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
-------------------- . . . ceterum censeo Carthaginem esse delendam.
posted
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.
-------------------- Bernd Schneider
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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.
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.
-------------------- Bernd Schneider
Registered: Mar 1999
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posted
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
-------------------- . . . ceterum censeo Carthaginem esse delendam.
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.
-------------------- "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 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.
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.
-------------------- "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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posted
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.
-------------------- . . . ceterum censeo Carthaginem esse delendam.
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).
-------------------- "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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Cartman
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posted
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.
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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.
-------------------- . . . ceterum censeo Carthaginem esse delendam.
posted
Somehow you're making this sum : mv + -mv = 2mv Which is of course bollocks.
mv + -mv = 0 as in my original example.
'nuff said.
-------------------- "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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posted
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.
-------------------- "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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