Rumor New Monorails Coming Soon?

Voxel

President of Progress City
Now calculate the force it would take to shear a grade 5 bolt of 3/8 diameter and take the other three out enough so the door falls off the train.
Assuming the manufacture actually placed the grade 5 bolt through it’s QA. I have seems bolts fail at 20% their “advertised” capabilities. Its rare but it happens.
 

mikejs78

Well-Known Member
Now calculate the force it would take to shear a grade 5 bolt of 3/8 diameter and take the other three out enough so the door falls off the train.
I never claimed to be an expert :). Was just pointing out that the force of an object moving at some velocity is higher than the same object at a lower velocity. E.g. an ECV moving at 5mph is very different than a scooter that just barely bumps or a very large individual simply leaning on the door ..
 

JoeCamel

Well-Known Member
I never claimed to be an expert :). Was just pointing out that the force of an object moving at some velocity is higher than the same object at a lower velocity. E.g. an ECV moving at 5mph is very different than a scooter that just barely bumps or a very large individual simply leaning on the door ..
Just for grins it is 7951.2 pounds of force. I can't believe WDW would use substandard bolts or purchase from an unreliable supplier but I can believe the bolts if threaded into the train could loosen or destroy the threads with all the opening and closing and pull out. They should have been torqued on inspection after the last incident and if they used the lowest cost supplier for critical parts I can't believe that either. But here we are with doors falling off.
Hit by an ECV? Maybe but there was something else going on here to weaken the connection that a door inspection should have shown.
 

Voxel

President of Progress City
Just for grins it is 7951.2 pounds of force. I can't believe WDW would use substandard bolts or purchase from an unreliable supplier but I can believe the bolts if threaded into the train could loosen or destroy the threads with all the opening and closing and pull out. They should have been torqued on inspection after the last incident and if they used the lowest cost supplier for critical parts I can't believe that either. But here we are with doors falling off.
Hit by an ECV? Maybe but there was something else going on here to weaken the connection that a door inspection should have shown.
It should be noted that bolts from great suppliers can fail. Look at the bolt failure on the SpaceX failure a few years ago.

I never meant to go off saying that this is what happened, just showcasing one of the many things that could have caused this :)
 

britain

Well-Known Member
The Original Post in this thread? Or the original post about the scooter? You have continue to have sharp hinting skills.

1542148506203.png
 

ChrisFL

Premium Member
Via twitter:

@[B]ScottGustin[/B] 2h2 hours ago

About the monorail door incident at Disney’s Grand Floridian earlier today: Multiple eyewitnesses tell me the monorail door came loose after a guest riding an ECV scooter collided with the door. As photo shows, crews are able to safely remove the door from the monorail.






......what we MAY be missing here in this whole jump to conclusions thing is whether or not photos were taken at a point where they decided it would be better to push the door out and remove it from the outside vs. trying to get it back inside and/or drive it back to the station in that condition.

In other words, the ECV crash may not have caused the door to completely fall off (or nearly compeletly) but instead just jarred it out of place enough where they had to decide how to remove it before driving it backstage.
 

LukeS7

Well-Known Member
This is where people underestimate the potential impact. Some ECV's can hit 5mph. The impact force of a moving object hitting a stationary object is F = (0.5 * m * v^2) ÷ d (where m is mass, v is velocity, and d is the distance the moving object will travel before it stops. Calculating d is another formula, so let's just say an ECV will travel 0.1m on collision before stopping, for the sake of argument. So for a 550lb ECV (250kg) and a velocity of 5mph (8 kph or 2.2 meters/sec) we get: F = (0.5 * 250 * 4.84) / 0.1, or 6,051 newtons, which is equivalent to 1360 pounds-force. Not a small amount of force.

If anyone is more versed in physics and math than I, please check my work.... I'm definitely not an expert here...
The equation you used is for centripetal force (where d is diameter, not distance), the equation that should be used is for kinetic energy, which is F = ma. Even if it reached its top speed (2.24m/s) within 1 second (acceleration of 2.24m/s^2), that would give us (and I'm using the highest weight of 800lb presented in this thread) F = (368.74) * (2.24) which comes out to 825.98 N. A boxer's punch can be up to 5,000 N, so I'd argue that without some sort of failure, this amount of force should not have taken the door off.

Edited to clear up some grammatical errors.

EDIT: Used the wrong equation the first time through as pointed out by GlacierGlacier, corrected it
 
Last edited:

GlacierGlacier

Well-Known Member
The equation you used is for centripetal force (used for something moving in an arc/circle), the equation that should be used is for kinetic energy, which is F = 1/2 * mv^2. Even if it reached its top speed (2.24m/s) in that short of a distance, that would give us approximately (and I'm using the highest weight of 800lb presented in this thread), we get F = 1/2 * (368.74) * (2.2)^2 which comes out to 910 N. A boxer's punch can be up to 5,000 N, so I'd argue that without some sort of failure, this amount of force should not have taken the door off.
No...

That's the equation for Kinetic Energy
KE = 1/2 mv^2
Kg m^2/s^2, which is Joules.

Force is Mass times Acceleration or kgm/s^2
F = ma

It's High School physics. I would know, because I learned this only a few years ago. In high school.

Regardless, all of this is pretty pointless. When it comes to real life, you're dealing with a ton more variables and forces.

Not to mention, it's pretty silly trying to see if the force of an overladen ECV could push open a monorail door when - according to that tweet's eyewitness reports - it can.

Regardless, at least the front didn't fall off.
 

LukeS7

Well-Known Member
No...

That's the equation for Kinetic Energy
KE = 1/2 mv^2
Kg m^2/s^2, which is Joules.

Force is Mass times Acceleration or kgm/s^2
F = ma

It's High School physics. I would know, because I learned this only a few years ago. In high school.

Regardless, all of this is pretty pointless. When it comes to real life, you're dealing with a ton more variables and forces.

Not to mention, it's pretty silly trying to see if the force of an overladen ECV could push open a monorail door when - according to that tweet's eyewitness reports - it can.

Regardless, at least the front didn't fall off.
Craaaaappp, yea, you're right.

Correct thing would be 368.74 x 2.24 = 825.98 N. And while it doesn't tell us if it can or can't push open the door, it does help point to a possible malfunction as it shouldn't be able to under normal conditions.
 

MickeyMinnieMom

Well-Known Member
Not to mention, it's pretty silly trying to see if the force of an overladen ECV could push open a monorail door when - according to that tweet's eyewitness reports - it can.
IF the eyewitness report is accurate it obviously CAN... I think the question is whether that's the result of improper maintenance or some other failure on Disney's part, or if these things are really designed and built such that an ECV can just knock out a door under ideal circumstances. The latter would seem absolutely insane to me -- WAY too risky to allow them unrestrained if all we need for the door to fly off is an out of control ECV.
 

networkpro

Well-Known Member
In the Parks
Yes
Also would like to add, reports are that a ECV rammed into it... I'm not 100% sure but it looks like the door wasn't on the platform side so idk if I'm buying that story....

Now calculate the force it would take to shear a grade 5 bolt of 3/8 diameter and take the other three out enough so the door falls off the train.

Easy if you'ver got the number of bolts and the number and thicknesses of the plates that are fastening it together

F ÷ (d x (t1+t2...)

but its not as simple of a construction as you've made it out to be. Each door section is attached by upper and lower arms.
 

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