AntiMatter Engine on Kickstarter (LOLWUT)

[PB666]

15 hours ago, Spaceception said:

Well, they only need 17 grams to get to .4 c.

That doesn't seem correct, not unless the space craft total weight was 100 grams, payload would have to be under 50 grams, 25 if you intent to stop at a destination. If you go by e = mc^2 and half the ships weight was antimatter the most you could get was 0.7c, that would mean a ship weight of 34g. Don't forget the ejecta is either light at 300MW/newton or some form of matter with an ISP in the 1e7 range. Not a ejecta would be going -y so there would loses there.

Is it possible to do a better job checking the sources math before we throw numbers like this out, i've seen highly optimistic quotes like this with the black hole drive that just don't pan out. A sense of realism here would also factor in the carrying weight of containment. 

[p1t1o]

12 hours ago, todofwar said:

If it's just antihydrogen I would assume that it would be a colorless gas like hydrogen. But the slow leaking in of gasses might cause constant annihilation at the surface, giving off a constant glow. I highly doubt any container would end up looking like that though. 

Matter/AM annihilation does not produce visible light. The only reason it would be glowing blue like that is if it is very hot, several thousand Kelvins hot. If your antimatter is very hot, it is also very high pressure, so you better hope that the magnetic fields in your penning trap are nice and strong (and symmetric).... (Also, if it is that hot from annihilation events, then it is also spewing gamma rays at quite a high rate, not to mention other high energy particles, so either way there are probably some health and safety concerns...)

 

Here is an image of a penning trap, I'm not sure how much stuff it hold but this one was used for an experiment designed to study individual ions, so probably not very much...

Edited April 11, 2016 by p1t1o

[Jovus]

The pretty blue glow of antimatter bottles (in movies) is provided by blue LEDs so you can see that the trap is working.

Because, you know, otherwise you might be unsure. Something else might have made off with the city block your storage container lives in.

[PB666]

Here is an image of the illustrated device.

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Popular mechanics quote  - .... 17 grams of antimatter on board, the craft would slowly accelerate toward the speed of light, topping out at about 40 percent. . . . . . estimated price tag of . . . .  . . . . or $1.7 trillion just for fuel costs. . (Edited for the sake of fair use)

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IN forbes the source of the PM sourcing article it says - n a paper presented at the 2003 Particle Accelerator Conference. . . . . .  .  a small antimatter interstellar probe that would only need 17 grams of antihydrogen . . . . . . .  touching a tenth of lightspeed . . .  . . (Edited for the sake of fair use)

A little more credible . . . . .ON my IPAD I see a different article that defines the weight of the sail as 100 kg of Uranium on carbon fiber.

http://www.iflscience.com/space/antimatter-propulsion-could-sail-stars

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antimatter-driven sail that could deliver a 10-kilogram (22-pound) probe  . . . .  . using 17 grams (0.6 ounces) of antihydrogen . . . . 100-kilogram (220 pounds) sail  . . . .made of carbon, and coated with depleted uranium. (Edited for the sake of fair use)

http://accelconf.web.cern.ch/accelconf/p03/PAPERS/FOAA005.PDF

It says that the Uranium 238 undergoes antimatter-triggered to 2 palladium 111 the kinetic energy of the release they claim is 14,000,000 m/s

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 FOAA005.PDF - The  velocity  of  the  fission  products  is  [13,900,000]   m/s  and  the mass is 1.85[e25]

The amount of hydrogen in 17 grams is equal to 6.0223E23 that is 1.02E25 atoms that would cause the release of that number of Uranium at 98%, basically 1E25
That means how much mass will be lost, It will lose ideally 3.97kg of depleted uranium (U238, plus trace 236 and 235).

If we take them at their word on the palladium post-fission velocities, the delta-v generated will be  14,000,000 * ln(110/106) = 514,000 m/s (0.0017c). Note: the article does not state what c it might reach. We could try other scenarios, lets say it releases instead of 1 uranium, 2 uranium at 7/10ths the velocity (0.0024c), 4 results in (0.0036c), 8 results in (0.0054c), we place the logical stopping point once we have depleted more than half the sail. That is 12.36 uranium lost per antimatter collision, this results in (0.0054c) beyond which the antimatter would cause the degradation of the sail. But lets just assume that all the uranium was lost and none of the carbon fiber was lost, lets assume that the carbon fiber and tether weights were zero, that means we have 100kg of depleted uranium to ablate. The dV produced is 0.022c. That is 1/20th the velocity stated in the cash-farming article and 1/5 the velocity of the forbes article.

Note that the following weight were not given in the article and not added to my calculations:
 

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FOAA005.PDF - (2)  solid  H2bar  crystal  storage  units,  (3)  an  antiproton  driven electrical power supply

In terms of the power supply, it would need to be nuclear (TNG) and it would need to produce enough power to store the gas in magnetically contained compartments.
The antimatter containment is a hand-waving engineering project so that its mass cannot be estimated.

In a perfect world were only that mass was lost the amount of thrust you can produce is N/300MW (i.e. N/300,000,000 the same value as the speed of light). If you have a 10kg payload and convert 17 grams of mass then you have 1.35E15 energy. You can benefit from mass loss via ejection, but only so much because you have to accelerate the mass. The actual dV produced is actually less than I stated because you have heat losses on the sheild and because the ejecta is not strait backwards (-y) there are |x| and |z| vectors.
 

Summary: Don't believe everything you read in a magazines, particularly articles pandering for money. . . .Caveot emptor.  Fact checking?

There is no magic bullet that gets us to the speed of light. If you convert mass the energy you get was defined 100 years ago everyone knows it, no one can forget it, its E = mc². If you have a ship that initially weight 110+ kg and you convert 0.017 kg of mass you cannot expect to get anywhere close to the speed of light, even if you shed 9/10ths of the ship mass in the process.

 

[AngelLestat]

I dont know exactly how kickstarter works, but it seems that you can receive money without the need to prove nothing.
So this open the oportunity for many scams, like the Triton Gills (kickstarter project) who receive almost 1 millon dollar when in fact is totally fake. 
http://www.iflscience.com/technology/artificial-gills-underwater-breathing-device-has-820000-funding-despite-being

This antimatter engine seems to have the same objective..  steal money, because at the end, they just can said.. it would work.. but we dont have fuel to test it. so..  thanks for your money.
 

[Spaceception]

Question, if you were in an orbit around the Sun, perhaps closer than Venus is, and had a magnetic sail several hundred meters wide, could you collect a decent amount of Antimatter? And if so, how long would it take to collect a few kilograms of it?

[HebaruSan]

18 minutes ago, Spaceception said:

Question, if you were in an orbit around the Sun, perhaps closer than Venus is, and had a magnetic sail several hundred meters wide, could you collect a decent amount of Antimatter? And if so, how long would it take to collect a few kilograms of it?

The stuff around Earth comes from "high-energy cosmic rays impacting Earth's atmosphere"; is there another source available to such a vessel, without a nearby atmosphere?

[PB666]

7 minutes ago, Spaceception said:

Question, if you were in an orbit around the Sun, perhaps closer than Venus is, and had a magnetic sail several hundred meters wide, could you collect a decent amount of Antimatter? And if so, how long would it take to collect a few kilograms of it?

For hydrogen (assuming you electrically neutralized protons) at the radius of mercury about the sun it would take 134,000 years to collect 1 kg across a meter squared. Antimatter is created in major solar flares, about 1/2 a kilogram, this is versus 1E9kg of hydrogen per second, and so the amount of antimatter production is less that 1/10¹²  The amount of matter production. IOW if you sat at Mercuries orbit collecting everything the flew by a meter squared in a year you would collect less that 10^-19 kg of antimatter. Making it 1 kg of antimatter would take  9 x 10¹⁶ joules of energy (with no manufacturing losses), about the daily human production of energy. 

http://www.nasa.gov/centers/goddard/news/topstory/2003/0903rhessi.html

So how if you had a sail about Mercury how big would it have to be to collect a kg of antimatter in a year. The sail would need to be a square 3 million by 3 million kilometers assuming that you had several decent solar flares in that year.