[Inspired by some annos to 'Negative Glider'. This one *is* an invention, though of dubious feasibility and utility...]
HBman and VFFman are still arguing over the results of their free-falling contest (qv). A re-match is arranged, this time in reverse. On the ground at the foot of the tower block are two identical standard HB catapaults, each capable of launching an object vertically at 98m/s. The challenge is to see who can reach the greatest height, without any additional propulsion, after this catapault launch.
"Aha!" cries the optimally- streamlined VFFman - "You have no chance of winning. If we were in a vacuum, we would both reach an altitude of 490 metres, decelerating to a standstill due to gravity before falling back down. As we are in air, we will both lose height due to air resistance but, since I am Optimally Streamlined, I will lose less than you and will win."
"We shall see." says HB man, climbing into his Super Slingshot Shot projectile. By good fortune, VFFman happens to weigh the same as HBman-plus-projectile. They each sit poised on their respective catapaults.
At the same moment, both catapaults fire, projecting the two contestants upward at 98m/s. VFFman, optimally streamlined, nevertheless experiences significant air resistance, slowing his ascent a little more than would gravity alone. HBman is almost as well streamlined but not quite, and begins to lag behind.
Suddenly, HBman opens a port in the nose of his projectile, allowing the air to rush past a variable-pitch turbine; the turbine is connected to a heavy spherical mass, which it sets spinning (and no, this isn't a gyroscope trick!).
"Aha!" cries VFFman "I am sure to win! HBman has foolishly traded some of his upward speed for rotational energy of that spherical mass! he is already falling further behind me!"
Seconds later, both of their speeds have dropped due to gravity and air resistance. HBman cunningly alters the pitch of his turbine, and now exploits it as a propellor, driven by the momentum of the spinning mass. As both contestants reach the apex of their trajectories (and as HBman's spherical mass finally slows to a standstill), HBman glances down to see that he has out-reached VFFman by just a few inches. The winner!
"Cheat!" cries VFFman - "what you have done violates energy conservation!" "Au contraire", replies HBman. "I merely exploited the fact that air resistance increases more-than-linearly with speed. At the bottom, when we were both moving fast, I traded some of my vertical motion for rotation of the spherical mass, losing vertical speed faster than you did. Sure, I lost some energy in the transfer, but not as much as *you* lost through air resistance at high speed. Then later, when we had both slowed a lot, I simply transferred the energy back from the rotating mass into vertical motion. Again, I lost energy in the transfer, but at these slower speeds I didn't waste much energy through air resistance. In effect" concludes HBman "I simply used an energy store to even-out my vertical speed, thereby wasting less energy through the disproportionately high velocities at the start of the contest. QED."
"Incidentally", adds HBman "you'll note that neither of us reached the theoretical maximum height of 490m - otherwise we would have violated all kinds of conservation laws."
"Bugger." mutters VFFman. "Outbaked again."-- Basepair, Mar 15 2005 He does indeed suffer losses in spinning up his mass, and again in converting its spin back into lift. *BUT* he is also spending less time travelling at a high velocity, and more time travelling at a slow velocity; this saves his total losses due to air resistance (as far as vertical motion goes), because air resistance is more than proportional to velocity. So, if his turbine/mass system were lossless (which it isn't) he would certainly win by evening-out his vertical speed during the ascent. The only question is whether the actual losses in the turbine/mass system would outweigh the gains through velocity- evening. I don't know if there is an absolute upper theoretical limit on turbines in the same way that there is on (say) a classical heat engine.
One more thing: at these velocities, it probably wouldn't work. But at higher launch velocities, the energy loss from air-resistance at the start of the flight will be **much** worse, and hence the advantage of storing the kinetic energy for use at later (slower) speeds will be proportionately greater. For any given turbine efficiency, there must be a launch-velocity above which it would give a net advantage....-- Basepair, Mar 15 2005 hmmm, I'm not so sure about this...So you're saying that the graph of wind-resistance vs velocity would be something akin to this
<attempts ascii graph>
<fails>
There's a curve in there that suggests there is more resistance at higher speeds than lower ones btw.
So, IF we use some of that force to speed up a turbine, we can release it later when there is less resistance to slow us down?
Holy Perpetual Motion HalfBaked man!
Sounds dodgy to me, more energy will be used to work the turbine than will be extracted from it - at whatever speed HB Man is going at. He will be slowed down more by the increased wind resistance as he is going fastest, and even though he's stored his energy, it will be a lossy conversion - both times. So say his turbine is 99% efficient, he looses 1% of his upward intertia as he powers it, then later, when he tries to use the spinning turbine to power himself for another couple of seconds, he's going to loose another 1% for the conversion back from rotational to vertical motion. This means (at best) he will only be able to use 98% of the energy he collected earlier on.
He might be able to use the turbine to slow his descent, and even power a small christmas tree for a while by converting the turbine's energy into electricty. But it is a lot of bother to go to in order to achieve a festive, but out-of-season lighting effect.-- zen_tom, Mar 15 2005 Sorry, nice try, but unfortunately like for the negative glider you hit the second law of thermodynamics.
[Una] there are no outside forces in a glider system. Lift is generated by converting energy, in the case of a glider gravitational potential energy, in the case of other planes chemical potential energy.-- scubadooper, Mar 15 2005 [zen tom] - the thing about the air- resistance curve is that it goes up steeper than linearly. In other words, if you double your speed, your wind- resistance is *more* than doubled (though not quite quadrupled, I think).
So, if you travel for (say) 10 seconds at 50m/s, you will lose less energy from air resistance than if you travel for 10 seconds decelerating smoothly from 100m/s to 0m/s, even though your *average* speed and *average* distance is the same.
What it comes down to is that steady speeds are more efficient (as regards losses to air resistance) than changing speeds. So, an energy-storage system that can smooth-out the velocity will give an advantage. In reality, losses in the conversion will indeed count against this advantage, but needn't completely offset it. All depends on velocities and efficiencies, rather than conservation laws.-- Basepair, Mar 15 2005 IIRC, wind resistance is proportional to the cube of velocity <Disclaimer: this is based on physics/maths lessons dating back to the mid-part of the second half of last century>-- AbsintheWithoutLeave, Mar 15 2005 [Absinthe] - many thanks. I didn't think it was as steep as cube, but if it is then so much the better. But for this to work in principle, it just needs to go more than linear (square or 1.5th power would be OK).-- Basepair, Mar 15 2005 Absinthe, air resistance is proportional to the square of the velocity for a range of velocities where drag coefficient can be held constant. (due to viscous and sonic effects, drag coefficient is not constant at very low and very high speeds).
Sorry, this is still bad science. Even with a perfectly lossless turbine, you'd break even at best.-- Freefall, Mar 15 2005 Freefall - thanks for the clarification on air resistance. And I accept that a 'real' turbine would be hopelessly lossy. *But* I still argue that a lossless turbine would make this work. A 'perfect' turbine (by definition) would allow a lossless exchange of vertical speed (k.e.) into rotational energy and vice versa, no? So, in theory the entire ascent could be made at a constant 50m/s, rather than at a speed which decreases linearly from 100m/s to 0m/ s (as it would for a normal projectile). Now, since air resistance increases as the square of speed, the total losses during the journey at 50m/s will be less than the total losses during the journey that starts at 100m/s and ends at 0m/s. Hence, less total energy is wasted in air resistance and the projectile must rise higher (since there's nowhere for energy to go ultimately except by being dissipated by air resistance).
If you're driving a car, you waste less energy in wind resistance if you drive at a constant 50mph than if you drive for the same time (and distance) acclerating steadily from 0 to 100mph, even though the average speed is the same. This is the equivalent argument turned vertically; we're just using the turbine to spread out the imparted kinetic energy of the launch so that velocity becomes more nearly constant during the ascent.
As I said, I am sure that at any realistic velocity, and with a turbine of any practicable efficiency, you'd lose out. But with a perfect turbine you'd win, getting closer to (though never reaching) the height which you would have reached without air resistance. No violation of energy conservation, just minimising air-losses by keeping closer to a uniform velocity.-- Basepair, Mar 15 2005 Very interesting idea [Basepair], and great annos.
This reminds me of a news video clip about a vertical rollercoaster. The car went straight up a tower, coasting, and the owner was bragging about how it was in freefall. The newsman had a tennis ball which did *not* float for the camera. The owner just ignored it and the newsman didn't argue with him. It took me a while to figure out that the car's wheels were acting as flywheels, keeping it moving up and then slowing its descent.
HBman's projectile is similar to that rollercoaster, except that the flywheel is spun-up after launch. And the spherical shape is a rather bad one for an energy-storage device. I'm going to take the principle off to a rotorcraft design, and see if I can add anything to this.
VFFman's projectile has a definite upper limit to its trajectory, streamline it how you will--purely ballistic. HBman may be said to be flying, even if rather poorly, by supporting his craft with moving air.
For example, and for less moving parts, HBman could used a winged craft, which upon vertical launch used its wings to do an immediate sweeping turn to the horizontal. Ideally, it would then be gliding at 98m/s and using its wings for lift and its momentum forward for energy to climb. Practically, it would lose energy on the turn, or be gliding at 98m/s, which is what HBman was planning to avoid. But a video of his possible paths, run backwards, could be compared to the negative glider, and show that as [Basepair] says in the idea, HBman can't go any higher than a ballistic path.
My best design for HBman is a rotorcraft, based upon various flying toys and too little sleep. The rotor is mounted on a frictionless bearing, with two perfectly-designed blades incorporating tip-weights and pitch control. After clearing the launcher, the rotor blades begin spinning up, and convert some of the forward motion to rotational energy. The blades then change pitch and shape to follow the ideal curve of lifting energy while giving the rotational energy back. (I know this is the original idea, but I've done rotorcraft in real life {such as it is} and I think this would be more efficient than a turbine and a spherical flywheel.) So, with perfectly-streamlined blades (ignoring the fact that I'm now putting spinning blades into the airstream) that convert all lift to energy and all energy to lift, we start the video camera and launch this puppy. When we run the video backwards, we see a roto-glider begin a slow drop, with its blades pitched for maximum lift, beginning to spin-up the craft. It passes through an curve of speed of rotation versus descent, then, when within only a few feet of the ground, the blades apparently malfunction horribly, converting all the built-up rotation into a mad smash into the ground at 98m/s. VFFman's craft, seen in the background, drops from the same height, falling faster, but reaches terminal velocity due to air resistance, and has nothing to cause a final burst of speed, and hits the ground at a slower speed, but gets there first.
Okay, that was odd, but to me it says that energy storage would be better, provided it was done ideally. But if rotor blades can have perfect efficiency, so can streamlining, which makes the whole thing pretty iffy.
Ideally, a perfectly designed craft would be a cylinder with a variable-geometry inlet and exit nozzle. Upon launch, the air hitting the front of the craft would channel down the frictionless inlet to a storage tank, slowing, compressing and heating, and slowing the craft. The compressed air would later be released out of the frictionless exit nozzle, boosting the craft as it expands and cools. Both nozzles would be varying geometry like mad. Looked at that way, I see HBman's craft as simply a rather odd way of going about making a pressure-recovery aft section.
I'm going to say that HBman would win for any launch speed greater than VFFman's terminal velocity, provided he has ideal energy to lift conversion. Sounds like a great contest for Junkyard Wars.-- baconbrain, Mar 16 2005 I think HB man could win. There are some red herrings in here which might cloud thinking about this.
Let us imagine that VFF and HB have frictionless go carts on a level surface. The gocarts have huge coiled springs on board, each containing the same amount of energy. VFF discharges his spring all at once, pushing off against a rock, and goes zooming away. HB discharges his spring little by little through a series of gears, powering his car along at 10 kph. VFF is sitting still in his cart somewhere out there at the end of his wild ride. Will HB pass him before his spring is exhausted?
It depends on how efficiently the HB cart can turn spring energy into motion. Essentially, the turbinepowered ball in the original example does the same thing - capturing catapult spring energy by way of air resistance. An efficient vehicle could win it for HB man.-- bungston, Mar 16 2005 [Baconbrain] - Many, many thanks - you have analysed this a lot more rigorously than I had, and your anno is very helpful.
Re "But if rotor blades can have perfect efficiency, so can streamlining, which makes the whole thing pretty iffy." - I agree, which does indeed make the whole thing a little moot. On the other hand, given slightly-less-than-perfect turbine efficiencies and slightly-less- than-perfect streamlining, my gut feeling is that, *at high enough launch velocities* the turbine system would still win, simply because the penalties of air resistance would have more impact on the not-quite-perfect streamlining. However, I can see converse arguments and I'm not sure which is right.
Re the use of a gliding path, I agree that we can never get higher than the in- vacuo-ballistic path under any circumstances, and I also agree that the gliding velocity would have to be high after an efficient turn, thereby not saving air-resistance losses. So, gliding doesn't seem to help.
I also agree regarding the design of the turbines - I can imagine that the spherical mass is not the best flywheel.
Your pressure-recovery system sounds better than the turbine system. I was thinking last night about having internal sprung weights which would lag behind the body of the craft on launch and then spring back during the later stages of flight to return the energy. But then I figured that these would be cheating, if the launch velocity of the catapault is constant, since they'd be drawing extra catapault energy to extend the springs. But your air-compression idea sounds like it could be done without cheating (as long as no nozzles open until the craft leaves the catapault; otherwise you'd again be drawing extra launch energy from the catapault, which would be sort of cheating).
[bungston] - many thanks for the clarifying anno. You're right in that this is effectively a vertical equivalent of a 'go cart' experiment, and your illustration makes the point that smoothing-out the speed will be more efficient overall.
Anyway, as a 'thought experiment' this has been grand! Many thanks to all annotators! (Wanders off to try to sell idea to Burt Rutan and/or former Supergun designers......)-- Basepair, Mar 16 2005 Great physics. I like it. Better than all these "backward time whirlpool bubble" ideas we get nowadays.-- sninctown, Nov 14 2005 Several dual stage versions are possible. For example, you could build a cannon which is lunched upwards. Compressed air is stored, which is later used to fire a projectile.
The body of the cannon would reach a reduced height, but the projectile could reach a greater height than that achievable even by frictionless projectile motion. Yes, it's cheating, but still, it introduces a slight greyness to //we can never get higher than the in- vacuo-ballistic path under any circumstances//. And no extra energy is stolen from the catapult.
What if the projectile were also a (smaller) cannon, etc.?
Basepair's internal sprung mass would be a stealthier way of cheating. It would have to spring back _before_ the projectile left the catapult, so that the internal mass is moving faster than 98m/s, and therefore the RMS velocity of the contraption exceeds 98m/s on launch.-- spidermother, Jan 30 2006 If this were not catapults, but spring powered helicopters, there is probably a way to calculate the ideal rotor speed to generate maximum height. It seems obvious to me that not all rotor speeds will attain the same height. VFFmans catapult launch corresponds to an arbitary rotor speed in the spring powered helicopter - not necessarily the best one.-- bungston, Jan 30 2006 random, halfbakery