News New Gondola Transportation - Disney Skyliner -

[nace888]

HM and Peoplemover board from synchronized moving platforms, so their effective speed is 0mph. Not relevant to this situation.

I'd actually completely forgot about those, and they're my favorite attractions. I meant the speed in general, not how fast the floors move with the vehicles. Either way, I know they'll figure it all out... Unless of course, they plan to use those types of floors on the gondolas?

 

[lazyboy97o]

It’s a shame Disney didn’t go with a reputable company with a history that spans more than a century and a large, international portfolio of projects to help them design this system.

 

[Lift Blog]

Its 11 on the line. Station speed is more like 1 or 2. Lift Blog would probably be able to tell us what normal station speed is.

0.25 m/s = 50 ft/min = 0.6 mph

 

[JoeCamel]

0.25 m/s = 50 ft/min = 0.6 mph

With multiple unload/ load areas not counting the extra handicapped load? seems like they have plenty of room to get off and on this system so it is not sequential load but multiple cabins being loaded at any time?

 

[Lensman]

I'd actually completely forgot about those, and they're my favorite attractions. I meant the speed in general, not how fast the floors move with the vehicles. Either way, I know they'll figure it all out... Unless of course, they plan to use those types of floors on the gondolas?

I decided to do some calculations on the possible speeds of the in-station gondolas.

The 10 passenger D-Line at Soelden runs at 6.5 meters per second and is estimated to have a throughput of 4,500 passengers per hour per direction. That's 450 gondolas per hour or a gondola every 8 seconds. At 6.5 meters per second that's a spacing of 52 meters. The concept art shows spacing that is much closer, but who knows if the concept art is meant to be accurate in that regard? @Lift Blog could tell us whether gondola cars can be spaced closer on the line than 52 meters with or without a decrease in stage length (distance between the pylons) or increase in rope diameter (up to what he mentions as the max of 64mm).

At any rate, going with 52 meters between cars, you could have a 20 to 1 ratio of speeds to get a 2.6 meter spacing. This is a theoretical maximum because it would leave ZERO space between cars. At this ratio the cars in the station would be crawling along at 4.9 / 20 = 0.245 meters per second = 9.6 inches per second or about twice the top speed of a Galapagos Tortoise.

OTOH I expect Disney to run shorter spacings on the line (@Lift Blog or others please confirm that this is possible) to get capacity up, so if they go with 40 meter spacing to maintain the 4,500 pph throughput, that would translate into a theoretical minimum one foot per second speed in the stations.

From that I guess we can speculate on how much a moving walkway would improve loading.

(Happy to have anyone check my engineering estimates!) As we often say on these boards, apologies for the crudity of my calculations!

Edited to add: I'm super happy that @Lift Blog's post, which crossed mine, confirms my crude calculations. I'm keeping mine above as a #humblebrag.

 

[GCTales]

Hang on. D23 was in July 2017, IAAPA was in September 2017.

What am I missing here? September was after July last time I checked.

Not saying KTP is correct and they discussed it last year, but it is possible. More likely though someone talked prior to the start of this year’s IAAPA perhaps?

Us IAAPA was sept 2017. Europe IAAPA just started. He specifically referred to European IAAPA.

 

[Smooth]

 

[lazyboy97o]

Us IAAPA was sept 2017. Europe IAAPA just started. He specifically referred to European IAAPA.

IAAPA Attractions Expo has been in November for years now. It was not in September in 2017. The Euro Attractions Show was in September 2017 in Berlin.

 

[Lift Blog]

 

[SLUSHIE]

Regardless of speed etc, I think this gondola will have significantly more time to load/unload than most simply because the stations are a bit longer than normal.

The cabin waits for no one. You better get in before the point of no return because the doors are going to close whether you are in it or not. Gondolas usually have a gate or pressure plate right were the doors close so that if a guest is still trying to get in and they trip the gate and the lift will stop. If a lift op is paying attention, they can slow the lift to give more time for that particular cabin to load before that happens.

The operating speed is also variable. There is a (*edit* turny thing) in the shack that can be used to finely adjust the speed. The exterior control panels just have preset slow/med/fast.

On the gondola I worked, during peak times we would have one 'Maze Master' that would make groups of 6 and then one of the load operators (up to 5 or so) would help that group load and then get back in line behind the other operators to get another group. There would also generally be a supervisor that would help out wherever needed and would generally be the one closest to the controls and would slow or stop the lift if needed. On unload you only need one person.

 

[GCTales]

IAAPA Attractions Expo has been in November for years now. It was not in September in 2017. The Euro Attractions Show was in September 2017 in Berlin.

http://www.iaapa.org/expos/euro-attractions-show/home
IAAPA EUROPE 2018 is going on now.

i. mistakenly said US IAAPA was in September. You are correct and I agree the US show is and has been in November with the IAAPA Europe (the EAS), which in September and is going on NOW. Doppelmeyer is presently exhibiting in Europe at IAAPA EAS.

Kenny the Pirate specifically referred to the European IAAPA / EAS show, NOT US IAAPA.

EDIT:
While the skyliner was announced at D23 in 2017 and the 2017 Europe IAPPA EAS was in September 2017 (meaning Doppelmeyer "could" have said some thing about it if they were there), the article was posted on the first day of the 2018 IAAPA EAS.t

2nd EDIT: Doppelmeyer does not appear on the version of the 2017 IAAPA EAS exhibitor list to which I have access and i could not find it in the online lists I found.

 

[nace888]

Forget that... I know what we can use the Gondola for!! Let's use them as PARK DECOR AND THEME USE!! The Sanderson Sisters can bewitch the cabins in the fall, and once in a blue moon jolt them ever so slightly, and for Christmas, we can decorate them! Let's install a multi-colored flood on the towers to shine down, and they can strobe or show colors.

EDIT: Now if only we can convince them to record audio for the Monorails... I can imagine them fussing on the Epcot beam..

Winnie: "Now look here... WE'RE RUNNING THE SHOW, and YOU'RE going to Epcot... You can taste all the MAHVELOUSH POTIONS we create at the SODA shop, and RUN AMOK!"
Sarah: "Heheheh, AMOK AMOK AMOK AMOK AMOK!"
Winnie: "SHUSH.. you idiots!"

 

[MisterPenguin]

Don't forget that all the loading stations with the exception of Riviera have the extra wheel for standing still loading, or, to double the loading ability, making keeping up with a faster line possible.

Anyway, to understand the correlation of line speed with loading speed it helps to use an extreme example:

Let's say that a gondola arrives every second. Now, if there were not the detachment to slow it down, and no extra gondolas in the station, then people in the gondolas would have less than half a second to jump out and those getting on would have less than half a second to jump in as the gondola whips around the wheel.

Now, let's fill the station with 60 gondolas (30 on one side unloading and 30 on the other side loading). Every second a gondola comes in, another gondola has to leave. If you follow that one gondola coming into the station, it has to wait for the 60 gondolas ahead of it to leave before it gets sent off. This gives it 30 seconds on the unload side and 30 seconds on the load side. (And if that's still too fast to load, it gets sent off empty).

Now, let's slow down the rope by a factor of 10 but keep the same distance (physical spacing) between gondolas. This means a gondola is entering the station once every 10 seconds instead of every second, and now, one has to leave once every 10 seconds. With 60 gondolas in the station, a gondola that arrives has to wait 10 minutes before it leaves. That gives unloading 5 minutes to unload and loading 5 minutes to load.

But sitting in the station 5 minutes waiting to leave is a long time. So, let's reduce the number of gondolas in the station by 10 to have just 6 of them. Arriving once every 10 seconds, a gondola will have to wait 60 seconds for the other 6 gondolas ahead of it to dispatch. This gives a gondola 30 seconds on the unload side and 30 seconds on the load side.

So, now we have the gondolas arriving every 10 seconds. Let's add more gondolas! We put another gondola in between every current gondola decreasing the physical spacing. So, now, every 10 seconds we have 2 gondolas arriving, which is one every 5 seconds. Which means a gondola has to leave every 5 seconds. So, when a gondola comes in and has 6 gondolas ahead of it that will dispatch once every 5 seconds, then that gondola will only be in the station for 30 seconds. This gives folks 15 seconds to unload and 15 seconds to load.

Now, the real math!

d = distance between gondolas
v = velocity of the rope
a = rate of arrival of the gondolas expressed in time between gondolas
g = number of gondolas in the station
l = time that a gondola spends in the station for unloading and reloading

a = d/v

So, if the distance between gondolas is 40m, and the velocity is close to 11 mph (5m/s), then the gondolas would be arriving once every 8 seconds. [40m / 5m/s = 8s]

l = a*g

So, if the gondolas are arriving every 8 seconds and there are 12 gondolas in the station, then an arriving gondola has to wait 8s * 12 = 96s to leave. This gives 48 seconds to unload and 48 seconds to load (maybe closer to 40 seconds each side since loading and unloading doesn't happen on the turn).

The full formula for amount of loading time is: l = dg/v

So, loading time is correlated to distance between gondolas, AND velocity of the rope, AND number of gondolas in the station.

 

[bhg469]

Don't forget that all the loading stations with the exception of Riviera have the extra wheel for standing still loading, or, to double the loading ability, making keeping up with a faster line possible.

Anyway, to understand the correlation of line speed with loading speed it helps to use an extreme example:

Let's say that a gondola arrives every second. Now, if there were not the detachment to slow it down, and no extra gondolas in the station, then people in the gondolas would have lest than half a second to jump out and those getting on would have less than half a second to jump in as the gondola whips around the wheel.

Now, let's fill the station with 60 gondolas (30 on one side unloading and 30 on the other side loading). Every second a gondola comes in, another gondola has to leave. If you follow that one gondola coming into the station, it has to wait for the 60 gondolas ahead of it to leave before it gets sent off. This gives it 30 seconds on the unload side and 30 seconds on the load side. (And if that's still too fast to load, it gets sent off empty).

Now, let's slow down the rope by a factor of 10 but keep the same distance (physical spacing) between gondolas. This means a gondola is entering the station once every 10 seconds instead of every second, and now, one has to leave once every 10 seconds. With 60 gondolas in the station, a gondola that arrives has to wait 10 minutes before it leaves. That gives unloading 5 minutes to unload and loading 5 minutes to load.

But sitting in the station 5 minutes waiting to leave is a long time. So, let's reduce the number of gondolas by 10 to have just 6 of them. Arriving once every 10 seconds, a gondola will have to wait 60 seconds for the other 6 gondolas ahead of it to dispatch. This gives a gondola 30 seconds on the unload side and 30 seconds on the load side.

So, now we have the gondolas arriving every 10 seconds. Let's add more gondolas! We put another gondola in between every current gondola decreasing the physical spacing. So, now, every 10 seconds we have 2 gondolas arriving, which is one every 5 seconds. Which means a gondola has to leave every 5 seconds. So, when a gondola comes in and has 6 gondolas ahead of it that will dispatch once every 5 seconds, then that gondola will only be in the station for 30 seconds. This gives folks 15 seconds to unload and 15 seconds to load.

Now, the real math!

d = distance between gondolas
v = velocity of the rope
a = rate of arrival of the gondolas expressed in time between gondolas
g = number of gondolas in the station
l = time that a gondola spends in the station for unloading and reloading

a = d/v

So, if the distance between gondolas is 40m, and the velocity is close to 11 mph (5m/s), then the gondolas would be arriving once every 8 seconds. [40m / 5m/s = 8s]

l = a*g

So, if the gondolas are arriving every 8 seconds and there are 12 gondolas in the station, then an arriving gondola has to wait 8s * 12 = 96s to leave. This gives 48 seconds to unload and 48 seconds to load (maybe closer to 40 seconds each side since loading and unloading doesn't happen on the turn).

The full formula for amount of loading time is: l = dg/v

So, loading time is correlated to distance between gondolas, AND velocity of the rope, AND number of gondolas in the station.

But will it blend?

 

[JoeCamel]

But will it blend?

Well yeah, you can't have three red ones in a row. It would break the thematic consistency.....

 

[G00fyDad]

Don't forget that all the loading stations with the exception of Riviera have the extra wheel for standing still loading, or, to double the loading ability, making keeping up with a faster line possible.

Anyway, to understand the correlation of line speed with loading speed it helps to use an extreme example:

Let's say that a gondola arrives every second. Now, if there were not the detachment to slow it down, and no extra gondolas in the station, then people in the gondolas would have lest than half a second to jump out and those getting on would have less than half a second to jump in as the gondola whips around the wheel.

Now, let's fill the station with 60 gondolas (30 on one side unloading and 30 on the other side loading). Every second a gondola comes in, another gondola has to leave. If you follow that one gondola coming into the station, it has to wait for the 60 gondolas ahead of it to leave before it gets sent off. This gives it 30 seconds on the unload side and 30 seconds on the load side. (And if that's still too fast to load, it gets sent off empty).

Now, let's slow down the rope by a factor of 10 but keep the same distance (physical spacing) between gondolas. This means a gondola is entering the station once every 10 seconds instead of every second, and now, one has to leave once every 10 seconds. With 60 gondolas in the station, a gondola that arrives has to wait 10 minutes before it leaves. That gives unloading 5 minutes to unload and loading 5 minutes to load.

But sitting in the station 5 minutes waiting to leave is a long time. So, let's reduce the number of gondolas by 10 to have just 6 of them. Arriving once every 10 seconds, a gondola will have to wait 60 seconds for the other 6 gondolas ahead of it to dispatch. This gives a gondola 30 seconds on the unload side and 30 seconds on the load side.

So, now we have the gondolas arriving every 10 seconds. Let's add more gondolas! We put another gondola in between every current gondola decreasing the physical spacing. So, now, every 10 seconds we have 2 gondolas arriving, which is one every 5 seconds. Which means a gondola has to leave every 5 seconds. So, when a gondola comes in and has 6 gondolas ahead of it that will dispatch once every 5 seconds, then that gondola will only be in the station for 30 seconds. This gives folks 15 seconds to unload and 15 seconds to load.

Now, the real math!

d = distance between gondolas
v = velocity of the rope
a = rate of arrival of the gondolas expressed in time between gondolas
g = number of gondolas in the station
l = time that a gondola spends in the station for unloading and reloading

a = d/v

So, if the distance between gondolas is 40m, and the velocity is close to 11 mph (5m/s), then the gondolas would be arriving once every 8 seconds. [40m / 5m/s = 8s]

l = a*g

So, if the gondolas are arriving every 8 seconds and there are 12 gondolas in the station, then an arriving gondola has to wait 8s * 12 = 96s to leave. This gives 48 seconds to unload and 48 seconds to load (maybe closer to 40 seconds each side since loading and unloading doesn't happen on the turn).

The full formula for amount of loading time is: l = dg/v

So, loading time is correlated to distance between gondolas, AND velocity of the rope, AND number of gondolas in the station.

Yeah, yeah, yeah, but where do you carry the 2 in that equation and what time is the train getting in to Albuquerque if it is travelling at 43mph?

 

[danlb_2000]

Don't forget that all the loading stations with the exception of Riviera have the extra wheel for standing still loading, or, to double the loading ability, making keeping up with a faster line possible.

Anyway, to understand the correlation of line speed with loading speed it helps to use an extreme example:

Let's say that a gondola arrives every second. Now, if there were not the detachment to slow it down, and no extra gondolas in the station, then people in the gondolas would have lest than half a second to jump out and those getting on would have less than half a second to jump in as the gondola whips around the wheel.

Now, let's fill the station with 60 gondolas (30 on one side unloading and 30 on the other side loading). Every second a gondola comes in, another gondola has to leave. If you follow that one gondola coming into the station, it has to wait for the 60 gondolas ahead of it to leave before it gets sent off. This gives it 30 seconds on the unload side and 30 seconds on the load side. (And if that's still too fast to load, it gets sent off empty).

Now, let's slow down the rope by a factor of 10 but keep the same distance (physical spacing) between gondolas. This means a gondola is entering the station once every 10 seconds instead of every second, and now, one has to leave once every 10 seconds. With 60 gondolas in the station, a gondola that arrives has to wait 10 minutes before it leaves. That gives unloading 5 minutes to unload and loading 5 minutes to load.

But sitting in the station 5 minutes waiting to leave is a long time. So, let's reduce the number of gondolas by 10 to have just 6 of them. Arriving once every 10 seconds, a gondola will have to wait 60 seconds for the other 6 gondolas ahead of it to dispatch. This gives a gondola 30 seconds on the unload side and 30 seconds on the load side.

So, now we have the gondolas arriving every 10 seconds. Let's add more gondolas! We put another gondola in between every current gondola decreasing the physical spacing. So, now, every 10 seconds we have 2 gondolas arriving, which is one every 5 seconds. Which means a gondola has to leave every 5 seconds. So, when a gondola comes in and has 6 gondolas ahead of it that will dispatch once every 5 seconds, then that gondola will only be in the station for 30 seconds. This gives folks 15 seconds to unload and 15 seconds to load.

Now, the real math!

d = distance between gondolas
v = velocity of the rope
a = rate of arrival of the gondolas expressed in time between gondolas
g = number of gondolas in the station
l = time that a gondola spends in the station for unloading and reloading

a = d/v

So, if the distance between gondolas is 40m, and the velocity is close to 11 mph (5m/s), then the gondolas would be arriving once every 8 seconds. [40m / 5m/s = 8s]

l = a*g

So, if the gondolas are arriving every 8 seconds and there are 12 gondolas in the station, then an arriving gondola has to wait 8s * 12 = 96s to leave. This gives 48 seconds to unload and 48 seconds to load (maybe closer to 40 seconds each side since loading and unloading doesn't happen on the turn).

The full formula for amount of loading time is: l = dg/v

So, loading time is correlated to distance between gondolas, AND velocity of the rope, AND number of gondolas in the station.

 

[NormC]

What... is the air-speed velocity of an unladen swallow?

 

[Ripken10]

What... is the air-speed velocity of an unladen swallow?

Is that an African or European Swallow?

 

[G00fyDad]

What... is the air-speed velocity of an unladen swallow?

I wish I had thought of that! Outstanding sir!