Yeah. The magnet quench flash boils a bunch of helium which is itself expensive, and presents a nice asphyxiation hazard as well. And then, assuming the quench damaged nothing, you have to set up the magnet again by getting the coils back down to superconducting temperatures… to get there, you end up boiling off a lot more helium. And then you have have to bring an engineer in to get the electrons spinning through the coil again and wait for the wobbles in the current to stabilize.

Or so I think. I work with NMR spectrometers and not MRIs, but it’s essentially the same technology.

Oh, I’m sure this’ll end well.

/s

So many solver solutions that day, either Z3 or Gauss-Jordan lol. I got a little obsessed about doing it without solvers or (god forbid) manually solving the system and eventually found a relatively simple way to find the intersection with just lines and planes:

1. Translate all hailstones and their velocities to a reference frame in which one stone is stationary at 0,0,0 (origin).
2. Take another arbitrary hailstone (A) and cross its (rereferenced) velocity and position vectors. This gives the normal vector of a plane containing the origin and the trajectory of A, both of which the thrown stone must intersect. So, the trajectory of the thrown stone lies in that plane somewhere.
3. Take two more arbitrary hailstones B and C and find the points and times that they intersect the plane. The thrown stone must strike B and C at those points, so those points are coordinates on the line representing the thrown stone. The velocity of the thrown stone is calculated by dividing the displacement between the two points by the difference of the time points of the intersections.
4. Use the velocity of the thrown stone and the time and position info the intersection of B or C to determine the position of the thrown stone at t = 0
5. Translate that position and velocity back to the original reference frame.

It’s a suboptimal solution in that it uses 4 hailstones instead of the theoretical minimum of 3, but was a lot easier to wrap my head around. Incidentally, it is not too hard to adapt the above algorithm to not need C (i.e., to use only 3 hailstones) by using line intersections. Such a solution is not much more complicated than what I gave and still has a simple geometric interpretation, but I’ll leave that as an exercise for the reader :)

My condolences, buddy.

You may be interested to know that these kinds of paper adhesives are usually intentionally designed so that the substrate (paper) tears before the adhesive does. This is meant to ensure robust packing and to give proof that the package has not been tampered with. Couple this with ever thinner and shittier substrates, and, well…

♪ How 'bout I do ANYWAY ♪