# Coaster Equations

Saturday, March 30, 2002 1:53 PM

Does anyone know where I can find a list of equations or something to find out the G force of a launch, the G force of a certain curve, etc. I am just completely bored and I was just wondering.

Glenn

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Rollercoasters are my life, I think about them more than I think about....wait, thats all I think about :)

Proud Owner of www.coasterkingdom.cjb.net

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Saturday, March 30, 2002 4:20 PM

I was waiting to see if anyone else posted but you can find formula's in David Bennett's book Roller Cosater. In it he interveiws Swartzkopf (sp?) and he provides some formulas. Good luck figuring them out though!

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It's his turn to feast, when you ride the Son of Beast.

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Saturday, March 30, 2002 4:40 PM
You'd have to be moderately good at mechanics, but you'd probably have to leave out some variables that would be too hard to calculate.

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Ride the New England Bush!

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Saturday, March 30, 2002 4:49 PM

G forces and end speed on a drop are actually pretty easy to find out, the formulas are simple. To calculate wind drag, friction of the track, etc.... more complex things, forget it, you need some more advanced forumulas. Unfotunately, I can't remember these formulas right now, but I did them in high school physics, so like I said... pretty simple stuff.

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Saturday, March 30, 2002 6:07 PM
The trick is, you have to understand mechanically what it is that is going on, and what you want to find out.
Definitions:
Force: The product of mass and acceleration.

1G: A force equal to 32 pounds per slug of mass, or 9.8 Newtons per kilogram. A force of 1G will accelerate any mass at 32 ft/sec/sec or 9.8 m/sec/sec.

Acceleration: The time rate of change of velocity.

Velocity: The time rate of change of displacement.

By doing successive integrations of acceleration you can express formulas for velocity and distance with respect to time under constant acceleration (which is what gravity gives you):
a = 32 ft/sec/sec.
v = at + v0 = 32t + v0 ft/sec
d = (1/2)at^2 + v0t + d0 = 16t^2 + v0t + d0 feet. (t=time in seconds)

If you know the acceleration in feet/sec/sec, you can divide by 32 to get the value in Gs.
For circular motion, acceleration is, if I remember correctly, equal to the square of the tangential velocity divided by the curve radius. If you know the angular velocity, then the tangential velocity is equal to the circumference multiplied by the rotation rate.

Make sense?

--Dave Althoff, Jr.
(edited for format)

*** This post was edited by RideMan on 4/1/2002. ***

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Saturday, March 30, 2002 6:46 PM
Are G's units of accel. or force?

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Sunday, March 31, 2002 1:48 AM
both
like if a ride has a launch with 2 Gs acceleration, and you weigh 120 pounds... then it would feel like a 240 pound weight was pushing you against the back of the seat
im not like a physics person or anything so im just guessing
i dunno if this is the right formula but its the one i use (for acceleration):
g = [v0 mph(5280/3600)]/(32.15(t sec))
g = (v0 mph*88)/[1929(t sec)]
i probably got it wrong though

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"It was sure nice of Mr. Krabs to give me a job here!" "And at \$50 an hour too! When I started, i had to pay Mr. Krabs \$100 an hour!"

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Sunday, March 31, 2002 6:41 AM

I was trying to figure this out too cause i love trying to design coasters. In a physics book I saw that the formula for centrepital force is velocity squared then divided by the radius of the curve (v^2 /r). I posted a topic about this a while back and only got one post though, http://www.coasterbuzz.com/forums/thread.asp?ForumID=11&TopicID=14282

As for accelaration its a lot easier. Gravitational Acceleration is 9.81 m/s/s (32.185 ft/s/s, or about 22mph faster each second) so for example a coaster launching from 0-66mph in 6 seconds it would have one half a G in acceleration. It would take one second to get to 11 mph which is half of 22mph.

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Millennium Force......It has its ups and downs.

I have 32 coasters in my track record.....I'm so proud of myself.

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Sunday, March 31, 2002 10:50 AM

Dave has the basic equations correct, but you need to correct all of them by adding the 1 G of gravity. This depends on the angle of the track to deternine the effect. For example, if you have a pull out at the bottom of a hill which has a radius that gives 3 G's of centripital acceleration just as the train becomes horizontal, you have the add the G of actual gravity to get a total acceleration of 4 Gs. Of cousrse you don't need to add that G of force to the force on the riders when the train is going straight down, since the train is accelerating along that axis. Then to make it really complicated, you have to include the fact that the train is constantly accelerating when going down and decelerating when going up, so the speed is only near constant when going around a level turn. And then if you are doing real world designing, you have to add in the effects of mechanical friction and aerodynamic drag. This is why modern coasters are designed on computer programs that are guarded pretty closely by the designers.

*** This post was edited by Jim Fisher on 3/31/2002. ***

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Monday, April 1, 2002 9:00 AM

ApolloAndy said:
"Are G's units of accel. or force?"

Wackokid has the right idea in saying that it is both. See, there is a special problem involved when you are dealing with forces. Force is defined by Newton's Second Law as the product of force and acceleration:

F = m * a

Think of it in terms of the units we use. Masses refer to how much stuff you have, and are measured in "slugs" or "kilograms". In fact, the Kilogram is the one SI unit which is still, to this very day, represented by a platinum-iridium slug kept in a vacuum jar in France. (All of the other SI units have been redefined in terms of various physical constants)

Forces refer to how much a thing weighs, or how hard you push on that thing, and as such forces are typically measured in pounds or in Newtons.

The inconvenient thing about all of this is that when we're talking about an amusement device, we have no real knowledge of what the mass is going to be, and because we don't know what the mass is, we don't know exactly what the force is. That's pretty darned inconvenient. The nice thing is that in Newton's Law of Universal Gravitation, (F = G((m1m2)/r)) the attractive force of gravity scales in a linear fashion with mass. So gravity supplies a force which is always proportional to mass (in fact, the force is 32 pounds per slug, or 9.8 Newtons per kilogram). That means that if we exert an unknown force which will produce a known acceleration, we can easily compare that force to gravity, because we know gravity scales! A "G" is a unit of COMPARISON. Since on amusement rides the mass is almost always held constant, as a practical matter, we're almost always measuring accelerations.

Now to get back to the hole that Jim Fisher fell into up there (undoubtedly accelerating at 1G until he hit the bottom), I should point out that he's right about adding 1G of normal Earth gravity to acceleration calculations. But one place where he missed is that spot where he talks about the train accelerating when it's going down and decellerating when it is going up. I disagree! In both cases, the train is accelerating, and it is accelerating in a direction towards the ground! When the train is pointing in the direction of the ground, it goes faster; when it is pointing away from the ground it goes slower...but the acceleration is the same either way: It's 1G downward! That's what all that Physics class stuff about changing from kinetic to potential energy is all about. If the coaster had no losses, you could easily determine the train speed based solely on its height anywhere on the course. Given an arbitrary height H in feet and a maximum height in feet Hmax, the train's speed at any point on the course would be:

V[mph] = 60/11 * sqr(Hmax - H)

Of course because of losses, it doesn't work out that nicely, but the point is, gravity is doing the work. Gravity is accelerating the train down the hill, and gravity slows it again as it comes up the hill. The force is the same at all points on the course. The effect of the force may vary because of track angle, but the force applied remains constant.

--Dave Althoff, Jr.

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