Fun with: Ion Engines

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Wyrm
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Post by Wyrm »

Kuroneko wrote:Addendum: since the question has attracted some interest, let's try to find the distance traveled as a function of time, assuming Newtonian mechanics . Let R = 1 - Ft/[Mu] be the mass fraction at t. Then v dt = -u log R dt = Mu²/F log R dR. Integrating, the distance traveled for zero initial velocity is D = [Mu²/F][Rlog(R)-R+1], where R is now the mass fraction at some particular t>0.
Cool! I didn't think of using the mass fraction when I was doing the same integration (the result looks similar). Damn thing looks a pain to invert, though. Ugh.

I assume that the mass fraction R is the ratio of initial mass at the beginning of the burn and final mass at the end. Am I correct?
Kuroneko wrote:This is true for a constant acceleration, but not a constant thrust rocket. If the exhaust velocity and thrust are constant, then the only relevant parameter is the mass fraction. Thus, taking a line segment OM for total mass, with subsegments OT for turnaround fuel, OP for total fuel, and PM for payload, we must have the proportion TM:OM :: PM:TM. Therefore, thus the amount of fuel burned before turnaround is the geometric mean of the initial mass and payload. That's why the acceleration and deceleration times in the calculations above were so skewed (93 years accelerating and 7 years decelerating for a 100-year fuel burn).
Yeah, well, I figured the constant thrust case would be a bit different. The point of my particular tirade was that flight time is minimized when you spend all your time with your engine on. No coasting.
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Post by Kuroneko »

Wyrm wrote:Cool! I didn't think of using the mass fraction when I was doing the same integration (the result looks similar). Damn thing looks a pain to invert, though. Ugh.
Let A = F/Mu and T = D/u for convenience, so that D = [Mu²/F][Rlog(R)-R+1] becomes AT-1 = R[log R - 1]. Let's substitute r = AT-1 and R = r/w, where w is to be determined. Then w = log(r/w) - 1, or e^w = r/ew. Hence w = W(r/e), where W is the Lambert-W function; back-substituting to R, we have t = [w-r]/[Aw], w = W(r/e). (Didn't we have a thread on the W function some time ago?)
Wyrm wrote:I assume that the mass fraction R is the ratio of initial mass at the beginning of the burn and final mass at the end. Am I correct?
Yes, but the other way--it's the mass remaining after time t to the initial mass. Hence the negative sign in v = -u log R or v = tanh(-u log R) in the relativistic case.
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Post by Sikon »

rhoenix wrote:Sikon, that was excellent, and deeply fascinating reading - I thank you.
You're welcome. :P

Ice rockets (and/or mass driver transports) may be like the supertankers of a space civilization, cheaply shipping many, many times their own mass in payload over distances such as millions of kilometers, refueling from natural space material. There aren't capable of very high velocities like nuclear pulse propulsion, but the potential economics are awesome, along with suitability starting from the beginning of space colonization.
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Post by Sikon »

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On a separate topic, regarding Orion, adding a little more to the other discussion referenced before seems worthwhile.

The VISTA ICF mentioned in the Lawrence Livermore National Laboratory paper is one hypothetical form of external nuclear pulsed propulsion, what I usually call an Orion variant. If one has heard of the Daedalus starship proposal, that's also ICF. ICF stands for inertial (not internal) confinement fusion. The theoretical concept involves a simpler equivalent of a miniature thermonuclear bomb but with an external driver pulse replacing the fission trigger (and bulky chemical explosives), allowing the efficient yield to be smaller since there is no longer a minimum critical mass with no fission used. Optionally, it may be scaled down to a very miniscule fraction of a kiloton per pulse. The fusionable fuel of each target (pulse unit) is momentarily compressed and heated from all sides by the inertia of an imploding capsule wall, imploding due to being blasted by the external driver array, igniting, and then exploding from the resulting fusion energy release.

Pure-fusion ICF is the most technologically difficult nuclear pulse propulsion system but has advantages if it did one day become plausible.

An Orion propulsion system with pulse units using A-bombs or regular thermonuclear bombs has some disadvantages. For example, it requires a relatively high minimum yield per pulse, for reasonable efficiency, with the critical mass of the fission triggers. Politically, there are the obvious political complications of using nuclear bombs, much more so than using small fuel capsules imploded and ignited by an external driver. But it is the system based most on existing technology, with an example of some missions readily possible with such well worked out some time ago.

Then there is mag-pulse Orion, e.g. this Mini-Mag Orion. It much reduces fission critical mass and reduces the pulse unit complexity compared to stand-alone nuclear bombs. Small plutonium capsules can be imploded by a large external driver system to greater compression, density, and lower resulting critical mass than with chemical explosive implosion, with a driver still requiring a lot less performance than that required for ICF fusion.

And there is antiproton-catalyzed microfission/fusion, as illustrated by the ICAN II concept. It is lesser in antimatter requirements than a pure-antimatter drive by a factor of enough millions to have a relatively good chance of being practical in the relative near-term foreseeable future. Again, it allows relatively low-yield pulses if desired.

For nuclear pulse propulsion in general, while starship concepts approach up to on the order of a million-sec specific impulse, that isn't even the goal for interplanetary transport, particularly not with the tradeoffs involved, and the average proposal for an interplanetary craft involves tens of thousands of sec Isp or less, along with usually <=~ 1g acceleration during periods of thrusting in deep space. That's substantial in today's perspective, compared to chemical rocket engines being a few hundred sec Isp; today's ion engines running off small solar panels like the ion engine of the DS1 probe accelerating it at around an 0.00002-g initial rate, taking a couple months for the first 1 km/s of delta v alone; and so on.

If politics prevented any nuclear-pulse-propulsion implementation, high-speed interplanetary transport market niches might be eventually satisfied by externally-powered propulsion, such as a sub-micron-thickness array of suitable high-temperature photovoltaic cells illuminated by a powerful laser at up to orders of magnitude greater intensity than natural sunlight. High efficiency is possible by matching the laser wavelength to the PV cells, unlike broad-spectrum sunlight. Such might obtain a power to weight ratio vastly above regular onboard power production, using instead the power beamed from large stationary installations millions of kilometers away. For example, while unlikely to provide minimum-cost bulk cargo transport, such might be like the Concorde of a space civilization for interplanetary passengers in a hurry.

However, by default, such tends to be relatively expensive per unit of energy involved, also with lesser delta-v capabilities compared to Orion variants.

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Topics like the delta-v to Pluto or the minimum delta-v to escape the solar system seem to have come up a lot in this thread.

Let's illustrate some of the complexity for calculating minimum delta-v to escape the solar system, plus some interesting possible "tricks."

In many cases, in order to do much calculation for a trip starting from earth orbit, one must consider earth's velocity around the sun, which is 29.8 km/s.

Let's first illustrate how that affects the delta-v to escape the solar system, neglecting techniques like gravity assists for now.

If one unrealistically neglected orbital velocity and incorrectly treated a spacecraft as starting stationary 1 AU (a distance of ~ 1.496E8 kilometers) from the sun, that would lead to calculations as follows:

F_gravity / M_ship = G * M_sun / (r^2), where G is the gravitational constant of 6.67E-11 N * m^2/kg^2, while M_sun is the Sun's mass of 1.99E30 kg; and r is the distance from the sun.

Force * distance equals work (energy), so one can evaluate the difference in gravitational potential energy of ship's mass being at a distance r_1 from the sun versus it being at a distance r_2 from the sun with the following integration:

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To find the amount equal to the kinetic energy for escape velocity from that location in the sun's gravitational field, set r_1 =~ 1.496E11 meters for 1 AU from the sun and r_2 to approach infinity. So the ship starting 1 AU from the sun needs 887 MJ of energy of kinetic energy for each 1 kilogram of its mass to depart the solar system, to obtain escape velocity from its starting position. Kinetic energy is KE = 0.5 * M * V^2, so that corresponds to ~ 42.1 km/s delta v.

Such is the nominal minimum delta-v for a ship to escape the solar system if starting 1 AU from the sun ... if the ship unrealistically starts as stationary with respect to the sun. To illustrate how the math works, if one instead substituted into the preceding integral a value of 6.96E8 meters for an imaginary ship starting on the surface of the sun, considering sun's radius, that nominal delta-v would become 618 km/s. Such suggests how there were no careless mistakes in derivation, as it appropriately matches the figure that is the sun's escape velocity.

The preceding math was fine in itself.

But is 42.1 km/s delta-v really the minimum involved in a real-world craft departing the solar system from 1 AU away from the sun? Not at all.

The math is only as good as the starting assumption, which is unrealistically treating the ship as starting from 0 km/s relative to the sun.

The problem is neglecting how a ship departing earth orbit or elsewhere would actually start with orbital velocity around the sun. No ship would really start just suspended in space stationary with respect to the sun while being 1 AU from it.

Let's now illustrate for a ship that starts in a solar orbit 1 AU from the sun (e.g. following several million kilometers behind earth). For an approximately circular orbit, it is currently orbiting the sun at 29.8 km/s. What happens if a burst of thrust from its engines gives 12.3 km/s delta v with that velocity in a directional parallel to the original orbital velocity, such that the ship reaches (29.8 + 12.3) km/s or thus 42.1 km/s relative to the sun? It eventually escapes the sun's gravity, departing the solar system.

Observe how different 12.3 km/s is from 42.1 km/s, illustrating the major effect of taking into account the starting orbital velocity.

Actually, even the 12.3 km/s figure is only valid for the ship starting in a solar orbit 1 AU from the sun without starting significantly within earth's gravity well, not for a ship in earth orbit, nor for an object starting on earth's surface.

Let's now illustrate for an object starting from earth's surface, such as a high-velocity mass-driver projectile that for some reason one wants fired at such velocity to leave the solar system. Assume it is launched in the right direction, which is quite important, as will be discussed later. One wants it to have 12.3 km/s additional velocity beyond earth's 29.8 km/s orbital velocity around the sun for 42.1 km/s total velocity relative to the sun.

That means it should have 12.3 km/s velocity relative to earth left over after escaping earth's gravitational field.

Earth's escape velocity that is 11.186 km/s corresponds to 62.56 MJ/kg being taken from the projectile by earth's gravitational field. If the projectile is to have 12.3 km/s velocity left over afterwards, it needs to have 75.6 MJ/kg left over afterwards (of kinetic energy relative to earth). So, aside from atmospheric drag losses, it needs to start with ~ 138.2 MJ/kg, which means 16.6 km/s velocity. One may note that is not the same as (12.3 + 11.2) km/s, not 23.5 km/s. If a projectile goes at 16.6 km/s just above earth's atmosphere, headed in the right direction, it can escape the solar system even after losses going out of earth's gravity well.

One can indirectly double-check that there weren't careless mistakes in the preceding calculations by comparing to the following, which implies the same:
C. Park & S.W. Bowen wrote:Ablation and deceleration characteristics are analyzed for a hemisphere-cylinder-shaped projectile protected by a graphite nose tip and launched vertically upward with a velocity in excess of 17 km/sec. It is shown that ablation and deceleration of the projectile are in tolerable ranges for a scheme in which such projectiles are packed with nuclear wastes and launched with a mass driver to dispose of nuclear wastes outside the solar system.
[...]
If the projectile survived the atmospheric flight and had a velocity when above the atmosphere greater than 16.6 km/s, it would escape the solar system, provided the launch was properly timed and aimed [e.g. launched vertically at dawn from the equator].
AIAA 1981-355

In the preceding, why does the launch have to occur at around a particular time to minimize velocity requirements?

It is because direction matters much, the right trajectory being practically necessary for this.

For example, suppose the vertically-launched projectile was fired at the worst time 12 hours different from the best time, so the velocity given to it by the launcher was directed opposite to earth's orbital velocity around the sun instead of parallel to it.

In that case, instead of eventually 12.3 km/s added to earth's 29.8 km/s orbital velocity for 42.1 km/s velocity, the situation would become 12.3 km/s subtracted from earth's orbital velocity. The projectile would drop to 17.5 km/s relative to the sun, moving slower than earth.

In that case, the projectile would end up in a highly elliptical orbit around the sun. Any elliptical solar orbit is like this:

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In the preceding, V_a is the aphelion velocity, which is the velocity at the greatest distance from the sun. In this case, that aphelion velocity is now 17.5 km/s. M_sun and G are the mass of the sun and the gravitational constant respectively again. R_a is the aphelion distance of 1 AU.

R_p is the new perihelion distance that is now going to be a lot closer to the sun than 1 AU. With respect to the sun, this projectile fired "backwards" in the wrong direction lost a lot of velocity compared to earth's orbital velocity.

Specifically, in this case, R_p =~ 3.12E10 m or 0.21 AU.

As the projectile "falls towards the sun" on its new orbit, it reaches a perihelion velocity of 83.9 km/s relative to the sun at its close point of 0.21 AU. That is a velocity subsequently lost as it moves back away from the sun later in its elliptical orbit, moving back out towards 1 AU aphelion.

As another illustration, one may look at dropping nuclear waste into the sun. Such is an idea sometimes suggested by individuals in the general public unfamiliar with orbital mechanics. To "drop into the sun" an object starting from earth's vicinity, it needs delta-v negating nearly all of earth's 30 km/s orbital velocity around the sun. Otherwise, the object merely ends up an elliptical orbit passing closer to the sun without intercepting the sun itself (the sun's radius being such that its surface is 7E5 kilometers from its center).

As previously implied, even escaping the solar system actually takes less delta-v. When starting at 30 km/s relative to the sun at 1 AU distance, it is easier to rise by 12 km/s to reach 42 km/s than to drop by about 30 km/s to reach about 0 km/s. That's why the preceding nuclear waste disposal paper suggested sending the projectiles on a trajectory escaping the solar system instead of sending them into the sun. (That's not to say that the trouble of either option is really at all necessary for waste disposal, just that the former is energetically easier than the latter).

Incidentally, although topics like orbit calculations are mainly covered in textbooks, one of the better online discussions is here.

Now, what about a craft starting in low earth orbit that is to escape the solar system? Its altitude is low enough that the situation is like starting from earth's surface aside from the 7.9 km/s LEO orbital velocity around earth. If it is to obtain solar system escape velocity, one again wants 16.6 km/s velocity (parallel to earth's orbital velocity) relative to earth, for the velocity when it is close to earth.

So the craft needs ~ 8.7 km/s delta-v if it does a quick engine burn parallel to its LEO orbital velocity around earth. That's for such being done at a time close to earth and in a direction parallel to earth's orbital velocity around the sun.

In contrast, if the craft did an engine burn of ~ 3.3 km/s to reach earth escape velocity from LEO orbit, drifted slowly several million kilometers out of earth's vicinity, thus becoming almost stationary with respect to earth and 29.8 km/s relative to the sun, then there would be an undesirable 12.3 km/s involved in reaching the solar system escape velocity at 1 AU of 42.1 km/s total velocity. The burn in LEO described previously is a better way, in part for reasons related to those involved in the Oberth maneuver described later.

Also, all of the discussion so far is neglecting the option of gravity assists, described later.

For the direct route without gravity assists, an example is that a 450-sec specific impulse (4400 m/s exhaust velocity v_e) chemical rocket starting in LEO would have an initial to final mass ratio of about 7.2 to 1 for escaping the solar system.

The calculations is straightforward as one knows the 8.7 km/s delta-v involved from the previous discussion, substituting it into the rocket equation that is change_in_velocity = exhaust_velocity * ln (m_initial/m_final), where ln is the natural log.

The rocket equation can be rewritten as m_initial / m_final = e^(∆v / v_e), where e =~ 2.71828.

However, rather than the "direct route" really being fully applicable to historical deep-space missions with chemical propulsion, the technique of gravity assists has been used instead. No historical missions have actually involved having to deliver 8.7 km/s delta v in LEO, only much less.

For example, the probe Voyager 1 launched back in 1977 is currently 15 billion kilometers from the sun, beyond the orbits of the planets, on the way out of the solar system, traveling at 17 km/s velocity relative to the sun. That was not obtained by it having had as much propellant mass as one would expect if not for gravity assists. Indeed, the delta-v capacity directly from its chemical propulsion was rather limited. But Voyager 1 used multiple gravity assists, taking advantage of how planets move fast around the sun while having gravitational forces tugging any object that passes close nearby. A little like bouncing off a giant moving billard ball but with a gentle gravitational tug swinging past a planet rather than destructive physical contact, gravity assists are useful for low-performance craft like chemical rockets; one description is the NASA educational page here, which includes an analogy:

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In such a manner, much less than 8.7 km/s delta-v beyond LEO is nominally sufficient for the amount of actual engine performance needed to leave the solar system or fly by locations in the outer solar system.

However, arranging repeated gravity assists after waiting for planets to get into the right locations for where one is headed is a slow process. For example, Voyager 1 did it over a period of years.

Because of the limited few km/s of delta-v deliverable with the gravitational fields of the planets and the time involved in going off to perform gravitational assists, such are less likely to be used as much with spacecraft having much higher performance engines. But they are useful if minimum delta-v for traveling to the outer solar system is the priority.

With enough gravity assists, one can eventually get anywhere in the solar system even with chemical propulsion, in the manner of some of NASA's historical space probes going as far as Neptune and beyond. However, such is very slow. (Such also requires a relatively substantial ratio of propellant to payload in LEO even just for getting on its initial trajectory despite the small number of km/s delta v involved, with the low Isp of chemical rocket engines). A vastly higher-performance propulsion system such as Orion variants and some other options would work much better if one doesn't want to spend years just getting to the outer solar system.

A different technique with some potential applications is the principle of the Oberth maneuver, for craft with engines having suitable thrust capabilities:

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Click on the above image for the source of its description, a PDF file that is quite a good presentation about a number of topics.

The above figures are just one illustration. The situation for an Oberth maneuver with the sun depends on the delta-v capabilities of the ship and on how close it flies to the sun.

The latter is restricted by the ship's delta-v for lowering its perihelion and by its ability to avoid getting fried, its heat resistance.

Under some assumptions, the Oberth maneuver it is quite helpful up to hundreds of km/s velocity left over after exiting the sun's gravity, for a ship with a fraction of that in delta-v provided by its engines and the ability for a close solar fly-by. That's too little to help a starship much, but it is a possibility for helping speed transport to the outer solar system and a bit beyond.

Compared to the more pedestrian gravity assist technique, the Oberth maneuver requires more delta-v to properly arrange, but it is applicable to a lot more km/s gain, with a solar fly-by able to give tens to hundreds of km/s effective extra delta-v under some assumptions.

Gravity assists have been important to historical chemical-propulsion space probes. In contrast, for example, higher-performance interplanetary Orion craft would tend to obtain proportionally less benefit from gravity assists, less likely to spend the time, since a small number of km/s matters less to them, but they could be more capable of using the Oberth maneuver for more "free" km/s gain if headed for the outer solar system, Kuiper Belt, or Oort Cloud.

The Oberth maneuver is curiously reminiscent of Star Trek IV's "slingshot around the sun," minus the unrealistic time-travel and warp drive.
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Ender
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Post by Ender »

TheLemur wrote:
You are full of shit. Neutron radiation will steadily break down your shielding by activating it.
Okay, just the fact that you used the word "break down" demonstrates that you have no idea how radiation shielding works.
Yeah, what the fuck do I know, I only work on a nuclear reactor every single day. Nitpick word choice all you want, it doesn't change the fact you are trying to bullshit your way through this. Same for Rickover and the guys who wrote that manual Howder linked to. Clearly they don't know as much as you do because they used that phrase in the RPM nd other nuclear guides.
It's not like a Star Trek shield that can be at 60% health or a video game shield with a certain number of hit points. Radiation shielding is deliberately manufactured to have a very low neutron-activation cross-section.
No shit. But when you are talking the flux from the constnat detonation of nuclear weapons for centuries like Orion requires, that doesn't make a damn bit of difference. Fuck, the shielding from the D1G test core in Saratoga Springs was so hot they encased it in several feet of concrete and you could still feel the heat when you pressed your hand against it. And that was just a few decades of high power operations.
Of particular concern is the cobalt 59 in the steel becoming cobalt 60, one of the most dangerous isotopes reactor operators deal with.
Okay, firstly cobalt-60 emits beta radiation, which is very easily blocked (you can block it with only a few millimeters of aluminium, iron or plastic).
That would be why it is not at all a concern then, as the very pipes holding the coolant would block it, right? And why Cobalt 60 is not one of the most commonly used gamma emitters, along with cesium 137? Afterall, beta emission is not coupled with gamma emission, right? Wait, thats wrong, and you are retarded.

1 Ci of Co60 at 1 meter gives you 1 Rem/hr of gamma radiation. This is one the most basic radiation dose thumbrules there is. It is a concern, and it is why we take precautions like portable shielding. The fact that Co-60 is a powerful gamma emitter is what makes it useful for sterlizing and nondestructive testing.

Decay shains show that beta emission is coupled with gamma emission. Whne the neutron converts to a proton and beta, after the beta is released the atom still has too much energy and emits a gamma as well. Now, I'm curious as to whether you never paid attention to your high school science teacher when they covered the chart of the isotopes, or if you are just a dishonest little shit. Given that you have been caught shifting the goalposts and strawmanning so fat, I'm going toward the latter.
Secondly, the vast majority of steel does not contain significant levels of cobalt (source).
So most steel doesn't. Whoop de fucking do. The alloys used in nuclear reactors and shielding does. Quit trying to weasel out of this.
That's why we don't use iron, cobalt,argon, flourine, xenon, cessium, or nitrogen in our shielding today, right?
We don't. You quite clearly have no clue whatsoever what you're talking about. [/quote]
Says the person trying to pass off google knowledge as better then school taught knowledge.
Argon and xenon are noble gases that are totally useless for shielding of any kind because they're GASES.
Do you even know how a fission reaction works? Argon 41 comes from the activation of argon 40 which is found in the poly (virgin and impregnated) used in secondary shielding, and in solution in the water used for shielding.

Xenon is intentionally used in because it acts as a poison, thermalizing neutrons. Way to prove you are just googling rather then have actual knowledge. In addition, Xenon 135 is a decay product from Iodine 135, which is a fission product of Uranium 235. U-235 is found as an impurity in fuel cell cladding.
Nitrogen is also a gas, but it can be fixed into various chemical compounds. It is not used in radiation shielding because it's not good for much (low density, low neutron absorption).
Nitrogen is in the water and the poly used in shielding. N-16 is the reason ELTs have to wait to do their samples on nuclear reactors, and one of the reasons they do shielding surveys.
Fluorine is a ridiculously reactive gas; in fact, it's the closest thing we have to a universal solvent. It will eat through glass, most plastics, most metals, and organic compounds. It is not used in shielding for the same reason as nitrogen.
Nitrogen 14 captures an alpha, forming Fluorine-18, which emits a beta+ (positron) and a neutron to decay to Oxygen-18
Cesium is another ridiculously reactive element that melts at a very low temperature and is too expensive to use for bulk shielding.
Xenon -135 decays to Cesium-135 9 hours later moron. As Xenon is intentionally placed in there to thermalize neutrons, you are ineffect intentionally using cesium as well.
Iron is used in radiation shielding, but it is fairly immune from activation as only 0.2% can transmute and the isotope it forms is a quick-decaying beta producer.
Odd that it makes up 37% of the radioactive CRUD (Crystal River Unidentified Deposits, got to love the Navy's accronyms) then, don't you think? I mean here you are declaring that you know better then people who live and work with these things daily, yet your google-fu doesn't match real life. Huh.
Cobalt is not used because it had a problem with neutron activation, as you pointed out, and it is also quite expensive.
Cobalt is used extensively in nuclear grade materials due to the hardness, strenght, and corrosion resistance it lends to the metals it is alloyed with.

I just want to add to what Howder pointed out earlier - you are debating nuclear physics with a nuke, rocketry with rocket scientist, and transit orbits with 2 physics grads.
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Post by Ender »

TheLemur wrote:
1) Wikipedia is not a valid source here
If you actually spent thirty seconds figuring out where the information came from, you'd realize it was from the US Bureau of Transportation Statistics, complete with a .gov website.

(direct link)
How dare I not do your job for you. I should clearly go and research and validate your arguments.
2) Look at the reference, the source for that information is stted as a year, not a publication, which means for all I know someone just made it up.
See above. The BTS may distort things, but they don't just make them up.
Yet I can go to wiki and do just that. Hence why it is not a valid source.
You compare their fuel efficiency to energy expended, which deliberately ignore the inefficiencies of biological processes. The stomach is only about 18% efficient at converting the energy in your food into energy you can use. So your dishonest 1.8 MJ should really be 10 MJ, which makes a car far more efficient.
Firstly, even if it was 10 MJ a car wouldn't be "far more efficient", since a car would use 8 MJ. [/quote]You are expending 10 MJ to move 70 kgs at a few mph, where as the car is expending 8 MJ to move 3000 kgs at sevaral times faster. Yet you claim that it not more efficient.

Second, I deliberately used a very generous estimate of 500 W total power (including waste heat), which is equivalent to around 430 kcal/hr;
this link gives 265 kcal/hr burned for walking at two MPH even when the guy weighs 120 kg.
How about we get a better source then a diet site before we go to "but but but I'm still right if you let me change a lot of things!" mm'kay?
4) you are shifting the goalposts by redefining efficient when it is convienent to you. Odd that for your arguements that non instant acceleration is inefficient you base it off speed, but when confronted here, you try to compare energy efficiency (though deliberatly exclude information to skew the answer)
Acceleration for human walking is damn close to instant; it's zero to full speed in less than a second.
Which does nothing to change the fact you are shiting between travel time and energy expended as the benchmark for efficiency whenever it suits you.
Please provide an example of information I've supposedly excluded and any redefinitions I've supposedly made.
Getting called out on neutron radiation, ingoring the gamma emission that accompanies beta emission, redefining the benchmark of efficiency being used here.
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Post by TheLemur »

Yet I can go to wiki and do just that. Hence why it is not a valid source.
Really? The BTS is a wiki where you can just move in and rewrite whatever you want? Wow!
where as the car is expending 8 MJ to move 3000 kgs at sevaral times faster. Yet you claim that it not more efficient.
Nice distortion of the word "efficient", considering that the vast, vast majority of that three metric tons is not used for anything other than pushing the vehicle forward.
How about we get a better source then a diet site before we go to "but but but I'm still right if you let me change a lot of things!" mm'kay?
Okay, the people at this site list 263 cal/hr for a 100 kg guy walking at 2 MPH.
Which does nothing to change the fact you are shiting between travel time and energy expended as the benchmark for efficiency whenever it suits you.
What? I was referring to acceleration time, not travel time. The two are totally different concepts.
redefining the benchmark of efficiency being used here.
I have calculated efficiency as joules expended per passenger-mile. Please name another calculation I have made that I have also labeled "efficiency".
Getting called out on neutron radiation,
I knew that neutron radiation would be a factor, and I specifically cited reasons why it could be handled: you can put cheap neutron absorbers in place and use low activation materials.
ingoring the gamma emission that accompanies beta emission,
Why would beta decayers have an exceptional tendency to produce gamma radiation? Tritium is a common beta decayer than can safely be used everywhere because all the radiation is blocked by a thin shield.
Yeah, what the fuck do I know, I only work on a nuclear reactor every single day.
Completely unsubstantiated claim of superior expertise. You tried to claim argon was used as a shielding material. Try again.
Nitpick word choice all you want, it doesn't change the fact you are trying to bullshit your way through this.
Don't blame me for you saying stupid things.
Same for Rickover and the guys who wrote that manual Howder linked to. Clearly they don't know as much as you do because they used that phrase in the RPM nd other nuclear guides.
If you can actually produce text from these guides showing that they used that phrase, then I will drop my claim. Otherwise, go pound sand.
But when you are talking the flux from the constnat detonation of nuclear weapons for centuries like Orion requires, that doesn't make a damn bit of difference.
Even assuming the material on the outside of the ship is so neutron-activated it glows with the heat, any radiation it releases will be absorbed by the other shielding further in, because the total amount of secondary radiation released from activation must be comparable to the primary radiation released by the bomb directly, and we already know shielding can protect against primary radiation.
you could still feel the heat when you pressed your hand against it.
What kind of idiot presses his hands against material that is literally hot from radioactive decay? If it really was that hot, all the radiation must have been absorbed well inside or nobody would dream of standing that close to it. Which was my point exactly.
And why Cobalt 60 is not one of the most commonly used gamma emitters, along with cesium 137?
Okay, I did check, and cobalt-60 is in fact a strong gamma emitter. Which still has nothing to do with my point.
Whne the neutron converts to a proton and beta, after the beta is released the atom still has too much energy and emits a gamma as well.
Except that a neutrino is also emitted which can carry off excess energy, and so while a gamma *might* be emitted, there is no reason why it *has* to be.
school science teacher when they covered the chart of the isotopes,
My high school science teachers never even tried to teach anybody how a single isotope decayed, and I never took a college course that included the characteristics of specific radioactive isotopes, so I didn't know off the top of my head that Co-60 was a strong gamma emitter. Are you just going to pound this one point all day?
The alloys used in nuclear reactors and shielding does.
Do you have any evidence whatsoever for this, especially considering that cobalt is a high activation material and the problems with it in conjunction with radiation are well known? And did you not realize that Orion ships are still *unbuilt* (gasp!), and that if Co-60 is a showstopper, the ship builders can simply change the steel they're using?
argon 40 which is found in the poly (virgin and impregnated)
Do you have any evidence that the shielding is impregnated with large quantities of argon? No? Thought not.
and in solution in the water used for shielding.
This is quite clearly *by accident*, because argon is used as an inert gas and will dissolve in water under pressure.
In addition, Xenon 135 is a decay product from Iodine 135, which is a fission product of Uranium 235. U-235 is found as an impurity in fuel cell cladding.
This is true. It has nothing whatsoever to do with your claim that xenon was deliberately being used for radiation shielding.
Xenon is intentionally used in because it acts as a poison, thermalizing neutrons.
Xenon thermalizes neutrons?! Firstly, xenon is horrible at thermalizing neutrons because it has a very high atomic mass and so does not slow down neutrons in each collision; that is why they use water/carbon and not heavy metals as moderators. Second, thermalizing neutrons will tend to speed *up* the reaction because cross sections for thermal neutrons are generally much larger than for fast neutrons, which is why moderators are used in the first place.
Nitrogen is in the water and the poly used in shielding.
Again, it is probably either there as a dissolved gas or impurity.
Nitrogen 14 captures an alpha, forming Fluorine-18, which emits a beta+ (positron) and a neutron to decay to Oxygen-18
This is true, but I sure as hell hope you aren't claiming F-18 is used for shielding, because it decays in ten minutes.
Xenon -135 decays to Cesium-135 9 hours later moron.
This again has nothing whatsoever to do with your claim, which is that these materials are used as *shielding*. I have no quarrel if you claim they appear in the reactor by accident or through nuclear reactions.
intentionally using cesium as well.
You have no clue what "intentionally" means, do you? Intentionally means you deliberately put it there to serve a purpose. If Cesium-135 or another dangerous isotope is produced by radioactive decay after neutron activation, it is quite clearly *not* going to serve a purpose because it's A), highly radioactive, B), produced in very small quantities, C), has the exact same chemical characteristics as the nonradioactive isotopes which can be brought in externally. The only counterexamples I can think of are fissile isotopes and isotopes which serve as radiation sources; Cesium-135 is not fissile and you claimed the stuff was used for shielding, not radioactivity.
Odd that it makes up 37% of the radioactive CRUD (Crystal River Unidentified Deposits, got to love the Navy's accronyms) then, don't you think?
That's because there's so damn *much* of it, because iron is cheap and fairly effective.
Cobalt is used extensively in nuclear grade materials due to the hardness, strenght, and corrosion resistance it lends to the metals it is alloyed with.
Then it is clearly not a very big problem, because if it was they would use metals which aren't as strong structurally but don't produce Co-60. Congratulations, you just shot yourself in the foot.
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Post by TheLemur »

Sikon, thank you for that excellent and well written explanation of orbital mechanics. Please continue!
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Post by Wyrm »

TheLemur wrote:Nice distortion of the word "efficient", considering that the vast, vast majority of that three metric tons is not used for anything other than pushing the vehicle forward.
Nice evasion, you little bitch. The engine aboard the car uses less energy to move more mass faster. In anyone's dictionary, that's a more efficient way to move. Except to you with your rubber definitions.
TheLemur wrote:I have calculated efficiency as joules expended per passenger-mile.
A lie. Most cars are capable of carrying more than a single passenger. Even with your bogus 1.8 MJ to walk a mile calculation, the average sedan (carrying 5 passengers) gets 1.6 MJ per passenger-mile (1.33 MJ per passenger-mile if you're carrying someone in the trunk), winning over your optimistic estimate of walkingn energy. Against a real estimate of energy taken in to energy utilized, cars soundly trounce walking.
TheLemur wrote:
Getting called out on neutron radiation,
I knew that neutron radiation would be a factor, and I specifically cited reasons why it could be handled: you can put cheap neutron absorbers in place and use low activation materials.
You are, of course, neglecting what you have to give up when you do this. In particular, not being able to use cobalt in your steel would result in weaker steel. That means your pusher plate has to be more massive in order to resist mechanical and thermal stresses (assuming that it could be built at all), which would of course degrade performance. Oops.
TheLemur wrote:Why would beta decayers have an exceptional tendency to produce gamma radiation? Tritium is a common beta decayer than can safely be used everywhere because all the radiation is blocked by a thin shield.
Cobalt-60 is not tritium. This should be obvious even to you. What makes you think that all decay chains have the same nuclear energy levels?
TheLemur wrote:
Yeah, what the fuck do I know, I only work on a nuclear reactor every single day.
Completely unsubstantiated claim of superior expertise. You tried to claim argon was used as a shielding material. Try again.
Now you're just projecting your own feelings of inadequacy onto Ender. It's pathetic.
TheLemur wrote:
But when you are talking the flux from the constnat detonation of nuclear weapons for centuries like Orion requires, that doesn't make a damn bit of difference.
Even assuming the material on the outside of the ship is so neutron-activated it glows with the heat, any radiation it releases will be absorbed by the other shielding further in, because the total amount of secondary radiation released from activation must be comparable to the primary radiation released by the bomb directly, and we already know shielding can protect against primary radiation.
See, even I know you're full of shit here. You think that shielding becomes radioactive one layer at a time. You think that shielding completely blocks radiation. It does neither. Shielding attenuates radiation in exponential proportion to its thickness. A fraction of the radiation always penetrates. Thus, the entire mass of the shielding is being exposed to some level of radiation — it's all becoming activated to one degree or another.

Take Ender's advice and stop pretending you know anything.
TheLemur wrote:
And why Cobalt 60 is not one of the most commonly used gamma emitters, along with cesium 137?
Okay, I did check, and cobalt-60 is in fact a strong gamma emitter. Which still has nothing to do with my point.
AHEM...
TheLemur, a liar, previously wrote:Okay, firstly cobalt-60 emits beta radiation, which is very easily blocked (you can block it with only a few millimeters of aluminium, iron or plastic).
The implication is clearly that you only need to put a few sheets of tinfoil between you and the cobalt, and you're protected from all the nastiness. So, yes, you lying little shit, that was your point.
TheLemur wrote:
Whne the neutron converts to a proton and beta, after the beta is released the atom still has too much energy and emits a gamma as well.
Except that a neutrino is also emitted which can carry off excess energy, and so while a gamma *might* be emitted, there is no reason why it *has* to be.
And the power of obscenely large numbers turns that "*might*" into a certainty that a good fraction of the cobalt-60 will emit gammas.

PS, I'll let Ender handle the rest of your bullshit.
TheLemur wrote:Sikon, thank you for that excellent and well written explanation of orbital mechanics. Please continue!
Don't pretend you understood any of it, dearheart.

-----

Alright. I got off my ass and applied the formula Kuroneko derived (and I almost derived) for the distance/mass ratio relation to the SMART-1 problem. Recall that specific impulse for the engine was 3100 s, thrust was 20 mN, mass was 367 kg and the distance to travel was 7.3 billion km. Kuroneko's formula is:

D = [Mu²/F][R log R - R + 1]

Letting A = F/Mu and T = D/u (as per Kuroneko) and rearranging, we get R[log R - 1] - (AT-1) = 0, so we find the root with respect to R. This turns out to be about 0.4102179408432, found The Hard Way™ (Newton's method doesn't seem to like this function — damn thing keeps going negative, then imaginary). This gives us, by t = (1-R)/A, a travel time of 10 years, five months. It also gives us a dry mass of 150.5 kg, which can actually accomodate the 29 kg SMART-1 Hall effect thruster. This same mass ratio gives us a delta-V of 27.07 km/s (a bit less than the crude estimate, but not horribly wrong).

Also, the plume power is Fu/2 = 303.8 W, and given the 60% efficiency of the engine, gives us a power consumption of 506.3 W. This can be supplied by a 551.23 W RTG (which will degrade over the lifetime of the mission to 506.3 W) which masses 110.25 kg (alpha = 200 kg/kW). So the mass of engine plus RTG is 139.25 kg, leavng us with 10.25 kg for structure. So we can build this craft using current power technologies if the SMART-1 thruster does indeed have the stated performance.

Does it? Well, not quite. I found out that the SMART-1 Hall effect thruster, in its implementation aboard SMART-1, had a specific impulse of 1640 s, but a thrust of 68 mN. It was the DS1 that had the 3100 s specific impulse and 20 mN, but on different power levels. Dammit.

Let's try the calculation again, this time with the off-the-shelf NSTAR (as flown on DS-1) and including RTGs.

The NSTAR ion engine (in high-power mode) has a specific impulse of 3100 s, a thrust of 92 mN, with a power consumption 2.5 kW at this power level. I was unable to find engine mass, but I assumed that the mass was proportional to its "performance", which I define here as the product of specific impulse and thrust ("performance" also happens to be proportional to exhaust power). For the Hall effect thruster in SMART-1, this ratio is (1640 s)(68 mN)/(29 kg) = 3.85 m/s, so I use the same performance/mass ratio for the NSTAR to get an engine mass of (3100 s)(92 mN)/(3.85 m/s) = 79 kg. Even though the NSTAR is an electrostatic and the SMART-1 was a Hall effect, this figure is probably close enough to reality.

The previous time calculation was done for an A = F/Mu of 1.7938088481e-9 s^{-1}. The thrust has increased, so to keep the time the same, simply increase starting mass by the same factor (4.6), and thus the initial (fueled) mass is 1,688.2 kg, and dry mass of 692.5 kg.

A couple of RTGs rated at 2.72 kW (the output would degrade to 2.5 kW by the time we arrive at Pluto) would mass approximately 544.2 kg. Together with the ion engine, this comes to 623.2 kg. This is under the dry weight of the mission, and leaves us with 69.3 kg for structure and mission packages.

We could build this craft today, with current technologies!
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Post by TheLemur »

Nice evasion, you little bitch. The engine aboard the car uses less energy to move more mass faster.
Didn't I just calculate that it used more energy? And most of the mass isn't serving any useful purpose. If you build an amazing one thousand ton vehicle that can go 80 MPH, get 25 MPG and carry five people and cargo, it's no better than a car is in an efficiency comparison because its useful payload is still the same even if the mass isn't.
A lie. Most cars are capable of carrying more than a single passenger.
So what? If you actually bothered to look at how the data was derived:

"For 1995 and subsequent years, highway passenger-miles were taken directly from Highway Statistics rather than derived from vehicle-miles and average occupancy, as is the case for 1960-1994."

They did take into account the fact that cars can carry more than one person. And even if they didn't, that still has nothing to do with how I was defining "efficiency". Try spending thirty seconds looking this up before jumping to conclusions.
cars soundly trounce walking.
Even going by your numbers, a margin of 10% is hardly a "trounce".
which would of course degrade performance. Oops.
How much would it degrade performance? 1%? 10%? Is it even significant? You don't know! Would they even use steel for a pusher plate? Steel is primarily used because it's very cheap, but that's hardly a concern for an interstellar spacecraft.
Cobalt-60 is not tritium. This should be obvious even to you. What makes you think that all decay chains have the same nuclear energy levels?
(sigh)

I provided tritium as an example to show that *not all* beta emitters produce significant levels of gamma, and so therefore you couldn't really assume that Co-60 did if you knew nothing about the isotope.
Now you're just projecting your own feelings of inadequacy onto Ender. It's pathetic.
Ooh, fancy psychoanalysis. Am I supposed to run away in fear now, or what?
You think that shielding becomes radioactive one layer at a time.
The outer layer will become the most activated, followed by the next outermost layer, and so on, each "layer" getting exponentially less activation (assuming the shielding is homogeneous). The vast, vast majority of activated material will be well on the outside of the shielding assuming the shielding has a decent absorption cross section because of the exponential decay, so there's going to be a lot of shielding still in between the activated material and the cargo for almost all cases.
Thus, the entire mass of the shielding is being exposed to some level of radiation — it's all becoming activated to one degree or another.
You're conveniently ignoring the *degree* to which the material becomes activated. Sure, the inner layers will get some degree of activation, but if the shielding is any good it won't be a concern because it will be too low to worry about.
The implication is clearly that you only need to put a few sheets of tinfoil between you and the cobalt, and you're protected from all the nastiness.
This is true, and I was wrong here because I didn't bother to look it up in advance (sound familiar?) This still has nothing to do with my main point, which is that radiation blocking is a solvable problem and will not be a showstopper for an Orion-type spaceship. You must really be getting desperate- this is what, the fifth time either you or Ender has attacked me over this one issue?
And the power of obscenely large numbers turns that "*might*" into a certainty that a good fraction of the cobalt-60 will emit gammas.
I was referring to beta-decay in general, not just Co-60.
Don't pretend you understood any of it, dearheart.
Understood it? I could have written most of it if I had several hours to spend.
So we can build this craft using current power technologies if the SMART-1 thruster does indeed have the stated performance.
I agree: you could build it, if there's no annoying stuff like gravity in the way and you could somehow fit all this stuff into your very small mass budget. New Horizons is going to Pluto in ten years solely on chemical propulsion (admittedly with a much larger rocket stage and Jupiter gravity assist, but then again you ignored gravity in your calculation).
We could build this craft today, with current technologies!
Yes, you could. However, it would not be able to transport any significant quantity of cargo or instruments because of the razor-thin margins, and it sure as hell would not be able to get to Pluto in two years, which was the original claim.
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Post by Wyrm »

TheLemur wrote:Didn't I just calculate that it used more energy? And most of the mass isn't serving any useful purpose.
Neglecting the fact that the frame is there to protect you in case of accidents, help provide a slipstream for the wind, and provide comfort and entertainment. That's hardly "not serving any useful purpose."
TheLemur wrote:"For 1995 and subsequent years, highway passenger-miles were taken directly from Highway Statistics rather than derived from vehicle-miles and average occupancy, as is the case for 1960-1994."
In other words, you don't know how the highway statistics were gotten. I can think of quite a few ways that that figure could be biased, or at the very least, not telling you what you think its telling you. Your figure is worthless.

And you're a liar, too:
Liar-boy previously wrote:Maybe not a bicycle, but it's sure as hell more efficient than a car or airplane. Automobiles and airplanes use roughly 4 MJ per passenger mile. (source) Assuming you're taking your time, you can walk two miles in one hour. In one hour, at a very high power consumption of 500 W, you would consume 1.8 MJ, versus 8 MJ for automobiles and airplanes.
So you inflated the figure as supposedly stated for automobiles. How the hell did 4 MJ per passenger mile for autos and aircraft become 8 MJ, which you used thereafter?

Further, you can always decrease efficiency by underutilizing what you're using, but in efficiency comparisions, that is not fair.

Also, I took the trouble to undress that link, underlined in your quote:

http://en.wikipedia.org/wiki/Amtrak#Amt ... sportation

Ie, it's WIKI-FAIL!
TheLemur wrote:And even if they didn't, that still has nothing to do with how I was defining "efficiency". Try spending thirty seconds looking this up before jumping to conclusions.
Given that you never state what you mean for a certain process to be "efficient", somehow expecting it to be obvious to all involved, you really shouldn't be surprised when others define it differently.

Or maybe you expected us to know because of you're telepathy. You need to go back to the Psi Corps.
TheLemur wrote:Even going by your numbers, a margin of 10% is hardly a "trounce".
Learn to read, moron. I call 10 MJ the realistic metabolic cost of a mile walking, not your bullshit 1.8 MJ figure.
TheLemur wrote:
which would of course degrade performance. Oops.
How much would it degrade performance? 1%? 10%? Is it even significant? You don't know!
And neither do you. But I'm not the one suggesting to use cobalt-free steel in the Orion pusher plate. You are. So I expect you to have a sound argument to back it up. And, fortunately, because of Debate Rule 6, I can.
TheLemur wrote:(sigh)

I provided tritium as an example to show that *not all* beta emitters produce significant levels of gamma, and so therefore you couldn't really assume that Co-60 did if you knew nothing about the isotope.
:roll: Why don't you fucking make sure before you start spouting your mouth off, you little twerp?
TheLemur wrote:
Now you're just projecting your own feelings of inadequacy onto Ender. It's pathetic.
Ooh, fancy psychoanalysis. Am I supposed to run away in fear now, or what?
You have plenty of choices. You can continue to accuse Ender of lying about his credentials, or you can ignore him and continue to spout your nonsense, or you can take his council and actually learn something. I prefer the latter route when I encounter people who outclass me in certain areas. Learning is actually a lot of fun, and it results in a much better outcome. C'mon, it'll grow on ya.
TheLemur wrote:
Thus, the entire mass of the shielding is being exposed to some level of radiation — it's all becoming activated to one degree or another.
You're conveniently ignoring the *degree* to which the material becomes activated.
No, you're conveniently ignoring the time-scale that the irradiation occurs, and that in space, you tend to reduce mass to the least that you can get away with, including the shielding. Over the time scale of centuries of operation, as would be needed to go interstellar on Orion, that's a lot of activation even after shielding. If the first x feet of shielding becomes dangerously radioactive in a decade of operation, then the next x feet will become the same in two decades of operation, the next x feet in four, and so on. A 600 year powered transit would require about 8 times the shielding than a decade-long powered transit time.

You're also ignoring another factor, radiation-induced embrittlement. This is caused by radiation particles recoiling off atoms, displacing them in the lattace and weakening it. Also, alpha and beta radiation, and in particular neutrons, cause the chemical makeup of the metal to change slowly through activation. That activated shielding can become physically hot means that enough disintegrations are occuring each second for enough of them to be absorbed by the material cause significant heat. Maintained activation radiation over the course of centuries and you might have a big problem. Embrittlement induced by activation can be staved off by the right materials (assuming that you accept the consequences of not using those materials), but embrittlement due to dislocations are inevidable. And remember this is not a static load this plate will be under.
TheLemur wrote:This is true, and I was wrong here because I didn't bother to look it up in advance (sound familiar?) This still has nothing to do with my main point, which is that radiation blocking is a solvable problem and will not be a showstopper for an Orion-type spaceship.
I don't remember Ender claiming that it would be a "showstopper". What he claimed was that the radiation shield would become of less and less utility because the shielding itself would become radioactive.
TheLemur wrote:
And the power of obscenely large numbers turns that "*might*" into a certainty that a good fraction of the cobalt-60 will emit gammas.
I was referring to beta-decay in general, not just Co-60.
So some isotopes beta decay to a stable ground state, some don't, needing to emit gammas. Thank you, Captain Obvious of the Interplanetary Space Corps.
TheLemur wrote:
Don't pretend you understood any of it, dearheart.
Understood it? I could have written most of it if I had several hours to spend.
The plain fact is that you didn't.
TheLemur wrote:
So we can build this craft using current power technologies if the SMART-1 thruster does indeed have the stated performance.
I agree: you could build it, if there's no annoying stuff like gravity in the way and you could somehow fit all this stuff into your very small mass budget. New Horizons is going to Pluto in ten years solely on chemical propulsion (admittedly with a much larger rocket stage and Jupiter gravity assist, but then again you ignored gravity in your calculation).
Ion engine-driven probes can use gravity assist, too. And as long as we're combining techniques, why not mount a chemical rocket to get us quickly out of the Earth's gravity well, then kick in the ion drive for a long period of constant thrust when we'd normally be doing nothing? I don't remember ever claiming that ion drive is teh uber; it has it's advantages and weaknesses.
TheLemur wrote:Yes, you could. However, it would not be able to transport any significant quantity of cargo or instruments because of the razor-thin margins, and it sure as hell would not be able to get to Pluto in two years, which was the original claim.
I didn't have to. The original (somewhat fanciful) ion engine system I proposed did the task in the required time, with no considerations for the payload. That was all I claimed, and I successfully showed that it could. I never claimed that the systems we currently have could make it within that time, and again no payload was included in the challenge. I only claimed that it could make it, and I successfully showed this.

I could reduce the propellant fraction to put in more instruments at the expense of time to Pluto. The calculation should be easy enough that you could do it, so I won't do it here. It means that I would have to spend some time coasting, and more carefully chosing my oribts, but I would still have plenty of delta-V to play with.
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Post by Kuroneko »

Wyrm, something is not quite right, but I believe the main point of your calculations still holds, since the figures aren't drastically incorrect and we wouldn't actually start from zero in a realistic situation.
Wyrm wrote:Recall that specific impulse for the engine was 3100 s, thrust was 20 mN, mass was 367 kg and the distance to travel was 7.3 billion km. ... D = [Mu²/F][R log R - R + 1] ... Letting A = F/Mu and T = D/u (as per Kuroneko) and rearranging, we get R[log R - 1] - (AT-1) = 0, so we find the root with respect to R. This turns out to be about 0.4102179408432, ... This gives us, by t = (1-R)/A, a travel time of 10 years, five months.
At this mass fraction, the distance is most of the way to Pluto at perihelion, but much smaller than 7.3e12m. For perihelion ([29.7-1]AU) flyby, the ship time would be eleven years; for aphelion ([49.5-1]AU), about thirteen and a half years. This does not, of course, count any special maneuvers or initial orbital velocity, which would be utilized in a real mission and would lower the time required or increase possible payload or both. In any case, for D = 7.30e12m, I calculate u = Ig = 3.04e4m/s, R = 0.231, t = 13.6yr.
Wyrm wrote:... found The Hard Way™ (Newton's method doesn't seem to like this function — damn thing keeps going negative, then imaginary).
Yes. The same problem occurs near -1/e for the Lambert W function. As mentioned before, the analytical solution is r = FD/Mu²-1, R = r/W(r/e); however, what I need to add here is that the non-principal branch that is real in [-1/e,0] should be used here. That is because increasing u (incidentally increasing power or keeping it constant by lowering F by the same factor) should gives us a lower mass fraction, so that the magnitude of W(r/e) should increase in [-1/e,0). Out of the two branches real on [-1/e,0), it is the non-principal one. The principal branch gives us a R>1 solution to the equation.
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