, hanging from an A-frame which I could easily have made from scaffolding poles. But she persuaded me that she would use it more in her room. So, going against the advice of the manufacturers, I decided to hang it from the ceiling.

In the UK, ceiling joists are generally 4" x 2" or about 100 mm high and 50 mm wide. I would not be happy hanging an aerial hoop from a single joist, but I reckon spreading the load across 3 joists would be enough, based on the experience of a friend who has done this sort of thing before. The idea would be to screw a piece of 6" x 2" (150 x 50 mm) timber, placed horizontally under the ceiling at right angles to the joists, into the centre-lines of the joists, and then hang the hoop from this new beam.

The first task is to find the joists. You can get a rough idea from tapping the ceiling or using a stud finder. From there, my friend has a trick involving a U-shaped piece of stiff wire (like a very thin croquet hoop) which you feed through a small hole in the ceiling, roughly between the joists, and then waggle it round until it hits the joist, and mark the position of the end that is still below the ceiling. Unfortunately, this didn't work because my joists have fibreglass insulation between them, which stops the free movement of the wire.  I resorted to guessing the location and alignment of the joists and then, along two parallel lines just under 150 mm apart that are destined to end up covered by the beam, poking a bradawl through the ceiling, noting where it goes through into thin air (or insulation) and where it encounters resistance due to the joist. By using the two parallel lines you can ascertain not only the position but also the exact alignment, so that you can be reasonably sure to drill into the centre line of each of the 3 joists (which by the way are usually 16" (about 400 mm) apart).

I bought M10 (10 mm) coach screws that were 150 mm long (50 mm for the new beam, 100 mm for the full height of the joist, and about 10 mm for the ceiling board, so we nearly reach the top of the joist). I made the beam 1100 mm long to cover the 3 joists and allow about 100 mm at each end.  I drilled 10 mm holes through the beam and 5 mm holes through the ceiling into the joists. I used 2 screws for each joist, about 80 mm apart. I did consider offsetting these so that any potential split in the joist didn't spread to the other hole, but decided that it was more important to be as close to the centre line as possible, given that I still couldn't be 100% certain of the position and state of each joist.

It's difficult to hold the beam in place to align the ceiling holes correctly. With the 6 holes drilled in the beam, I drilled one hole through to the centre joist, screwed the beam into place with one screw, and marked the positions of the other 5 holes. It was then possible to loosen the screw and rotate the beam to uncover the 5 marked holes and drill them (even with the holes uncovered, you need a longer than average 5 mm drill to get right up to the top of the joist). Then move the beam back into place and put in the other 5 screws. Tightening up the 6 screws with a 17 mm socket is quite a workout - at least 100 quarter-turns (with some elbow grease) for each one!

Now to a refinement: I wanted to make sure that the load would be applied equally to the 3 joists, especially if it was a heavy load.

This little bit of applied maths shows that putting a piece of 1/8"" (3 mm) plywood packing between the beam and the middle joist should help to equalise a large load of half a tonne. The hand-waving argument is that making the beam bend will push up on the middle joist, reducing the net downward force from the load on that joist and transferring it to the outer joists.

And here's the result...

Fundamental Theorem of Arithmetic (update)

This is an update on my 2019 post on the subject. This new proof is even simpler and (hopefully) better explained.

Here's a simple, fairly non-algebraic and not particularly rigorous proof of the Fundamental Theorem of Arithmetic.  The theorem states that any natural number >1 can be uniquely expressed (apart from re-ordering) as a product of primes.  For example, 140 = 2 x 2 x 5 x 7 and there is no other set of primes that multiply together to make 140.

It is obvious by construction that any natural number can be expressed as some product of primes.  Now, if the Fundamental Theorem of Arithmetic is false, then, for some such natural number, there is another set of primes with the same product.  We can divide out any prime that appears in both sets, so that we end up with two sets of primes with equal products and no primes in common.

Now take the largest prime P (looking over both sets).  We need to show that P does not divide into the product of primes in the other set.  We do this by induction, building up the product in the other set.  We note that P does not divide into the first factor in the other set because that is prime (and because it is smaller than P).  Then if P does not divide into the product so far, we show that it won't divide into that product multiplied by the next factor.

Let R be the current product, and N the next factor. So we know that P does not divide into R, and we want to show that P does not divide into NR.

Instead of worrying about N itself, we look for the smallest number k between 2 and P-1 inclusive for which P divides into kR. If we can prove there is no such k, we have also proved that P does not divide into NR, since we know that N is between 2 and P-1.

So suppose such a k does exist.  Now look at the largest multiple of k that is less than P. Call this sk.  Then P divides skR.  But P also divides PR (obviously), so P divides (P-sk)R.  But P-sk <= k (otherwise sk wasn't the largest multiple). And P-sk is not equal to k (otherwise P = k(s+1) and P would not be prime) so p-sk < k. So k wasn't the smallest after all.  So there is no such k, including in particular the next element N in our product.  So P does not divide into NR.  Continuing this argument until we have reached the last element in the product, we have shown that P does not divide into the other product of primes, so the theorem is true.