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CommentAuthoraltlenny
• CommentTimeMay 22nd 2014 edited

Hi all,

I've already done one of those posts and I really feel the urge to come back to discussing the fundamental ways of how physics (and more specifically dynamics) can be applied to roller coasters, which simplifications can be made and how big the error will be due to those simplifications. As many of you know I'm the developer of FVD++ which aims to simulate roller coaster dynamics dynamically via user inputs in order to generate highly realistic tracks by means of mathematical input. Therefore a potent yet simple to compute mathematical model has to be constructed that represents the behaviour of real roller coasters as closely as possible.

## 1. A (simple) Physical Model for a Coaster Train

Let's look at a roller coaster train shall we.

As you can see any real train has a specific length (denoted by l), a height (denoted by h) and a width (denoted by w). Most of the time l will be much larger than b an h, but you can also find examples of coasters where l is more or less equal to w or h (see Gerstlauer) or where w is indeed quite significant (see 4D Coasters or Wingriders). But for now let us look at a very simple model of a train, I call it the stick-on-track model.

In this model the train length l and width w shrinks to zero. Furthermore we let the red stick be of no mass while the whole mass of the real train the stick model should represent is compressed in the blue cross which is widely known as the center of mass. That point is somewhere above the center of the rails (for non inverted coasters). Let us denote the height of the center of mass above the center point between the rails as h_m (as height of the center of mass). The green point in the above picture shall be called the heartpoint (which will later be extruded to a heartline). Similarly h_h denotes the height of that point. Be aware that h_m will usually NOT be the same as h_h, and it is not the same for the human body either (h_m will be around the height of your belly button).

With those terms in mind let us see how we can calculate the velocity of such a stick-train at any point during a track. I will do this by adding up all the (relevant) energies that can be found during a ride. At the first glance this is rather easy.

We can see that the total energy of a train is a sum over the kinetic energy and the potential energy at the current point in time. In a perfect world this total energy would be constant over the whole track (coasters with unlimited length YAY), but as the train races through the track it constantly loses energy mostly due to friction and air resistance. That is why the total energy E will decrease during a track and the only way to increase E is to either feed kinetic energy into the system (e.g. by launching the train) or to feed potential energy into the system (by lifts). Be aware that the h used in the formula for the potential energy is the total height of the center of mass above a global zero level.

Okay let's look a bit deeper into the kinetic energy because some of you that had physics in school might be wondering what the second (cyan) part of that formula stands for. This formula expresses the energy that is contained in any rotation of a body with a spatial extend. This is the energy that is saved during any spinning or rolling motion of your coaster. My stick-on-track train has no spatial extend and that's why that train will indeed NOT have any of this energy, which is why the stick-train will be faster in rotating elements than a real train. However, except for trains with significant height or width (again wingriders or 4D coasters) this effect won't be too noticable and I have yet to experiment with NL2's physics engine to verify whether such a simulation is done for NL2 or not.

A far bigger error due to the stick-train model in most cases is the absence of a train length. Let us see why this is a problem.

As you can see a the center of mass of a long train is as the arithmetic mean of all center of masses of the vehicles in valleys higher than the center of mass of a stick-train without any spatial extend. This will lead to a miscalculation of E_pot which will lead to a miscalculation of E_kin = E - E_pot. This will remain a problem for math building because you can't calculate the velocity of a track at point A if you don't know the curvature of the track beyond point A, and to calculate the curvature of the track beyond point A you need to calculate point A first.
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CommentAuthoraltlenny
• CommentTimeMay 22nd 2014 edited

There are ways around this problem such as extrapolating the current curvature of the track or iteratively recalculating the velocity based on past results, but for the sake of our sanity let us just come back to our beloved stick-trains.

## 2) Modelling the Coaster Track

With the knowledge of chapter 1 we can calculate (or at least we can estimate) efficiently how fast a coaster train will be going at any point during a track by only knowing the position of the track and the total energy of the train at that position. But how is it possible to calculate the shape of the track at any position by knowing the velocity, the orientation of the track, the normal and lateral acceleration and the angular velocity around the heartline? Our best friend here is the superposition principle, stating that for all linear systems the total response of the system is the sum over all partial responses. That principle allows us to look at the reponse for the normal acceleration, the response for the lateral acceleration and the response for the angular velocity seperately first, and then just add all those responses up.

But before we start, let me rant about the use of the word force in the context of coasters. Those are no forces goddammit, they are accelerations measured in multiples of earth's gravitational acceleration g which is around 9.81 meters per second squared.

Now let us start with the easiest response, that of the lateral acceleration. Let me draw you a circle.

This beautiful top view shows the radius of a (more or less perfect) circular track segment which is traversed with the velocity v. We can calculate the lateral acceleration a_l a rider experiences during that segment by utilizing the centrifugal acceleration (whereas the track has to withstand the corresponding centripetal force which in turn is the centripetal acceleration multiplied with the mass of the train). However, it would not be wise to use the resulting radius of our track as a measure of its curvature because for small accelerations this value can get arbitrarily big (and even infinity if the lateral acceleration is zero, which it most and usually too oftenly is when math building a track). A better but less intuitively accessible value is the angluar velocity denoted by a small omega. This value describes the magnitude of the angle the train has to turn in a second for a rider to experience the lateral acceleration.

The normal force is similar, but not exactly the same. Let me remind you of the stick-train introduced earlier.

As you can see here the radius for the center of mass, heartline and track respectively are different. At the first glance this is no problem because we are interested in the angluar velocity just like before. But due to the different radii a rider's head will have a different velocity (and thus normal acceleration) than her feet. This is the point where we have to agree upon on the fact that the train's velocity is only accurately described at its center of mass (the only point our stick-train exists anyway) while the rider wants to experience the normal acceleration at the heartpoint. If we keep in mind that he angular velocity for every point on the train must be the same, we can solve the problem by hopefully correctly applied math.

I've already talked about this part of coaster dynamics in another post but my last paragraph should have summarized it well enough for now.

Let us see what the rollspeed has left over for us to talk about. Well, similarly to the normal force we have to agree on the fact that any rolling motion of the train should be performed around the heartline (duh, that's what the whole principle of the heartline is for).

The velocity of the train is still measured at the center of mass, which means that every rolling motion around the heartline introduces a displacement of the center of mass which in turn results in a velocity in that direction. This velocity due to the angular velocity from the rolling motion has to be considered as part of the actual velocity of our train. If we split up the velocity into a part that is dependent on the rollspeed and a part that is independent from it, we can calculate how fast the heartline has to move forward to stay in synch with the train's center of mass. Similarly we can calculate how fast the wheels of the train have to move on the track (and then how fast they have to spin etc).

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CommentAuthoraltlenny
• CommentTimeMay 22nd 2014 edited

## 3. Calculating what seat to choose AKA Why the Back Row is Always Most Fun

In chapter 2 we looked at how a track can be shaped and calculated by only specifying accelerations and roll rate around the heartline. This is possible because according to chapter 1 we can calcu... accurately estimate the velocity of a train at any given position along the track. Now suppose we have designed some awesome new Inverted Spinning Mack Launched Giga layout with 42 inversions (and you even remembered putting a heartline roll at the end) in FVD++, import it in NoLimits just to see that for your new 16-car train design with 12 seats per row the accelerations reach insane levels in the back left seat. Clearly it's time for you to look up my e-mail address and write an angry letter because you designed the coaster perfectly with all acceleration limits in mind. Well, shame on you because I will link you to this chapter.

### 3.1. Rotations AKA Being Torn Apart

All amusement park rides introduce some kind of motion to your body. General motion comes in two flavors. Linear motion moves you along a trajectory in space. There was this guy called Sir Isaac Newton that found out that changing this trajectory (either its direction or magnitude) is only possible by applying some kind of acceleration to you as a passenger, and this acceleration will be the result of a force acting on you. The second flavor is rotational motion around your center of mass. Just like with linear motion changing rotational motion requires a angular acceleration which is the result of torque acting against the rotational inertia of a body. With pure rotational momentum you won't be able to get around too much in space, you'd need some portion of linear motion to do that. The interesting thing about rotational motion is that all points that don't lie exactly on a rotating bodie's center of mass will have a linear motion due to the body rotating. Moreover that linear motion will constantly change it's trajectory (thus all points will feel an acceleration, the centrifugal acceleration) and the sum over all linear motions of all points of that body has to be exactly zero for a purely rotating body. This means if you spin fast enough, your body will literally be torn apart by the centrifugal acceleration your outer body parts will experience.

Don't try this at home, you will explode.

Let's talk about what kind of rotations can be found while riding conventional roller coasters (the non-spinning, non 4D variety). At this point it is be handy to define the terms yaw, pitch and roll angles/motion. In roller coasters one can find all three kinds of rotational motions, all with their own effect on the different seats in a coaster car. Yaw motion is most abundantly experienced during flat turns but also during typcial coaster segments superposed to other motions. The seat position inside the car may greatly change the way a passenger experiences those flat turn, a good classic example would be the good old mouse wingrider (who doesn't know that one). Let us say that the seats on a wingrider train have a 2.5m horizontal displacement from the center of rails.

To make things ridiculous let us build a flat right turn for that wingrider mouse coaster with a radius of exactly 2.5m. This would look something like this.

The train will pass through that element with a linear velocity of 5m/s. We immmeditaly notice that the passengers in the right seats would not move at all because they would sit exactly on the center of the circular turn. Thus they would just sit there while not being accelerated at all during the turn. On the left side of the train the passengers would move with a velocity of 10m/s through a circle with 5m radius (I leave the needed calculations to get to those values open to you). As a consequence the left passengers would feel about 2g lateral acceleration (the middle of the train would experience about 1g lateral). The rule of thumb for any mouse coaster (especially mouse wingriders) would be to always sit on the seats that are as far away from the turn's center as possible. The physical explanation is quite straightforward. For a car with velocity v and width w riding through a turn with radius r the acceleration in the three positions middle of the car, inner seat and outer seat (displacement +-w/2) are the following.
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CommentAuthoraltlenny
• CommentTimeMay 22nd 2014 edited

Pitch motion is similar, except that I don't know of any coasters that stack steats on top of each other. However, the human body has the interesting property that it's height is larger than it's width or length. So while riding through hills and valleys your head will experience different accelerations than your feet just because they will move along circles with different radii. Those differences won't be significant as long as the radii are much larger than the height of your body (same formulas as above apply, if w/2 is significantly smaller than r you can just ignore its influence). Here's an example (and another one) for some unique acceleration differences between your head and your feet on a roller coaster due to pitch motion.

The really interesting stuff happens when doing Robb Alvey's favourite thing... twisting (or rolling). Those are the rotations on a roller coaster that are performed around the heartline. As a consequence, everything that doesn't lie exactly on the heartline will experience additional accelerations to be part of the twistyness.

And as always the accelerations depend on how fast the coaster is twisting and how far away you are from the heartline. This is the reason why you can't use wingrider trains on Intimidator 305 without killing a bunch of people in the process (I bet even 4 seats per row would be pretty painful on that one). The interesting part about the additional accelerations due to roll motion is that you can't really eliminate them by changing the position of the heartline or introducing some kind of additional curvature to the track.

Let's take a zero-g roll as an example. In case the zero-g roll is designed symmetrically riders in the left seats will be accelerated to the left and riders in the right seats will be accelerated to the right (always away from the center of rotation). If you design a zero-g roll for the left seats, the right seats will experience double the acceleration due to roll motion. This fact is one of the reasons why the perfect zero-g roll does not exist. This is especially true if you think about the fact that your body takes up space and so only exactly one point of your body can have exactly zero g at any given time when you are rotating. Reducing the occuring accelerations is only possible by decreasing the roll rate of an element, either by decreasing the velocity of the train during that element or by decreasing the twistyness of the element. Finally a perfect zero-g stall or airtime hill is less impossible than the perfect zero-g roll, but a drop tower beats all of those.

### 3.2. Front Row or Back Row?

For some reason even the most general of general public has some kind of understanding that it matters in which row you sit in a coaster train. But the most common explanation for this ("The back row is faster") is utter crazytalk. If the back row were indeed faster than any other row, it wouldn't be the back row for too long, so there has to be different explanation. Let us first agree that each car and thus each row of a coaster train has the same velocity at every point in time (else the coaster is broken). The crucial point is that the different cars and rows are at different positions along the track at the same time. Therefore the experience on all elements can vary widely across rows. This leads to the famous effect of the front half of a train pulling the back half of the train over the crest of a lifthill (or any hill for that matter).

And that is why you should ride in the back.
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CommentAuthoraltlenny
• CommentTimeMay 22nd 2014 edited

I encourage you to ask questions about coaster physics here to start a discussion and if you have any idea on how I can extend or improve this introductory post please drop a comment below.

A short note to the mods, I will whisper some posts to myself because I plan on both keeping this introductory text up to date and extending it in the near future. I hope you agree that it would be very helpful to have those kinds of posts as close to each other as possible.

Changelog:
22.05.2014: First version
13.05.2015: Renamed initial version to chapters 1 and 2, added chapter 3
• CommentAuthorbigjoe97
• CommentTimeMay 22nd 2014

Will it be possible to show the center of mass in the next release of FVD?
• CommentAuthorMilBee
• CommentTimeMay 22nd 2014

I really never realized how many calculations it takes to build a coaster. Computers do it all now but it's still cool to see it explained like this. Great lesson, Lenny!
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CommentAuthorskyasaurus
• CommentTimeMay 24th 2014

really nice to see an in-depth and organized analysis of questions i've had for a long time about the complications of using point-mass approximation for coaster trains. thank you!
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CommentAuthoraltlenny
• CommentTimeMay 13th 2015 edited

Update: Please refer to the initial posts! Again, I encourage you to start a discussion on the physics of roller coasters here!

13.05.2015: Renamed initial version to chapters 1 and 2, added chapter 3

Greetings, Lenny
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CommentAuthorInstant Mix
• CommentTimeMay 13th 2015 edited

This makes me cry, in both happiness and sadness. I need to go over kinematics at some point.

The fact you made FVD++ totally fascinates me as I'm studying computational physics and to be fair making a program similar to this would be, quite frankly, amazing.

Here's a question I had that I had to scribble some stuff down for in order to understand : is the mass of a coaster vehicle negligable?

Someone brought this to my attention when they mentioned that adding dummies did nothing and my instantaneous reply was "of course they do, they add mass and therefore kinetic energy, course the thing will go faster".
I thought a bit more about it and then realised that the force of gravity is obviously going to scale with M, therefore it'll do almost nothing apart from potentially decrease the effects of drag? I also thought about wether it'd decrease the amount of energy lost via friction , but then realised that $F_{k} = \mu F_{n}$ , and obviously $F_{n}$ is going to be equal to the modified gravity equation..

so honestly the only thing I can think that extra mass does is reduce the effects of drag ( as $F_{d}^{M} < F_{n}^{m}$ , they rely on area rather than masS ).. why on earth do they bother putting in water filled dummies then?
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CommentAuthorMorganFan
• CommentTimeMay 13th 2015 edited

Probably some testing regulation made up by someone who was too lazy to read up on physics.
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CommentAuthoraltlenny
• CommentTimeMay 14th 2015

^^Have you thought about the increased structural stress on both the trains and the track/supports?
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CommentAuthorDC High Heat
• CommentTimeMay 15th 2015 edited

Ever see the difference between speed in a test run and that of a run with people? Water dummies make for more realistic test conditions and also help rides to not stall. Maybe they won't be much faster toward the beginning of the ride, but at the end the decrease in the effects of drag makes a difference.
1.
Oh dear god why is there education here.
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CommentAuthorInstant Mix
• CommentTimeMay 15th 2015 edited

DC, that's the whole point of my question though - the extra weight has absolutely NO effect on speed of the car. The added weight shouldn't have that much of an affect on structural stress as the cars themselves are going to be considerably heavier than the people riding them, I don't want to imagine how heavy the dive machine cars are.

See DC, the fact rollercoasters run on gravity makes the whole situation considerably easier.
At the top of a drop, or say the lift hill, they'll have kinetic energy from the lifthill ( assuming they aren't stationary, we can make it move without forcing it but that's quantum mechanics and I do NOT want to see the HUP again ) and potential energy from being x distance away from the ground.
The equation for kinetic energy of a body is $E_{k} = \frac{1}{2}MV^{2}$, and the potential energy for an object H metres from the surface of earth is $E_{p} = MGH$. By combining the two, you'll get some equation like $E_{t} = M_{car}gH + \frac{1}{2}M_{car}V_{lifthill}^{2}$. This is the potential energy of the car from the ground PLUS whatever energy the lift hill chain is putting into it. Now, the energy increase of the train going down the slope will be whatever the displacement in X is, so let's say there's a 50 metre drop. There'll be an increase in energy ( decrease in potential) of $50M_{car}G \textrm{Joules}$. This change in potential is EXACTLY* (disregarding effects of friction, drag etc) how much kinetic energy the car will now have, this being $\frac{1}{2}M_{car}V^{2}$, so we can equate the two terms so that we have a new equation , this being $\frac{1}{2}M_{car}V^{2} = 50 M_{car}G$.
What do you notice about the equation?
Both sides have a common factor of $M_{car}$ , meaning that the mass has absolutely no effect on it's speed, because it balances out. Regardless, we'll solve it for V.
$\frac{1}{2}M_{car}V^{2} = 50 M_{car}G$.
$M_{car}V^{2} = 2 ( 50 M_{car}G )$.
$V^{2} = 2 (50 G)$.
$V= \sqrt{ 2 (dx G)} \quad \textrm{where} \quad dx = 50$.
Which shows that for any height displacement, dx, the car will be at least travelling at $2(dx G)$ - regardless of mass!
Obviously DC, i've thought that having the dummies in speeds up the car ( or having a longer train, thanks RCT ), but physically it shouldn't as far as I can see.

Looking into it a bit more a user friendly version of the equation would be $V= \sqrt{ 2 (H - dx) G)} \quad \textrm{where} \quad H = \textrm{Highest point of Coaster} , dx = \textrm{Height above ground}$.

Altlenny, please correct me if i'm wrong here
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CommentAuthoraltlenny
• CommentTimeMay 15th 2015

Posted By: Instant MixI don't want to imagine how heavy the dive machine cars are.

Yeah, but also consider how heavy 30 people are (3 rows with 10 people each). The answer is about 2.4 tons, or about 1.4 small cars (thanks Wolfram Alpha). Even if those 2.4 tons were 5%-10% of the total train's mass, it would matter, and I don't really think each dive car weighs 8-16 tons...

As far as drag / air resistance is concerned, it makes a difference too as you pointed out. And again even if the difference is small, you should test that case before actual people go on the ride.

Greetings, Lenny
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CommentAuthormkj3322
• CommentTimeMay 15th 2015

It's pretty neat being able to recognize a lot of these concepts from physics 1 and physics 2.
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CommentAuthorRK
• CommentTimeMay 15th 2015 edited

^^ Yeah I was about to write the same yesterday. But I decided not to comment due to the lack of knowledge. Because everything you said is correct for the ideal case, so when theres no friction etc. So it could be that if you add the terms of friction and so on the mass has indeed influences the speed of the train.

First thing that came in my mind was the rolling resistance, which says F(r) = c(r) * F(n). F(n) the Normal Force is influenced by the mass (for a piece of track parallel to the ground: F(n) = F(g) = m * g ) and F(r) the Resistance Force is in the opposite direction the train is moving. So a bigger mass should actually slow the train down ( So this doesn't help to proof that the train should be faster). But again other forces should speed the train up when the mass is bigger e.g. the air resistance would be lower.

So I don't know exactly why a train is faster with bigger mass, but it is for sure.
And your guess that
Posted By: Instant Mixthe only thing I can think that extra mass does is reduce the effects of drag
is mine too.
• CommentAuthorJetPulse
• CommentTimeMay 15th 2015

Posted By: RKFirst thing that came in my mind was the rolling resistance, which says F(r) = c(r) * F(n). F(n) the Normal Force is influenced by the mass (for a piece of track parallel to the ground: F(n) = F(g) = m * g ) and F(r) the Resistance Force is in the opposite direction the train is moving. So a bigger mass should actually slow the train down.

You forgot that you are calculating force and not acceleration. Heavier trains will generate more rolling resistance in terms of force, but it also takes more force to slow them down. Also, that equation isn't for rolling resistance, it's for friction. Rolling resistance is proportional to the amount of deflection in the track and wheel due to the contact forces between the two (a little more complicated than the simple friction model shown above). This deformation is proportional to normal force, but not linearly because the more it deforms, the larger the contact area becomes and then the more force it takes to create additional deformation. Thus, a heavier train will create more rolling resistance, but the increase in rolling resistance force is smaller than the increase in mass thus the total acceleration effects are smaller for heavier trains (making them faster).

Likewise for air resistance. The total force for a full and empty train is pretty similar. But with a larger mass, the acceleration due to air resistance is smaller on heavier trains.

Instant Mix, what you've written is true for frictionless vacuums, but according to that simple model a rollercoaster would never slow down. Thus, it is missing every force that acts against the train's motion (which is where a train's weight comes into play).

Might help to think of it this way: it's not that heavier trains travel faster, its that they don't slow down as fast. The forces acting against the forward motion of heavier and lighter trains are similar, but since a=F/m, larger mass means less acceleration.
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CommentAuthorskyasaurus
• CommentTimeMay 15th 2015

remember that all of these equations are just approximations for extremely complex micro and macroscopic phenomena. normal force (and thus friction) increases by m times a small constant, whereas momentum and kinetic/potential energy increase by m unmodified by a constant. also, a large part of the friction is drag, which stays basically the same as m increases. yay differential equations
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CommentAuthorKingRCT3
• CommentTimeMay 15th 2015

I know that manufacturers design their rides and estimate their speed/forces based on calculus for sure, but also from previous experiences. There is a lot of diferences between the ideal and simplidfied situations and reality.
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CommentAuthorInstant Mix
• CommentTimeMay 15th 2015 edited

@ JetPulse: Obviously, hence why i mentioned that I was ignoring the effects of drag and resistance , but was trying to clear up the point that heavier trains do not travel faster than lighter ones. Of course there, as drag is directly related to surface area it is independent of mass, it'll have less of an affect on heavier masses.

I like these sort of discussions, i'm beign a total physicist right now
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CommentAuthorRK
• CommentTimeMay 15th 2015 edited

^^^^ Thank you for your explanation to look at the accelerations and not the forces. That absolutely makes sense.

But I'm not sure about the rolling resistance thing, though.
I thought rolling resistance is the frictional force, because in german it's called 'Rollreibung' or 'Rollwiderstand' which is the equation I used. I just looked up the corresponding english wikipedia article (because I don't know all the english terms for technical purposes) which's title says 'rolling resistance' (which would actually be the direct translation of 'Rollwiderstand') and the same equation is in there too.

^ This kind of physics is not completely new to me but I never really used it actually. I love this kinds of discussions too :)

| Ah ok. Didn't get you meant the coefficient.
v
• CommentAuthorJetPulse
• CommentTimeMay 15th 2015 edited

Just saying, heavier trains do travel faster. It's not because of point kinematics (that's mass independent as you've shown) but instead its because of mass dependent energy losses. As DC said, they won't be different at the beginning but energy loss differences have a noticeable effect by the end of the ride.

Posted By: altlenny^^Have you thought about the increased structural stress on both the trains and the track/supports?

As a structural engineer, the structural integrity of a ride should be well above loads experienced even with overly obese people in every seat of the train. The reality is that if you're going to get structural failure, it's going to be in fatigue over years of service. Since all the structural loads are cyclic, I would assume design stresses are below the fatigue endurance limit for steel which about 30% of the yield stress. They might use higher values for columns that only experience nearly pure compression where fatigue cracking isn't an issue. But still, if you notice anything structurally different between empty and full trains (other than a little more temporary deflection as the train rolls over) you've got bigger problems.

RK, the wiki article states that Crr (coefficient of rolling resistance) is dependent on many things (and not a constant). One of the things load.
2.
Now I know next to nothing about physics, yet I've been very interested in this thread.

To throw a small monkey wrench in the situation, how is it that on fair (meaning identical, or nearly identical length) racing roller coasters, heavier trains tend to beat the lighter trains? Is that only a result of momentum? (Because I do know momentum is dependent on mass). Just wondering :)
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