by John Ananny
Answer: ROUGH CALCULATION
At the start of the ride, assume the coaster car to be at rest. For remainder of these calculations, assume unit mass for car plus riders and work in metric. Use the standard gravitational acceleration value of 9.80665 m/s2 as one G. Relevant formulas include (1) the exchange of gravitational potential energy for kinetic energy (i.e., mgh = (mv2)/2) as well as (2) the centripetal acceleration of a body moving through a curve of particular radius (i.e., a = (v2)/R). Assume frictionless operation throughout. Acceleration magnitudes are unrealistic and unreasonable for humans, so this "design" is intended to be read as a wildly unworkable draft by an over-caffeinated intern, not as something that has actually been built. Dimensional precision varies wildly, perhaps also suggestive of someone with more enthusiasm than experience. :-)
After the initial 40 m drop, the car has a speed of 28.01 m/s. Per the "let the fun begin!" note and the lack of dimensioning of the curve, the acceleration corresponding to the initial drop is uninteresting. It exists to let the car build up some initial speed.
The extraction content starts now.
The first loop is meant to be read as a clothoid, with a radius at the top of the loop smaller than the radius at the bottom of the loop. The bottom and top of the loop are dimensioned, but the intervening smoothly varying curve is not, which is meant to draw focus to those two points only.
Moving through a radius of 4.444 m at a speed of 28.01 m/s gives a centripetal acceleration of 176.54 m/s2, or 18.00 G, yielding the letter R. By the top of the 15 m loop, enough kinetic energy has been converted to gravitational potential energy to bring the speed of the car down to 22.14 m/s. Moving through a radius of 3.333 m at a speed of 22.14 m/s gives a centripetal acceleration of 147.11 m/s2, or 15.00 G, yielding the letter O.
Once the car has returned to the bottom of the loop, it is back to its original entry speed of 28.01 m/s courtesy of frictionless operation.
The second loop is wildly ridiculous, having the same circular profile at both bottom and top. When solution letters pair up such that the second letter is earlier enough in the alphabet relative to the first one, it is possible to choose a loop height such that the bottom and top loop radii are the same. This seemed like fun (for certain values of "fun"), and allowed for use of "(TYP)" markings on the drawings, which might lead some solvers to conclude incorrectly that such loops yield one letter rather than two.
Moving through a radius of 3.81 ms at a speed of 28.01 m/s gives a centripetal acceleration of 205.92 m/s2, or 21.00 G, yielding the letter U. By the top of the 26.667 m loop, the car has slowed to 16.17 m/s, and moving through a radius of 3.81 m gives 7.00 G, yielding the letter G.
The third loop has the same shape as the second. Entry speed of 28.01 m/s and a radius of 10 m gives 8.00 G, yielding the letter H. By the top of the 25 m loop, speed has been reduced to 17.15 m/s, which into a radius of 10 m gives 3.00 G, yielding the letter C.
Producing a calculation yielding one G only is a bit tricky, since actual coaster loops need to stay well above this value (or else things start falling out of the car). Instead, we use a short section labelled "freefall" to cue a value of 1 G, yielding the letter A. This section also introduces an additional 2 m drop, which is assumed to go entirely to the kinetic energy of the car, increasing its speed slightly to 28.70 m/s.
The fourth loop has an entry speed of 28.70 m/s and a radius of 7 m, yielding 12.00 G and the letter L. By the top of the 31.5 m loop, the car has slowed to 14.35 m/s, and moving through a radius of 7 m gives 3.00 G and the letter C.
The fifth loop has an entry speed of 28.70 m/s and a radius of 4 m, yielding 21.00 G and the letter U. By the top of the 18 m loop, the car has slowed to 21.70 m/s, and moving through a radius of 4 m gives 12.00 G and the letter L.
A second 2 m freefall section gives the letter A, and increases the speed of the car slightly to 29.38 m/s. (At this point, solvers failing to adjust the speed of the car may get into the weeds, but they'll have plenty of letters already.)
The sixth loop has an entry speed of 29.38 m/s and a radius of 4.4 m, yielding 20.00 G and the letter T. By the top of the 24.2 m loop, the car has slowed to 19.71 m/s, and moving through a radius of 4.4 m gives 9.00 G and the letter I.
Given the closeness of the final two letters, another clothoid is used as the seventh and final loop. An entry speed of 29.38 m/s into 5.867 m yields 15.00 G and the letter O. By the top of the 15 m loop, the car has slowed to 23.85 m/s, and moving through a radius of 4.143 m gives 14.00 G and the letter N.
Assembling the pairs of letters in order of travel and treating the two freefall sections as the letter A gives the final answer, RO-UG-HC-A-LC-UL-A-TI-ON, or ROUGH CALCULATION.
The "Geez, kid" that starts the sticky-note flavortext is meant to hint conversion of the acceleration values to multiples of one G.