quote:Originally posted by Identity Crisis: Nope. When the exhaust is redirected the change of momentum is 2p not just p.
Okay, here's the deal.
1. Behold the Flare Rocket in deep space:
2. Behold the Flare Rocket engage engines for one second. Note the exhaust particles flying out of the back.
3. Behold the Flare Rocket three seconds later, when mystical forces have stopped it. (The engaging of engines followed by mystical stopping is a "jump".)
4. Behold as the Guardian places a mostly flat object behind the ship.
Now, if we engage engines, the ship will proceed forward. Meanwhile, the exhaust will also push the flat thing backwards.
5. Behold as Guardian attaches the flat thing to the ship with rigid beams and girders that hold it firmly in place.
6. The attempt to engage engines now will result in the following:
An accelerative force, F, will be imparted on the spaceframe by the impulse reaction of the engines. The mass leaving the back of the ship in the form of high-speed exhaust will impart a force on the spaceframe roughly equal to the mass times the acceleration.
The exhaust shall travel to the right. The ship shall thus be thrown to the left.
Upon departing the aftmost part of the ship, the exhaust products will quickly collide with the new addition, the flat device. Because the flat device is now connected to the spaceframe, the exhaust products will impart F onto the flat device, and thus the spaceframe. The force will tend to direct the flat device (and thus the spaceframe) to the right.
F - F = 0 (zero)
Now, if you had a forcefield which was somehow able to impart a new momentum force upon each individual particle of exhaust as if the forcefield were a collection of mini-trampolines, and also be able to reflect this force against the spaceframe, you might have something.
However, it would also have to impart lateral force in order to prevent the exhaust products from bouncing against the impulse drive from whence they came, or new exhaust products from bouncing into the old and ruining the whole effect.
The same problem holds true for the straw example. Let's say that Geordi could put big metal tubes (or a forcefield) up against the impulse drive. These would be shaped like a C, with one open end at the impulse drive and the other open end facing forward.
Many lateral forces would be imparted on the spaceframe of the ship (as delivered by the "C" via its connections to the spaceframe), but the net acceleration on the spacecraft would still be zero, unless the exhaust products were accelerated through the tube out the front, as by some sort of redirect of the space-time driver coils.
If the driver coils were not so engaged, the most one could hope for out of the "C" would be that the newer exhaust products could push against the old ones, producing a column of exhaust products for the Enterprise to push against, much as if you were working with thrust within an atmosphere (atmospheres alter the rules somewhat, such as the "column of air" the Harrier jump-jets get to use for lift-off).
However, this would still produce forward thrust, at least until the exhaust products bounced around in the tube and again caused a net acceleration of zero.
G2k
-------------------- . . . ceterum censeo Carthaginem esse delendam.
posted
Sorry I had to delete the attempts at pictures to go along with the "Behold . . . " parts above, but UBB was being a pest, and wouldn't even allow me to make an ANSI-looking picture of the deal. Grr.
G2k
-------------------- . . . ceterum censeo Carthaginem esse delendam.
posted
So, if you exhaled (through your mouth) in space, your lungs would thrust the air upward. So, would you fly downward, even though your throat/mouth redirects the air to the front? Seems to me that you would fly backward, opposite the direction in which the air finally leaves your body.
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quote:Originally posted by TSN: So, if you exhaled (through your mouth) in space, your lungs would thrust the air upward. So, would you fly downward, even though your throat/mouth redirects the air to the front? Seems to me that you would fly backward, opposite the direction in which the air finally leaves your body.
That would be a right-angle situation, and not a 180 of the "exhaust" like we were talking about.
However, if such a circumstance were possible, you'd probably simply engage in an unstable pinwheel, for the most part, since the gases being ejected from your lungs would impart much of their force on the back of your throat where it curves toward the mouth and nose, resulting in a rearward acceleration of your head and neck.
I don't think you'd care, though, since you'd be more concerned with the various gases boiling out of your blood in the low pressure of space.
The important concept here, yet again, is that it's not a matter of which way the exhaust products end up going. As soon as they exit your mouth (or, really, just stop touching you or affecting something that will), they could hang a left, or stop and form into a human-looking gaseous anomaly and dance the macarena. It wouldn't matter, because they would not be imparting that force to you or your "spaceframe".
-------------------- . . . ceterum censeo Carthaginem esse delendam.
posted
Exactly. You move in the opposite of the direction the gas is going when it leaves your body. And the ship goes in the opposite of the direction the gas is going when it leaves the ship. And since the forcefield is part of the ship, the gas leaves the ship when it leaves the forcefield.
So what if the forcefield curves the gas 180 degrees, and your throat only covers 90 degrees? Is there some sort of limit? At 115 degrees, the angle has an effect, but 116 will do nothing? I don't think so. Try pointing your face at your feet and then exhaling. Now your throat is curving the gas 180 degrees. And you'll fly upward.
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posted
A real world example of diverting thrust is the Harrier jump-jet. Its engine has four curved exhaust nozzles that direct the thrust at a variety of angles, from straight back to straight down and even slightly forward of straight down (allowing the plane to hover and eek backwards). I suppose you could have a "forcefield engine nozzle" that would perform the same function on a starship.
-------------------- "Well, I mean, it's generally understood that, of all of the people in the world, Mike Nelson is the best." -- ULTRA MAGNUS, steadfast in curmudgeon
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posted
Okay, look. I'm sorry if I haven't explained this basic physics concept sufficiently well to you previously. Here's my final attempt.
(Forgive my utter lack of artistry)
Thrust is achieved by the detonation of some whatever within the bell-shaped area. Forward acceleration is achieved by the momentum transfer of the exhaust against the bell-shaped area. Unused exhaust products exit immediately from the rear. (Note the fact that this is a terribly inefficient engine due to the amount of unused exhaust products (unlike the engines aboard the space shuttle, which achieve efficiencies above 95%), and thus Joshua Bell's concept from page 1 of this thread might almost work.)
Let's simplify. Suppose there were four big springs in the back of the ship, each launching a big heavy cannonball backward, as opposed to millions of particles. This is still a reaction engine, so it still follows the argument, and makes for easier demonstration. It won't be as exact, but the idea might get across better.
Let's assume that the total momentum that the cannonballs cause on the ship is 4p. Therefore each cannonball is moving with a momentum of 1p (assuming perfect momentum transfer of the cannonball-spring system, equal mass, et cetera).
Now, attached to the ship is a big flat thing that will redirect the exhaust cannonballs, preventing them from continuing aft.
Each cannonball, during launch, has transferred 1p to the ship. Now, each cannonball's 1p is coming to a rapid end against the big flat thing, thereby causing a reverse acceleration of 4p on the ship.
The net acceleration on the ship, therefore, will be zero.
But, but, what about a tube?
No, you're still screwed. Even if a rigid curved tube is suddenly attached to the ship somehow, it still isn't going to help.
Look at the lower tube. When the ball collides with the rear inner surface, it will transfer part of its energy in the 45-degree blow. This means that the ball will now be moving downward with .7071 of its original velocity (the cosine of 45 degrees).
Thus, about 30 percent of the momentum will now be redirected to the ship's spaceframe in the direction indicated by the upper arrow.
Then, blammo, a second collision with the inside of the tube. This one is at what we'll call a 30 degree angle. The ball will maintain .866 of its velocity, meaning about 15 percent of the momentum will transfer to the tube, in roughly the middle arrow's direction.
Then, something of a gentle nudge for the third collision, at what we'll call a 15 degree angle, transferring only about 5 percent of the momentum in the rough direction of the arrow. (The ball retains about 96 percent of its velocity in this collision.)
Now, the ball will be exiting toward the front of the craft with about 50% of its original velocity. According to your theory, this suggests that (for all four balls) the rocket should slow by about 50%, assuming it was still moving forward with whatever velocity that 4p from earlier created before we slapped the tubes on to redirect the exhaust products.
The problem with your theory is that the exhaust products have nothing to do with it. It is the collisions with the spaceframe that produce any result.
So, let's find out what really happened. Let us assume that the cannonballs have a mass of one kg (yeah, I know, but it's just to make the math easy), and that their original velocity is 1000 m/s.
p = m * v p = 1kg * 1000m/s
p = 1000 newtons
The first collision therefore transfers about 300 newtons to the spaceframe in the upward direction of the first arrow. The second collision transfers about 100 newtons in the middle arrow direction, and the final collision transfers a whopping 30 newtons in the lower arrow direction.
That'll work out to about 276.6 newtons of force in the direction of the long arrow by geometric vector addition. Multiply that by about 4, and you get the total rearward momentum of all four cannonballs, about 1100 newtons backward.
In other words, you accelerated by 4p, then got 1.1p back out of the tube system. This means that you've still accelerated forward by 2.9p.
Naturally, this is an imprecise calculation of what you might expect from some sort of tube that redirected the exhaust products forward from the rear end of a Galaxy Class starship impulse drive, but you see the problem you're facing now, right?
No matter what you do, the best you can hope for is *zero acceleration*, not negative (i.e. reverse) acceleration.
Even if you add in the pressure of additional exhaust products pushing the old ones out of the tube faster, there will still be no way you can expect to get the exhaust products flying out of the tube with greater momentum than what they first had, because the momentum has to come from somewhere. That's asking for a free lunch, and physics just isn't a welfare state.
posted
This would seem to suggest that, if you did exhale in space, you'd fly in the direction of the top of your head.
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posted
Identity Crisis already got the collision thing exactly right with his momentum analysis. In kinetic gas theory, one usually assumes the collisions will be elastic, so that energy is preserved in addition to momentum. In reality, this assumption actually holds pretty well, but we don't need energy conservation. We only need momentum conservation.
Momentum is always conserved *for the entire system in isolation*. Here our system is the ship and the exhaust (in vacuum, so there's nothing to thrust against and no air resistance). Initially, they are at rest wrt the camera, so momentum p(system)=p(ship)+p(gas) is zero.
Regardless of what the gas does within the ship (it could stop for tea at 10-fwd for five minutes if it wanted), eventually it exits the vessel. When it does, it carries momentum p(gas)=m(gas)v(gas). The other half of our system, the ship, therefore MUST carry the momentum p(ship)=-m(gas)v(gas), no matter what. The sum of momenta, p(system)=p(ship)+p(gas), MUST remain zero.
And don't even think about ranting about the gas hitting the walls at odd angles. You CAN'T lose momentum in collisions. You can lose mechanical energy into heat and vibrations and whatnot if the collision is inelastic, but you can't lose momentum. You can *redistribute* it from the gas to the ship and back for all week if you want, you can send some of it aboard a shuttlecraft to Risa and then bring it back, but eventually the gas exits and carries p(gas), which must exactly match p(ship), no matter what route the gas took.
All the physical, chemical or theological interactions between the gas and the ship are irrelevant - the gas is not even obligated to hit the engine nozzle walls at all, even though that's how most rocket engines tend to work. The absence or presence of an easily recognizable coupling between the gas and the ship will not alter the fact that momentum for the system MUST be conserved, or God Himself comes down and whacks you silly.
So in the end, the ship IS moving at momentum p(ship)=m(gas)v(gas) opposite the direction of v(gas). Of course, when you write out p(ship)=m(ship)v(ship) and equate that with -m(gas)(v(gas), you find out that v(ship)=-v(gas)m(gas)/m(ship) is pretty darn small, because m(gas)/m(ship) is bound to be low. Unless your v(gas) is very high, which it certainly is in a Trek impulse engine. (In fact, it's probably so high that one has to evoke Einstein to correct the momenta, but that's another discussion altogether.)
posted
PS about the transporter-based drive system: recycling a chemical propellant would of course be futile, but if the propellant is simply an inert substance being heated by a separate fusion reactor, or accelerated by EM fields, then recycling is no problem. You don't expend propellant, you only expend the fuel that fires up your reactor or your EM fields. (And if you already defy the conservation of momentum, I'm sure you can rig up a perpetual motion machine to give you the required energy without the need for a fuel!)
I'd be happier believing in a momentum-conserving transporter, though. But the momenta are still immense, since even though the masses of the people being moved are small, the distances and velocity differences are extreme.
quote:Originally posted by Timo: Identity Crisis already got the collision thing exactly right with his momentum analysis.
I hope so. I did do a physics degree, it was a few years ago now but this is fairly basic stuff.
quote:(And if you already defy the conservation of momentum, I'm sure you can rig up a perpetual motion machine to give you the required energy without the need for a fuel!)
A PM device would give so many tech possibilities that it would redefine most of what we think we know about Trek science, so let's assume that there is no PM device and that transporters somehow conserver momentum.
quote:I'd be happier believing in a momentum-conserving transporter, though. But the momenta are still immense, since even though the masses of the people being moved are small, the distances and velocity differences are extreme.
Momentum for an object in an orbit is given by p = mrw (where m is the mass of the orbiting object, r is the radius of the orbit and w is the angular velocity).
For a ship in geostationery orbit over a beam down site the initial angular momentum is p1 = (m(ship) + m(party)) * r(orbit) * w
When the party beam down they end up with momentum p2 = m(party) * r(planet) * w
So the ship will now have momentum p3 = p1 - p2.
But ths ship wants to remain above the landing site so must adjust it's momentum by firing thrusters. How big an adjustment must it make? Enough to make it's final momentum m(ship) * r(orbit) * w
Fell free to work out the following for yourselves, I can't be arsed to type it out fully.
As Rp is less than Ro the change in momentum required from the thrusters is negative, which is what we expect as the party lost momentum when it moved to the planet's surface and so the ship must have gained momentum.
Ro is the killer. For Earth (Ro - Rp) is 35,787 km. If the ship drops down to a lower orbit for transport ops then the momentum change can be reduced considerably.
-------------------- "My theories appal you, my heresies outrage you, I never answer letters and you don't like my tie." - The Doctor
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quote:Originally posted by Timo: Identity Crisis already got the collision thing exactly right with his momentum analysis.
Oh, yes, of course, quite right, I did forget that there's a witch doctor in Engineering.
I do apologize, but I find it incomprehensible that I've found myself in this argument about basic physics, especially since there's someone here who claims to have a degree in it.
I'm really very sorry, but I cannot accept the notion that space vessels are Voyager-esque devices which can magick momentum out of nowhere.
And don't get me wrong, you argued your point quite well, insofar as the very convincing misuse of the notion of the conservation of momentum. It's just that it isn't how you describe.
Momentum is well-conserved in my model without resorting to magic, declarations, et cetera, and is well in keeping with modern technological applications of the concept (reference: NASA shuttle engines and other hypothetical space propulsion technologies). At no point did I rant, and at no point did I suggest that momentum is lost. Indeed, quite the contrary, it is opposing arguments which suggest that matter is magically gained.
I'm afraid I really just can't accept that notion. Were it within our power, I would suggest that we construct a space vessel. First, I'll accelerate towards the sun, and then turn the ship around and fire the propulsion system to stop the ship. Then, I'll get out. At this point, you'll accelerate toward the sun, and lower this goofy C-shaped reverse thrust system over the engines and try to stop.
I do hope you decide to jettison the silly thing and turn the ship around to decelerate, or else I'll have to break in to the ISS to get home again.
Again, so sorry,
G2k
-------------------- . . . ceterum censeo Carthaginem esse delendam.
posted
I think you have got the wrong starting point. Thrust reversers *work*. I've seen thrust being reversed, and a jet taxi caboose first under the power of such reversed thrust. Sure, the process is wasteful, but mainly because of all the turbulence and thermodynamic losses that wouldn't be present if we dealt with ideal gas and laminar flow. This would not work if a thrust reverser merely killed forward thrust.
So there *must* be something wrong with your analysis. What could it be?
Let's try trace again what happens to a gas particle in a rocket engine. First, it obtains kinetic energy from an exothermic reaction (chemical, nuclear) or is given the energy by some other means (EM fields, laser or solar heating). It has a nonzero momentum p now, too.
Now, the particle moves out through the open nozzle. It didn't deliver momentum to the ship by any visible mechanism. Did it violate conservation of momentum? Not from its own POV. It was its own system. The wider system of ship and particle conserved momentum as well, because the particle didn't gain its p by stealing it from the ship - it got the p from the chemical/other reaction, which wasn't necessarily coupled to the ship.
So statistically only half of the particles matter at this stage - namely those that impact the front wall of the rocket burn chamber and thus deliver a momentum to it that has at least some component pointing toward the bow of the ship. When they deliver this momentum, they also lose some of their velocity.
Now introduce the thrust reverser. Which particles will impact that one? Not all the ones that hit the front wall - they are doomed to be recycled for another round of reactions within the chamber, since they don't have the velocity. But many of them will. However, the ones that previously escaped scot free through the nozzle opening will now play the crucial role - essentially, the reverser plate will be the "chamber front wall" for them, except that it is the back wall and results in opposite movement.
With the reverser plate in place, each particle could suffer multiple collisions during its journeys. Eventually, however, the particles that escape the ship have logged in more collisions with the reverser plate than with the front wall on the average (remember that half the particles initially head out for the reverser plate without touching the front wall at all), whereas the particles that do not yet escape are just cancelling out each other.
When one does the summing of momenta over this entire system, and remembers to include the "creation" of momentum out of "nothing" due to the chemical/other reaction, the sum should come up zero and the ship should move as hoped.
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