**PART 1: STACKED 8s**

NOTE 1: The following assumes you care more about the G forces exerted on riders everywhere on the track than you care about the overall, average ride ratings. A few brutal G spikes on an otherwise benign ride won't have much effect on the overall average ratings, so won't change the ride's attractiveness to any noticeable degree. But if you'd rather not send peeps home with souvenir cervical collars, then you want to eliminate those spikes. That's what this whole thing is about in the specific case of Flying Turns coasters. With a few tweaks, this info will also be applicable to other types of coasters.

NOTE 2: Due to real life, what was originally intended to be a 1-post-covers-everything instead has to be broken up into sections, so here's Part 1.

INTRO

As I have recently discovered by pouring through old pics and Google maps, a lot of the old wooden Flying Turns, as well as the modern reproduction at Knoebels, had/have stacked figure-8s. You don't see this so much on the modern steel bobsled coasters. So naturally I wanted to put stacked 8s in my rides. And then depression set in.

THE PROBLEM

Actually, there are 3 problems. While it's a no-brainer to make perfectly stacked 8s using the default minimum 12m track section turned 90^ x 3 for a 270^ turn, then going 270^ the other way, this causes 2 problems and those lead to a 3rd problem:

- The abrupt change from the hardest possible turn in 1 direction to the hardest possible turn in the other causes horrific lateral Gs which spike fear and nausea, thus decreasing excitement because the peeps are worried about survival and getting puked on by the peep in the seat in front of them instead of enjoying the ride. This really mauls ride prestige, which is the dominant factor on getting peeps to ride the thing at all. It also keeps the overall fear higher than it should be, so many families will avoid the ride even though it's family-capable.
- The minimum possible downslope in such a minimum-sized stacked 8 is 4^ per track section. Anything less means the peeps get a support in the face going under the crossover. A 4^ downslope per 12m of length prolonged through every track section of a stacked 8, let alone 2 or more stacked 8s directly above each other, means that the train accelerates all the way down, thus increasing the lateral Gs at each change of direction and making everything less fun.
- Avoiding the above problems means putting some straightaways between the turns of stacked 8s, so the transitions between the directions of turns aren't so harsh. These straightaways mean the downslope can be reduced as well, so the problem of increasing lateral Gs throughout the stacked 8s can be avoided. But unless you do some geometry and trigonometry, you won't be able to get your stacked 8s to line up directly above each other. And math is hard for me, even though I have a degree in engineering. I had to google up all the formulæ below because I've killed a LOT of braincells since high school I feel I'm not alone in this so hopefully this refresher will save some time and frustration for others.

GOAL #1: CROSSINGS AT 60^

The goal is to make something like the pic below, all perfectly stacked however far down you want to go, with a crossing angle of 60^ which approximately matches the Knoebels ride and many older, long-dead examples, without making peeps wish they were dead. This stacked 8 doesn't accelerate the train enough to make the lateral Gs exceed 1.5 while keeping the excitement in the 6-7 range throughout and nausea negligible. But you can't make this without math.

THE MATH

There are tons of potential variables in a stacked 8 so the 1st thing to do is eliminate as many of them as possible. To do that, I decided to make the turns out of 3 track sections all the same length and degree of turn, and 2 equal sections in the straightaway (of lengths not necessarily equal to the arc lengths). To get a 60^ crossing angle in the middle of the 8, I needed a 120^ gap in the circle of the turn at each end, so with 3 sections of track per turn that's 80^ of turn per section. This leads to a minimum downslope per section of 3^ to avoid face-smashers (as opposed to head-choppers) at the crossovers, which keeps the speed essentially constant throughout and minimizes lateral Gs. Of course, it's impossible to get exactly 80^ of turn and 3^ of downslope but come as close as you can, which will be about 0.05^ on average (angle snap MUST BE OFF).

With these initial conditions, plus the restriction that all track section lengths must be integers, and whether or not you keep the built-in track restrictions of sections having to be 12-24m long, this reduces the problem to the following variables:

A = length in meters of the curved sections of track (3 per turn at 80^ of turn each)

S = length in meters of the straight sections of track (2 per straightaway)

R = radius of the turn

C = length in meters of the chord of the turn's circle left open by the 240^ arc of the 3 curved sections

With a 60^ crossing angle, S > A, so the purpose of the math is to derive a formula where S is a function of A { in general terms, S =

*f*(A) }. Then you just plug in interger values of A and see if you can get nearly integer values for S, so you can build such a thing within the 1m track section length restriction imposed by the game. To do this, you have to relate S and A together via the intermediary variables R and C. It goes like this...

1st, you have to determine R. This is based on the fraction of 360^ covered by the arc of length A:

A = 2πR * (80^/360^)

R = A/(2π) * (360^/80^)

R = 0.716 * A

Because you can relate S to C via simple trig, the next step is to relate C to R because we already have R in terms of A, so this will give S in terms of A. Bisecting C and drawing a line from C's midpoint to the center of the curve, we have a right triangle with a 30^ angle between C and the curve's radius. Thus:

cos30^ = (0.5 * C) / R

C = 2 * R * cos30^

C = 2 * (0.716 * A) * cos30^

C = 1.24 * A

Again with C bisected and looking the other way towards where the tracks cross, we have another right triangle with a 60^ angle between C and S. Thus:

cos60^ = (0.5 * C) / S

S = (0.5 * C) / cos60^

S = (0.5 * 1.24 * A) / cos60^

S = 1.24 * A <--- what we were looking for, and C = S, which is interesting to note for later.

Because both A and S must be intergers due to PC only allowing 1m track length intervals, we need solutions to S = 1.24 * A where A is an integer and S is as close to an integer as possible. Plugging in integer values for A between 12 and 24 (assuming you're using the default track limits) gives the following table of good results (the others have S being too far off an integer value):

A = 12, S = 14.88 <--- VERY close to S = 15, which I used in the above pics. Good enough.

A = 13, S = 16.12 <--- no worse than the above but a tad bigger, use S = 16m.

A = 17, S = 21.08 <--- closer to perfect but biigger still, use S = 21m.

A = 21, S = 26.04 <--- closest fit but biggest, suitable for very high speeds. Use 4x S sections of 13m each per straightaway as S can't exceed 24m nor be less than 12m.

DOWNSLOPE of 60^ STACKED 8s

The less downslope you have per track segment, the less speed the train gains through the stacked 8, which is a good thing because stacked 8s add up to a lot of total length going downhill and you don't want the train to accelerate too much through it all or G forces will become not fun. The bigger A and S are, the less downslope you need for the minimum safe clearance at the crossover. And the less downslope you need, the better the above values for A and S fit to stacking multiple 8s directly above each other.

This last is a VERY important point. All the above calculations assume all track segments are in the same horizontal plane (IOW, zero downslope). In PC, the length you set for a track segment is constant along the length of that segment, NOT in the horizontal plane. Thus, the steeper the downslope of a track segment, the shorter its projection on the horizontal plane, and the less well the above "best" values for A and S will stack above each other through multiple 8s. So, if you want the above math to be of any use to you, always go for the minimum possible downslope. But never use less than 2^ downslope with default friction or the train will decelerate, which the peeps find boring and, if prolonged enough, will cause the train to stop.

Anyway, for the stacked 60^ crossing stacked 8s pictured here, I used A = 12 and S = 15. The entry track into this was horizontal (IOW, 0^ downslope, so the 1st turn needed a 4^ downslope to get under it safely. But from there on, it was 3^. In the pic below, the selected section of track as 4^ downslope, all the rest 3^. If your entry is into the stacked 8s is going down at all, you can do the whole thing at 3^ downslope. And because the entry speed determines the speed and thus excitement throughout the stacked 8s, you'll probably be coming into them steeper than the 8s are themselves, to get the speed you want.

GOAL #2: CROSSINGS AT 90^

If all you do is 12m and 90^ turn per segment, then you're crossing at 90^, need 4^ downslope, and have a not-fun ride in terms of lateral Gs getting worse through the stacked 8s. But suppose you WANT 90^ crossings for aesthetic reasons. This is how to get that without breaking the necks of your peeps. You'll end up with something like this:

90^ CROSSING MATH

The math is the same BUT, because we're doing 270^ of turn at each end instead of 240^, things are greatly simplified because R = S instead of R = C, enabling us to skip dealing with C at all. And because S < A this time, it's easier to do things as A =

*f*(S). At the bottom line:

A = 1.57 * S

And because S < A in this case, A =

*f*(S), so the table of good values for S and A is based on plugging in integer values of S and trying to get close to integer values of A. We thus have:

R = S = 14, A = 21.98 <--- IOW, A = 22m, used for the pic above.

Nothing else stacks properly due to rounding errors.

DOWNSLOPE OF 90^ CROSSINGS

Because this is a rather large turn, a 2^ downslope gives safe clearance provided the track above is sloped at least that much. This is the minimum possible downslope ever, resulting in absolutely constant speed throughout however many 8s you want to stack. Anything less results in a slowing coaster. So, if using 2^ downslope, you have to be sure your entry speed is what you want. About 35mph entry speed keeps lateral Gs about 1.5.

GOAL #3: SMOOTHING

Regardless of crossing angle, when making 8s you want to stack directly above each other, ONLY smooth banking, NEVER "all". Throughout stacked 8s, the turns should be banked the default max of 45^ in the appropriate direction. Build the whole thing that way, then go back and smooth the transitions. Select the whole straightaway and then the 1st section of curve on each end of it. Hit "Smooth Banking" 3 times. Done.

And that's all for now. Next time, I plan to delve into stacked zig-zags, such as on the Knoebels ride.