No, I'm not accusing Bernd of being inconsistent! Instead, he has some new commentary in the "Inconsistencies" section of Ex Astris Scientia, and of course it deserves the usual level of careful scrutiny.
In the department of "I always had the nagging feeling there was something wrong but I couldn't put a finger on it", the new pages point out the inconsistencies in navigation and propulsion, weapons and tactics, and some tech miscellany. Lots of good points there.
In the first category, though, Bernd says that a solid or forcefield-based jet reverser attached to the impulse engines would not alter the net force on the ship. This isn't true, as the end result will still be that the exhaust jet is directed forward, thus transferring momentum in that direction and giving the ship momentum in the opposite direction. The internal structure of the engine does not matter - internal forces will cancel out, save for the one opposite the final exhaust direction.
In the second one, I found no nits to pick. Darn.
In the Miscellany category, I'd suggest that the oddly shaped "forcefields" are based on the same technology as the "shields". That is, their effect is based on gravitons trapped in a subspace field. The abruptness of the forcefield effect in the direction normal to the field surface would then be easier to explain - an effect is only created in the region where there are trapped gravitons, and that effect doesn't obey the 1/r^2 rule any more than the other gravitic tech (say, deck plating) of Star Trek does. As for the ability of even a single generator to create an oddly shaped subspace field to suspend the gravitons in... I doubt subspace follows 1/r^2, either.
On the same vein, I gather that "magnetic" boots are based on gravitic tech, too. They just use a catchy name that has alternate meanings. If any technology in Trek is cheap, reliable and easily miniaturized, gravitics is (by necessity, because artificial gravity always has to be present in even the most unlikely locations for production reasons).
As for stairs as an alternative to turbolifts, Kirk's old ship had plenty. There was that vertical ladderway easily accessible from the curved corridor set, and often seen used. And ST2 showed the crew operating efficiently without turbolift access anywhere below the top three decks. Either there was a big staircase somewhere (as Probert originally intended), or then using the stairs wasn't all that hard after all.
The E-D crew seldom resorted to vertical Jeffries tubes as an alternative to turbolifts. There just wasn't an obvious need to go from one deck to another during a crisis: most of the heroes would sit on the bridge, while LaForge and perhaps Data would sit at Engineering, and nameless professionals would man the torp bays or the phaser control rooms and so forth, so the heroes needn't go there.
Vertical access was only needed for LaForge to access crucial damaged systems - and those systems all seemed clustered around the Jeffries tubes anyway (which is a separate yet very annoying nit). If the crew had to resort to crawling, it was not because the turbolifts were hiccuping - it was because the whole ship, lifts and doorways and corridors and the unseen staircases alike, was locked up by malfunctions or malevolently used security protocols.
Voyager had a greater problem with this: a small number of competent crew had to access vital systems distributed on multiple decks, and they were *always* crawling around. The "locals" just couldn't handle the crises. And there were indeed episodes where people were trapped on a certain deck by what looked like a simple turbolift failure or general power failure.
Timo Saloniemi
Posted by CaptainMike (Member # 709) on :
Is there an official EAS decision as to whether or not bunnies are soft?
Posted by Timo (Member # 245) on :
"Canonical or semi-official bunnies?"
"Gee, I don't kn... Aaaaaggggghhh!"
A king has to know a thing or two...
(But it's still a good security measure to carry a grenade with you in case the bunny turns out not to be so soft after all.)
Timo Saloniemi
Posted by Guardian 2000 (Member # 743) on :
quote:Originally posted by Timo: In the first category, though, Bernd says that a solid or forcefield-based jet reverser attached to the impulse engines would not alter the net force on the ship. This isn't true, as the end result will still be that the exhaust jet is directed forward, thus transferring momentum in that direction and giving the ship momentum in the opposite direction. The internal structure of the engine does not matter - internal forces will cancel out, save for the one opposite the final exhaust direction.
Final exhaust direction is irrelevant. Rockets, jets, et cetera all operate by imparting force on the spaceframe of the vessel, be that a shuttle, aircraft, or what-have-you. The exhaust does not "push against" anything to produce thrust, though that is a common misconception.
(It's kinda accurate in reference to hovercraft, where the engine exhaust is contained and provides a cushion of air, but that's not what we're talking about.)
Some sort of physical device built to reflect the exhaust toward the front of the vessel would, by default, have to be connected to the vessel. Therefore, the net acceleration on the starship spaceframe would be zero.
The only way such a system would work would be if some sort of device could grab the exhaust products, redirect them forward, and in the process cause the exhaust products to impart a new force on the spaceframe, providing an accelerative force against the ship (albeit in a bass-ackwards direction).
G2k
Posted by Treknophyle (Member # 509) on :
Referencing the 'reverse exhaust' discussion.
1- A close look at the ventral surface of the E-e's impulse bays shows 4 rectangular vents - which I interpret to be reverse-direction thrust ports.
2- However - I must disagree with the argument that a force field-based thrust reveral nozzle would be ineffective: a- The plasma would be vented forward at high speed. b- There is a verifiable degree of 'momentum feedback' with shields (if a meteor of bolt of energy hits the external shields, the ship as a whole is shaken. I assume that the momentum is transferred back to the generators (which therefore must be braced and welded down).
Posted by Guardian 2000 (Member # 743) on :
quote:Originally posted by Treknophyle: Referencing the 'reverse exhaust' discussion.
1- A close look at the ventral surface of the E-e's impulse bays shows 4 rectangular vents - which I interpret to be reverse-direction thrust ports.
2- However - I must disagree with the argument that a force field-based thrust reveral nozzle would be ineffective: a- The plasma would be vented forward at high speed. b- There is a verifiable degree of 'momentum feedback' with shields (if a meteor of bolt of energy hits the external shields, the ship as a whole is shaken. I assume that the momentum is transferred back to the generators (which therefore must be braced and welded down).
If there are forward-facing ports near the shuttlebay, they could be used as reverse thrusters. However, they would either have to have their own power source, or else the IPS reactors would (A) have separate outlets for the thrust, closing off the rearward ones or (B) the reactors themselves would have to turn to blast in a new direction.
As for the "vented forward" argument, the principle is that first, the fusion reaction exhaust would have pushed against the front part of the reactor, exiting out of the tail and producing forward thrust.
Any system which causes the exhaust to reverse direction (even a forcefield) will be doing nothing more than "bouncing" the exhaust toward the front of the ship, canceling out the forward thrust and bringing the thrust total down to zero. Just because the exhaust products are now headed forward does not mean that the ship must now move backward.
If you need confirmation, take a nasty little cheap-o plastic toy car . . . something light, but big enough to tape a balloon to. Take the balloon (blown up with air), tape the bottom of it to the car, and then let go and watch the car fly.
Now, attach something big and flat to the back of the car. It doesn't matter what it is . . . it can be a big flat piece of cardboard, a sail/parachute, or a nifty something designed to redirect the airflow backward. As long as you do your job well, the car isn't going to move.
The only way it would move the way you want it to would be for you to make something to catch all of the air out of the balloon, hold it, and then blow it out toward the front of the car. But, it would just be easier to turn the original balloon around, or the whole car.
G2k
Posted by David Templar (Member # 580) on :
quote:Originally posted by CaptainMike: Is there an official EAS decision as to whether or not bunnies are soft?
Damnit, you stole my line! Posted by Joshua Bell (Member # 327) on :
quote:Originally posted by Guardian 2000: Any system which causes the exhaust to reverse direction (even a forcefield) will be doing nothing more than "bouncing" the exhaust toward the front of the ship, canceling out the forward thrust and bringing the thrust total down to zero. Just because the exhaust products are now headed forward does not mean that the ship must now move backward.
Hogwash.
There are two ways to look at the problem, both of which are equivalent and give the same answer.
-----
First, consider the ship and any assorted engines and thrust deflectors as a black box. The only things coming out of the black box is a stream of particles moving faster than the ship in the direction of travel. To conserve energy and momentum, the black box must be accelerating in the opposite direction. That's basic physics.
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Second, consider the transfer of momentum at the level of particles, starting in the fusion drive. The fusion engine is imparting energy to the particles along random orientations which imparts no net momentum to the ship. Some of the particles bounce off the sides of the engine, but this happens in a symmetric fashion so there is no net transfer of energy (some of the particles hit the top accelerating the ship upwards and the particles reflect downwards; some of the particles hit the bottom, accelerating the ship downwards and the particles reflect upwards). Some of the particles hit the front of the engine, accelerating the ship forwards while the particles reflect backwards. The rest of the particles escape from the engine without a (net) transfering energy to the ship (i.e. they do, but it is offset by the particles that started off going forwards).
The net result of this is that there is a stream of particles leaving the engine to the reverse while the ship is given a forwards momentum. Note that all of the particles that leave the engine must have a negative direction - and all particles must leave the engine eventually or it will clog up.
Now this stream of particles impacts a thrust reverser and reflects forwards. This imparts a negative momentum to the ship. Note that ALL of the particles leaving the engine impact the thrust reverser and transfer negative momentum, while only particles which impacted the front of the engine imparted positive momentum. Thus, there is a net negative momentum imparted to the ship.
If this were somehow balanced (i.e. the same number of particles impact the front of the engine as the impact the back) the ship would still experience a negative momentum increase. For example, if the initial momentum imparted to the particles was not omnidirectional (imagine you're running an ion drive which acts like a microscopic rail gun) then just imparting positive momentum to the particles (i.e. towards the front of the engine) necessarily imparts negative momentum to the ship. The particles then bounce off the front of the engine (a bit of positive momentum), then off the thrust diverter (a bit of negative momentum). Since the latter two particle bounces cancel each other out, you are left with the initial negative momentum from the particle generation.
Both of these viewpoints (black box & watching particles) match, which is what you'd expect.
------
Your car/balloon/cardboard example is flawed. First, in most cases, the balloon is going to have only just enough energy to move the car without the cardboard in place (i.e. given air resistance and friction, it's not imparting 10x the energy to get the car going, probably only around 2x). Secondly, the air bouncing off the cardboard primarily goes off to the sides; it is not a unidirectional thrust diverter. More than likely, this reduces the available energy from 2x to 0.5x and thus the car doesn't move.
If you used a styrofoam cup as the diverter to direct the flow unidirectionally, I bet it would work just fine.
-----
Lastly, jet airliners use thrust diverters to slow the plane upon landing. This is practical evidence that the darn things do work.
Posted by Guardian 2000 (Member # 743) on :
quote:Originally posted by Joshua Bell: Hogwash.
(Snicker)
I'll just let you take it up with Sir Isaac Newton, Albert Einstein, Stephen Hawking, and the others with whom you are in disagreement.
G2k
Posted by Guardian 2000 (Member # 743) on :
quote:Originally posted by Joshua Bell: Lastly, jet airliners use thrust diverters to slow the plane upon landing. This is practical evidence that the darn things do work.
No, this is evidence that they needed a way to reduce the forward thrust from, say, %10 to %5 without stalling the engine.
The thrust diverters do not slow the plane, they divert the forward thrust that is keeping it from slowing. Hence the name "thrust diverters".
G2k
Posted by TSN (Member # 31) on :
The car+balloon+cardboard example is also flawed because of the cardboard itself. You're acting like the impulse engines just have a big flat forcefield behind them that the exhaust bounces off of. That's just silly. The forcefield should be shaped like a tube that comes out from the impulse engine and curves around to be pointing to the front of the ship. It would be no different than if a giant curved metal tube came out and redirected the exhaust to the front of the ship. Take your balloon and stick a straw into the opening, seal it, and make the straw curve around to point in the other direction. Then see which way it flies.
Posted by CaptainMike (Member # 709) on :
assuming of course we understand the way a forcefield works. maybe its more a scattering field than a 'wall'
Posted by Identity Crisis (Member # 67) on :
quote:Originally posted by Guardian 2000: Any system which causes the exhaust to reverse direction (even a forcefield) will be doing nothing more than "bouncing" the exhaust toward the front of the ship, canceling out the forward thrust and bringing the thrust total down to zero. Just because the exhaust products are now headed forward does not mean that the ship must now move backward.
Nope. When the exhaust is redirected the change of momentum is 2p not just p.
A particle heading aft with momentum p hits a forcefield (we'll assume that the collision is perfectly elastic) and bounces off with momentum -p.
The total change of momentum (from p to -p) is thus 2p.
So the ship gains forwards momentum of p when it first generates the particle and backwards momentum of 2p when it bounces the particle.
p-2p = -p.
The ship now has a total backwards momentum of p.
Of course, no collision is 100% elastic and the redirected particle can't be heading directly back the way it came (otherwise it would go straight back into the impulse drive) so the ship's backwards momentum would be less p but, giving Starfleet engineers some credit, that's only a change of magnitude not of direction.
Posted by Timo (Member # 245) on :
Now that this confusion is out of the way, we might consider the possibilities offered by the transporter. That machine doesn't seem to preserve momentum or angular momentum - after all, it can handle surface-to-orbit transport just as fine as a ship-to-ship one.
Let's start easy. What about rigging a rocket engine inside the ship, and having the jet of propellant be transported back to the fuel tanks after it has been spat out by the rocket? An exhaustless drive that consumes no propellant. You only need the fuel that gives you energy for the rocket and the transporter - say, a fission or fusion reactor or something.
Or, if the transporter can give the required momentum and angular momentum to somebody who is transported from ship to surface, I guess it could also give such momentum to a jet of gas. No need for actual engines - just transport a puff of gas that will shove the ship in the desired direction, then grab that gas again and give it another dose of momentum, again and again and again until your ship moves at relativistic speeds... Your "engine nozzle" could be a hollow sphere at the heart of your ship, with the puffs of gas hitting the inner walls of the sphere.
And so forth and so forth. A machine that doesn't conserve momentum is a pretty darn impressive propulsion system, even if it can only handle a small amount of matter at a time. You don't need to transport an entire starship to get it moving...
Or does the transporter in fact conserve momentum? If so, where is it dumped? Some sort of huge flywheels which rev up when somebody beams up, and are tapped for momentum when somebody beams down? Heating and cooling of something by fiddling with the momenta of the molecules? That's pretty impressive, too. And a great weapon or way to get free energy, or then a horrid source of environmental damage if the transporter dumps the momentum on nearby planets and stuff.
Next week on Flare Forum: artificial gravity and how it causes 6.7-Richter earthquakes in Newton's grave...
Timo Saloniemi
Posted by Identity Crisis (Member # 67) on :
I'd guess that the transporter (and thus the ship) absorbs/supplies the required momentum. After all the change in momemntum for six humanoid masses going from standard orbit to planetary surface could be transfered to a starship sized mass without much effect on its orbital velocity.
If this works out then your transporter drive idea is a dud.
I think your rocket idea is dodgy as well. It would only work if the nature of the fuel wasn't changed. If you ignite the fuel (to release the energy needed to accelerate it to exhaust velocities) then the fuel will no longer be the same after use as before so transporting it back to the fuel tank would be pointless.
Posted by Guardian 2000 (Member # 743) on :
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
Posted by Guardian 2000 (Member # 743) on :
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
Posted by TSN (Member # 31) on :
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.
Posted by Guardian 2000 (Member # 743) on :
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".
Posted by TSN (Member # 31) on :
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.
Posted by MrNeutron (Member # 524) on :
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.
Posted by Guardian 2000 (Member # 743) on :
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.
G2k
[ February 17, 2002, 22:21: Message edited by: Guardian 2000 ]
Posted by TSN (Member # 31) on :
This would seem to suggest that, if you did exhale in space, you'd fly in the direction of the top of your head.
Posted by Guardian 2000 (Member # 743) on :
quote:Originally posted by TSN: This would seem to suggest that, if you did exhale in space, you'd fly in the direction of the top of your head.
You "must... find... brain... think...".
G2k
Posted by Timo (Member # 245) on :
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.)
Timo Saloniemi
Posted by Timo (Member # 245) on :
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.
Timo Saloniemi
Posted by Identity Crisis (Member # 67) on :
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.
Posted by Guardian 2000 (Member # 743) on :
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
Posted by Timo (Member # 245) on :
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.
Does that sound plausible now?
Timo Saloniemi
Posted by Shik (Member # 343) on :
*pop*
OW! My fissure of Rolondo!!
Posted by Guardian 2000 (Member # 743) on :
quote:Originally posted by Timo: 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.
False analogy. First, atmospheres introduce new variables into the situation, since you can get some rebound off of the surrounding air (or even the ground in a Harrier or hovercraft). Second, you're talking about a turbofan gas turbine engine, not a direct reaction engine or a rocket . . . same basic principles, to be sure, but significantly different execution than in, say, a space shuttle main engine. Ninety percent of the thrust of a turbofan engine comes not from the combustion of the fuel and air, but from the massive fan at the front of the engine nacelle which is driven by the gas turbine part of the engine.
Thrust reversers on an aircraft are predominately non-propulsive, though (and I am correcting an earlier statement after further research) can provide some effect, thanks in large part to the atmospheric effects.
quote: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.
And above is the problem with your notion, wherein you presume low efficiency for the engine. Modern chemical rocket engines, such as the shuttle main engines, use a multistage combustion approach, leading to excellent combustion efficiency (99%) and excellent thrust efficiency.
The impulse engines of a Galaxy Class starship, according to the TNG:TM, produce thrust by channeling the exhaust products from the spherical fusion reactor in an aftward direction, then through an accelerator, and finally more thrust is obtained via the spacetime driver coils . . . all of this is translated to the spaceframe by whatever technobabble means.
We are not given the efficiency of this engine configuration, but I'd expect it to be a helluva lot better than 50%.
The remainder of your argument is based on the low engine efficiency presumption and therefore will only hold true in that case.
G2k
Posted by Bernd (Member # 6) on :
This has become more complicated than I expected. At first, my thoughts were exactly the same as those of Guardian 2000 (zero acceleration is the best that could be achieved), with the momentum conservation in mind, but then several new problems showed up.
The prerequisites for my considerations are: 1. The exhaust particles don't interact with each other, not even if they are trapped and accumulated between the engine and a possible reflector. 2. The particles don't interact with a medium, be it air or occasional atoms in space. 3. All momentum transfers are elastic. 4. It is a stationary random process, meaning that fluctuations in speeds, rates and directions average out over time. This allows us to look at a momentum instead of a differential momentum. Strictly speaking, we would have to look at a differential momentum though, if we take into account that the ship is still being accelerated while the gases are not as soon as they have left the ship (and thereby the common system), but we may postulate that the acceleration is negligible over a short enough time. 5. We use only classical physics. 6. We don't take combustion efficiency into consideration. What counts is the momentum the accelerated particles exert on the ship, no matter how they have been accelerated.
My starting point was that, if
system momentum = sum of the particle momentum vectors + ship momentum = 0
it would not matter what happens inside the rocket engine. We could simply add the momentum vectors of the particles across a sphere around the ship to determine the momentum of the ship. The consequence is that the final direction of the exhaust products would ultimately determine the direction of the ship's momentum. But how does this statement get along with Guardian 2000's pipe where the exhaust moves in the same direction as the ship?
Next, I considered the argument that a good deal of the exhaust gas would not be coupled to the ship's spaceframe, if we imagine a rocket drive where gases are mixed and exploded in a burning chamber. But this problem wouldn't exist in an ion drive where particles are accelerated through an electromagnetic field, for instance. I'm actually not sure as to how much of the exhaust products actually propel a fighter jet or a rocket, but I don't need to care about that because there exist concepts where all particles contribute to it. So, ideally, it's really a matter of measuring the momentum of all particles that leave the engine.
The redirection tube is something like a little efficient rocket engine in this respect, but why does the ship still have a momentum forward, if the final momentum of the particles goes forward too? What do we have to add or subtract to get the momentum conservation right again?
And what do thrust diverters actually do? It seems they are really used to slow down a plane. Moreover, the translation of the German word is actually "thrust reversers".
Although I still believe that no reflector could reverse the thrust, I'm confused.
Posted by Joshua Bell (Member # 327) on :
quote:Originally posted by Bernd: system momentum = sum of the particle momentum vectors + ship momentum = 0
That's it exactly. Now go ask a 10-year-old what happens if you attach a bent tube to the end of a balloon glued to a toy car. The 10 year old will get it right, since she's not thinking about momentum transfers. She's just thinking "the air goes this way so the car goes that way". This is one case where common sense is correct.
Or watch a Harrier take off or land via VTOL, as someone pointed out previously. Or, even better, try it yourself.
But how does this statement get along with Guardian 2000's pipe where the exhaust moves in the same direction as the ship?
The ship is still slowing. Imagine that the ship is at rest (relative to something, obviously) and the engines crank up with the funky hoses attached. The ship will start moving backwards. If you imagine what would happen on the filmstrip before t=0 the ship will have forwards momentum but be slowing due to the forwards exhaust.
Again - just try it. Go out, buy a balloon, a toy car, and a bendy straw.
Posted by Bernd (Member # 6) on :
quote:Or watch a Harrier take off or land via VTOL, as someone pointed out previously. Or, even better, try it yourself.
Wouldn't it be a bit exaggerated to join the Royal Air Force only for that purpose?
quote:The ship is still slowing. Imagine that the ship is at rest (relative to something, obviously) and the engines crank up with the funky hoses attached. The ship will start moving backwards.
I would like to see the correct momentums for that case which would have to sum up to zero. Viewed from far away, as I already suggested, it would be quite clear, but the effects inside the tubes are still a mystery to me.
Posted by Guardian 2000 (Member # 743) on :
Actually, as it stands now, there wouldn't appear to be much difference of opinion.
I was presuming a vessel with a sublight impulse reaction drive system of very high efficiency . . . thus leading any thrust reversing system to at best end up doing squat, even at maximum possible efficiency.
The others were thinking about a system with less than 50% efficiency, meaning that an extremely efficient thrust reverser might be able to slow the ship ever so slightly.
It basically comes down to how efficient you think an impulse engine is.
G2k
Posted by Identity Crisis (Member # 67) on :
quote:Originally posted by Guardian 2000: I was presuming a vessel with a sublight impulse reaction drive system of very high efficiency
Really? Please tell us how your are defining efficiency in this case. Because conservation of momentum shows that the thrust reverser works perfectly regardless of what happens inside the engine itself.
Posted by Guardian 2000 (Member # 743) on :
quote:Originally posted by Identity Crisis:
quote:Originally posted by Guardian 2000: I was presuming a vessel with a sublight impulse reaction drive system of very high efficiency
Really? Please tell us how your are defining efficiency in this case. Because conservation of momentum shows that the thrust reverser works perfectly regardless of what happens inside the engine itself.
"Efficiency" in this instance refers to the percentage of drive products that actually get used for thrust, energy imparting to the spaceframe in the appropriate direction, instead of floating off and doing nothing.
My cannonball example vessel, for instance, had roughly one hundred percent efficiency, and thrust reversers were not useful. The example ship for the argument to the contrary had a ship which slung most drive products out of the ship in a more wasteful fashion, and it was these drive products which caused reverse thrust to be possible, since they only ever imparted energy to the reverser.
[ March 06, 2002, 20:52: Message edited by: Guardian 2000 ]
Posted by Identity Crisis (Member # 67) on :
quote:Originally posted by Guardian 2000: "Efficiency" in this instance refers to the percentage of drive products that actually get used for thrust, energy imparting to the spaceframe in the appropriate direction, instead of floating off and doing nothing.
My cannonball example vessel, for instance, had roughly one hundred percent efficiency, and thrust reversers were not useful. The example ship for the argument to the contrary had a ship which slung most drive products out of the ship in a more wasteful fashion, and it was these drive products which caused reverse thrust to be possible, since they only ever imparted energy to the reverser.
Rubbish. My simple conservation of momentum example works with a single 'cannonball' exhaust. This whole thing about particles being generated in all directions is a huge red herring and has nothing to do with the problem in hand.
It's very simple so let's go through it one more time.
1. The ship is at rest relative to some external viewpoint. (Both ship and viewpoint may be moving at some constant velocity, but in a frictionless medium such as a vacuum this is irrelevant). In this frame of reference the ship has zero momentum and thus the whole system has a total momentum of zero.
2. Ship generates a single cannonball exhaust particle of mass m and velocity v moving in the -x direction. Particle has momentum of -mv so the ship must have a momentum of mv giving the whole system a total momentum of zero.
Before: SHIP zero(x) zero(y) PARTICLE zero(x) zero(y) TOTAL FOR SYSTEM zero(x) zero(y)
After: SHIP mv(x) zero(y) PARTICLE -mv(x) zero(y) TOTAL FOR SYSTEM zero(x) zero(y)
3. Cannonball particle strikes a 45 degree plate rigidly attached to the ship (In 'reality' a force field) and bounces off in the +y direction. If the collision is perfectly elastic then kinetic energy is preserved and the particle now has velocity v in the +y direction. The particle now has zero momentum in the x direction and mv in the y direction. To conserve momentum the ship must have gained -mv in the x direction and -mv in the y direction.
Before: SHIP mv(x) zero(y) PARTICLE -mv(x) zero(y) TOTAL FOR SYSTEM zero(x) zero(y)
After: SHIP zero(x) -mv(y) PARTICLE zero(x) mv(y) TOTAL FOR SYSTEM zero(x) zero(y)
4. Cannonball particle strikes another 45 degree plate rigidly attached to the ship and bounces off in the +x direction. The particle now has zero momentum in the y direction and mv in the x direction. To conserve momentum the ship must have gained -mv in the x direction and mv in the y direction.
Before: SHIP zero(x) -mv(y) PARTICLE zero(x) mv(y) TOTAL FOR SYSTEM zero(x) zero(y)
After: SHIP -mv(x) zero(y) PARTICLE mv(x) zero(y) TOTAL FOR SYSTEM zero(x) zero(y)
Gosh, what's this? The ship now has momentum of -mv in the x direction. If it was moving with any velocity before the process started it is now moving slower, and if it was stationery it is now moving backwards.
Even if the collisions are not perfectly elastic it still works. Assume that a massive 75% of the particles kinetic energy is lost to waste heat at each collision. By conservation of energy this means that 1/2 m v2 ^2 = 1/4 * 1/2 m v1 ^2. Cancel the common terms and v2 ^2 = 1/4 v1 ^2 so v2 = 1/2 v1. So after the collision the particle is moving at half the speed it was before. So the first collision is now:
Before: SHIP mv(x) zero(y) PARTICLE -mv(x) zero(y) TOTAL FOR SYSTEM zero(x) zero(y)
After: SHIP zero(x) -0.5mv(y) PARTICLE zero(x) 0.5mv(y) TOTAL FOR SYSTEM zero(x) zero(y)
and the second collision is
Before: SHIP zero(x) -0.5mv(y) PARTICLE zero(x) 0.5mv(y) TOTAL FOR SYSTEM zero(x) zero(y)
After: SHIP -0.25mv(x) zero(y) PARTICLE 0.25mv(x) zero(y) TOTAL FOR SYSTEM zero(x) zero(y)
So even with 75% of the particles energy lost to waste heat at each collision the ship still ends up with negative momentum.
Obviously two 45 degree collisions is a simplification but the principle remains the same no matter what the details are: if the exhaust particle ends up going forwards then the ship has gained negative momentum.
Now please let this thread die.
Posted by Guardian 2000 (Member # 743) on :
This thread has been bothering me for over a year. Though the conceptual visualization I engaged in still seemed to be spot-on, the inherent messiness of simple-text equations and the confusion over elastic collisions just had my eyes glazing over to the point that I simply wasn't sure. I ended up never replying . . . in part because my head hurt, and in part because I didn't want to end up in a state of war.
This topic, however, was recently revisited at the Subspace Comms Network (link). And as I returned to the problem, and the thread, I saw what I'd missed so long ago. (I could smack myself for missing it, too.)
Using a simplified version of the "bendy straw" reverse thrust concept, we came to the following diagram (I've long since lost the ones I made for the thread):
1. Now, in that last message, IC argued in section #3 that "If the collision is perfectly elastic then kinetic energy is preserved and the particle now has velocity v in the +y direction."
The problem is that he failed to follow conservation of anything. By giving the cannonball ("particle") a pre-collision velocity of V and a post-collision velocity of V (or, more specifically, a momentum of mv in both cases), he increased the overall momentum of the system. Why? Because he gave the ship a momentum of mv in the opposite direction. That's 2mv to come out of the collision. Instead of the ship and cannonball sharing a set amount of momentum, it has been artificially increased at the moment of collision.
The conservation of momentum is not limited to the cannonball, but to the entire system . . . the respective kinetic energies of both the vehicle AND the cannonball.
2. In section #4, he once again gave the collision an artificial doubling. The end result of this is that the cannonball departed the ship with the exact same amount of momentum it originally had, and yet the ship also had that same amount.
There is no such thing as a free lunch. If momentum could be extracted from nothing in this manner, there would be one. The concept behind the conservation of momentum is that, for the system, mv(before) = mv (after) . . . not for just one component of that system, as IC described.
Posted by Identity Crisis (Member # 67) on :
"You wouldn't let it lie..."
Momentum is conserved at every step.
Momentum is a vector quantity not a scalar one. You have to add it according to vector rules.
But even according to your simplistic system of scalar addition the momentum is conserved. 2[mv] goes into the system (mv for the ship and -mv for the ship, both in the x-axis) and 2[mv] comes out of the system (-mv for the ship and mv for the particle, but now both in the y-axis).
No extra momentum is being created at any stage other than when the particle is first created in the reactor.
The only scalar quantity available to us is kinetic energy. Can you give any step in the first (perfectly elastic) process where the KE of the particle is not the same as in all other steps? No, it is always 1/2 m v^2 because the speed of the particle is always [v] (the velocity changed with every step because the direction changes, but the speed stays the same).
Posted by Cartmaniac (Member # 256) on :
quote:False analogy. First, atmospheres introduce new variables into the situation, since you can get some rebound off of the surrounding air (or even the ground in a Harrier or hovercraft).
Rebound doesn't account for more than one percent of the Harrier's total upward momentum (due to the position of the exhaust nozzles) and can be neglected.
quote:Second, you're talking about a turbofan gas turbine engine, not a direct reaction engine or a rocket . . . same basic principles, to be sure, but significantly different execution than in, say, a space shuttle main engine. Ninety percent of the thrust of a turbofan engine comes not from the combustion of the fuel and air, but from the massive fan at the front of the engine nacelle which is driven by the gas turbine part of the engine.
The "fuck"? Dude, we're talking literally TONS of exhaust gas moving at twice the speed of SOUND here. That produces a LOT more thrust than the bypass airflow generated by the fan itself.
Posted by Guardian 2000 (Member # 743) on :
quote:Originally posted by Identity Crisis: "You wouldn't let it lie..."
� That is correct. �
quote:Momentum is a vector quantity not a scalar one. You have to add it according to vector rules.
� Yes, yes, a magnitude and a direction. That's all well and good, but it won't allow for a doubling of the momentum . . . how you add it up is irrelevant compared to that. �
quote:But even according to your simplistic system of scalar addition the momentum is conserved. 2[mv] goes into the system (mv for the ship and -mv for the ship, both in the x-axis) and 2[mv] comes out of the system (-mv for the ship and mv for the particle, but now both in the y-axis).
� That's zero for the ship resulting in 2mv! What are you even talking about? � Look, just forget the tube for a moment. I think the axes are confusing you. If you put a plate at the back of the ship and have it attached to the ship, then when the ejected cannonball hits it the cannonball and the ship will stop relative to one another. In effect, it cancels the momentum of each individual participant, and overall momentum is conserved. � The principle with the tube is the same. The y-axis components of the hits (one "down", one "up") cancel one another out, leaving us with no y-ward momentum. We're left with the two x-axis hits. � In your example, each one of these produces mv on the ship on the x-axis . . . one stops the ship, the other sends it flying backward. If two 90 degree-trajectory-change collisions can produce this effect, why not a single 180 degree collision, as with the plate? � The answer is, they can't. Otherwise, four 45 degree-trajectory-change collisions would propel the ship backwards at 4 times the rate . . . but that doesn't happen. � The first collision (your step #3) should not have caused 1mv(-x) and 1mv(-y). The collision should've caused 1/2mv(-y) and 1/2mv(-x). (In other words, the force on the ship would've driven her both backward and downward simultaneously.)
Why? See below:
Your version of events is colored red and purple. My version is in blue and green. I'll try to use the proper lingo in explaining the image.
The ball, when it strikes the angled wall (or another ball), transfers force along the line connecting their centers (the diagonal blue line). We can take the original momentum of the ball (which is 1mv(-x)) and render it as two vectors . . . the one along that center-line, and a perpendicular one (the green one going up and to the left which, since this is to be a friction-free event, won't experience any change in magnitude).
Meanwhile, like the swinging ball desktop toy, momentum will be transferred to the ball as well, along that centerline. From the wall, that's the green line going up and to the right.
The result is the same for the ball . . . it will head upward along the green line with 1mv(y).
For the ship, however, the result is quite different. Isolating the two wall-hits for the moment, we find that they both result in .7071mv to the ship in their respective diagonal directions, or 1mv(-x) along the x axis. The other 1mv of the occasion was along the y axis, but it effectively cancels itself out during the process of changing the cannonball's course (remember, there's no such thing as negative momentum).
Since the ship had a momentum of 1mv(x) from ejecting the ball in the first place, the 1mv(-x) results in a final velocity change of zero.
That means that if the ship were moving with any velocity beforehand, it is now going to continue moving with that same velocity. If it were stationary, it will remain so.
Now, I'll grant that there may be an error on my part somewhere. Despite the fact that I have an intuitive grasp of (and ability to visualize) some of the more exotic physics issues even beyond my mathematical understanding, it remains a curious irony that certain issues of mechanics just give me a headache. (On the other hand, I kick ass at pool, so go figure.) But, if I have made an error, I can't see (or visualize) it.
Posted by Identity Crisis (Member # 67) on :
Somehow you're making this sum : mv + -mv = 2mv Which is of course bollocks.
mv + -mv = 0 as in my original example.
'nuff said.
Posted by Guardian 2000 (Member # 743) on :
quote:Originally posted by Identity Crisis: [QB] Somehow you're making this sum : mv + -mv = 2mv Which is of course bollocks.
Funny . . . I explicitly did not make that sum. On the other hand, it does resemble your conclusion in regards to the final momentum of the spaceship.
It appears that you're somehow trying to suggest that I engaged in a variation of the artificial momentum doubling you seem to have employed.
Please explain.
Posted by Identity Crisis (Member # 67) on :
You claim that I double the momentum of the ship.
As at every stage the momentum of the particle, the ship and the system in my example is given as eiether mv, -mv or 0 there is no way that I can have done so. So i had to reach the conclusion that you had made such an error when trying to follow my maths.
Please say explicitly in which step I doubled the momentum of which element. Because I don't see any step where any of the three elements (ship, particle, whole system) goes from mv to 2mv or from -mv to -2mv.
Posted by Identity Crisis (Member # 67) on :
Oh, I see. You are confused.
You say "remember, there's no such thing as negative momentum".
That's a potentially confusing statement.
Momentum is a vector quantity as we agree.
Momentum of mv in any direction is cancelled out by momentum of mv in the opposite direction.
You get this right when you apply it to the momentums created along the y-axis. And you get it right when you apply it to the momentum of the particle along the x-axis. But for some reason, you get it wrong when you apply it to the momentum of the ship along the x-axis.
Here are three simple statements, please write your versions of the same in the same terms.
* When the particle is created the ship gains momentum of magnitude mv in direction x. * When the particle strikes the first barrier the ship gains a momentum of magnitude mv in the direction -x. * When the particle strikes the second barrier the ship gains a momentum of magnitude mv in the direction -x.
BTW, this is _exactly_ the same as if the particle had hit and bounced off a barrier in the y plane. (i.e. a 180deg collision). The only reason I made it two 90deg collisions was because people were complaining about the exhaust particles going back into the impulse drive.
Posted by Guardian 2000 (Member # 743) on :
quote:Originally posted by Identity Crisis: Oh, I see. You are confused.
Perhaps, but I know not how.
Correction . . . I do now, more or less. See below: my original comments and my new italicized versions.
I know I'm confused about one thing, though it's quite unrelated to all of this. What the hell is "maths"? You use that term often.
(Snipping my way down to the good stuff)
quote:BTW, this is _exactly_ the same as if the particle had hit and bounced off a barrier in the y plane. (i.e. a 180deg collision). The only reason I made it two 90deg collisions was because people were complaining about the exhaust particles going back into the impulse drive.
I find that incomprehensible.
Point One:
In the above, the top section represents a standard rocket firing. Scene 1 (so as not to be confused with your steps) shows the rocket just sitting there. In Scene 2, the red particle is ejected from the rear of the rocket, and the rocket flies forward (blue).
Below the green line, we see a new version of events. The rocket has a flat plate doohickey attached. Scenes 1 and 2 are identical.
In Scene 3, the particle has collided with the plate. This is my version of what happens . . . note how the two objects are no longer mobile. Their momentums have cancelled.
Scene 4 shows your version of what happens, as judged by your statements. The particle has rebounded off of the plate and the rocket is now moving backwards (at least until they collide again, as is about to occur . . . but, if the plate rebound occurred and produced reverse ship motion, then the particle should rebound off the rocket just as well, and we'd end up with a bouncing-back-and-forth action).
And now that I have the visual and a proper analogy in mind, I see that your version is actually correct, I think. The analogy would be to one of those swinging ball thingies. If you pick up one ball from each side and drop them, they both "bounce" as opposed to stopping entirely.
Funny . . . I knew that. I actually used that very concept on one of my pages several months ago, too. But, in reference to the ship example, I missed it. Must've been the headaches from all of this.
This leads us straight into what was originally going to be:
Point Two:
Here we have two particles . . . one red, one yellow . . . striking a stationary object (or more precisely, stationary to our frame of reference) at different times. Due to the angles of the strikes (with the momentum transfer direction identified by similar colors to the striking particle), the net effect on the object would be the same.
And here we come to the reason why I was disagreeing with you. And in case this is too hard to read, pretend the rest of this is in italics:
It was all based on my ship-plate concept that was wonky. Because I was thinking that the momentum would cancel and the two would become stationary, it struck me as absurd to conclude that two 90 degree collisions could produce rearward motion where one 180 had failed.
As near as I can figure, I must've had an inelastic collision on the brain in reference to the 180, instead of an elastic one.
In any case, as you can see in the pic above, the yellow (180) particle smacks the stationary doohickey and gets reflected back. The total momentum on the doohickey would be 2mv, since it has to both stop the particle and send it back on its way as any good elastic collision should produce. When I was thinking the wrong thing about the ship for whatever reason, I was thinking ("in effect" only, not quite consciously) that this should only be 1mv against the doohickey.
Hence my disagreement with you over the 1mv leftward and 1mv downward in relation to the ship.
On the other hand, I'm still confused by something. The bit about "two 90 degree collisions could produce rearward motion where one 180 had failed" was only part of it.
Earlier in the thread, I used this pic:
Now first, we have the red particle smacking into the gray angled wall. We know the actual transfer should occur along the blue (and green) diagonal line perpendicular to the wall.
If that is so, then we still cannot have -1mv acting on the wall on the x plane, since the only way to get the particle to deflect upward with 1mv would be for the particle and wall to transfer .7071mv between one another along that perpendicular line. Then the vectors add up, as shown by the green lines. That would still leave the wall (or ship) with .5mv and .5mv along the horizontal and vertical axes (the blue lines). For a second collision that sent the particle right-ward, we'd still only get a total of 1mv against the ship.
(Incidentally, a collision simulator I was using seems to agree with this view, insofar as it shows a .5-.5 momentum transfer from the light particle against a stationary wall.)
To my mind, we're left with two possibilities:
1. The momentum transfers involved in getting the particle deflected upward count toward the total, meaning that the two-hit thrust reverser wouldn't work because a whole 1mv gets effectively wasted by the act of getting the particle up and out of the way so it won't hit the ship on its way out.
2. If two 90 degree collisions are to be the same as a single 180 (i.e. no wasted effort as suggested above), then by damn the ship will get 2mv leftward somehow-or-other. The only way I can imagine this working within the confines of my diagram would be for us to simply add the two .7071mv diagonal lines into a single 1.414mv diagonal, which would then require 1mv left and 1mv down for the ship. However, that doesn't strike me as a proper maneuver off-hand . . . though it sounds similar to geometric vector addition.
Given my "Before After" diagram, I must assume that such a maneuver is the way to go. But now I'm just curious about how to think about it.
Suggestions?
Posted by Identity Crisis (Member # 67) on :
Maths is short for mathematics. Surely you know that? Oh yeah, for some reason people on the other side of the Atlantic drop the final 's'. Dunno why.
Your diagram with the rocket shows why a flat plate, although giving simple maths, confuses people. They see the particle hitting the rocket again and that just seems wrong. If rocket exhaust goes into a closed box attached to the rocket then no thrust has been generated - for all intents and purposes the exhaust hasn't left the ship. So in my exampe I added my two 45deg plates in order to make the exhaust end up going forwards but not back into the ship. (This was a simplified case, other people had previously drawn a more rounded tube with many collisions. Bugger doing the maths on that).
You say "Incidentally, a collision simulator I was using seems to agree with this view, insofar as it shows a .5-.5 momentum transfer from the light particle against a stationary wall."
But in this case the wall is not stationary. The wall has equal but opposite momentum to the particle.
This is the case with the desktop balls where you set both of the end balls swinging at once.
Would it help if I gave you the matrix tansformation to convert a vector in x,y co-ordinates into a vector in your rotated 45deg co-ordinate system?
I am having trouble following your diagram. I'm also getting pissed off with trying to write maths in ascii. I may write it out longhand and scan it. Quicker than fiddling about with any of the software I've got installed here.
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