# High Speed Transit Help

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1. Jun 5, 2015

### LaLiLuLeLo

Hi everyone,

I am an aspiring writer, and I have come here seeking some insight/a second opinion. I recently decided to work on my one novel, a science fiction story, and unlike some other sci-fi writers, I want to make my book as scientifically plausible as I can, without relying on something like this to explain the tech...

Me: In the future, people can do THIS!

Reader: And how does that work?

Me: Umm...err...SCIENCE!

In any case, I've been hard at work looking at theories, crunching numbers, and of course, making the physics plausible. So far, I've come up with ideas for various scientific advances within the story's world, such as warp travel, artificial gravity, and in-space resource conservation, to name a few. However, today, I was working on another key aspect of the tech, and something didn't seem quite right. I did the research for what sort of math I'd need to use to find the answers I was looking for, but either A) I put the wrong figures in, or B) I am right, though I'm not sure that I am.

Anyways, sorry to ramble a little bit. So here's the issue...I've made plans for there to be a public transit system. A high speed train, in essence.

THE FACTS
+Travels in an environment that has gravity a little less than that on Earth (5.882 m/s2), though most humans have grown up in this environment, so this is their "normal" gravity.
+The trains travel through a tube with little air within, to allow the least amount of air resistance while enough air to breathe for a few minutes should someone accidentally (or stupidly) fall/jump off the train (in either case, there's emergency terminals with oxygen tanks, if anyone's curious).
+The trains, depending on the line, either make more or fewer stops, with the latter being more express routes. For example...

Line A: 997.793 km in 20 minutes
Line B: 402.336 km in 10 minutes
Line C: 2993.38 km in 1 hour

These three are probably the most prominent train lines, so feel free to use them as a reference.

+Maglev
+In terms of train weight, standard trains in our modern time are fine as a reference point. For sake of example, I will use the TGV Atlantique as my reference point. Unlike the Atlantique, however, there's also some room for cargo. This puts the train's weight at a range from 444 t (completely empty) to 509 t (completely full, including cargo). The computer on-board can compensate for whatever weight the train ends up being.
+Train paths will be relatively straight and on a flat surface, aside from a bit of a curvature to the planet (it is a good bit smaller than Earth, so that may play a role.
+Travelling underground in order to save time/decrease distance is not possible in this environment.

THE GOAL
-Find out the best mix of acceleration and deceleration so that the passengers won't end up uncomfortable, or worse, DEAD. While fighter pilots are trained to deal with many degrees of G-forces, the general populace of the story's world is not. Plus, I imagine it would be bad for PAYING CUSTOMERS to get WHIPLASH every time they took the train. Every time I did the math, it seemed as though I did something wrong, or that I used the wrong formula. I have some experience with geometry and physics (both of which have proved immensely helpful), but I admit I'm not confident of the accuracy of my figures. Some help to ensure I'm on the right track (pun not intended) would be greatly appreciated. Feel free to ask for any more info about the conditions, and I'll post them in a reply. Thank you in advance!

BONUS: If you can figure out a way for the train to make the same distance of trip at a faster speed without endangering the passengers, that would also be appreciated.

-Lalilulelo

2. Jun 5, 2015

### phinds

Let me see if I have this right ... you want us to help you with some unspecified calculations which you do not care to post?

3. Jun 5, 2015

### Staff: Mentor

You get the fastest trip if you constantly accelerate with the maximal acceleration you want to allow to the mid-point, then decelerate again with the same magnitude of acceleration. Limiting trains to a lower top cruise speed can be advisable in terms of friction and safety concerns.

Apart from that, see phind's post.

4. Jun 5, 2015

### LaLiLuLeLo

@phinds My apologies. Here's one of the calculations I used. I admit, I'm not an expert in physics, but I did the best with what I had. In addition, there seems to be an issue with the copy and paste function on this forum, crashing the page each time I try. As a result, I'll have to recreate it from scratch. For this example, we'll use the aforementioned "Line A."

V=997.793 km/20 min=49.88965 km/min

Converting to km/h gives us 2993.379 km/h
in m/s, we're looking at 831.494 m/s

Vf=Vi+(a*t)

@mfb Thanks for the idea to have it reach max halfway through. Solving for a given the 10 minute halfway point.

831.494=0+(a*10)

a=83.1494 m/s2

In addition, I found a formula for G-Forces on this site, which is not something I learned a lot about in class. Here's where things got fuzzy for me during the day. Still, it helped with the whole curvature-of-the-planet hurdle.

v2/r

83.14942/952822=0.00726 G's

...And just in case my hunch for what I did wrong is true...

83.1494/952822=0.0000873 G's

Yeah, I get the hunch I'm doing something wrong. Like I said, G-forces weren't really covered for me, and admittedly some rust may have set in. Regardless, the help is appreciated!

5. Jun 5, 2015

### Staff: Mentor

I think you're going about this the hard way. An airliner here on Earth accelerates from zero to about 90 m/s in around 30 seconds (all ballpark figures for simplicity). This is an acceleration of 3 m/s2, or just under 1/3 g. Assuming this is around where you'd want your passengers comfort level to be, you could simply set your maximum acceleration to 1/3 of 5.9 m/s2.

Note that the formula for finding the acceleration is: Vf = Vo + At, where Vf is the final velocity, Vo (pronounced "v-naught") is the initial velocity, A is the acceleration, and t is time. V2/r is the centripetal acceleration (aka radial acceleration) of a rotating object, not the linear acceleration.

Last edited: Jun 5, 2015
6. Jun 5, 2015

### Staff: Mentor

Now, would the acceleration be appropriate for your train paths?

Your average velocity for line A is about 50 km/min, or about 800 m/s (I'm using 800 instead of 833 to keep the math simple). How long will it take for the train to reach 800 m/s?

Using the equation from my post above: 800 = 0 + 2t, simplifying to 800 = 2t, 400 = t.
So it will take 400 seconds to reach 800 m/s at an acceleration of 2 m/s2. 400 seconds is about 6.67 minutes.

Since you'll only be traveling at 800 m/s for about 1/3 of the trip, the rest spent accelerating, you'll need to change your top speed if you want to have a 20 minute trip.

7. Jun 6, 2015

### Staff: Mentor

You are mixing minutes and seconds here. Also, you are mixing peak velocity and average velocity. The right acceleration is a bit above 2m/s2, probably tolerable.

v2/r is a formula for acceleration on circular tracks. You don't have that (the curvature of your planet is still negligible at that speed).
Also, you used the (wrong) acceleration value here instead of a velocity.

8. Jun 6, 2015

### Rippetherocker

Taking Line A of 1000km in 20min most comfortable would be constant acceleration and deceleration. So cover 500km in 10min then decelerate.
S= ½at² => 500km= ½a(10min)²
500000m= ½a(600sec)²
So acceleration is around = 2.77m/s²
That is doable for line A.
Note that as line distance reduces acceleration required increases. Case in point line B where acceleration will be around 4.77m/s²