Mach Effect progress

[GIThruster]

I wonder if we could try to estimate density of advanced civilisations (colonised planets, more likely) with the data on Universe's expansion rate..

If we assumed that the universe's acceleration in expansion was due exclusively to M-E use, we could indeed calculate the degree of use. I'm not entirely sure whether we would get the degree of use at this time, or if we'd get the degree of use in the history of the universe. I think it's the former, since we measure acceleration of expansion in current time, but I'll leave that to others who know better.

[TallDave]

I wonder if we could try to estimate density of advanced civilisations (colonised planets, more likely) with the data on Universe's expansion rate..

If we assumed that the universe's acceleration in expansion was due exclusively to M-E use, we could indeed calculate the degree of use. I'm not entirely sure whether we would get the degree of use at this time, or if we'd get the degree of use in the history of the universe. I think it's the former, since we measure acceleration of expansion in current time, but I'll leave that to others who know better.

As I understand it the force is constant, thus the expansion is always accelerating (except of course for that deceleration after early cosmic inflation).

Now, if you could find localized variations in the cosmological constant, well then...

[GIThruster]

I'm still not comfortable with the conclusions 93143 is drawing. They seem intuitively wrong. For instance, where they true, electric motors would have this same conservation issue. Like an M-E thruster strapped to a flywheel, or in the case of a spacecraft, like a linear motor, the electric motor has electrical energy in, and force out. If the conclusion is that at any efficiency, an M-E thruster strapped to a flywheel will necessarily go over-unity, then this holds for all electrical motors as well, and we know this isn't true.

I think the key may be what is missing from the analysis is that in the case of an M-E thruster on a flywheel, you don't really get constant force out. You get constant force acting across a distance--both as the flywheel turns or the rocket accelerates through space--and force times distance is not just force. It's kinetic energy. The only instance where force is not acting across a distance is when there's no motion--the force is inadequate to cause acceleration, so KE sums to zero.

The reason we use the term "thrust efficiency" is that we want a value where we can calculate based upon an arbitrary mass being accelerated, so this gives us a handy value, but in the real world examples, there is always a mass, that determines the acceleration and hence the distance the force acts across.

I don't know if 93143 has accounted for this in his equation, but just saying, I have a nagging feeling something very basic is missing from the analysis. I don't think M-E thrusters can go overunity unless they're driven into wormhole territory.

[Stoney3K]

Please keep your terminology in mind here. Kinetic energy is something that is already stored in a moving object (e.g. flywheel or spacecraft).

Force times distance is work, which is the change of total kinetic energy in the system. This is consistent with Newton's Second Law of motion.

After all, there would be no single object in the universe with zero kinetic energy. It would have to be standing still (in what reference frame, please) and its temperature would need to be at absolute zero.

My example does not deal with HOW the thrusters do what they do, nor does it need to. It only matters that they do it. They're black boxes. Once you've supplied them with power and gotten thrust out, it's all Newtonian all the way.

I think what you're missing here is an important piece of phyiscs. Thrust by itself does not mean power.

In essence, if you have a device that generates thrust, but is pushing against nothing, it would not effect any work on an object, and therefore generate zero power. It would also, in theory, require zero power to function.

M/E thrusters, if they prove to work as advertised, would never give out more power than the amount of power that is put into them in the first place. For a thrust generating device, computing the power for it is simply the amount of work exerted by that thrust each second, or the amount of thrust multiplied by the velocity resulting from that thrust.

So if you have a train that is already moving and stick an M/E thruster on the back, it won't do any good if it doesn't accelerate the train. Replace 'train' with 'piston', 'paddlewheel', 'drive belt' or your favorite prime mover, and I think you get the picture.

Even if an M/E thruster is reactionless, it doesn't mean it ignores the law of conservation of energy. 1 Joule of energy in, means 1 Newton-meter of work out on the other end. Nothing more. So if you spin a flywheel with that beautifully crafted M/E power plant, you'd at most get the same amount of power out of it as you would need to power the M/E thruster in the first place.

Say you've attached a flywheel to the driveshaft, with tangentially-oriented Mach-effect thrusters arranged uniformly around the edge so as to spin it without generating a bending moment on the driveshaft. As we've just established, these thrusters need to counter the torque induced by the generator, so they need to supply 159 kN·m in total. Let's say that the proponents are right, and that 1 N/W is feasible. So we'll give the thrusters a thrust efficiency of 1 N/W.

Let the radius of the flywheel be 1 m.

The required torque is then reached at a total thrust of 159 kN. At a thrust efficiency of 1 N/W, operation of the thrusters requires 159 kW in total.

Any questions?

What you're missing here is that the M/E thrusters exert work along the circumference of the flywheel. Let's roll it out flat:

Radius of flywheel = 1m
Circumference of flywheel = (2pi * radius) = 6,28m
Velocity along M/E's trajectory: (circumference * RPM/60): 62,8 m/s

... hey, wait, does that number look vagely familiar already?

Force exerted by M/E's in total: 159 kN

Power required by M/E's: 159 * 62.8 = Exactly 10MW.

So in the ideal situation, there's nothing left. What goes in, must come out.

[93143]

It isn't me that's confused.

[Stoney3K, all you've done is calculate the work done by the M-E thruster, not the input power requirement. It's no surprise that it matches my own calculation of the work done by the M-E thruster.]

Consider two cases.

1) A Mach-effect thruster on a test stand, wired to a stationary power supply.

2) A Mach-effect thruster on a trolley (with brakes) moving at a constant speed, wired to a stationary power supply via a long extension cord.

Supply the same amount of power to both of these thrusters. Do they generate the same amount of thrust?

Of course they do, because of Lorentz invariance (or even the Copernican principle). It shouldn't matter what the thruster's velocity state is. And keep in mind that while mechanical power depends on the observer's reference frame, force does not.


Conventional prime movers (like electric motors) always react against something, and the relative velocity between the mover and the something determines how much power is required to exert a given force against that something. The same is true of a Mach-effect thruster, but the something is nonlocal and its velocity state relative to the thruster is not immediately obvious, though given a thrust efficiency you can calculate an upper bound. THIS is the part you're missing. Even though I explained it already.

You're talking as though the M-E thruster is reacting against something in the laboratory frame of reference, which is entirely without physical justification.

...

The reason an electric motor needs more power to deliver the same torque at a higher speed is because the armature is reacting against the stator, which is in the lab frame. It has nothing to do with the relative velocities of the armature and the power supply.

Try another thought experiment. Take a portable light bulb socket with a 60 W bulb in it, and plug the other end of the cord into the wall. How much power does it draw? 60 W, of course. Now wave it around. Does it draw more power? Of course not. Why not? Because unlike the electric motor armature, the operating principle of the light bulb doesn't have anything to do with the laboratory frame of reference, so there is no mechanism by which its power consumption could change based on its velocity in the lab frame.

[Of course there will be a bit of difference due to inductive effects in the cord and such, but those will be very minor.]

A Mach-effect thruster is the same. Its principle of operation doesn't reference the lab frame in any way. Otherwise your minimum power requirements for propulsion on a space mission would be determined by how fast you were moving relative to the spaceport.

[GIThruster]

Numbers first this time. Say you've got a 12-pole generator, efficiency 100% for the sake of argument. Your rotational speed is 600 rpm, or 62.8 rad/s. Your power out is 10 MW. Therefore the torque applied by the generator to the driveshaft is 159 kN·m, or 117,387 ft·lb. This torque must be countered by something in order to maintain the speed of the driveshaft and keep the system producing electrical power.

Say you've attached a flywheel to the driveshaft, with tangentially-oriented Mach-effect thrusters arranged uniformly around the edge so as to spin it without generating a bending moment on the driveshaft. As we've just established, these thrusters need to counter the torque induced by the generator, so they need to supply 159 kN·m in total. Let's say that the proponents are right, and that 1 N/W is feasible. So we'll give the thrusters a thrust efficiency of 1 N/W.

Let the radius of the flywheel be 1 m.

The required torque is then reached at a total thrust of 159 kN. At a thrust efficiency of 1 N/W, operation of the thrusters requires 159 kW in total.

Any questions?

Lots!

Isn't there a way to calculate backward and find a critical thrust efficiency below which you don't have an overunity condition? I still don't understand how this could not be so.

BTW, I'm the guy responsible for the whole "thrust efficiency" notion and all the misunderstandings it has created. It was about 5 years ago I was talking with Paul about wanting to see a real "applications" paper at STAIF, which if you'll recall, is was the Space Technologies and APPLICATIONS International Forum. There just weren't any applications papers being presented in the Advanced Concepts forum, and so we conspired together and the result was Paul's WarpStar paper. The point was, to find a watermark above which we could safely say that the technology would enable human spaceflight.

For satellite station keeping, things most commonly used are Hall thrusters and they produce about 50uN/W. Paul used Andrew's math model to extrapolate what frequency MLT would be needed for 1N/W, found this is a perfectly reasonable goal. (I think it was something like 500Mhz as opposed to something unreasonable like a Thz figure--ceramics don't have ionic response in the Thz range so we can't go there with evolutionary development of current embodiment.)

So here's my question, it still appears to me with a low enough thrust efficiency, you will not have an overunity condition. Isn't there a way to work the same calculation you just did backward and solve for that critical thrust efficiency figure?

[KitemanSA]

Here's an amusing thought: what if the reason for the slowing of the expansion of the universe is that millions of civilizations are out there flying around trillions of ships using ME drives harvesting momentum from the universe?

Harvest??? I thought this thing was supposed to EXCHANGE momentum with the universe, similar to a car exchanging momentum with the Earth. Harvest? What an odd notion.

[Stoney3K]

It isn't me that's confused.

[Stoney3K, all you've done is calculate the work done by the M-E thruster, not the input power requirement. It's no surprise that it matches my own calculation of the work done by the M-E thruster.]

Consider two cases.

1) A Mach-effect thruster on a test stand, wired to a stationary power supply.

2) A Mach-effect thruster on a trolley (with brakes) moving at a constant speed, wired to a stationary power supply via a long extension cord.

Supply the same amount of power to both of these thrusters. Do they generate the same amount of thrust?

They do. But static thrust does not equate to energy.

If I place my feet on the floor I will exert a static force (just like thrust) on said floor because of my weight acting on it. However, I'm not doing any work on it (the floor doesn't budge), since neither me, nor the floor are moving as a result of the gravity acting on my body.

If I'm climbing a set of stairs, however, I will need to effect work on my body (through my muscles) because I'm going up, and need to convert kinetic energy into potential energy for each step I take.

That's the difference between the first and second situation. In the first situation, the theoretical energy requirement would be zero, much like me standing on the floor or an object leaning against a wall, since the M/E thruster is not effecting work on anything. That is, if you have an ideal thruster -- a practical one would still require power to counteract internal losses.

In the second, provided the brakes (drag) are set to such a point that motion is constant, the theoretical energy requirement of the M/E thruster would be just as much as is required to keep the vehicle moving. In essence, that would be the thrust times the vehicle's speed.

This is also the work done on the vehicle by the M/E thruster each second (power), and also the amount of heat developed in the brakes needed to slow the thing down. Net energy balance remains zero.

However way you turn it, the ideal M/E thuster would be reactionless with zero losses. It would never be able to do any more work than the Joules you throw at it, unless there's some glittery magic unicorn physics behind it.

Also, keep in mind that the definition of static thrust here is a different one than the number specified on aircraft engines. On aircraft engines, the number can vary because of test circumstances and the airframe being used. A theoretical jet engine with zero mass running at maximum power will still effect zero thrust, because it has no mass that is willing to move.

You're talking as though the M-E thruster is reacting against something in the laboratory frame of reference, which is entirely without physical justification.

It is not. The M/E thruster is acting against two objects here, because of Newton's Third Law. The first, obviously, is the target you want to move, the other, being the object that receives the equal and opposite reactive force, is outside the laboratory frame of reference. Hence the definition of reactionless.

The analogy of a car going down a road also holds here: If that car is moving, the engine not only exert a force on the car to move it forward, but the car will also exert a force on the road. If there is no grip, the car cannot exert any force on the road and it won't go anywhere (it will just skid or wheel-spin).

An M/E thruster will operate on the same idea, but the 'road' is beyond the scope of our research. It will still have a limited efficiency ('grip') with which it can exert a force on that 'road', and thus generate thrust.

[CaptainBeowulf]

Kiteman - agreed, harvest is an incorrect word, exchange is better. In this example - take it with a grain of salt, it's not meant entirely seriously - M-E ships would be the dark matter momentum gets exchanged to, resulting in changes in the rate of expansion of the universe/changes to the cosmological constant.

[Carl White]

Let's leave the M-E device out of the picture for a moment and try something else:

Take our flywheel, rotating at 600 rpm = 62.8 m/s at the rim.

Say there's a liquid-oxygen/hydrogen rocket attached to the rim of the flywheel (okay, there should be at least two for balance but let's ignore that).

1g of H2 and 8g of O2 burned each second produce 142 J each second.

The hydrogen and oxygen are piped up to the rim of the flywheel at a rate of 9g each second. So 9g have to be accelerated to 62.8 m/s each second:

energy used = 0.009kg * 62.8 m/s * 62.8m = 35J consumed each second.

Net energy = 142 - 35 = 107J.

Now, the flywheel is producing 10MW so the torque is 159000 N.m = 159000 J.

So we need 159000 J input = 13.4 kg of fuel each second.

10 MW each second = 10 M joules.

If we crack water at a 50% efficiency, that would be 284 J/g x 13.4 kg = 3.81 M J each second.

3.81 M J used each second = requirment of 3.81 MW.

Input = 3.81MW
Output = 10MW

Perpetuum mobile? Clearly something doesn't add up here. Where are the errors?

[Stoney3K]

You're wrong on the 159 kNm torque equating to energy here.

In a rotational system, power (energy per second) equates to torque times the angular velocity of the flywheel. Which should hold for either end of the energy balance:

Power output = 10MW = 10.000.000 J/s
angular velocity: 600RPM = 10 RPS = 62.8 rad/s

Required torque: 10.000.000 / 62.8 = 159 kNm (we have already estabilished that).

For the O2-H2 rocket, you will have to consider the RADIAL (not angular) velocity as you already did. However, what you missed is that the torque does not equate to energy, instead:

Radial velocity along rocket's trajectory: 62,8 m/s

Force required to keep everything spinning at a constant speed: 159 kN (flywheel radius is 1m)

ENERGY required to keep everything running in 1 second: 159.000 * 62,8 = 10.000.000 J.

Remember, the force is acting in a tangential direction to the flywheel. That's where most of the math errors come from: Work is only calculated with the distance vector along the effective axis of motion.

159 kN with a flywheel radius of 1m does not mean an energy input of 159 kJ. The motion vector is pointing at right angles to the radius of the flywheel, so that does NOT work.

The correct way to calculate this is to integrate the motion vector over time along the path of motion in 1 second:

P = integral(0, 1s) F(t) * v(t) dt

Now, fortunately for us, the path of motion is circular, so this relates in polar coordinates to:

P = integral(0, 2pi) F(theta) * r v(theta) d-theta

With 'theta' being the angle along the flywheel and r being the flywheel's radius.

Simplifying this a bit more, since r is 1, and both F and v are constant, and in line, we get:

P = integral(0, 2pi) F * v d(theta) = 159.000 * 10 * 6,28 = 10 MW.

It's simpler to imagine if you have an infintely long piece of rail with the M/E thruster or H2/O2 rocket pushing a carriage forward, and a pair of brakes (or generators) slowing it down to the point where motion is constant.

OK, finishing this up, 10MW of power input from the H2/O2 rocket means the fuel flow will need to be:

10.000.000 / 142 * 0,009 = 633 kg per second.

If you want to crack that amount from water at 10% efficiency, you will need:

1,42 MJ/kg * 633 kg/s = 900 MW.

So I suspect that's not a good practical perpetual motion device after all. :p

[GIThruster]

Kiteman - agreed, harvest is an incorrect word, exchange is better. In this example - take it with a grain of salt, it's not meant entirely seriously - M-E ships would be the dark matter momentum gets exchanged to, resulting in changes in the rate of expansion of the universe/changes to the cosmological constant.

Actually "harvest" is a pretty good word since it reminds us that the power into the device is not entirely electrical, which is why it can appear to go overunity when in fact it does not.

Also note, according to Mach's Principle, the primary reaction mass is the farthest mass in the universe, or what Mach calls Far Off Active Mass or FOAM.

[TallDave]

Thrust by itself does not mean power.

Why not? You can always turn one into the other.

However, I'm not doing any work on it (the floor doesn't budge), since neither me, nor the floor are moving as a result of the gravity acting on my body.

That's only because you've reached an equilibrium state in which the floor prevents you from moving, your tissues hold each other together, etc. But you can get power from gravity; hydropower generally works on this principle. It's actually the cheapest way to get power.

It will certainly be very interesting to see what the upper bounds of N/W efficiency actually are, assuming the effect is real.

Waiting, waiting, waiting. We'll know the Singularity is here when empiricism doesn't take so damned long. Maybe this willl help.

[Betruger]

Don't need singularity if we can reach some less extraordinary interim like biomedical aging equilibrium. From then on the waiting is a formality and not a race against time anymore. From that day on the world will have not just a sequence of March's and Woodwards, and Feynmans and Einsteins, contributing for only for a few decades before exiting stage left for their successors to pick up where they left off, but a cumulative.. ever growing population of people investigating.

This is where the "paradigm paralysis" argument crops up (a few people camping at the top) and .. I think it's not quite accurate. If you have effectively indefinite lifespan, indefinite time ahead to develop anything, then it's not so crushingly urgent for anyone to "have the floor" so to speak. You have instead the luxury of waiting and seeing if their hunch was right and if their work delivers. (Not just as far as research goes, but also socially. What's to be concerned about with town hall meetings if you have everything you need - Mr Fusion + Drexler, etc?)

Unlike now where it's basically a crime to not pursue only the most promising avenues (scientific but also political etc) because it wastes that precious commodity - time. Which is curious in an ironic and almost comical way, given how adverse are people in general to admitting that aging must be cured. Ironic because the driving force behind all the conflicts of the world, at all scales, all come down to that same thing: everyone wanting a piece.

And indefinite lifespan pretty much hands you that on a platter. Provided you have patience. But that virtue ought to come to people much more easily, naturally, once aging is cured.

[TallDave]

It would certainly help, though there's still the Coefficient of Impatience.

Honestly though, aging is such a horrifyingly complex problem to solve. We've barely made the slightest dent. I think the one may only come with the other.