$$e^{\Delta v/v_{exhaust}} - 1=\frac{mass_{fuel}}{mass_{structure}}$$ $$\text{let } k=e^{\Delta v/v_{exhaust}} - 1 \implies mass_{fuel}= k \times mass_{structure} $$ According to SMAD (Space Mission Analysis and Design book), overall tank weight is $1.25\times (0.1\times mass_{fuel})$ (the meaning behind those values is 10% of propellant and 25% extra of the weight for PMDs and hardware).

According to this assumption,

$$ mass_{fuel}= k \times (mass_{structure-tank}+mass_{tank})$$ $$ mass_{fuel}= k \times (mass_{structure-tank}+0.125 \times mass_{fuel})$$ $$ mass_{fuel}\times (1-0.125k)=k \times mass_{structure-tank}$$ As $k$ can reach $50$, $mass_{fuel}$ would become -ve which should not be so. I would be grateful if you could tell me what exactly I have missed. Thanks!

  • $\begingroup$ If I'm understanding correctly, the 1.25(0.1 x mf) formulation determines what k is; you can't vary k while keeping the 1.25 and 0.1 constant. (Not sure why tank weight isn't just 0.125mf; is there something else being let out?) $\endgroup$ Mar 14, 2018 at 18:17
  • $\begingroup$ @RussellBorogove $k=e^{\Delta v/v_{exhaust}}-1$, so k is a constant. Tank weight is 0.125 mf. $\endgroup$
    – Infi
    Mar 15, 2018 at 2:52

1 Answer 1


K value is too large. In most cases major $\Delta V$ is provided by gravity assists. So if you consider $\Delta V$ performed just by the propulsion systems, $k$ value would decrease drastically. An excerpt from SMAD $-$ $$$$ enter image description here

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.