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Thread: Rollercoaster problem

  1. #1 Rollercoaster problem 
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    Hello Guys,
    I'm doing a research for my graduation of a rollercoaster.
    They want to replace the electric powered launch to pneumatics, hydraulics or steam.
    And my assignment is to investigate which is the best alternative to use in new rollercoasters.
    I have a problem with the calculation of a launch of a rollercoaster train.
    I have to calculate the acceleration and force needed to launch a rollercoaster train.
    The train will launch from 0-100 km/h in 3 seconds.
    The tip I got from my teacher was I had to look to the impulse theorum.
    Also they want the train gets the launch over a hill instead of a slow constant velocity.
    Can somebody help me? I can't find a similar problem anywhere on the net or anything.

    The data:
    Mass train: 22720 kilograms
    Height Hill: 35 meters
    Alpha hill: 30 degrees

    Thanks!

    greetz Lesly Spronck
    Graduate student Mechanical Engineering
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  2. #2  
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    Hi Lexus, welcome to the forum. Are you able to attach a diagram? I am struggling to understand what is required.
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  3. #3  
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    Hello Jilan,
    I was trying to attach a diagram but the link is stuck when I want to upload the image.
    Do you have mail then I will send it to you via e-mail.

    greetz
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  4. #4  
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    Sorry, no not a public one. What is the alpha hill?
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  5. #5  
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    The angle of the hill what the train has to overcome.
    You have to imagine the train will be launched like a lift hill but then with a higher speed.
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  6. #6  
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    Lexus, are you allowed to launch it on the flat. I would assume that it would be flat so the people could get on?

    Edit, Lexus if you can't and you have to accelerate up the hill it doesn't matter as the impulse theorum gets you there pretty easily. Of course I cannot tell you how to do it as it's your project. But I can help you get there. How far have you got so far?
    Last edited by Jilan; 08-04-2014 at 08:19 PM.
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  7. #7  
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    Yes but that is what I mean.
    The people get on a flat platform and then the train will go a little forward on the beginning of the hill and there will the train be launched.

    Well the impulse theorum stands J = m x v.
    M stands for mass and v for velocity.
    The velocity will be in m/s so the velocity is 27.78 m/s at the end of the launch.
    J= 22720 x 27.78
    J= 631161.6 kg dot m/s

    The force needed to create this impuls is the next formula.
    F= 1/delta t x J
    delta t is the time needed to overcome the impuls
    F= 1/3 x 631161.6
    F=210387.2 N

    Is this correct? My teacher gave me the tip I have to look on the principle of taking air out of a balloon because it's pneumatics and this is kind of the same principle.
    With pneumatics you fill a buffer and then you release all of the air out of the buffer.
    My problem is how do I solve this and calculate the buffer needed to accelerate my train?
    I really want to explain it to you properly so I can be helped, I'm really stuck.
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  8. #8  
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    Hi, I see you are given v, I was trying to calculate it! Yes it looks like that should be the force, it pretty big isn't it ? That going to be one big balloon. Do you know any equations associated with pneumatics?
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  9. #9  
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    Yes that's true, but these installations cost in the million euros.
    In case there's a hydraulic system of the Kingda Ka that costs 5 million euros so this force can be this big.
    But according to you this is correct what I have calculated?

    Well I have searched for formulas for calculating the volume of the compressor and this is:
    Vc= (Vb x fmax x (pU - PL) x Tcimax)/0.25 x Tbi x pci)

    Vb=buffercapacity(l)
    Vc=compressorcapacity(l/s)
    pci=pressure intake compressor(bar)
    Ticimax=maximale intake temperature of the compressor(K)
    Tbi=temperature of the accumulated air in the buffer(K)
    (p_U-p_L )=pressure difference between in use and not in use(bar)
    fmax=maximum frequency (1 cycle every 30 seconds)

    Before I can do this I need the volume of the buffer.
    To get this I need F=p x A
    I want the pressure to be 20 bar.
    This comes to a surface of:
    210387.2=20x10^5 x A

    A=0.1052 m^2
    With this A I can calculate the volume.

    A= pi x D^2
    0.1052 = pi x D^2

    D= 0.366 m

    With this diameter I can calculate the volume of the buffer right?
    I take a height of 40 meters for example and this comes to a volume of:
    V=A x H
    V = pi/4 x D^2 x 40
    V= 4.207 m^3 = 4207 l

    Is this all correct or am I missing something?
    I think the height of 40 meters is very much to my opinion.
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  10. #10  
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    It looks like you are making excellent progress with this project. So far I follow everything you have done. Just a couple of questions at this stage I have.
    A) does the mass of the train include the mass of the people?

    B) is there a particular reason you have settled on 20 bar?

    C) how long will the piston need to be to stay in contact with the train for 3 seconds?

    I have never designed a roller coaster before.
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  11. #11  
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    A) Yes the mass is included. I have 20 passengers with each a mass of 146 kilograms.
    This mass is a standard in designing these rollercoasters.
    B) No, I have to pick a pressure so I don't know if this pressure is big enough to accelerate such kind of mass with this velocity.
    C) The piston need to stay in contact till the train is over the hill and the distance of the hill is 50 meters. So the piston needs to be about 40 to 45 meters.

    But my question is do I have the calculations correct so far according to you?
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  12. #12  
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    OK, I think you need to look into how far the piston moves in 3 seconds to get the length it needs to be. Then perhaps consider the area it needs to be so it doesn't snap, (Look up the tensile strength of steel). That will then tell you what the pressure might need to be.
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  13. #13  
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    I think your calculations look fine by the way.
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  14. #14  
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    OK, so how do I do that? With sigma = F/A?
    So when say the length is 40 meters, I have to multiply this with the diameter and so I have the A?
    And then divide the force with this A?
    If this is lower than the tensile strength the installation complies?
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  15. #15  
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    Yes, there is that element to consider, but A is the cross sectional area of the piston. I was more thinking of the buckling that might occur in a long piston. I should have said the elasticity rather than the tensile strength I have just discovered.
    Buckling - Wikipedia, the free encyclopedia
    I don't think the piston would need to extend as far as 40m if it only has to supply the impulse for 3s.
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  16. #16  
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    Sorry for the late reply. Ok I'm gonna try that, I went to the teacher and it could be solved with impuls but also an easier way with dynamics.
    But the rest of the steps should be the same. The distance it needs to overcome for the acceleration in 3 seconds is 42 meters so this must be the minimum right?
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  17. #17  
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    Applying Fxd = mgh I get 37m so 42m gives you some headroom.
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  18. #18  
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    When I calculate your formula I get:
    F x d = mgh
    210387.2 x 0.366 = 22720 x 9.81 x h
    H= 0.345
    This can't be right.
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  19. #19  
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    We already have H, it's 35m. It's D we are calculating.
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  20. #20  
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    Allright then it's correct.
    But is this D the diameter of the cilinder or is this the length of it?
    So if this is 37 meters the impulse has been overcome?
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  21. #21  
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    The length of it.
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  22. #22  
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    Allright.
    I've been to my teacher and I can't solve this problem with the impulse theorum.
    This will be an differential problem and I don't have had this in my class.
    Now he has an other way to solve this problem.
    With the energy theorum:
    Mass = 22720 kgs
    G=9.81 m/s^2
    Height = 22.5 meters
    Velocity at the end = 100 km/hour = 27.78 m/s

    The kinetic energy will be:
    Ekin = 1/2 x m x v^2=
    Ekin= 1/2 x 22720 x 27.78 = 8766834.6 J

    The potential energy will be:
    Epot= m x g x h
    Epot= 22720 x 9.81 x 22.5 = 5014872 J

    Total energy = 13781706.6 J
    The power that is needed to get the acceleration with 3 seconds:
    P= E/t
    P= 13781706.6 / 3 = 4593 kW.

    This power isn't this way too much?
    Or can I use this energy in the formula of the energy in the formula of the energycapacity and then the diameter of the cilinder getting out of this?
    The formula E= 1e5 x p x pie x D^2 x H (ln(p-1)/p)?
    I found this formula Opslag van Energie.
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  23. #23  
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    LOL, is this roller coaster not built yet? The power looks fine to me. It is pretty big, but it's only for a few seconds. (Out factory draws this level 24/7 from the grid). This would mean the force would need to be 13781707/42 = 328kN. This is 50% higher than you arrived at on post #7. Didn't you already have a good equation for the force of a compressor?
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  24. #24  
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    No I was busy doing the research of the different alternatives and the essay.
    Now I have to calculate the force and power to make a choice out of these alternatives.
    Yes that's correct but the teacher said I have to calculate this way because this principle is different than a simple dynamic problem.
    This is sort of like emptying a ballon and this force will be lower when the buffer deflate.
    So this 328 kN I have to use to calculate the cilinder and compressor like the posts before?
    The 4593 kW is not the power for the compressor?
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  25. #25  
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    Ow and by the way you derive the energy and the distance?
    But the formule is P= F x s right?
    So I have to derive the power with the distance right?
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  26. #26  
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    Excuse me.
    Isn't it P = F x v?
    So when you need F you need P / v?
    The force = 4593/ 27.78 = 165.3 kN?
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  27. #27  
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    The energy is the force multiplied by the distance. I took the energy from your post #22 and the distance from your post #16. You cannot use the 27.78 as this is a final velocity (not a constant).
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  28. #28  
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    Allright.
    And so this is the force I need to calculate the compressor like before?
    Do you have a reference where this formula coming from?
    I need the sources where I get it from.
    Do you also have experience in steam calculations?
    I also need to calculate the steam installation and I have little experience with this.
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  29. #29  
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    Lexus, the energy of one joule is the work done by a force of one Newton moving through a distance of one metre.
    Work (physics) - Wikipedia, the free encyclopedia
    I have no experience with steam calculations. Perhaps this site might be useful for the calculations involved.
    Engineering Calculator | TLV
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  30. #30  
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    Ok so concluded this force from the work is needed for the compressor?
    These calculations only refer to flow ratings through pipes etcetera.
    I need to design a steam installation like the pneumatic and hydraulic system.
    So I need to calculate the boiler, buffer and cilinder to achieve this launch.
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  31. #31  
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    Didn't you do the cylinder already in post #9? I still don't know why you picked 20 bar though. Is it some industry standard?
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  32. #32  
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    Yes but when the force changes the cilinder size will also change.
    I now picked 40 bars because this is indeed an industry standard.
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  33. #33  
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    So what bit are you stuck on?
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  34. #34  
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    Well the pneumatic part is now complete and I think this is correct.
    Now I only have to calculate the costs of the components of the system and the maintenance of the installation.
    Then I also have to calculate the hydraulic and steam systems.
    Don't you have any experience with hydraulics?
    I have a system with pumps, accumulators, motors and a winch drum that spins a cable around it with big speed.
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  35. #35  
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    None at all. It all sounds very interesting though! What is the winch cable used for?
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  36. #36  
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    It's about the same as pneumatics only with hydraulic fluid.
    The installation for hydraulics there are reservoirs, pumps, accumulators, hydraulic motors and a winch needed to launch a rollercoaster.
    This cable winch is attached to a catch car that launches the train.
    This winch will wrap up very quickly with the motors that are connected to the accumulators.
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  37. #37  
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    I'm back again and went back to my teacher.
    He said I have to use the power needed I calculated with the energy so the 4593 kW.
    With help of a diagram of a compressor in stages I have to read the compressor needed with the accumulator.
    He showed me but I can't find the diagram anywhere I thought I saw it on wikipedia but I lose it.
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  38. #38  
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    Lexus, there seems to be many different types of accumulators and compressors. If you know what you need you could try contacting a company that make them? Most people are happy to help because one day you might be their customer,buying their products!
    Hydraulic Accumulator Basics
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