X- The Fastest Coaster with Inversions
DrachenFire04 said:
Yah that is true, if all conditions are the same a train going down a 60 degree and 20 degree drop will have the same final speed
Actually, I would disagree.
The Normal Force of a train going down a 60 degree slope is less than the Normal Force of a train going down a 20 degree slope. Thus, the smaller Normal force * the same coefficient of friction results in a smaller loss of energy. . . therefore, more energy goes into kinetic energy. The speed is faster on a steeper drop.
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~~~ Maddy ~~~
*** This post was edited by Chernabog on 11/6/2002. ***
Chernabog: the equation you're using for friction applies to one thing sliding on top of another (like a wood block on a desk). It does not apply to something rolling or to wind resistance. I know wind resistance is proportional to speed, but I'm not really sure about rolling.
A steeper slope will result in a faster final velocity because the train travels less distance to get down the hill. If the force of friction is constant or directly proportional to the speed (I still can't figure out which it is) then *the integral of the force of friction over the distance travelled* (i.e. loss of energy) will be greater for a shallow slope in both cases.
If the friction for a rolling wheel is proportional to the normal force than you can use Chernabog's formula to determine the force of friction (which will be smaller for a steeper slope) and intergrate it over the track length (which will also be smaller for a steeper slope.
Yes, I'm an engineer.
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ApolloAndy said:
Chernabog: the equation you're using for friction applies to one thing sliding on top of another (like a wood block on a desk). It does not apply to something rolling or to wind resistance. I know wind resistance is proportional to speed, but I'm not really sure about rolling.
A steeper slope will result in a faster final velocity because the train travels less distance to get down the hill. If the force of friction is constant or directly proportional to the speed (I still can't figure out which it is) then *the integral of the force of friction over the distance travelled* (i.e. loss of energy) will be greater for a shallow slope in both cases.
If the friction for a rolling wheel is proportional to the normal force than you can use Chernabog's formula to determine the force of friction (which will be smaller for a steeper slope) and intergrate it over the track length (which will also be smaller for a steeper slope.
Yes, I'm an engineer.
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Better than I could ever hope to put it. For sanity's sake, I sometimes just assume all of the losses in a system occur in one place. In this case, I imagine there to be no friction between the wheels and the track - and all the friction occurs as the shaft slides against it's bearings. So, its similar in concept, but I'd bet that fun rotational things screw around with the math a bit.
I was just about to edit my post to include the information about distance, even, because I neglected to explain how a 'force' of friction could immediately result in an energy loss. Thanks for the better explanation.
I'm an engineer in training.
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~~~ Maddy ~~~
*** This post was edited by Chernabog on 11/6/2002. ***
*** This post was edited by Chernabog on 11/6/2002. ***
And a feather and a penny should hit the ground at the same time!
Wrong!
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Charles Nungester.
Is it about coasters or friends? I say both!
I thought that if an object was rolling then you could still apply the equation F=(mu)*(normal force). Don't they have coefficients of friction for rolling objects? Of course, when you think about it, F=ma, so therefore ma=(mu)*(normal force) and normal force would be a function of mass for a roller coaster train. Thus two objects that have the same coefficient of friction but different weights, end up having virtually the same deceleration. However, other things such as wind resistance do play a vital role in weight, and momentum can play a part in how the train acts after it drops down the first hill. I guess the equation you could use for air-resistance is F=1/2*(Drag Coefficent)*(rho)*(velocity)^2*area, then plug in your F=ma, solve for acceleration and thus different masses have different decelerations due to air resistance.
Then again, its midnight and I'm tired so who knows what I'm saying! Anyway, another engineering student here =)
Charles Nungester said:
And a feather and a penny should hit the ground at the same time!
Wrong!
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Charles Nungester.
Is it about coasters or friends? I say both!
In a perfect vacuum they do, but basically this statement points out clearly what the differences are between theory and the real world.
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Werner Stengel: Ich sprich, denke und traüme Achterbahn!
I guess I shoud've put a disclamer on my post about it all being theoretical. Chernabog, I find your comment insulting and unecessary.
I am not able to calculate the friction and wind resistance as I do not know the properties of Intamin's wheel assemblies or the drag coefficient of their trains plus any other additional factors. In a perfect vacuum and in freefall, two objects dropped from the same height will always hit the ground simultaneously. Real world situtaion...a trainful of hefty ACE members will not go significanly faster than a trainful of horse racing jockeys. The added mass does help overcome some of the wind resistence and affects friction. I'm tired of seeing people beleive that a train that is twice as heavy will go twice as fast.
-Seth
ucdaap42 said:
I guess I shoud've put a disclamer on my post about it all being theoretical. Chernabog, I find your comment insulting and unecessary.
I am not able to calculate the friction and wind resistance as I do not know the properties of Intamin's wheel assemblies or the drag coefficient of their trains plus any other additional factors. In a perfect vacuum and in freefall, two objects dropped from the same height will always hit the ground simultaneously. Real world situtaion...a trainful of hefty ACE members will not go significanly faster than a trainful of horse racing jockeys. The added mass does help overcome some of the wind resistence and affects friction. I'm tired of seeing people beleive that a train that is twice as heavy will go twice as fast.
-Seth
Fair enough. My first post was a bit curt.
I mentor robotics teams in my spare time, and one of the most prevalent errors I see among students on these teams is trying to apply elementary physics to their siuations. ...stuff like saying that traction is directly related to weight, or similarly, that pushing an 800 lbs. object can be equated to exerting 800 lbs. of force.
So, I tend to get frustrated a bit more quickly than most when it comes to people making really bad, generalizing statements about the behavior of objects in the real world.
I never said the differences would be significant, but there *will* be differences. Like you, I wouldn't even hazard a guess as to the actual losses in a coasters chassis - so I just assume it all to be in one place, and I don't even care to guess what that number is. I think that because riders comprise a small proportion of the weight of a roller coaster train, it's safe to say that they wouldn't have quite a large effect. But, to hazard a guess and say a B&M Hypercoaster train is heavier than an Arrow hyper train - the weight difference may be significant enough to notice a difference (engineering differences and resulting efficiencies aside).
Again, I apologize for being nasty :)
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~~~ Maddy ~~~
Derrick Whitsett said:
Mr. Freeze or Batman&Robin:The Chiller are the fastest coasterwith inversions. They both of them have inversion and reach around 7omph
Did you read the whole thread. X goes 76 or should go 76. SOB goes almost 78.
Charles Nungester said:
The trains turned out heavier than anticipated. It has been clocked at 82mph
Design speed was for 78mph (Main drop)
Euhm...
Colossos at Heide-Park, Germany has had a top speed of 89.5 mph before the trims on the first drop were added !!!
It was designed to go 75 mph during normal conditions, but when the track was wet and the temperature was good it hit 80, 85 and even 89.5 mph !!!That ride has a 170 feet tall drop. Son of Beast has a 215 ft. drop I think and only goes 82 ? :s
Well, Colossos does have another track design and has other train,s but 45 feet height difference and the lowest goes 7.5 mph faster....
170' can hit 89.5 MPH? I'm finding that hard to believe, even in the ideal situation. Yeah. Someone correct me if I'm wrong, but even in the perfectly ideal situation, you're not going to get faster than 73 MPH from a 170' drop.
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You must be this dumb to ride Viper. -SFGAdv.
Belgian, Not making a slam or nothing, Just a suggestion.
Please research a little more before making such a far off statement. A good place to do it would be www.rcdb.com Type in Collosos in the search and it has all the design stats.
Thank you.
Chuck, who can readily agree that coasters can increase a mph or two for various reasons, The only way they are going to change 10+mph. is with a rocket on the tail.
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Charles Nungester.
Is it about coasters or friends? I say both!
Perhaps you're confusing Colossos with Colossus? Monstrous wooden coaster vs. delicate Intamin looper?
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*** This post was edited by Chernabog on 11/7/2002. ***
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