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How do manufacturers tend to calculate theoretical throughput?

Matt N

Strata Poster
Hi guys. Whenever parks are sold a ride by a manufacturer, the manufacturer usually calculates a theoretical throughput, which gives an indication of the amount of people a ride could theoretically take in an hour. But I was interested to know; how do the manufacturers calculate this figure? Do they do it using the duration of the longest block section as an average dispatch time or something, or does it take into account other factors, too?
 

bmac

Giga Poster
Hi guys. Whenever parks are sold a ride by a manufacturer, the manufacturer usually calculates a theoretical throughput, which gives an indication of the amount of people a ride could theoretically take in an hour. But I was interested to know; how do the manufacturers calculate this figure? Do they do it using the duration of the longest block section as an average dispatch time or something, or does it take into account other factors, too?

Most manufacturers take into account cycle time, and assume all available units will be cycled through in that theoretical hour.

It's how you end up with wild ones like El Toro and Intimidator 305 getting 1500 riders per hour despite neither of them being able to break 40 trains per hour.
 

oriolat2

Giga Poster
Most of them are calculated considering perfect operating conditions, as in:
- having all units available
- filling all available seats
- no stacking, thus hitting interval consistently
- no slow guests or other hiccups (guests not fitting, etc.), which slow operations

On top of that, these throughputs are calculated on paper, prior to testing. What happens is that after testing most of these rides suffer minor/major tweakings to fix unforeseen ride behaviour (train cycling too fast/too slow), which can then affect the originl hourly ride capacity.

Another thing is that some manufacturers tend to be oversell ride throughput to make their product seem more attractive/competitive to potential buyers. There are some notable examples, especially flatrides (Zamperla, looking at you!) or smaller coaster models (wild mices, spinners or SRII), which are FAR from their advertised capacity.

The only manufacturer that seems to give accurate theoretical hourly capacities is B&M because... you know, B&M.
 
Cycle time is only one factor of many that determines throughput.
The design of the blocking system gives the starting point to calculate throughput as it provides the absolute maximum possible performance for a given ride. Once the ride is built, the positions of the sensors that control and monitor the block system may be different to the original design for any number of reasons, this almost always lenghthens the block time reducing throughput.

By far the biggest factor that takes all the work done by the engineers during design and testing and promptly craps all over it is the design of the station.
 

Matt N

Strata Poster
I know this sounds like quite a crude way of calculating it, but would a basic formula for theoretical throughput on a certain number of trains be to time how long there is between the dispatches of one particular train (so essentially, the full cycle time and the time in the station), use that to calculate how many times this individual train goes round in one hour, and then multiply that by the number of trains on the track?

For instance, Smiler at Alton Towers is said by Gerstlauer to have a theoretical throughput of 1,200pph on 5 trains. Would the theoretical for each number of trains be as simple as; 1 train = 240pph, 2 trains = 480pph, 3 trains = 720pph, 4 trains = 960pph, 5 trains = 1,200pph? And Wicker Man in the same park has a theoretical of 952pph on 3 trains; would it be 1 train = 317pph, 2 trains = 635pph, 3 trains = 952pph? Or is it a bit more nuanced than that?
 
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JoshC.

Giga Poster
Trying to calculate it with multi cars is more complicated, as I imagine it depends on the spacing between blocks. For Smiler's theoretical throughput, it might be 1200pph on 5 trains, but that takes into account the fact that there's 4 other trains on the track and the speed at which it slows down on blocks, etc will therefore be taken into account. On 1 train, it would be able to go pass the blocks at the maximum speed for those sections, which may will be quicker than the speed it goes on 5 trains. That in turn could cut the ride time time, and mean that the theoretical throughput on 1 train is slightly higher than 1/5 of the 5 train throughput.

Smiler may not be the best example to see how that could happen (and I don't know much about it). But Saw you can see how this can happen. On the final block section (before the dive loop), the car goes through that much faster when the ride is running less than 8 cars. I don't think it's actually run on 8 cars this year, so if you've been on Saw this year, that final section will feel much faster than, say, an average ride a couple of years back.
We're talking tiny amounts of time in practice, half second or smaller possibly. But over an hour, that can add up.

But still, yes, the method you suggest Matt is a crude, vague but ultimately reasonably good way of calculating a theoretical throughput on a given number of trains, if you know the maximum theoretical throughput.
 

Matt N

Strata Poster
Would the method I suggested be more reliable on a coaster that has fewer trains holding more guests as opposed to lots more, smaller cars?

On an unrelated note; with regard to Saw, is that why the ride blitzes through that final dive loop at what feels like a somewhat excessive pace?
 
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Matt N

Strata Poster
Sorry for double posting, but I was looking through Vekoma’s website earlier when I found their info page on the Bermuda Blitz model (Lech Coaster): https://www.vekoma.com/sitdown-thrill-coasters/bermuda-blitz

In terms of the listed theoretical throughput for the Bermuda Blitz, Vekoma states that the throughput on 1, 2 & 3 trains is as follows;
  • 500pph on 1 train
  • 1,000pph on 2 trains
  • 1,200pph on 3 trains
You might notice that the throughput does double from 500pph to 1,000pph when the number of trains increases from 1 to 2, but the rule of 500pph per train does not continue, because the throughput does not increase to 1,500pph when the number of trains increases from 2 to 3.

So my question is; would I be correct in saying that the method I suggested above would be accurate while there is a gap between the final train leaving the station and the first train re-entering the station, but becomes inaccurate once you have trains stacking outside the station? Meaning that there is a certain maximum number of trains for which my suggested rule applies, which would depend strongly on the ride and how it’s composed?

Sorry to badger you all with so many questions; I’ll admit that the science behind things like throughput is really starting to interest me!
 
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JoshC.

Giga Poster
So my question is; would I be correct in saying that the method I suggested above would be accurate while there is a gap between the final train leaving the station and the first train re-entering the station, but becomes inaccurate once you have trains stacking outside the station?
In short, yes. Your method is crude, but works in theory in crude situations where there's a decent amount of space between vehicles.

Meaning that there is a certain maximum number of trains for which my suggested rule applies, which would depend strongly on the ride and how it’s composed?
A maximum number of trains, no. The ride could have any number of trains, and your linear suggestion would work if the ride had a block system which was suitably spaced out. The thing is, rides with a large number of vehicles don't have such nice spacing between blocks.
So yes, it depends much more on the ride and how it's composed, how the blocks are laid out, etc. As Undead Creature says, this is why the block system is important, arguably more important, than cycle time.

As you may well be aware, the maximum number of vehicles you can have on a coaster is one less than the number of block sections. That's because a vehicle cannot enter a new block section until the vehicle in front has cleared it.
Look at something like Colossus at Thorpe. That has 3 block sections: Station, the Lift Hill, and the Brakes at the end of the ride. So the maximum number of trains it can have is 2.
Smiler has more: Station, Brakes before Lift Hill 1, Lift Hill 1, Brakes before Lift Hill 2, Lift Hill 2 and Brake run at end of the ride. That's 6, so maximum number of cars would be 5.

I think really it's best understood with an example with real numbers. Let's look at Lech in closer detail.

- Lech seats 20 people per train. To hit 500pph, you would need to send 25 (full) trains per hour. That's a dispatch every 2mins 24sec.
-One 1 train, it seems to take 1min 43sec for a train to run from station to station (based off this video).
-This means that in the theoretical throughput, Vekoma allow 41secs for a dispatch.

Lech has 4 block sections: Station, Lift Hill, End Brakes 1 (immediately after the ride) and End Brakes 2 (after End Brakes 1, just before the Station) .
On 2 trains (call them Train 1 and Train 2), you have the following situation: You send Train 1, and Train 2 rolls into the station shortly after. You can send Train 2 at any point after Train 1 has 'cleared' the Lift Hill, in theory. However, you ideally want to send it so that by the time Train 2 reaches the top of the Lift Hill, Train 1 has cleared End Brakes 1. In other words, Train 1 needs to be in End Brakes 2.

Of course, you can dispatch Train 2 at any point. However, this either means the lift hill goes slower than the optimal speed, or Train 2 stops on the Lift Hill. In either case, an earlier dispatch could mean you have the contradictory-sounding situation that the train ultimately ends up leaving the Lift Hill later than if you dispatched it at the 'right' time.

On 3 trains (Train 1, Train 2 and Train 3), it becomes trickier. You start off with Train 1 in Station, Train 2 in End Brakes 2 and Train 3 in End Brakes 1. Train 1 is dispatched and hits the lift hill, allowing Train 2 to roll into the station, which means Train 1 can go into End Brakes 2.
Train 1 then leaves the lift hill and goes about the ride. You only want to dispatch Train 2, however, when you know it can clear the Lift Hill. For that to happen, you need End Brakes 2 to be available for it. That can only happen when Train 1 is in End Brakes 1, which in turn can only happen when Train 3 is in the station. But that can only happen when Train 2 is dispatched.
This means there's only a small window of time when it is optimal to dispatch Train 2.
And this whole process repeats.

So anyways, let's look at some numbers again, again using this video:
-Time taken to go from Station dispatch to bottom of lift hill: 14secs
-Time taken to go from bottom of Lift Hill to top of Lift Hill: 22secs
-Time taken to go from top of Lift Hill to End Brakes 1: 47secs
-Time taken to go from End Brakes 1 to End Brakes 2: 10secs
-Time taken to go from End Brakes 2 to Station: 10 secs
(The brake times will be a few second longer with multiple trains as they will have to include stopping the train and getting the train moving between the brake sections, whereas the video, with 1 train only, they just fly through. This is an important caveat to the multi-train case!!).

So 1 train is easy, in theory. Train takes 1min 43secs to go round the course, you take 41secs to unload/loads train and do safety checks. Boom, 25 trains sent an hour, 500pph.

Two trains again is fairly straightforward.
-0secs: Dispatch Train 1 from Station. Train 2 sat on End Brakes 2.
-14secs: Train 1 at bottom of Lift Hill, Train 2 moves forward.
-24secs: Train 2 in station
-36secs: Train 1 at top of Lift Hill and goes through ride.
-1min 5secs: Train 2 in station for 41secs, can now be dispatched (as Lift Hill is clear).
-1min 23secs: Train 1 at End Brakes 1.
-1min 33secs: Train 1 at End Brakes 2. This is the earliest moment that Train 2 can leave the Lift Hill.
-1min 43secs: Train 1 returns to Station.
-1min 46secs: Train 2 at top of Lift Hill, and as End Brakes 2 is clear, moves on.
Rinse and repeat.
In other words, the addition of the second train doesn't slow down the allowed time for dispatching a train, because all relevant blocks are cleared when needed.
So each train is sent 25 times an hour. 1000pph.

Now let's look at 3 trains. I'll go for a bit longer to show what is happening:
-0sec. Dispatch Train 1 from Station, Train 2 sat on End Brakes 2. Train 3 sat on End Brakes 1.
-14secs. Train 1 at bottom of Lift Hill, Train 2 moves forward. Train 3 still on End Brakes 1.
-24secs: Train 2 in Station. Train 3 can move forward.
-34secs. Train 3 is in End Brakes 1. Train 1 is clear to leave the Lift Hill and go through the ride when possible.
-36secs: Train 1 at top of Lift Hill, and leaves as End Brakes 1 is clear..
-1min 5secs: Train 2 in Station for 41secs, can now be dispatched (as Lift Hill is clear).
-1min 19secs: Train 2 at bottom of Lift Hill, Train 3 moves forward.
-1min 23secs: Train 1 at End Brakes 1.
-1min 29secs: Train 3 in Station. This is the earliest Train 1 is allowed to move from End Brakes 1 to End Brakes 2. As it is in End Brakes 1, it moves immediately.
-1min 39secs: Train 1 at End Brakes 2. This is the earliest moment that Train 2 can leave the Lift Hill.
-1min 41secs: Train 2 at top of Lift Hill, and as End Brakes 2 is clear, moves on.

-2min 10secs: Train 3 in Station for 41secs, can now be dispatched (as Lift Hill is clear).
-2min 24secs: Train 3 at bottom of Lift Hill, Train 1 can move forward.
-2min 28secs. Train 2 at End Brakes 1.
-2min 34secs: Train 1 in a Station. This is the earliest Train 2 is allowed to move from End Brakes 1 to End Brakes 2. As it is in End Brakes 1, it moves immediately.
-2min 44secs: Train 2 at End Brakes 2. This is the earliest moment that Train 3 can leave the Lift Hill.
-2min 46secs: Train 3 at top of Lift Hill, and as End Brakes 2 is clear, moves on.

-3min 15secs: Train 1 in Station for 41secs, can now be dispatched (as Lift Hill is clear).
-3min 29secs: Train 1 at bottom of Lift Hill, Train 2 moves forward.
-3min 33secs: Train 3 hits End Brakes 2.
-3min 39secs: Train 2 is in Station. This is the earliest Train 3 is allowed to move from End Brakes 1 to End Brakes 2. As it is in End Brakes 1, it moves immediately.
-3min 49secs: Train 3 at End Brakes 2. This is the earliest moment Train 1 can leave the Lift Hill.
-3min 51secs: Train 1 at top of Lift Hill, and as End Brakes 2 is clear, moves on.


The sections in bold are the most important things. If everything runs like this, you have a period of 2 seconds of wiggle room. BUT, they don't, because this doesn't take into account the time it takes when a train stops and restarts. Say you then have to add an extra 5 seconds to move between End Brakes 1 and End Brakes 2, and an extra 5 second between End Brakes 2 and the Station, that's an extra 10 seconds which needs to be accounted for.

That can be achieved by, say, slowing the lift hill speed down, or giving more time in the Station for a dispatch. But in any case, it slows the time in which you can dispatch a train in theory. All that time means that instead of being able to send each train 25 times in an hour (as you can, in theory, with 1 or 2 trains), you can only send each train 20 times an hour, in theory.

So that means achieving 1500pph (the 500pph achieved on 1 train multiplied by 3) becomes an impossibility, and it brings the theoretical throughput to 1200pph.

Obviously this all goes down the drain when you add in guests, checking bars, etc. I'd imagine that with the set up, 3 trains literally would not get a meaningful higher throughput than 2 trains, except in exceptional circumstances.
In other words, Lech is not a ride that's really designed to run 3 trains efficiently or effectively.

---

Okay, that's a lot of waffle. Hopefully it makes sense and the maths adds up. Though it very well might not. But in any case, that should give an insight into why theoretical throughputs work the way they do.

The important thing to note is that rides with a sense of uniform spacing between blocks have a better chance at being closer to the theoretical throughputs in practice (as long as they have well-designed stations, efficient staff, etc). Rides with LOTS of vehicles have added complications if their block sections are awkwardly spaced. Something like Smiler, for example, never did much better on 5 trains compared to 4 trains because the block system, along with the operations, meant it simply never worked. Saw is similar: the different in practice between 7 cars and 8 cars in minimal.

Anyway, to give clear, short answers to your questions:
Would the method I suggested be more reliable on a coaster that has fewer trains holding more guests as opposed to lots more, smaller cars?
Yes, if a coaster with a lot more cars has an awkward block set up. But as a simple, crude method, it's not bad for get ball park figures.
On an unrelated note; with regard to Saw, is that why the ride blitzes through that final dive loop at what feels like a somewhat excessive pace?
Pretty much, yes.
(Saw is also a difficult case to look at given the dual dispatch, etc. My advice would be to try not to understand the throughput of that!)
So my question is; would I be correct in saying that the method I suggested above would be accurate while there is a gap between the final train leaving the station and the first train re-entering the station, but becomes inaccurate once you have trains stacking outside the station?
To reiterate (since I waffled): It's more accurate in the first situation you suggest. It becomes less accurate in the latter, but still okay in the grand scheme of things.
Meaning that there is a certain maximum number of trains for which my suggested rule applies, which would depend strongly on the ride and how it’s composed?
The more important thing is the design of the ride and its block system, and how well it can handle its given number of trains. The number of trains is less important.
 

Furiustobaco

Mega Poster
^^^

Intamin calculate their model's throughput differently. Basically any throughput number on their website should be divided by two as usually it is a bunch of bull**it. Then divide by two again to account for very unavoidable downtime.
 

Matt N

Strata Poster
^^^

Intamin calculate their model's throughput differently. Basically any throughput number on their website should be divided by two as usually it is a bunch of bull**it. Then divide by two again to account for very unavoidable downtime.
Interestingly, though, some Intamin throughputs seem to be quite accurate or sometimes even modest at times, so only some are overstated! For instance, Thirteen’s throughput is said by Intamin to be 1,100pph, but it got 1,260pph in its first hour of operation, and it got as high as 1,440pph during its opening year!
In short, yes. Your method is crude, but works in theory in crude situations where there's a decent amount of space between vehicles.


A maximum number of trains, no. The ride could have any number of trains, and your linear suggestion would work if the ride had a block system which was suitably spaced out. The thing is, rides with a large number of vehicles don't have such nice spacing between blocks.
So yes, it depends much more on the ride and how it's composed, how the blocks are laid out, etc. As Undead Creature says, this is why the block system is important, arguably more important, than cycle time.

As you may well be aware, the maximum number of vehicles you can have on a coaster is one less than the number of block sections. That's because a vehicle cannot enter a new block section until the vehicle in front has cleared it.
Look at something like Colossus at Thorpe. That has 3 block sections: Station, the Lift Hill, and the Brakes at the end of the ride. So the maximum number of trains it can have is 2.
Smiler has more: Station, Brakes before Lift Hill 1, Lift Hill 1, Brakes before Lift Hill 2, Lift Hill 2 and Brake run at end of the ride. That's 6, so maximum number of cars would be 5.

I think really it's best understood with an example with real numbers. Let's look at Lech in closer detail.

- Lech seats 20 people per train. To hit 500pph, you would need to send 25 (full) trains per hour. That's a dispatch every 2mins 24sec.
-One 1 train, it seems to take 1min 43sec for a train to run from station to station (based off this video).
-This means that in the theoretical throughput, Vekoma allow 41secs for a dispatch.

Lech has 4 block sections: Station, Lift Hill, End Brakes 1 (immediately after the ride) and End Brakes 2 (after End Brakes 1, just before the Station) .
On 2 trains (call them Train 1 and Train 2), you have the following situation: You send Train 1, and Train 2 rolls into the station shortly after. You can send Train 2 at any point after Train 1 has 'cleared' the Lift Hill, in theory. However, you ideally want to send it so that by the time Train 2 reaches the top of the Lift Hill, Train 1 has cleared End Brakes 1. In other words, Train 1 needs to be in End Brakes 2.

Of course, you can dispatch Train 2 at any point. However, this either means the lift hill goes slower than the optimal speed, or Train 2 stops on the Lift Hill. In either case, an earlier dispatch could mean you have the contradictory-sounding situation that the train ultimately ends up leaving the Lift Hill later than if you dispatched it at the 'right' time.

On 3 trains (Train 1, Train 2 and Train 3), it becomes trickier. You start off with Train 1 in Station, Train 2 in End Brakes 2 and Train 3 in End Brakes 1. Train 1 is dispatched and hits the lift hill, allowing Train 2 to roll into the station, which means Train 1 can go into End Brakes 2.
Train 1 then leaves the lift hill and goes about the ride. You only want to dispatch Train 2, however, when you know it can clear the Lift Hill. For that to happen, you need End Brakes 2 to be available for it. That can only happen when Train 1 is in End Brakes 1, which in turn can only happen when Train 3 is in the station. But that can only happen when Train 2 is dispatched.
This means there's only a small window of time when it is optimal to dispatch Train 2.
And this whole process repeats.

So anyways, let's look at some numbers again, again using this video:
-Time taken to go from Station dispatch to bottom of lift hill: 14secs
-Time taken to go from bottom of Lift Hill to top of Lift Hill: 22secs
-Time taken to go from top of Lift Hill to End Brakes 1: 47secs
-Time taken to go from End Brakes 1 to End Brakes 2: 10secs
-Time taken to go from End Brakes 2 to Station: 10 secs
(The brake times will be a few second longer with multiple trains as they will have to include stopping the train and getting the train moving between the brake sections, whereas the video, with 1 train only, they just fly through. This is an important caveat to the multi-train case!!).

So 1 train is easy, in theory. Train takes 1min 43secs to go round the course, you take 41secs to unload/loads train and do safety checks. Boom, 25 trains sent an hour, 500pph.

Two trains again is fairly straightforward.
-0secs: Dispatch Train 1 from Station. Train 2 sat on End Brakes 2.
-14secs: Train 1 at bottom of Lift Hill, Train 2 moves forward.
-24secs: Train 2 in station
-36secs: Train 1 at top of Lift Hill and goes through ride.
-1min 5secs: Train 2 in station for 41secs, can now be dispatched (as Lift Hill is clear).
-1min 23secs: Train 1 at End Brakes 1.
-1min 33secs: Train 1 at End Brakes 2. This is the earliest moment that Train 2 can leave the Lift Hill.
-1min 43secs: Train 1 returns to Station.
-1min 46secs: Train 2 at top of Lift Hill, and as End Brakes 2 is clear, moves on.
Rinse and repeat.
In other words, the addition of the second train doesn't slow down the allowed time for dispatching a train, because all relevant blocks are cleared when needed.
So each train is sent 25 times an hour. 1000pph.

Now let's look at 3 trains. I'll go for a bit longer to show what is happening:
-0sec. Dispatch Train 1 from Station, Train 2 sat on End Brakes 2. Train 3 sat on End Brakes 1.
-14secs. Train 1 at bottom of Lift Hill, Train 2 moves forward. Train 3 still on End Brakes 1.
-24secs: Train 2 in Station. Train 3 can move forward.
-34secs. Train 3 is in End Brakes 1. Train 1 is clear to leave the Lift Hill and go through the ride when possible.
-36secs: Train 1 at top of Lift Hill, and leaves as End Brakes 1 is clear..
-1min 5secs: Train 2 in Station for 41secs, can now be dispatched (as Lift Hill is clear).
-1min 19secs: Train 2 at bottom of Lift Hill, Train 3 moves forward.
-1min 23secs: Train 1 at End Brakes 1.
-1min 29secs: Train 3 in Station. This is the earliest Train 1 is allowed to move from End Brakes 1 to End Brakes 2. As it is in End Brakes 1, it moves immediately.
-1min 39secs: Train 1 at End Brakes 2. This is the earliest moment that Train 2 can leave the Lift Hill.
-1min 41secs: Train 2 at top of Lift Hill, and as End Brakes 2 is clear, moves on.

-2min 10secs: Train 3 in Station for 41secs, can now be dispatched (as Lift Hill is clear).
-2min 24secs: Train 3 at bottom of Lift Hill, Train 1 can move forward.
-2min 28secs. Train 2 at End Brakes 1.
-2min 34secs: Train 1 in a Station. This is the earliest Train 2 is allowed to move from End Brakes 1 to End Brakes 2. As it is in End Brakes 1, it moves immediately.
-2min 44secs: Train 2 at End Brakes 2. This is the earliest moment that Train 3 can leave the Lift Hill.
-2min 46secs: Train 3 at top of Lift Hill, and as End Brakes 2 is clear, moves on.

-3min 15secs: Train 1 in Station for 41secs, can now be dispatched (as Lift Hill is clear).
-3min 29secs: Train 1 at bottom of Lift Hill, Train 2 moves forward.
-3min 33secs: Train 3 hits End Brakes 2.
-3min 39secs: Train 2 is in Station. This is the earliest Train 3 is allowed to move from End Brakes 1 to End Brakes 2. As it is in End Brakes 1, it moves immediately.
-3min 49secs: Train 3 at End Brakes 2. This is the earliest moment Train 1 can leave the Lift Hill.
-3min 51secs: Train 1 at top of Lift Hill, and as End Brakes 2 is clear, moves on.


The sections in bold are the most important things. If everything runs like this, you have a period of 2 seconds of wiggle room. BUT, they don't, because this doesn't take into account the time it takes when a train stops and restarts. Say you then have to add an extra 5 seconds to move between End Brakes 1 and End Brakes 2, and an extra 5 second between End Brakes 2 and the Station, that's an extra 10 seconds which needs to be accounted for.

That can be achieved by, say, slowing the lift hill speed down, or giving more time in the Station for a dispatch. But in any case, it slows the time in which you can dispatch a train in theory. All that time means that instead of being able to send each train 25 times in an hour (as you can, in theory, with 1 or 2 trains), you can only send each train 20 times an hour, in theory.

So that means achieving 1500pph (the 500pph achieved on 1 train multiplied by 3) becomes an impossibility, and it brings the theoretical throughput to 1200pph.

Obviously this all goes down the drain when you add in guests, checking bars, etc. I'd imagine that with the set up, 3 trains literally would not get a meaningful higher throughput than 2 trains, except in exceptional circumstances.
In other words, Lech is not a ride that's really designed to run 3 trains efficiently or effectively.

---

Okay, that's a lot of waffle. Hopefully it makes sense and the maths adds up. Though it very well might not. But in any case, that should give an insight into why theoretical throughputs work the way they do.

The important thing to note is that rides with a sense of uniform spacing between blocks have a better chance at being closer to the theoretical throughputs in practice (as long as they have well-designed stations, efficient staff, etc). Rides with LOTS of vehicles have added complications if their block sections are awkwardly spaced. Something like Smiler, for example, never did much better on 5 trains compared to 4 trains because the block system, along with the operations, meant it simply never worked. Saw is similar: the different in practice between 7 cars and 8 cars in minimal.

Anyway, to give clear, short answers to your questions:

Yes, if a coaster with a lot more cars has an awkward block set up. But as a simple, crude method, it's not bad for get ball park figures.

Pretty much, yes.
(Saw is also a difficult case to look at given the dual dispatch, etc. My advice would be to try not to understand the throughput of that!)

To reiterate (since I waffled): It's more accurate in the first situation you suggest. It becomes less accurate in the latter, but still okay in the grand scheme of things.

The more important thing is the design of the ride and its block system, and how well it can handle its given number of trains. The number of trains is less important.
Oh wow; thanks for the in-depth post @JoshC.! It makes a lot more sense now!

I don’t know about you, but I find the science of throughputs absolutely fascinating, I’ll admit!
 

Gazza

Giga Poster
It’s literally as simple as calculating dispatch interval. It’s not a linear relationship with the number of trains, as as undeadcreature said depends on the physical design of the ride and station.

This goes for any type of ride. You only need to measure at one point.



Eg a water slide. 2 person raft every 20 seconds = 360 per hour.

Omnimover. 2 person buggy every 3 seconds = 2400 per hour.

Top Spin. 38 seats. Cycles every 5 mins = 456 per hour.

B&M Coaster. 32 Seats. Dispatch every 90 secs = 1280 per hour.



Adding extra trains only helps insofar as the amount it can reduce your dispatch intervals.



For example, if you have a long coaster and the ops are twiddling their thumbs between trains, adding a 2nd train could near on double capacity if they can cleanly load and dispatch without stacking in that spare time.



On the other hand, some very short coasters there’s minimal benefit. Eg on Jet Rescue at Sea World Australia, the whole thing is like 30 seconds long so they have barely started loading 2nd train by the time the first train is back, so they perhaps gain a 30 second head start on a 90 second process. It’s impossible not to stack due to the shortness of the ride, so in the end the head start in loading means a 2nd train only nets an extra 20-30% extra capacity.



Stacking complicates the process and eats into capacity.



For example on DC Rivals, if the crews work fast the station is clear and the train 1 is going up the lift, so train 2 simply cruises smoothly off the main brake straight around into the station and crews can get to work loading immediately.



If a train has stacked, then capacity is reduced because the train comes to a full stop, and the brakes open and the train has to get moving again by gravity off the main brake, and wastes time crawling at a slower speed into the station…A double whammy because the slow rolling wastes time, and simultaneously gives the crew less time to load the next train. Think of it like stop start traffic congestion on a highway.



Double stations like Storm Runner or Air help because you are in effect overlapping the loading and unloading part of the sequence. A train can always enter the station immediately and commence unloading, and the double-up gives ops extra time to load, which is a bit more forgiving and means they can comfortably complete the loading process and be ready to dispatch as soon as the track is clear. Net effect these combined factors reduce dispatch interval.


The number of seats on a train does not necessarily correlate with capacity.
Eg a Eurofighter has only 8 seats, but because you’re only dealing with 8 humans at once, and they are seated in a compact block, it should theoretically be a quick process to load, check and dispatch.
Same goes for some wild mouses, so quick and easy to check that they can even have cars rolling through non stop.

However, with a long train with 32/36 seats, yes more people are going out in each dispatch, but now you are dealing with 32 humans on the platform, and it can be like herding cats. Longer physical distance for ops to check and walk.
If a guest has a problem at the other end of a train or has a phone in their pocket then it takes longer to get to them and deal with it.
So the extra seats comes with the flipside of a longer loading time and the only way to offset that is to have heaps of ops on the platform.



Some newer B&Ms like the Gigas and Banshee have 3 trains without an MCBR, but they cheat a bit by making that final brake looooooong, and splitting into separate blocks, so a train has cleared the main brake so another train can enter the circuit, but the length of the sloped brake block keeps it from stacking behind the train ahead. Operatiionally it’s much better because stopping a train on the MCBR usually means an evac, but stopping on the long brake has no such consequences.

Net result is a coaster where it’s easier to run three trains by having a means of keeping them spaced evenly time wise.
 
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Eyebrows

Mega Poster
Interestingly, though, some Intamin throughputs seem to be quite accurate or sometimes even modest at times, so only some are overstated! For instance, Thirteen’s throughput is said by Intamin to be 1,100pph, but it got 1,260pph in its first hour of operation, and it got as high as 1,440pph during its opening year!
Intamin just sort of looks at the ride and takes a blind guess
 
Lech has 4 block sections: Station, Lift Hill, End Brakes 1 (immediately after the ride) and End Brakes 2 (after End Brakes 1, just before the Station) .
On 2 trains (call them Train 1 and Train 2), you have the following situation: You send Train 1, and Train 2 rolls into the station shortly after. You can send Train 2 at any point after Train 1 has 'cleared' the Lift Hill, in theory. However, you ideally want to send it so that by the time Train 2 reaches the top of the Lift Hill, Train 1 has cleared End Brakes 1. In other words, Train 1 needs to be in End Brakes 2.

Man knows his shtuff.

Although if I was in charge, with 2 trains I would wait till train 1 is clearing the trim station before dispatching train 2. This gives the operators the most time possible to process the train in the station before it has to go without causing the returning train to stop in the waiting block.

Single train operation is actually suprisingly difficult to pull off efficiently. In order to get the best throughput you have to turn the train around in the station as quick as humanly possible.
 
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