Author Topic: Thin Film Isotope Nuclear Engine Rocket (TFINER)  (Read 5690 times)

Offline Twark_Main

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Re: Thin Film Isotope Nuclear Engine Rocket (TFINER)
« Reply #20 on: 05/30/2025 09:31 pm »
In general I agree but alas the sharing mechanisms don't allow for editing right now, and sharing via screenshot makes looking for text difficult, and this forum doesn't have good support for Latex equations.

Any ideas how to solve all of these problems?

Ahh, I understand now.  Yeah here I was naively thinking they had implemented some sort of sane and disability accessible text fallback when you copy the math equations.  ::) ::)

Thanks.


Here's an edited down version of TFINER giving possible mass flow and force numbers for the TFINER proposal

https://chatgpt.com/share/683a17fd-b63c-8013-b1c9-0e3305af7728

I really appreciate you making the effort. Thanks again.

As far as fact-checking, I would double-check:

   * Is it just assuming the same decay energy (48 MeV) for all alpha particles?  Are there substantial differences in energy between the original decay and the subsequent daughter isotope decay energies?

   * Naturally there's going to be some self-absorption in the reverse direction in the 10 micron Thorium film. I'm guessing they chose 10 microns because any thicker and this becomes too much of an issue. It would be interesting to see if you can get ChatGPT to run a First Principle (vs fudge-based) estimate of those losses, including the spherical trig.

   * Similarly with the forward absorber, I expect they chose 50 microns not because it absorbs all forward alpha particles, but because any thicker and you lose more to the extra mass than you gain from the extra absorption.  I wonder if it can construct a simple model to estimate the "break even" point, and what the resulting efficiency might be.

  * The other possibility is that they chose 50 microns because of gauge issues, and the material simply isn't available in thinner sizes.

   * Maybe it can give suggestions into what material might be chosen for the absorber. I'm assuming it's just a bog standard kapton film, but it would be interesting to see if ChatGPT can come up with any other possible materials that might have been chosen.

« Last Edit: 05/30/2025 09:41 pm by Twark_Main »

Online InterestedEngineer

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Re: Thin Film Isotope Nuclear Engine Rocket (TFINER)
« Reply #21 on: 05/31/2025 02:49 am »

   * Similarly with the forward absorber, I expect they chose 50 microns not because it absorbs all forward alpha particles, but because any thicker and you lose more to the extra mass than you gain from the extra absorption.  I wonder if it can construct a simple model to estimate the "break even" point, and what the resulting efficiency might be.


you can calculate the absorption thickness needed from https://physics.nist.gov/PhysRefData/Star/Text/ASTAR.html which gives the range in g/cm2 units.

then to get thickness:

Thickness (μm) = 10,000 × (Range in g/cm²) ÷ (Material density in g/cm³)

which for Mylar:

Thickness = 10,000 × 0.00475 ÷ 1.39 ≈ 34.2 μm

for the entire 225m2 of the panels:

Mass (kg) = Area (m²) × Thickness (m) × Density (kg/m³)

Mass = 225 × (34.2e-6) × 1390 ≈ 10.7 kg


source:  ChatGPT - I got it to put equations in plaintext format so a little easier to post here directly.
« Last Edit: 05/31/2025 03:02 am by InterestedEngineer »

Online InterestedEngineer

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Re: Thin Film Isotope Nuclear Engine Rocket (TFINER)
« Reply #22 on: 05/31/2025 03:01 am »


   * Naturally there's going to be some self-absorption in the reverse direction in the 10 micron Thorium film. I'm guessing they chose 10 microns because any thicker and this becomes too much of an issue. It would be interesting to see if you can get ChatGPT to run a First Principle (vs fudge-based) estimate of those losses, including the spherical trig.



Thickness (μm) = 10,000 × (Range in g/cm²) ÷ (Material density in g/cm³)

Range for Thorium is about 0.01843g/cm2  at 5.5MeV - a guess, they don't have TH-228 on the list. but all the high density metals are in the same ballpark.

Density is 11.7g/cm3

so that makes 15.75μm.   Since it's a radioactive decay chain we'd have to do this for each decay product, which is too tedious for now, but suffice it to say there's some but not much self-absorption in the reverse direction.  The average is 5μm which is 1/3 the absorption length.

Not to mention a geometry problem, as particles that go at 45 degrees go through 1.4x the material.

still, seems to be in the ballpark of "that'll work".
« Last Edit: 05/31/2025 03:03 am by InterestedEngineer »

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