Maglev frictionless catapult system for interstellar travel

[TakenItSeriously]

Assume we adapted our interstellar space ship to use a maglev propulsion system (like those used for high speed trains) only over tracks surrounded by a sealed tube in order to create a perfect vacuum so that we travel in a frictionless state.

 

Assume ideal tracks that encircle the Earth at the equator and are always perfectly level and straight.

 

Wouldn't supplying a steady state power supply create a steady state acceleration given the frictionless state?

 

Theoretically, what prevents us from accelerating the ship to close to light speed over a period of years before releasing the ship to coast towards a destination star system light years away at say, *22% of the speed of light without using any of the ships resources for propulsion.

 

It seems like it's too easy but I can't think of any reasons that would prevent it from working, at least theoretically.

 

Thanks.

 

*changed from 75% thanks to Swansots pointing out of an error where centripetal g-forces would be too high to stay on the tracks at that speed

Edited June 11, 2016 by TakenItSeriously

[swansont]

What's the centripetal acceleration involved in moving at 0.75c at the radius of the earth?

[TakenItSeriously]

 

Nice catch, I think the g-force at 75% C is somewhere in the twenties so I'm off the rails and it's a bad example.

 

I originally had the g-force capped at 2 g's (centripetal force)for a couple of reasons. Any thing higher would be too uncomfortable for people to deal with over extended periods of time since g-forces at this scale would change over periods of months.

 

Also I could deal with a 2 g force mechanically by using a four rail track that had a par of rails above and a pair below specifically to keep the ship from derailing, when centripetal forces first kicked in.

 

So if we only accelerate the ship to 2 g's then, the initial speed for our ship is 22% the speed of light which is still a pretty respectable speed that we get for free so to speak.

 

Edit to add:

Also note that as the circumference of the track gets larger the initial speed we start out at also climbs. In that respect it's like a super collider for spaceships instead of particles. (For acceleration, not collisions)

 

I came up with an idea of creating a maglev track that was in orbit and spinning at orbital speeds, then we would require a counter acceleration so that accelerating our ship didn't slow down the track which would cause it to collapse, but that seemed too unstable.

 

Also I realized that the instant we left a vacuum, our ship would hit the atmosphere at a ridiculous speed that would first obliterate us then burn us up like flash paper.

 

Better to choose a planet without an atmosphere such as the moon but the sacrifice is our initial speed drops to around 11% the speed of light.

Edited June 11, 2016 by TakenItSeriously

[Danijel Gorupec]

Imperial percentages!

 

How did you compute this 22%? (Compute the percentage more carefully and I guarantee you the 'aha moment' you will not forget.)

[swansont]

Nice catch, I think the g-force at 75% C is somewhere in the twenties so I'm off the rails and it's a bad example.

 

 

20? You are a few orders of magnitude short. 20g's at 6400 km is ~36,000 m/s, which is 0.012% of c. I don't see where you're getting .22c giving 2g's. Did you forget to square the speed? a_c = v²/r

[Danijel Gorupec]

Even more striking... at 22% of the light speed the g-force would be some seventy million g... gee.

 

(Now I wonder how much my result is off if proper, relativistic, computation is used instead. Can anyone make and educated guess, please?)

[Janus]

So if we only accelerate the ship to 2 g's then, the initial speed for our ship is 22% the speed of light which is still a pretty respectable speed that we get for free so to speak.

For "free"? Even at 100% efficiency, the cost of getting a mass up to 22% of c would be $73,000,000 per kg(At 12 cents per kwh). And such a system couldn't come even close to achieving that kind of efficiency, even in operational costs.

[TakenItSeriously]

For "free"? Even at 100% efficiency, the cost of getting a mass up to 22% of c would be $73,000,000 per kg(At 12 cents per kwh). And such a system couldn't come even close to achieving that kind of efficiency, even in operational costs.

By free, I meant free from the perspective of the intestellar travelers who have serious payload budget issues, especially if the vast majority of their payload must be fuel. So if an intestellar space crew can get all of their speed right from the grid it's free relative to their payload budget.

 

Economies of scale are a very complicated issue that goes well beyond the scope of this thread. This is why I tried to keep it to what was theoretically possible, though a first order approximation of what might be practical is fine. Beyond that the speculation grows exponentially.

 

That electricity bill of $73 mil/Kg seems a bit high. Did you account for removing drag?

 

 

Even more striking... at 22% of the light speed the g-force would be some seventy million g... gee.

 

(Now I wonder how much my result is off if proper, relativistic, computation is used instead. Can anyone make and educated guess, please?)

It depends on the time frame you are choosing. Let's assume 1 g of constant acceleration since that's a popular choice for interstellar travel as well as a practical number for the acceleration of a maglev train in a vacuum. I didn't run the calculations myself, since math is not my strong suit but when looking at others numbers, I think the timeframe for reaching 22% C is roughly on the order of a year or two. Edited June 11, 2016 by TakenItSeriously

[swansont]

It depends on the time frame you are choosing. Let's assume 1 g of constant acceleration since that's a popular choice for interstellar travel as well as a practical number for the acceleration of a maglev train in a vacuum. I didn't run the calculations myself, since math is not my strong suit but when looking at others numbers, I think the timeframe for reaching 22% C is roughly on the order of a year or two.

The centripetal acceleration is a function of speed and radius. You do not get to choose it independently.

 

That electricity bill of $73 mil/Kg seems a bit high. Did you account for removing drag?

Easy enough to check. What's the kinetic energy of a 1 kg object at that speed?

 

Just in my head, it's around 2 x 10^14 J. A kwh is 3.6 x 10^5 J. So you need ~6 x 10^8 kwh. At $0.12 per kwh, that's $72 million.

[TakenItSeriously]

Imperial percentages!

 

How did you compute this 22%? (Compute the percentage more carefully and I guarantee you the 'aha moment' you will not forget.)

The centripetal acceleration is a function of speed and radius. You do not get to choose it independently.

I set the limit at 2 g's of centripetal force for practical matters of manned space travel. Another words, at 2 g's we cut the power and let the ship coast in a straight line away from the planet*.

 

 

Treat the earth like a centrifuge and calculate the revolutions per second to find the speed based on the circumference of the Earth.

 

Radius of the Earth: r = 6,371,000

Circumference of the Earth: C = 2πr

Centripetal force: g-force = 2 * 9.8m/s²

 

The equation for g-forces of a centrifuge:

g-force = 0.000001118r(RPS)²

or

RPS = SQRT(g-force/0.000001118r)

 

Convert revolutions/s to m/s

Speed = C * RPS

 

So

Speed = 2π * 6,371,000 * SQRT[2 * 9.8/(0.000001118 * 6,371,000)]

 

Note that at 2 g's centripetal force away from the planet and considering the 1g of gravity towards the planet, people inside the ship should experience something like anti-gravity or a 1g force radiating from the planet. Thus I used a four rail track with two below and 2 above to suspend the ship between the maglev rails.

 

*I did address the problem of hitting the atmosphere at this speed in another post, where we could build the catapult on the moon rather than try to maintain a vacuum all the way to outer space on the earth. For the radius of the moon, I calculated the speed at 2 g's centripetal force to be around 11% of c.

 

 

 

.

Edited June 12, 2016 by TakenItSeriously

[Ken Fabian]

For "free"? Even at 100% efficiency, the cost of getting a mass up to 22% of c would be $73,000,000 per kg(At 12 cents per kwh). And such a system couldn't come even close to achieving that kind of efficiency, even in operational costs.

 

The price of building those bridges, with or without PV or wind turbines to supply power, would add something to the costs too I would imagine1

 

As a thought experiment it's mildly interesting. As a physics exam question - how much energy to reach x% of c, what centripetal acceleration - it may have some application. I'm not convinced interstellar travel is a reasonable goal, let alone an achievable one for a human civilisation, even if it makes for engaging fiction.

[TakenItSeriously]

The price of building those bridges, with or without PV or wind turbines to supply power, would add something to the costs too I would imagine1

 

As a thought experiment it's mildly interesting. As a physics exam question - how much energy to reach x% of c, what centripetal acceleration - it may have some application. I'm not convinced interstellar travel is a reasonable goal, let alone an achievable one for a human civilisation, even if it makes for engaging fiction.

Well some people think of solving interstellar travel as more than mildly interesting but to each their own.

[Strange]

Well some people think of solving interstellar travel as more than mildly interesting but to each their own.

 

 

After reading the previous posts, do you still think of this idea as a "solution"? If so, I think you need to work through the calculations again.

[TakenItSeriously]

After reading the previous posts, do you still think of this idea as a "solution"? If so, I think you need to work through the calculations again.

I ran the calculations and find that at the point of experiencing 2 g's of centripetal force or a net force of 1g pointing away from the planet, the ship is moving at 66,344,000 m/s

 

Perhaps I made a mistake. Can you point to the post you were referring to?

[Janus]

I ran the calculations and find that at the point of experiencing 2 g's of centripetal force or a net force of 1g pointing away from the planet, the ship is moving at 66,344,000 m/s

 

Perhaps I made a mistake. Can you point to the post you were referring to?

Ac = v^2/r

For a net 1 g force at the surface of the Earth:

r=6378000m

Ac= 19.6 m/s

 

solving for v gives

 

[math]v = \sqrt{19.6(6378000}[/math]= 11,181 m/s* or 0.0037% of the speed of light.

 

 

 

* It is also exactly equal to the escape velocity from the surface of the Earth, or put another way, it would just enough to get you into an independent orbit around the Sun.

[swansont]

 

The equation for g-forces of a centrifuge:

g-force = 0.000001118r(RPS)²

 

 

Not sure where you got this, bit it's wrong. By a lot.

 

1 meter radius at 1 revolution per second should give 39.4 m/s^2 of acceleration (~4 g's). Your equation says 0.000001118 in whatever units it's supposed to have. So that's where ~6 orders of magnitude of error come into it.

[TakenItSeriously]

Not sure where you got this, bit it's wrong. By a lot.

 

1 meter radius at 1 revolution per second should give 39.4 m/s^2 of acceleration (~4 g's). Your equation says 0.000001118 in whatever units it's supposed to have. So that's where ~6 orders of magnitude of error come into it.

Yeah, I noticed that from Janus's post.

 

I admit that I did take a big Liberty with my usage of the formula only because it was the closest my searches were able to find for some reason. I likely wasn't using the correct terminology which might be related to my using the term "centrifugal force" which seems to now have a different usage than what I was taught in high school physics around 1980.

 

Any way, the formula was for calculating the g-force for a centrifuge. I thought that the situation was equivalent enough that it should still work but clearly, I was wrong about that.

 

I have been trying to understand why modeling it like a centrifuge could be so wrong though. I noticed the .000001118 term as well, for which no explanation was given so I thought maybe it was a conversion factor of some kind. The fact that the answer is 11181 seems to just be an odd coincidence.

 

I started trying permutations of different units but then I noticed the radius factor was inverted. So clearly there is something more fundamentally wrong with trying to use centrifuge model than I can understand.

 

Note that the centrifuge formula does use angular velocity instead of velocity but I figured that could be handled by multiplying revolutions by the circumference.

 

Edit to add:

I'm beginning to wonder if that formula was some kind of curve fit equation with a huge fudge factor

 

BTW, 6 orders of magnitude off makes me laugh at myself. Talk about biggest blunders!

Edited June 12, 2016 by TakenItSeriously

[Janus]

Note that the centrifuge formula does use angular velocity instead of velocity but I figured that could be handled by multiplying revolutions by the circumference.

Using angular velocity the formula is

 

[math]A_c = \omega^2 r[/math]

 

where [math]\omega[/math] is the angular velocity measured in radians/sec

 

solving for [math]\omega[/math] gives:

 

[math]\omega = \sqrt{\frac{A_c}{r}}[/math]

 

v is then just

 

[math]v = \omega r[/math]

[TakenItSeriously]

Using angular velocity the formula is

[math]A_c = \omega^2 r[/math]

where [math]\omega[/math] is the angular velocity measured in radians/sec

solving for [math]\omega[/math] gives:

[math]\omega = \sqrt{\frac{A_c}{r}}[/math]

v is then just

[math]v = \omega r[/math]

Actually the formula used (revolutions/min)² which I changed to (Revolutions/sec)² to fit the units for g of m/s².

 

I figured the distance traveled per revolution is just the circumference or 2πr

Edited June 12, 2016 by TakenItSeriously

[Janus]

Actually the formula used (revolutions/min)² which I changed to (Revolutions/sec)² to fit the units for g of m/s².

 

I figured the distance traveled per revolution is just the circumference or 2πr

The conversion is where you likely made the error. Your answer is ~3600 times larger than mine and since the Conversion factor between rev/min and rev/sec is 60, and you are squaring...

[Ken Fabian]

Well some people think of solving interstellar travel as more than mildly interesting but to each their own.

 

An actual feasible means of doing so would be better than mildly interesting and I suppose if somehow it was relatively quick and easy to visit other stars it would get a lot more interesting. I have yet to see any proposals that are feasible and many, like your equatorial slingshot, don't offer anything practical.

 

I remain dubious that, in the absence of a quick and easy (relatively speaking) means, there is any compelling reason to go interstellar. Scientific curiosity is in my opinion worthwhile but I don't accept that interstellar colonisation is - projects of that scale run on expectations of tangible returns and historically, successful colonies had a sound economic basis, as well as proceeding on the back of existing, economically viable technologies - not built from scratch. Curiosity - and I suppose vanity/PR projects for morale and inspirition - can be indulged when it doesn't cost too much, but as things stand interstellar travel is such a long shot that it has to be reasonably classed as impossible. Aiming high is fine and good, but exploring possibilities for doing the impossible should not divert significant resources from goals more immediate and compelling.

 

As long as it remains extraordinarily difficult and expensive and/or likely to take longer than human civilisation has so far existed to get there then practical priority means those resources should be put to use nearer to our home world.

[Janus]

I set the limit at 2 g's of centripetal force for practical matters of manned space travel. Another words, at 2 g's we cut the power and let the ship coast in a straight line away from the planet*.

 

 

Treat the earth like a centrifuge and calculate the revolutions per second to find the speed based on the circumference of the Earth.

 

Radius of the Earth: r = 6,371,000

Circumference of the Earth: C = 2πr

Centripetal force: g-force = 2 * 9.8m/s²

 

The equation for g-forces of a centrifuge:

g-force = 0.000001118r(RPS)²

or

RPS = SQRT(g-force/0.000001118r)

Okay, I think I found the source of your equation, it is this site.

 

http://clinfield.com/2012/07/how-to-convert-centrifuge-rpm-to-rcf-or-g-force/

 

Now here's whee you went wrong:

 

First off, in this equation r is measured in millimeters not meters.

 

Secondly, as you said, it uses RPM for the angular velocity.

 

So to convert to a radius in measured in meters you have to multiply the right hand of the equation by 1000 (1000 millimeters to the meter)

 

which gives:

 

g-force = 0.001118r(RPM)²

 

Now if you want to use RPS rather than RPM you have to note that R/min = R/60sec, so to convert from measuring in RPM to RPS you have to do this:

 

g-force = 0.001118r(RPSx60)²

 

60²=3600 so this converts to

 

g-force = 4.0248r(RPS)²

[TakenItSeriously]

Okay, I think I found the source of your equation, it is this site.http://clinfield.com/2012/07/how-to-convert-centrifuge-rpm-to-rcf-or-g-force/Now here's whee you went wrong:First off, in this equation r is measured in millimeters not meters.Secondly, as you said, it uses RPM for the angular velocity.So to convert to a radius in measured in meters you have to multiply the right hand of the equation by 1000 (1000 millimeters to the meter)which gives:g-force = 0.001118r(RPM)²Now if you want to use RPS rather than RPM you have to note that R/min = R/60sec, so to convert from measuring in RPM to RPS you have to do this:g-force = 0.001118r(RPSx60)²60²=3600 so this converts tog-force = 4.0248r(RPS)²

Now that makes sense, thanks for the help!