All Star Nathan
New Member
So much for sarcasm and wit.
Lets see you try and measure financial peril.:drevil:
Lets see you try and measure financial peril.:drevil: 
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Fastest Ride at WDW
- Thread starter ArchiDanDisney
- Start date
Mikejakester
Active Member
:lol:
:lol: I know you were just being sarcastic. I was just pulling your hair bud...
:lol: I know you were just being sarcastic. I was just pulling your hair bud...
All Star Nathan
New Member
I know, I know. All is fair in love and forums.
WDW John
Member
So if you and the whale exited the plane at the exact same time the whale would hit the ground first? Or you would?
Also, I'm confused, are we now saying that dropping 13 stories in 2 seconds has no speed? I am accelerating to the ground at a higher rate than gravity would pull me on it's own. Seems fast. Isn't that what we are talking about here?
Michael Darling
New Member
Mass (or for the folks who have a hard time thinking of mass and weight as two different things, weight) has absolutely nothing to do with the velocity of an object in a true freefall.
A terminal velocity is the maximum velocity an object in a freefall can reach due to the resistance of air on the object.
A vacuum is the only place you can truely have an absolutely pure freefall. Anywhere there is air you will have wind resistance which will vary with both mass and surface area. This has absolutely nothing to do with the Tower of Terror because the wind resistance in such a small system can be considered negligable.
The elevators are pulled downward from a standstill (remember that at the very top of their up/down motion their velocity is zero) with an initial acceleration of about 11.76M/s/s (that's about 1.2G- forgive me if the acceleration is off, that's what I remember figuring out one day when somebody asked me). That's to say that after one second of that acceleration (assuming it's constant, which it isn't on the ride) the car would be moving toward the ground at 11.76M/s.
Yes, when you are moving toward the ground you are initially accelerating toward it... but at the moment you begin to slow you are actually accelerating away from the earth. (consult your nearest physics book for a brief lesson in basic kinematics to fully understand more)
Sorry for being so crass in my first post, but look who I hang around with...
A terminal velocity is the maximum velocity an object in a freefall can reach due to the resistance of air on the object.
A vacuum is the only place you can truely have an absolutely pure freefall. Anywhere there is air you will have wind resistance which will vary with both mass and surface area. This has absolutely nothing to do with the Tower of Terror because the wind resistance in such a small system can be considered negligable.
The elevators are pulled downward from a standstill (remember that at the very top of their up/down motion their velocity is zero) with an initial acceleration of about 11.76M/s/s (that's about 1.2G- forgive me if the acceleration is off, that's what I remember figuring out one day when somebody asked me). That's to say that after one second of that acceleration (assuming it's constant, which it isn't on the ride) the car would be moving toward the ground at 11.76M/s.
Yes, when you are moving toward the ground you are initially accelerating toward it... but at the moment you begin to slow you are actually accelerating away from the earth. (consult your nearest physics book for a brief lesson in basic kinematics to fully understand more)
Sorry for being so crass in my first post, but look who I hang around with...
Mikejakester
Active Member
to find out the speed of the elavator you can use this equation.
v=V(o) + a(t)
V= Final Velocity
V(o)= Incial velocity
a= Aceleration (freefall, {9.81})
t= Time.
let say the elevator is droped for 3 seconds.
v=?
V(o)= 11.76m/s^2 (acording to michael)
a= 9.81
t=3 s.
v= 11.76m/s^2+ (9.81m/s^2)(3s)
the seconds cancel out and we are left with m/s which is velocity
v=41.19 m/s or 92.15 mph
so the speed at the bottom of the tower IF IT WERE TO be DROP WITH NO BREAKS FOR 3 SECONDS. would be 92.15 mph. But I don't think it would reach that. becuase it's starts decelerating around the 2 second mark to go back up. If anyone has more info... or if something is wrong. Please tell me. we learn from our mistakes. :king: :wave:
v=V(o) + a(t)
V= Final Velocity
V(o)= Incial velocity
a= Aceleration (freefall, {9.81})
t= Time.
let say the elevator is droped for 3 seconds.
v=?
V(o)= 11.76m/s^2 (acording to michael)
a= 9.81
t=3 s.
v= 11.76m/s^2+ (9.81m/s^2)(3s)
the seconds cancel out and we are left with m/s which is velocity
v=41.19 m/s or 92.15 mph
so the speed at the bottom of the tower IF IT WERE TO be DROP WITH NO BREAKS FOR 3 SECONDS. would be 92.15 mph. But I don't think it would reach that. becuase it's starts decelerating around the 2 second mark to go back up. If anyone has more info... or if something is wrong. Please tell me. we learn from our mistakes.
:king: :wave:Mikejakester
Active Member
But now that i think about it the fastest ride would be the plane that I take to get to orlando.
Originally posted by Mikejakester
to find out the speed of the elavator you can use this equation.
v=V(o) + a(t)
V= Final Velocity
V(o)= Incial velocity
a= Aceleration (freefall, {9.81})
t= Time.
let say the elevator is droped for 3 seconds.
v=?
V(o)= 11.76m/s^2 (acording to michael)
a= 9.81
t=3 s.
v= 11.76m/s^2+ (9.81m/s^2)(3s)
the seconds cancel out and we are left with m/s which is velocity
v=41.19 m/s or 92.15 mph
so the speed at the bottom of the tower IF IT WERE TO be DROP WITH NO BREAKS FOR 3 SECONDS. would be 92.15 mph. But I don't think it would reach that. becuase it's starts decelerating around the 2 second mark to go back up. If anyone has more info... or if something is wrong. Please tell me. we learn from our mistakes.
aw!!! i love physics
Mikejakester
Active Member
you remind me of my Physics Teacher. :lol: :lol: :lol: :hammer:
WDW John
Member
OK, I seem to remember way back when ToT opened the claim was that you would drop 13 stories in 2 seconds. Like I said it's what I remember from about 10 years ago when there was one drop all the way from the top (of the shaft), but I could be wrong.
If that were true the actual speed of the fall would be more than 92.15 MPH (based on a 3 second drop). However, by using the equation you provided I come up with a SLOWER speed, which is contradictory. You can't move an equal distance in less time at a slower speed (just to clarify, I mean on Earth, at WDW
). What's up with that? I also noticed that your equation says "a= Acceleration (freefall, {9.81})". The problem here is that we aren't in a freefall on ToT, we are being pulled down at a higher rate than gravity alone would provide.
PS: You guys need to lighten up...
Mikejakester
Active Member
What do you mean by lighten up?
i don't seem to understand where you comming from, but I never talked about distance. I just figured out the Velocity. I don't know where you got the distance. But I would love to the see what you did with the calculations BTW. But anyway...
Lets try it with 2 seconds.
You said it your self, the elevator is pulled down by what MICHAEL said at 11. 76 Meters/ Second So thats your inicial Speed.
So you have
Final Velocity=(inicial Velocity) + (aceleration of Gravity)(time)
Final velocity= (11.76) + (9.81)(2)
final Velocity = (11.76) + ( 19.62)
Final Velocity = 31.38 m/s or 70.21 mph
Now you know that the elevator at the bottom of the shaft before it stops can reach a velocity of 70.21 mph with an incial speed of 11.76 m/s or 26.31 mph
In order to find the distance travel during that time you use this equation.
(Final velocity)^2 = (inicial velocity )^2 + 2(gravity) ( distance)
We found out that the final velocity is 31.38 m/s on the problem above. So now we solve for distance.
(31.38)^2=( 11.76)^2 + 2(9.81) (distance)
Solve for distance
984.70= 138.29 + (19.62)(distance)
846.41= (19.62) (distance)
43.14 meters= distance traveled.
Now this is the distance traveled of the elevator. To find out how many stories that is we first have to estimate how tall each story is. From my house I can tell my ceiling is about 10 ft high. Thats about 3 meters tall. so we do this...
Divide 43.14 by 3 meter for each floor gives us
43.14/3= 14.38 Stories high... This number is on and off depending on how tall the stories of the actual building is. But it's more or less. and it's acurate to your 13 stories high.. take away the friction by the wheels and you'll get 13 stories high.
Man I love Physics! Hahahahah
:lol: I really hope that anwsers your question. I really can't see it any other way... Please if you have anymore questions feel free! I love this stuff!
:slurp:
:hammer: :sohappy: :wave: Lighten up pal! :wave:
Velocity= 70.21 MPH
Distance traveled =43.14 meters or 141.54 ft. or aproximatly 13 stories ( 14 .38 stories high)
Time = 2 seconds.
TOT!!!
WDW John
Member
I know you didn't mention distance (earlier) but that is why I was confused. When you figured the drop at 3 seconds the speed was 92.15 MPH. At 2 seconds it comes out to 70.21 MPH. My confusion is that the distance doesn't change, so with my feeble brain I figured that the rate would have to be higher in a 2 second drop than in a 3 second drop. Now that I've thought about it more I understand that the equation doesn't care about distance at all.
I did the same thing you did with the 2 second drop and came up with the 70 MPH figure. Thus, my confusion.
OK, here's my next confusion:
I thought that velocity would be m/s and acceleration would be m/s^2. Why is initial velocity 11.76m/s^2? Isn't that an acceleration? If that is the case would the seconds still cancel out? If the seconds don't cancel out does that change the Final Velocity?
I always have questions. So far this is my last one. At this point are we hitting 70.21 MPH and then slamming into the basement floor? We haven't accounted for braking at all. If the fall truly is 13 floors in 2 seconds that 2 seconds must include braking time, right? If so, how does that affect the final velocity above?
trendymagic
Member
Well I know according to Disney that it is Test Track, but with all this math... ToT wins it for me. In a close second is that car cleared for takeoff on World Drive.
WDW John
Member
OK, after staring at this a little longer I'm figuring that the 11.76m/s^2 is a typo and it's just supposed to be 11.76m/s, right? The seconds that are cancelling out are the a and the t, I think.
That would be just about the same as gravity, but supposedly ToT PULLS you to the ground faster than gravity would.
Also, I'm curious now 'cause I remember the feeling that there may be multiple accelerations, or, we accelerate down at one rate briefly, then the acceleration increases briefly, then again, then brake. Anyone else get that feeling ever?