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Why we don't use cyclotron for ion thrusters?

If specific impulse is directly related to exhaust velocity, would a ion post-accelerator improve the Isp of a propulsion system?Which one of these Isp's for the Dawn spacecraft is wrong?Could a ship take off from Earth using ion thrusters?Would an ionocraft have better or worse performance in the upper atmosphere?Could Earth's electric grid be used to power rocket launches via cable?Can ion thrusters be scaled up?Could chemical energy quicken ion propulsion?Propellant Settling for RCS ThrustersCould an ion engine use an asteroid as an in-situ resource for fuel?Why can't we use charged fine metal dust instead of plasma for ion thruster?Why are Ion Thrusters so energy hungry?Are ion/plasma thrusters sufficient for propulsion from Earth?

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We know that the advantage of thruster is high efficiency and 'everlasting' propulsion. Which It can not provide a sudden huge propulsion during launch like chemical fuel is its biggest problem. I wonder if we can using cyclotron to accelerate ions to get a higher speed to provide a higher propulsion in the ion thruster. In this way, it gonna has a wider range of application.And this improvement seems feasible...

rockets propulsion

share|improve this question

edited May 18 at 15:02

Organic Marble

64.7k4180274

asked May 18 at 14:03

Hanzhi ZhangHanzhi Zhang

163

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  I don't know the answer to your question, but I do know that a cyclotron needs big magnets--big massive magnets--and a linac does not need big magnets...
  $endgroup$
  – Solomon Slow
  May 18 at 16:34
  

  
  
  
  


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  @SolomonSlow yeah but I think linac also needs very high voltage which means more electricity and longer distance which means larger volume of the engine.
  $endgroup$
  – Hanzhi Zhang
  May 18 at 17:32
  

  
  
  
  


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  @HanzhiZhang a cyclotron is a bit like a linac rolled up. They both use high voltage RF acceleration gaps of hundreds of kilovolts. The magnet is used to send the particles through the same RF gaps over and over, while a regular linac strings separate RF gaps in a long line. Slightly related space.stackexchange.com/a/33576/12102
  $endgroup$
  – uhoh
  May 18 at 17:38
  
  

  
  
  
  

add a comment | 

2

$begingroup$

We know that the advantage of thruster is high efficiency and 'everlasting' propulsion. Which It can not provide a sudden huge propulsion during launch like chemical fuel is its biggest problem. I wonder if we can using cyclotron to accelerate ions to get a higher speed to provide a higher propulsion in the ion thruster. In this way, it gonna has a wider range of application.And this improvement seems feasible...

rockets propulsion

share|improve this question

edited May 18 at 15:02

Organic Marble

64.7k4180274

asked May 18 at 14:03

Hanzhi ZhangHanzhi Zhang

163

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't know the answer to your question, but I do know that a cyclotron needs big magnets--big massive magnets--and a linac does not need big magnets...
  $endgroup$
  – Solomon Slow
  May 18 at 16:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SolomonSlow yeah but I think linac also needs very high voltage which means more electricity and longer distance which means larger volume of the engine.
  $endgroup$
  – Hanzhi Zhang
  May 18 at 17:32
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @HanzhiZhang a cyclotron is a bit like a linac rolled up. They both use high voltage RF acceleration gaps of hundreds of kilovolts. The magnet is used to send the particles through the same RF gaps over and over, while a regular linac strings separate RF gaps in a long line. Slightly related space.stackexchange.com/a/33576/12102
  $endgroup$
  – uhoh
  May 18 at 17:38
  
  

  
  
  
  

add a comment | 

$begingroup$

We know that the advantage of thruster is high efficiency and 'everlasting' propulsion. Which It can not provide a sudden huge propulsion during launch like chemical fuel is its biggest problem. I wonder if we can using cyclotron to accelerate ions to get a higher speed to provide a higher propulsion in the ion thruster. In this way, it gonna has a wider range of application.And this improvement seems feasible...

rockets propulsion

share|improve this question

edited May 18 at 15:02

Organic Marble

64.7k4180274

asked May 18 at 14:03

Hanzhi ZhangHanzhi Zhang

163

$endgroup$

share|improve this question

edited May 18 at 15:02

Organic Marble

64.7k4180274

asked May 18 at 14:03

Hanzhi ZhangHanzhi Zhang

163

share|improve this question

edited May 18 at 15:02

Organic Marble

64.7k4180274

asked May 18 at 14:03

Hanzhi ZhangHanzhi Zhang

163

edited May 18 at 15:02

Organic Marble

64.7k4180274

edited May 18 at 15:02

Organic Marble

64.7k4180274

asked May 18 at 14:03

Hanzhi ZhangHanzhi Zhang

163

asked May 18 at 14:03

Hanzhi ZhangHanzhi Zhang

163

1 Answer
1

active

oldest

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4

$begingroup$

The energy required would be enormous to see a big difference, even ignoring the other points of mass of the magnets and the fact that you probably can't accelerate that much mass at a time.

The largest cyclotrons had radii of about 4m and magnetic fields around 2T. Let's use those numbers as a starting point. The energy per particle in a cyclotron is:

$$E = fracq^2B^2R^22m$$

where $q$ is the particle's charge, $B$ the magnetic field, $R$ the radius of the cyclotron, and $m$ the particle's mass. For a fully ionized Helium atom and the above cyclotron we get an energy per particle of about $1.5 times 10^-10$ J.

That sounds pretty decent. But now let's make some assumptions. The Dawn spacecraft had 500Kg of fuel and burned for 2000 days, for a rate of $3 mu g /sec$, which translates into $4.5 times 10^17$ atoms per second. To get a similar amount of mass out of our cyclotron engine, we'd need about 45 MW of power, assuming absolutely no losses.

Now, let's instead say we want a much more reasonable 45KW cyclotron. That would mean we'd be using mass at a rate of $3 ng$ per second. What kind of momentum change would this give us? Assuming (big assumption) that all particles are exiting at the same velocity, we'd get a momentum change of 0.7 kg m/s. Not bad, but remember that force is $F = fracdpdt$, with everything constant this is about 0.7N of thrust. At those power levels, ion engines are comparable, less mass, and more flight proven.

Now, these back of the envelope calculations did show that the cyclotron would be far more efficient, but the mass of the magnets alone would nullify this, ignoring power requirements, given the thrust you end up with anyway.

share|improve this answer

answered May 18 at 17:58

Michael StachowskyMichael Stachowsky

1,228211

$endgroup$

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  Emm...I still don't know why the mass of magnets would nullify...because the magnet must be very huge or...?
  $endgroup$
  – Hanzhi Zhang
  May 18 at 18:36
  

  
  
  
  


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  Yes, basically. Cyclotrons weigh in the order of tons. You'd maybe use about 10x less fuel but the engine would weigh more than 10x as much!
  $endgroup$
  – Michael Stachowsky
  May 18 at 18:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see! If I wanna do calculation about what the mass of magnet i need , what equation should be applied to get the detail of the magnet? Thx
  $endgroup$
  – Hanzhi Zhang
  May 18 at 18:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That I'm afraid I do not know. Perhaps start with material density and magnet size
  $endgroup$
  – Michael Stachowsky
  May 18 at 18:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It would be great to calculate how small a cyclotron could be that provided the same specific impulse as Dawn, or say 10x better. I think for helium (not xenon) it would be a lot smaller than the ones mentioned here. I went to look up Dawn's Isp but got stuck: What is the mass-specific impulse (Isp) of the ion engine used by the Dawn spacecraft? Which of these is wrong?
  $endgroup$
  – uhoh
  May 18 at 19:04
  

  
  
  
  

 | 
show 5 more comments

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4

$begingroup$

The energy required would be enormous to see a big difference, even ignoring the other points of mass of the magnets and the fact that you probably can't accelerate that much mass at a time.

The largest cyclotrons had radii of about 4m and magnetic fields around 2T. Let's use those numbers as a starting point. The energy per particle in a cyclotron is:

$$E = fracq^2B^2R^22m$$

where $q$ is the particle's charge, $B$ the magnetic field, $R$ the radius of the cyclotron, and $m$ the particle's mass. For a fully ionized Helium atom and the above cyclotron we get an energy per particle of about $1.5 times 10^-10$ J.

That sounds pretty decent. But now let's make some assumptions. The Dawn spacecraft had 500Kg of fuel and burned for 2000 days, for a rate of $3 mu g /sec$, which translates into $4.5 times 10^17$ atoms per second. To get a similar amount of mass out of our cyclotron engine, we'd need about 45 MW of power, assuming absolutely no losses.

Now, let's instead say we want a much more reasonable 45KW cyclotron. That would mean we'd be using mass at a rate of $3 ng$ per second. What kind of momentum change would this give us? Assuming (big assumption) that all particles are exiting at the same velocity, we'd get a momentum change of 0.7 kg m/s. Not bad, but remember that force is $F = fracdpdt$, with everything constant this is about 0.7N of thrust. At those power levels, ion engines are comparable, less mass, and more flight proven.

Now, these back of the envelope calculations did show that the cyclotron would be far more efficient, but the mass of the magnets alone would nullify this, ignoring power requirements, given the thrust you end up with anyway.

share|improve this answer

answered May 18 at 17:58

Michael StachowskyMichael Stachowsky

1,228211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Emm...I still don't know why the mass of magnets would nullify...because the magnet must be very huge or...?
  $endgroup$
  – Hanzhi Zhang
  May 18 at 18:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, basically. Cyclotrons weigh in the order of tons. You'd maybe use about 10x less fuel but the engine would weigh more than 10x as much!
  $endgroup$
  – Michael Stachowsky
  May 18 at 18:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see! If I wanna do calculation about what the mass of magnet i need , what equation should be applied to get the detail of the magnet? Thx
  $endgroup$
  – Hanzhi Zhang
  May 18 at 18:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That I'm afraid I do not know. Perhaps start with material density and magnet size
  $endgroup$
  – Michael Stachowsky
  May 18 at 18:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It would be great to calculate how small a cyclotron could be that provided the same specific impulse as Dawn, or say 10x better. I think for helium (not xenon) it would be a lot smaller than the ones mentioned here. I went to look up Dawn's Isp but got stuck: What is the mass-specific impulse (Isp) of the ion engine used by the Dawn spacecraft? Which of these is wrong?
  $endgroup$
  – uhoh
  May 18 at 19:04
  

  
  
  
  

 | 
show 5 more comments

4

$begingroup$

The energy required would be enormous to see a big difference, even ignoring the other points of mass of the magnets and the fact that you probably can't accelerate that much mass at a time.

The largest cyclotrons had radii of about 4m and magnetic fields around 2T. Let's use those numbers as a starting point. The energy per particle in a cyclotron is:

$$E = fracq^2B^2R^22m$$

where $q$ is the particle's charge, $B$ the magnetic field, $R$ the radius of the cyclotron, and $m$ the particle's mass. For a fully ionized Helium atom and the above cyclotron we get an energy per particle of about $1.5 times 10^-10$ J.

That sounds pretty decent. But now let's make some assumptions. The Dawn spacecraft had 500Kg of fuel and burned for 2000 days, for a rate of $3 mu g /sec$, which translates into $4.5 times 10^17$ atoms per second. To get a similar amount of mass out of our cyclotron engine, we'd need about 45 MW of power, assuming absolutely no losses.

Now, let's instead say we want a much more reasonable 45KW cyclotron. That would mean we'd be using mass at a rate of $3 ng$ per second. What kind of momentum change would this give us? Assuming (big assumption) that all particles are exiting at the same velocity, we'd get a momentum change of 0.7 kg m/s. Not bad, but remember that force is $F = fracdpdt$, with everything constant this is about 0.7N of thrust. At those power levels, ion engines are comparable, less mass, and more flight proven.

Now, these back of the envelope calculations did show that the cyclotron would be far more efficient, but the mass of the magnets alone would nullify this, ignoring power requirements, given the thrust you end up with anyway.

share|improve this answer

answered May 18 at 17:58

Michael StachowskyMichael Stachowsky

1,228211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Emm...I still don't know why the mass of magnets would nullify...because the magnet must be very huge or...?
  $endgroup$
  – Hanzhi Zhang
  May 18 at 18:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, basically. Cyclotrons weigh in the order of tons. You'd maybe use about 10x less fuel but the engine would weigh more than 10x as much!
  $endgroup$
  – Michael Stachowsky
  May 18 at 18:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see! If I wanna do calculation about what the mass of magnet i need , what equation should be applied to get the detail of the magnet? Thx
  $endgroup$
  – Hanzhi Zhang
  May 18 at 18:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That I'm afraid I do not know. Perhaps start with material density and magnet size
  $endgroup$
  – Michael Stachowsky
  May 18 at 18:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It would be great to calculate how small a cyclotron could be that provided the same specific impulse as Dawn, or say 10x better. I think for helium (not xenon) it would be a lot smaller than the ones mentioned here. I went to look up Dawn's Isp but got stuck: What is the mass-specific impulse (Isp) of the ion engine used by the Dawn spacecraft? Which of these is wrong?
  $endgroup$
  – uhoh
  May 18 at 19:04
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

The energy required would be enormous to see a big difference, even ignoring the other points of mass of the magnets and the fact that you probably can't accelerate that much mass at a time.

The largest cyclotrons had radii of about 4m and magnetic fields around 2T. Let's use those numbers as a starting point. The energy per particle in a cyclotron is:

$$E = fracq^2B^2R^22m$$

where $q$ is the particle's charge, $B$ the magnetic field, $R$ the radius of the cyclotron, and $m$ the particle's mass. For a fully ionized Helium atom and the above cyclotron we get an energy per particle of about $1.5 times 10^-10$ J.

That sounds pretty decent. But now let's make some assumptions. The Dawn spacecraft had 500Kg of fuel and burned for 2000 days, for a rate of $3 mu g /sec$, which translates into $4.5 times 10^17$ atoms per second. To get a similar amount of mass out of our cyclotron engine, we'd need about 45 MW of power, assuming absolutely no losses.

Now, let's instead say we want a much more reasonable 45KW cyclotron. That would mean we'd be using mass at a rate of $3 ng$ per second. What kind of momentum change would this give us? Assuming (big assumption) that all particles are exiting at the same velocity, we'd get a momentum change of 0.7 kg m/s. Not bad, but remember that force is $F = fracdpdt$, with everything constant this is about 0.7N of thrust. At those power levels, ion engines are comparable, less mass, and more flight proven.

Now, these back of the envelope calculations did show that the cyclotron would be far more efficient, but the mass of the magnets alone would nullify this, ignoring power requirements, given the thrust you end up with anyway.

share|improve this answer

answered May 18 at 17:58

Michael StachowskyMichael Stachowsky

1,228211

$endgroup$

share|improve this answer

answered May 18 at 17:58

Michael StachowskyMichael Stachowsky

1,228211

answered May 18 at 17:58

Michael StachowskyMichael Stachowsky

1,228211

answered May 18 at 17:58

Michael StachowskyMichael Stachowsky

1,228211