Calculating Robert Zubrin's Dipole Drive Mass Flow

[Sebastiaan]

I'm trying to understand calculation made Robert Zubrin in his presentation on Dipole Drive

source:

In here he made the following statement

"A Coulomb of protons has a mass of 0.011 milligrams. If the jet power is 400 W, and the potential difference is 64 V, so the proton current will be 6.25 A, and have a mass flow of 0.0652 mg/s"

My question is, how did he derive to "a mass flow of 0.0652 mg/s"?

 

[Nugatory]

My question is, how did he derive to "a mass flow of 0.0652 mg/s"?

I haven't checked to see if the calculation is right, but the idea is pretty simple. We're given the power and the voltage, so can calculate the current, and hence the amount of charge per second.

Be aware that getting this calculation right doesn't necessarily validate the rest of the video.

 

[Ibix]

I haven't checked to see if the calculation is right, but the idea is pretty simple. We're given the power and the voltage, so can calculate the current, and hence the amount of charge per second.

If the obvious computation is used then the maths, at least as reproduced by Sebastiaan, is wrong. A current of 6.25A is 6.25C/s. The given mass of a Coulomb of protons is 0.011mg, which means 6.25×0.011 mg/s, which I make 0.06875 mg/s.

Either there's a typo or there's something more sophisticated going on that I can't think of right now (I can't watch the video at the moment).

 

[mfb]

The 11µg are probably the result of poor rounding, the actual value is 10.4 µg and multiplied by 6.25 we get 65.25 µg.

Edit: Oops, µg.

 

[mjc123]

That's 10.4 µg.

 

[Sebastiaan]

If the obvious computation is used then the maths, at least as reproduced by Sebastiaan, is wrong. A current of 6.25A is 6.25C/s. The given mass of a Coulomb of protons is 0.011mg, which means 6.25×0.011 mg/s, which I make 0.06875 mg/s.

That exactly what I though. So I though I was missing something

 

[Sebastiaan]

Either there's a typo or there's something more sophisticated going on that I can't think of right now (I can't watch the video at the moment).

the text can be found at https://www.centauri-dreams.org/2018/06/29/the-dipole-drive-a-new-concept-for-space-propulsion/

 

[Sebastiaan]

The 11µg are probably the result of poor rounding, the actual value is 10.4 µg and multiplied by 6.25 we get 65.25 µg.
Edit: Oops, µg.

I guess you mean 0.0104396 mg * 6.25 A = 0.06524803 mg/s. Thanks, now it makes sense

Still the question why did he round 0.0104396 down to 0.011 ? Seems to me this is a mistake which will result in confusion