Cedar Point says Top Thrill Dragster is done, sort of

I was attracted by the physics. Allow me to wildly speculate.

My understanding is the hydraulic system actually still used compressed air to store energy. The hydraulics were simply method of delivering that energy to the train. The ideal gas law tells us that PV = nRT so as the air decompresses (volume goes up), the pressure has an inverse relationship. So I would conclude that the hydraulic launch has the largest acceleration at the beginning, though I'm not sure by how much.

My understanding of LSM's is that they apply a constant acceleration to the train.

If the launch tracks are the same length and the exit velocity is the same and the hydraulic launch falls off in force, then the average acceleration from the LSM must be higher, but the peak acceleration from the LSM must be lower (If I'm thinking about this correctly).

That sounds plausible to me. But no matter what, going from 0 to 120 in the amount of track available is going to be Just Fine. Those folks who tell you it makes a major difference are the same ones who swore using markers on the edges of their CDs mattered.

We don't have to speculate. It's science, and we know two variables: The end speed necessary to climb over the top, assuming trains of a similar weight, 120 mph, and the length of the launch, 560-ish feet. (This is, coincidentally, the distance it takes my car to go 0-60 in 4.8 seconds, which is 0.57 G's.)

So with that in mind, there are some things we know.

• Average velocity isn't going to change, because it's measured as (final velocity + initial velocity) / 2. (120+0) / 2 for both launches.
• Acceleration is calculated as Δvelocity / time, and we don't know time.
• Similarly, G's are measured as Δvelocity / (time * gravitational constant), so again, we don't know the time.

In order for the time to be different, the acceleration would therefore have to vary.

Let's go back to hydraulic motors. It seems these things have a language all their own, but the short story is that torque can be constant or variable, depending on the type of motor. I don't know which kind Dragster is, but I imagine we could find out. Given Dave's diagram based on the patent in the PointBuzz explainer, I'm guessing that the torque is constant.

Electric motors can also vary with all kinds of precision. If you've ever seen a graph of measured Tesla output, the torque is constant when you floor it to a certain speed then drops off gradually as the car goes faster. So if the LSM version delivers constant torque, and we can't know that yet, then we can assume it would feel exactly the same.

There is one more variable that we haven't talked about, and that's the weight of the train and people. The old trains were 15 tons, and I'm pretty certain the replacements will not be. Now we have to figure mass into an equation somewhere. The power required to move the train depends on mass and time.

• Power = [mass * (change in velocity/time) * displacement] / time

Now we can plug in numbers and play around to at least use time as a proxy to the G's and acceleration. The shorter the time, the more you will "feel it." And again, I will contend that because the displacement is so short, I'm not convinced that one method over another will matter, especially if they both deliver constant torque. A lighter train with the same amount of power applied will reach the target velocity in less time.

Check my math and physics here... I haven't thought about any of this in a very long time, and my fact checking was casual at best.

I think you forgot to carry a 1 somewhere. 😉

I'm not sure you're using torque correctly here. Torque is defined as force*lever length and is the rotational equivalent of a linear force. An LSM doesn't have a rotational element. I'm not entirely sure why torque is used as a measure for car performance, but it's basically how "hard" the engine turns the driveshaft. Maybe "force" is what you want here?

Jeff:

Average velocity isn't going to change, because it's measured as (final velocity + initial velocity) / 2. (120+0) / 2 for both launches.

I don't think this is true (We just covered the mean value theorem in calculus yesterday, so this is fresh in my brain). Your calculation assumes a linear acceleration, but we're assuming they are not linear. The acceleration curves are different and I assume one of them spikes and trails off, giving it a higher average velocity. When I look at the patent, I see an expansion of pressurized air causing pressure (and force) to decrease over the course of the launch, and that's kind of the point.

And I don't think the train size/weight is relevant unless it affects exit velocity. If the exit velocities are the same, mass should cancel anytime it appears in force because of Newton's third law (the same mass will also be the thing resisting the acceleration due to that force).

Now, do I think the launch experience will be noticeably different? No. (Callback to all the people that swore MF went 100 mph in certain conditions). But I think it will take longer to get up to the same speed.

Jeff:

We don't have to speculate.

I had to look up the definition to confirm I wasn't having a senior moment. All we have is a teaser video and some aerial photos. Isn't the rest is speculation?

Torque is used for car performance because the twisting the engine is doing to the driveshaft eventually makes it to the wheels, and the torque on the wheel is converted to a linear force via friction between the tire and road to accelerate the car forward. So higher torque means a greater ability to accelerate.

Interestingly, a vehicle with a stepped transmission will have a different 0-30 time (quicker) than it’s 30-60 time, which is quicker still than it’s 60-90 time. That’s because while the power source—be it an IC engine or an electric motor—has a max power it can operate at, the transmission takes that power and spreads it out over a wide range by trading torque for speed (mechanical power is torque * rotational speed). So in the lower ranges, you have more torque to the wheels and thus will accelerate quicker, whereas in the upper ranges you have less torque but will travel faster.

ApolloAndy:

PV = nRT

Is this a first for CoasterBuzz? I have not seen the ideal gas law referenced here before.

And always remember that the vice president is over the treasurer!

This is new. A ride sucks before we've seen a POV or even a layout.

Jeff:

We don't have to speculate. It's science, and we know two variables: The end speed necessary to climb over the top, assuming trains of a similar weight, 120 mph, and the length of the launch, 560-ish feet.

Don't speculate, then you speculate the length of the launch track? That variable is hardly known and is pure speculation right now given the work being done.

Keeping with the car analogy, you can have two cars at a drag strip... One being a high torque, low horsepower car and the other car being an inverse with a low torque, high horsepower engine racing over a given track length. Both in theory would cross the finish at the exact same time. However, they will feel different of course. One will punch you back in your seat off the line but loose it's power at higher speeds, while the other will kick off the line slowly but catch up down the track.

Both will feel drastically different, of course. At the end of the day, they still hit that same top speed in "X" amount of time. We have not seen LSM yet match the acceleration capability over a given distance as we have with hydraulic launches. This is just fact.

Dragster (0-120 in 4s) gives us 1.37g average acceleration. (Hydraulic)

Storm Runner (0-72 in 2s) at 1.64g (Hydraulic)

Xcelerator (0-82 in 2.3s) at 1.64g (Hydraulic)

Kingda Ka (0-128 in 3.5s) at 1.67g (Hydraulic)

Red Force (0-112 in 5s) is 1.02g (LSM)

Maverick (0-70 in 3s) is 1.06g (LSM)

Blue Fire (0-72 in 2.5s) is 1.13g (LSM)

Thunderbird (0-60 in 3.5s) is 0.78g (LSM)

Those are some examples today and is quite a clear indication of what we have seen, up to this point. Not to say they don't have some new tech to deliver higher acceleration than what has been done yet, but again, until we see it...

Going from 1.02g to 1.64g is a big jump... That is doing a 0-60 in your car a full second faster.

My Camaro will do 0-60 in 4.0s or 0.68g :)

I guess force is really what I'm after, and not torque, because torque is force with a circular vector. But if we wanted to be ridiculous, isn't an LSM conceptually similar to an "unrolled" electric motor? Meh, irrelevant probably.

ApolloAndy:

Your calculation assumes a linear acceleration...

It does.

When I look at the patent, I see an expansion of pressurized air causing pressure (and force) to decrease over the course of the launch, and that's kind of the point.

Perhaps, but what I learned is that pressure is not the same thing as flow. The text I found says, "While some people use the terms interchangeably, they actually refer to different aspects of hydraulic system operation/performance. Hydraulic flow refers to the movement of the hydraulic fluid through the system. Hydraulic pressure refers to the force of that flow when it encounters resistance." Again, it depends on if the motor is constant or variable. I mean, we all know a linear piston in a hydraulic system makes an elevator move at a constant speed. Turning that into rotation on a ride like this, the train keeps moving faster, so the speed has to increase, meaning more flow. That's why I'm not convinced that the acceleration isn't fairly linear. I'd like to know.

And I don't think the train size/weight is relevant unless it affects exit velocity.

That would depend on where the exit velocity is achieved, no? I suppose for efficiency, you would to spread out the force applied, otherwise you have to reach a higher speed to account for whatever you lose in friction and air resistance before going vertical.

SteveWoA:
Don't speculate, then you speculate the length of the launch track?

It seems pretty reasonable to speculate given the space we're looking at that the launch track isn't going to get much longer unless it launches from inside the station. Even if that's the case, we're talking about 80 feet, which gets you to 640 feet total.

Now that I'm thinking about it though, the old ride didn't use all 560 feet to launch the train. Some of that track is where the catch car had to slow down, and while it's been a few years, I recall feeling the end of the acceleration pretty far uptrack, like maybe at the entrance tunnel. If that's the case, that's almost 200 feet or space to stop the catch car. If the actual acceleration happened inside of the remaining 360 feet of track, then I will concede that an LSM launch using all of that space would feel fundamentally different.

I know you don't like his channel, but ElToroRyan did a couple of videos recently showing the station demolition with some measurements and speculation about increased launch track length, relocation of transfer tables, etc. But his physics is not great. If I were his physics teacher, I'd give him a B at best. ;)

Jeff:

a linear piston in a hydraulic system makes an elevator move at a constant speed

Presumably the elevator is not powered by compressed air, but by a pump with a constant flow speed.

Jeff:

the train keeps moving faster, so the speed has to increase,

Constant or decreasing acceleration (borne from constant or decreasing pressure) will still result in an increase in speed.

Jeff:

Now that I'm thinking about it though, the old ride didn't use all 560 feet to launch the train. Some of that track is where the catch car had to slow down, and while it's been a few years, I recall feeling the end of the acceleration pretty far uptrack, like maybe at the entrance tunnel. If that's the case, that's almost 200 feet or space to stop the catch car. If the actual acceleration happened inside of the remaining 360 feet of track, then I will concede that an LSM launch using all of that space would feel fundamentally different.

You are definitely correct there. I believe the sled even disconnects a little further up track than the tunnel. Looking at the POV closely, I notice the sled brakes begin about mid-way on the white track segment just prior to that tunnel. Using the google, I get an estimate of just ~275ft from the launch area up to this point using the lamp post and such as reference points from the POV.

Using this information, that means the launch is actually 0-120 in about 3.13s (or about 1.75g) to cover that 275ft of track.

So with Dragster 2.0, using the transfer track area as the start of the launch (and using the entire length of the available launch track), you have nearly double the track length at ~550ft. If they happen to modify even further to launch from where the station area would be, they could gain even more distance (~600ft) as previously mentioned.

Using the 550ft reference, this would effectively reduce the 0-120 time down to 6.25s or so.

This is of course ignoring the fact that Dragster 2.0 may not even need to hit the same top speed, given they could use that extra ~200-225ft of track... While it may not be a huge difference, they won't have to overcome the drag and friction for that period of time when the train is loosing speed prior to the top hat while the sled de-accelerates.

It will be interesting none-the-less once we get the real details here. But I suppose it's fun to guesstimate in the meantime.

Assuming similar aerodynamics and friction, you still need to hit 120 mph to climb the tower. Once you go vertical, gravity is the only force acting on the train, and gravity is constant. I don't imagine that aerodynamics or friction are going to be radically different enough to require a different launch speed.

SteveWoA:

But I suppose it's fun to guesstimate in the meantime.

I guess you haven't been over to Pointbuzz yet...

I'm bad at math, but isn't this a f=m*a thing? The weight of the train would factor into how fast it needs to go to clear the tower, right? And aren't Intamin trains notoriously heavy? If i'm bastardizing physics 101 please correct me.

It's a kinetic energy thing, so T = 0.5*m*V².

You can actually calculate how much energy the train loses to various forms of friction if you know (or assume) the mass.

If you ignore friction, then you can calculate the velocity required to climb 420 ft against gravity, since gravitational potential energy is U = m*g*h. So setting that equal to the previous equation gives V = sqrt(2*g*h). For 420 ft then, V = 112.1 mph. But if we assume a Dragster train launches to 120 mph, and if we assume that this is just enough to clear the tower, then we plug mass back into the first equation to figure out the energy loss.

Found some mass estimates from a YouTube clip, so a train at 13,600 kg loaded with 18 adults having an average mass of 77.75 kg gives a total mass of 15,000 kg. That mass at 120 mph has 21.58 MJ of energy. But we said that you only need 112.1 mph to ascend 420 ft with no friction, so that equates to 18.83 MJ of energy for the same train. Therefore, a train loses 2.75 MJ of energy from the moment it reaches top speed to the moment it reaches the top of the tower.

What you come to realize is that a lighter train would actually need to be launched faster in order to clear the same tower. We see the opposite of this when the park launches an empty train that rolls back, when a fully loaded train would've cleared. If both trains reach 120 mph, the empty train has less kinetic energy, so if we assume the same energy loss during the climb (which isn't quite true, but that's for a later post), then the lighter train won't have enough leftover kinetic energy to make it to the top.

Jeff:

Assuming similar aerodynamics and friction, you still need to hit 120 mph to climb the tower... I don't imagine that aerodynamics or friction are going to be radically different enough to require a different launch speed.

To clarify, what I was trying to say is that the extra ~200ft at the end of the launch (from where the train hits maximum speed the sled disengages and begins to slow down) to the base of the tower, the train is losing some speed “X” due to drag/friction where the train is just coasting. So if the train hits a maximum speed of 120mph, let’s assume the train speed drops something like 3 mph by the time it actually hits the base of the tower and begins the ascent upward.

So my point being, if the LSM’s end up being installed that extra 200ft right up to the base of the tower utilizing the entire launch track length, technically, the LSM’s only have to launch to say 117mph since you don’t have to overcome those losses in that section of track before the tower. So the speed at the very end of the straight launch track would be the exact same between the two versions, ignoring train differences and all that right now, but again the train won’t have the deal with that ~200ft of loss of speed due the drag/friction.

Just generalizing of course, and perhaps maybe such losses are such a small parasitic difference that it hardly matters. But you do have 72 wheels in contact with the track, plus the drag at 120mph for that second or so at the end of the launch track that I would think that your speed would drop a little (few mph?). But maybe again it’s negligible and can be ignored for such a short duration of time.

And don't forget about the newly advertised* ability to put LSMs on the vertical part and continue the acceleration upwards.

So yeah... all speculation at this point.

* "new" meaning Mr. Freeze, apparently.