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A question: "Lateral force"


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Some people might think this is a nitpicky question, and I suppose it is, but still it's kind of a gray area I've never seen discussed. A lot of people refer to the force you feel during lightning-quick twists like on Maverick and I305 as "lateral force", but is that really what it is? Lateral force is the force that pushes you side to side, like on the switchbacks of a wild mouse. Coasters like the Voyage are designed with some of the turns intentionally under-banked so you feel a combination of lateral and positive Gs. But on Maverick, all of the turns are fully banked so you're not really feeling laterals at all, it's all positive. Is the force of the transitions between the turns really lateral force, or is it something else? Pretend for a moment that you weren't strapped in, during the transitions you wouldn't be pushed to the side of the car, you'd be totally ejected out away from the train, right? So it seems to me like it must be something different. Is it worth making a distinction between the two?

 

A question for physics nerds and overly analytical people like me, I guess.

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It's called a "moment" or "torque", and will be equal to lateral forces with a magnitude proportional to the distance from the axis of rotation. That it, it will be unevenly distributed about your body. In a heartline twist, for example, your head and feet will feel forces in opposite directions, with the strongest forces at the extremes of your body. Your body will also experience 'centrifugal' forces, a reaction to centripetal forces holding you in place (from the restraints or the rest of your body), so you will also experience (mostly vertical, compared to your body) forces outward from the axis of rotation.

 

A lateral force from an unbanked or underbanked turn, like on thunderbolt @ kennywood will produce lateral forces evenly distributed across your body, with no vertical component.

 

Edits: some corrections

Edited by Fooz
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Moment, for engineers, is short for "moment of force", which does mean "torque", but moment in physics is the component of distance from the axis to the perpendicular force. For engineering, they are the same, so it probably wasn't worth it to put that in.

 

I am an engineering student, btw. Lots of this is fresh in my head but if someone reading this is more knowledgable than me please correct me!

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But on Maverick, all of the turns are fully banked so you're not really feeling laterals at all, it's all positive. Is the force of the transitions between the turns really lateral force, or is it something else?

 

 

 

It is rotational force. Wicked Twister is a prime example of this. Anyone who's ever rode it or other Inverted impulses know what I'm talking about.

 

Most of the laterals are eliminated with heart-line transitions, replaced by rotational force (a.k.a. torque: as Fooz explained).

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This topic is interesting to me. If a track were banked 90 degrees and curving at the same time (if I305's first curve was 90 degree banked), would that be completely positive g force? You would be moving in a circular direction, like laterals, but banked and curving in a way that the track only curves upwards, like a loop on its side. What would this do?

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Hmmmmm, if you simplify it reference ground and gravity etc; the concept of lateral, positive, and negative g's should remain the same regardless if you're in a banked or unbanked turn because you're still being pulled in that direction whether you're upside down, right side up, or sideways.

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The force of gravity would still be acting on your body, producing 'lateral' G's on your body in the direction of the ground. Flat turns are banked so that the component of the force of gravity producing lateral g's in one direction are exactly counteracted by the lateral g's produced by the 'unbanked' component of the turn producing traditional laterals in the opposite direction.

 

Turns like I305's flat turns already produce only positive g's with this method.

Edited by Fooz
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This topic is interesting to me. If a track were banked 90 degrees and curving at the same time (if I305's first curve was 90 degree banked), would that be completely positive g force? You would be moving in a circular direction, like laterals, but banked and curving in a way that the track only curves upwards, like a loop on its side. What would this do?

 

 

Sounds similar to this.

www.themeparkreview.com/forum/files/img_5824__copy_.jpg

 

 

About the "loop on it's side".

 

Pay attention to the banking change, to compensate & reduce laterals.

upload.wikimedia.org/wikipedia/commons/7/7e/Chang_(Six_Flags_Kentucky_Kingdom)_02.jpg

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Hmmmmm, if you simplify it reference ground and gravity etc; the concept of lateral, positive, and negative g's should remain the same regardless if you're in a banked or unbanked turn because you're still being pulled in that direction whether you're upside down, right side up, or sideways.

What kind of forces does Outlaw Run's wave turn produce? There is no lateral movement, but gravity pulls you towards the ground, which is your right side, which seems like laterals.

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^Negative or Zero Gs.

 

The momentum is moving vertical, against the force of gravity. It only seems lateral due to it being 90* banked; therefore pushing riders to their side instead of the usual upwards "out of your seat" sensation.

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In the matter of complex shapes like incline loops and overbanks, the changing vertical momentum produces more forces to be balanced. The banking is selected so that the components in lateral directions cancel each other exactly, so that all forces the rider experiences are vertical.

 

For outlaw run's wave turn, the force produced by the changing vertical momentum is designed to equal the force of gravity, so that both of these forces neutralize each other. The remaining force is the lateral "bow" of the hill, which produces a change in lateral momentum, creates (sideways) forces that are oriented vertically along your (sideways) body.

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In the matter of complex shapes like incline loops and overbanks, the changing vertical momentum produces more forces to be balanced. The banking is selected so that the components in lateral directions cancel each other exactly, so that all forces the rider experiences are vertical.

 

For outlaw run's wave turn, the force produced by the changing vertical momentum is designed to equal the force of gravity, so that both of these forces neutralize each other. The remaining force is the lateral "bow" of the hill, which produces a change in lateral momentum, creates (sideways) forces that are oriented vertically along your (sideways) body.

That makes sense, thanks. As for Goliath's stall under the lift, is that positive or negative? The approaching negative g's rotate into positive, but you are also suspended in the air upside down, producing what seems like a combination of the two. (Well, obviously we haven't ridden it, but that's what it seems like it will do.)

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Zero Gravity, hence the name " zero gravity stall".

 

 

The track inverts while still rising, creating positive G's. However, the force of gravity is also at the same time acting against this. For a brief moment, the opposing forces will meet at a near equilibrium, creating a sense of zero gravity.

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Zero Gravity, hence the name " zero gravity stall".

 

 

The track inverts while still rising, creating positive G's. However, the force of gravity is also at the same time acting against this. For a brief moment, the opposing forces will meet at a near equilibrium, creating a sense of zero gravity.

 

Yes, gravity is counteracting the force of the changing vertical momentum (the colloquial 'centrifugal force') presumably at 0 or close to 0 g's.

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It's called a "moment" or "torque", and will be equal to lateral forces with a magnitude proportional to the distance from the axis of rotation. That it, it will be unevenly distributed about your body. In a heartline twist, for example, your head and feet will feel forces in opposite directions, with the strongest forces at the extremes of your body. Your body will also experience 'centrifugal' forces, a reaction to centripetal forces holding you in place (from the restraints or the rest of your body), so you will also experience (mostly vertical, compared to your body) forces outward from the axis of rotation.

 

A lateral force from an unbanked or underbanked turn, like on thunderbolt @ kennywood will produce lateral forces evenly distributed across your body, with no vertical component.

 

Edits: some corrections

 

Thanks for the answer, this makes a lot of sense. So they really are two different things, because the way they're experienced is completely different. Your body in a heartline twist is actually trying to go in two directions at once... that's kind of crazy to think about.

 

For the record, it doesn't bother me if people want to call it "lateral force", but I do think it's important that people are aware there's a difference. Knowledge rocks.

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^What you're referring to is called a red out: http://en.wikipedia.org/wiki/Redout. It is the opposite of the gray out or black out that comes with positive Gs. This article explains how well people can tolerate G forces: http://en.wikipedia.org/wiki/G-force#Vertical_axis_g-force. Basically, we are much less suited to tolerate negative Gs than positive, which is why you see the most intense rides hitting 4.5+ positive Gs (some actually exceed 6 briefly), whereas the strongest negative ones are about -1.5 (I believe, certainly no higher than -2).

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A lateral force from an unbanked or underbanked turn, like on thunderbolt @ kennywood will produce lateral forces evenly distributed across your body, with no vertical component.

 

Ahhhh my favorite the Kennywood Thunderbolt. Just the thought makes me warm and fuzzy inside. Not only are these excellent lateral forces (including on the last turnaround, when the ride is running extra fast!), but you get to enjoy them in a classic manner; no seat dividers and it's crush time for your riding partner!

 

Legend at Holiday World also has great lateral forces. You're encumbered a bit with the seat divider, but still a favorite of mine.

 

And you can even enjoy pseudo-classic lateral time on a classic steel coaster such as Mind Bender at SFOG. The individual lap bars sort of have the seat division built in, but not enough to keep you from sliding over that hard plastic foam seat and squishing your riding partner! Coaster heaven right there! Coaster perfection...

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Am I the only one here who feels like they experience negative G's on the transitions between Maverick's first few turns? It's almost in the same sense as a barrel roll. Maybe I've kept my body still while riding it, but I never felt as much of a lateral shift during those transitions as I've felt actual airtime.

 

 

Edit: And apparently not...that was explained on the last page

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^What you're referring to is called a red out: http://en.wikipedia.org/wiki/Redout. It is the opposite of the gray out or black out that comes with positive Gs. This article explains how well people can tolerate G forces: http://en.wikipedia.org/wiki/G-force#Vertical_axis_g-force. Basically, we are much less suited to tolerate negative Gs than positive, which is why you see the most intense rides hitting 4.5+ positive Gs (some actually exceed 6 briefly), whereas the strongest negative ones are about -1.5 (I believe, certainly no higher than -2).

 

1 G is normal gravity, 0 G is floating. "Extreme ejector air" is still lower magnitude than -1, probably less than -0.5 . Limits are mainly restraint comfort, and the physics that limit total airtime when not inverted. Rides that do hang you upside down never increase forces substantially above that of gravity.

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I've been wondering about this, specifically how orientation should affect calculated forces.

 

As an example, a barrel roll taken at a slow speed. (Think Tempesto, or Hydra) As you go through the inversion you're always experiencing 1 G but the angle changes, so you'll take 1 G of lateral force when you're body is parallel to the ground. When you're upside down you're experiencing -1 G.

 

So then do we consider a ride that just holds you upside down to be providing negative Gs?

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I've been wondering about this, specifically how orientation should affect calculated forces.

 

As an example, a barrel roll taken at a slow speed. (Think Tempesto, or Hydra) As you go through the inversion you're always experiencing 1 G but the angle changes, so you'll take 1 G of lateral force when you're body is parallel to the ground. When you're upside down you're experiencing -1 G.

 

So then do we consider a ride that just holds you upside down to be providing negative Gs?

 

It took me a few moment to understand your question, but it basically comes down to your choice of frame of reference.

 

For example, take a (slow) barrel roll and an airtime hill.

 

While both elements provide negative G from the rides point of view, only the airtime hill provides negative G (or reduced positive) with respect to the earth. Apart from the rotational forces and velocity, your body shouldn't be able to tell the difference between the barrel roll and an airtime hill providing -1 G.

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