Fun Physics problems with coasters

coasterbabez, the few posts I've seen have been all but meaningful. One had to do with meshing SOB or something, whatever that is.

...and perhaps some people on this board want to know the answer, and the means to find it. Especially if they want to get a job it this field.

Joe E. said:
"How about a tough Question.

What is the Average Speed of the Millennium Force from point A, located at the top of the lift to point B, located at the first break?

I am willing to bet that the Force WASTE all other (complete circuit) coasters in this category.

To do this we need the distance between B and A and the time it takes the train to travel between A and B. I can talk with Pythagorus and get the length of the lift, but I don't know the length of the brake run and Station, nor the time it takes for the train to get from A to B."



Well, based on the MF QuickTime that CP put out, the ride lasts 1:45 seconds, but that's just before the station. Let's assume that from the end of the brake run to station is 150 feet (guesstimate). The Distance from the base of the lift to the top is roughly 424 feet. This gives a track length from top of the hill to the first brake run at approximately 6021 feet. The time from top of lift to end of brake run on the POV is 1:15.

This yeilds an average running speed of MF of ~54.73 MPH while under the influence of gravity.

Assuming the ride is 1:52 seconds long (looks about that way until the station hits)...it's averaging 40.14 MPH for the entire course, including the slow roll into the unload station.

Jman
NICE JOB. Quite Clever, I never thought of using the POV as a timer. Obviously the only way to figure out it exactly is to have the exact measurements, but ride time's do vary so I would say 55 MPH is a great estimate. That is killer, most coaster's have trouble attaining that as their TOP speed.

If I get around to it, I will do this. Since I have POV's for other coaster's, I will time them and estimate the brake and approach runs. *** This post was edited by Joe E. on 1/15/2001. ***
Although, it does go much faster now since that was way back in April.
Yes, that's for an empty train when it's cold. I'm sure a full train at 4PM in July would probably up the avg. speed to about 56 or 57.

Jman
I take AP bio. Physics make my brain hurt... owww...
I actually love to see these kind of problems. I am not only infatuated with riding coasters, but with the physics and forces behind them. If I could do it all again, knowing what I know now, I would pursue a career in coaster design.

Now, do another problem....and show your work! That is the fun part.
Jeff's avatar
Yes, that original POV of The Force is really slow. I was going to post it on Guide to The Point, but I'm going to wait until I render it as a shorter clip, adjusting for the difference in time.

Anyone know what that time (lift crest to brakes) is these days?

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Jeff
Webmaster/Admin - CoasterBuzz.com
I am really impressed. You can probably help me with something I have wanted to know. Explain the logistics involved in getting the trains on Nessie to go through the inter-locking loops at the same time.
What would be really neat is if we could find the speed after the final overbanked turn into the station. That baby still is probably cooking at least 60.
87mph.., next question please.
Here is Magnum's Average Speed

Time A-B 1:20 seconds (B being right after the final tunnel)
Track Length, 5160
- Lift 346 FT
- Other
APX 4200 FT (Wild estimate, gave it less than it is probably so it may appear faster)

Average Speed, (A-B) APX 31 MPH
Hyper, HA.

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Joe
Beyond Post, Giga Post *** This post was edited by Joe E. on 1/18/2001. ***
Lorrie.
There really is nothing to it assuming you don't take into account the difference in the weight of the passengers wich could slow down or speed a train up slightly.

I have not been on this coaster so i do not know the approximate times or the layout.
You would just need to send one train and use a stop watch when the train goes through the first loop start the timer. Stop the timer when the train goes through the second loop.

The time that is on the stopwatch would be the time in wich you would have to wait after dispatching one train to dispatching another.

On the Orient Express at WOF this is extremely hard to do since the difference in time i would guess to be about 20 seconds at the most. So basically you would have to dispatch a train and you would have twenty seconds to get another into the station and dispatch it. This would be impossible if you had to load people on to it.
Unless you were to give a rewride to the second train.
However I believe the Orient Express does not have a brake in between the loops so you would have to bypass a safety feature to allow two trains to be in the same section of track at the same time. I do not know if Loch Ness Moster is this way or not.

But basically The quick answer is you have to dispatch another train at the same time interval that it takes the train to travel in between the two loops.

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#1 Steel-Incredible Hulk
#1 Wood--Timber Wolf
Loch Ness Monster makes the matching, interlocking loops fairly easy, as the first loop is at the beginning of the ride (after the second drop) and the second loop is at the end of the ride, two blocks later. In fact, I believe the ride is timed out so that if they hit interval AND are running three trains, two trains will match up in the loops every time. Thing is, like most parks, BGW doesn't usually make interval.....

--Dave Althoff, Jr.
Well in that case LNM would beable too but as for Orient Express the two loops are after the second drop and the third drop.

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#1 Steel-Incredible Hulk
#1 Wood--Timber Wolf

smoothncomfy said:
"I have a physics problem for you all and it goes like this: you have a roller coaster with a height of 240 ft. Calculate the speed at the end of the drop if the mass of the train is 2000kg."


I didn't read this whole thread so this might have been stated, but because it is a free fall then the mass has no affect at how fast it falls, what causes there to be varying speed throughout masses, is when it has a slope greater than 0 and less than 90, the reason is a little phisical rule called "horizontal force."
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What is life without geniuses?
Now, now, coastergenius...
If you're gonna get technical, then get technical ACCURATELY. :)
When the velocity of the dropping object is affected only by gravity, the mass cancels out of the equation and therefore does not affect the top speed. The reason that mass makes a difference is because the mass influences the total energy of the system. There are multiple resistive forces which remove energy from the train. Some of these resistive forces are independent of mass, such as wind resistance and gravity, while there are other resistive forces such as friction and trim brakes which ARE related to mass. More mass means more total energy, which means more force is required to slow the train. Since some of the forces are independent of mass and therefore are not increased with the mass, the losses will effectively vary with the mass of the train.

Whew! Does that make sense?

--Dave Althoff, Jr.
the work-energy theory

Initial gravitational Potential Energy=Final Kinetic Energy

PE=mgh KE=1/2mv^2
PE=KE
m=2000kg g=9.8m/s^2 h=240ft=72m
solve for v.

Jman said:



The Distance from the base of the lift to the top is roughly 424 feet.

It's longer than that isn't it, it goes 300 feet above the ground, at a 45 degree ancle isn't that 600 feet right their. and the curve over at the top is very large. 624 sounds better to me.
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Get wrapped in the coils of Viper at SFGAm.
Since it is a 45 degree angle, it makes a perfect right triangle. I know the lift is not a triangle but we are estimating. We use the Pathagereon theorem to do this. Since MF dose not start all the way on the ground, I used 305 feet, assuming that it starts about 5 feet of the ground, which is still pretty low compared to other coasters

305*2 X 305*2 = 186050. Rooted =431.33 feet

When I figured out Magnum lift, I assumed the it was 30 degrees and hence would have to use the SIN or COS functions.

To find the leghth of the Parabola at the top, we would need some crazy measuremts, like the distance between when the 45 degree angle stops on the upside, and when the drop hits 45 degree's

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