Disney is advertising a new ride - the hollywood Tower of Terror, or some such thing.
The thing, it seems includes a thirteen-story free fall. If the star is near-instantaneous, as is the stop, what are the G-forces endured by the forces, and wouldn’t it be physically dangerous to endure them ?
I’ve ridden Disney World’s version (one of the best ride ever made, imo) and although the ride does both begin dropping and cease stopping very quickly, it’s really quite gentle (e.g. you won’t get banged up). As for the Gs endured, if I recall correctly, it’s about 1.3, nothing close to being “physically dangerous”.
I can’t wait to try out DisneyLand’s version, I’ll be down there in just two weeks.
stopping=dropping
Well, everything except the stop is at 1 G. That’s what free-fall means.
Thing is, it’s not actually a free fall. The elevator is actually being pulled down, thus increasing forces.
I know that free fall is 1G, the high Gs happen on start and finish, with finish being far bigger, and causing more damage. Which made me quiet surprised to see a 1.3 G force.
No, no, no. I’m quite certain you’re wrong. Standing still is 1G, by definition of “G.” Freefall (if it is truly freefall, which DuderDude2 says it isn’t) is 0G. You don’t feel any forces while you’re in freefall. However, http://www.ultimaterollercoaster.com/thrillrides/towerterror/ says that it is in freefall for 6.5 seconds, and pulls 4.5 G’s.
It’s possible my 1.3 statistic is incorrect, but I know for certain the elevator does not reach speeds of “100 mph”, as the source you listed denoted. I’ve read that the top speed is 43mph (sorry no cite, going from memory). Nor did the ride cost a mere “16 million”, it was in excessive of “100 million”.
Ah-ha. http://tower-of-terror.com/facts.html
The 1.3 stat is about halfway down. However, I advice anyone who has not ridden the ride to not check out the side, as spoilers are abound (and it’d truely be a shame to spoil this attraction).
That doesn’t mean anything. If I drop a ball it starts moving instantaneously and as others correctly said will have a net force of 0 a it accelerates downward.
If you said that a ride reached a specific velocity near-instantaneously that would be a different matte.r
To be entirely pedagogical, a net force of 0 means it is not accelerating at all. I’m not really sure what you’re trying to say here.
Stopping ‘neary instantaneously’ from freefall is pavement-pizza, as in what happens when you jump off 13th story. Presumably they brake you at the bottom, and it’s impossible to measure the G force without knowing how long that takes.
This isn’t correct - when in freefall, your body is experiencing a downward force due to gravity of 9.8 newtons per kilogram of body mass (we’ll ignore air resistance). If it didn’t, you’d fall at a perfectly constant speed rather than accelerating at 9.8 m/s[sup]2[/sup]. Your body doesn’t feel any resistive force from the ground any more, but that doesn’t mean gravity has switched off.
Sorry, just to check, you mean ‘feel no forces’ is incorrect, not 0G is incorrect? In a way ‘feel no forces’ IS correct: gravity certainly is acting, but you can’t FEEL it (except for the rushing air).
No, I mean 0G is incorrect, almost by definition. You are accelerating downwards at 1G and must therefore be experiencing the commensurate force. It is true to say that we don’t experience the effects of that force directly, as we do when standing on the ground, but this doesn’t mean it isn’t acting.
When you’re standing on the ground, the earth is pulling you down with a force of 1G and the ground pushes up on the soles of your feet at 1G. The net force is zero Gs, so you don’t accelerate.
In free fall, the earth pulling you down with a force of 1G*, and there is no opposing force, so you accelerate toward the ground.
I think everybody understands that, but is expressing it differently. When people say “Freefall is 0 G,” they are referring to that lack of an opposing force on the soles of your feet (or butt, or wherever). They don’t mean that gravity is no longer pulling.
*And yes, you are also pulling the earth up to you with the same amount of force.
Suppose you’re standing on a platform ‘rising’ at 10m/s in deep space. You feel 1G. If you drop a ball, it ‘falls’ to the platform as if you were on earth. If it accelerates more, you’re ‘pulled’ to the platform more, your knees are more likely to buckle, balls ‘fall’ faster. This is, say, 2G. If it’s not accelerating, you feel 0G. Balls stay when you let go of them. Your knees are under no strain at all.
If you’re in free fall, it feels exactly like 0G above. If you’re on the ground, it feels like 1G above. If you’re on a heavier planet, or a lift going up, it feels like 2G. Hence the terminology.
Certainly whenever you’re on the earth you have gravity acting on you, but “g force” I thought normally refered to the relative force, ie. how crushed you were getting instead.
No, you would feel zero G. The platform is moving at a constant velocity of 10 m/s. You are moving at a constant velocity of 10 m/s. The ball is moving at a constant velocity of 10 m/s. Nothing is being accelerated.
To answer the other half of this question: Is it dangerous to pull Gs like this?
According to a study funded by the army to determine the forces that could be endured while piloting the UH-60 Black Hawk and the AH-64 Apache through acceleration and deceleration, A force of 5 Gs applied for 2 to 3 seconds is usually harmless, but the same force applied for 5 to 6 seconds can cause blackout or unconsciousness. Forces experienced by test pilots in the 60s and 70s experienced forces of about 15Gs when ejecting from planes. During the .2 second duration, the effect was also harmless, but if it was applied for 2 seconds, unconsciousness would occur. The force of a crash landing can reach 40 Gs in a fraction of a second, and you can still walk away from it. Needless to say, such a force sustained over 2-3 seconds would be fatal. As for body position, the body doesn’t tolerate high G forces as well in the Anterior-Posterior (Head to Foot) axis as well as it does along the Dorsal-Ventral (Front to Back) axis due to the increase in arterial pressure along the A-P axis. Hope this helps.
If you substitute “9.8 m/s^2” for the “10m/s”, the poster’s “platform-in-space” analogy is correct - I have a feeling this was a simple oversight. So, to re-state, with terms correct (instead of just gainsaying a typo and clouding the waters):
*Suppose you’re standing on a platform ‘rising’ at 9.8m/s^2 in deep space. You feel 1G. If you drop a ball, it ‘falls’ to the platform as if you were on earth. If it accelerates more, you’re ‘pulled’ to the platform more, your knees are more likely to buckle, balls ‘fall’ faster. This is, say, 2G. If it’s not accelerating, you feel 0G. Balls stay when you let go of them. Your knees are under no strain at all.
If you’re in free fall, it feels exactly like 0G above. If you’re on the ground, it feels like 1G above. If you’re on a heavier planet, or a lift going up, it feels like 2G. Hence the terminology.
Certainly whenever you’re on the earth you have gravity acting on you, but “g force” I thought normally refered to the relative force, ie. how crushed you were getting instead.*
So, Dead Badger - even though you’re always affected by earth’s gravity when you’re in it (gravity is never revoked), the discussion revolves around how many G’s of acceleration (times your mass) are transmitted through your feet as force. When you’re on flat, non-moving ground, you experience 1G * m, which is your “weight”; when you’re in the Tower of Terror, the ground is accelerating at the same rate you are. So even though you’re accelerating at 1G, the force you experience through your feet is the difference in accelerations between you and your frame of reference (the floor, 0 m/s^2) times your mass (zero, total), so you feel “weightless” and experience (from your perspective, in the Tower frame of reference) zero G.