I've been working on a tool to calculate delta-v, and I got a little stuck. Specifically, how do you calculate specific impulse of a rocket when you have multiple engines with different specific impulse and thrust firing at the same time (example: 4x RS-25 + 2 SRBs)? Assume constant thrust, no atmosphere, no gravity. Stoplookin9 Hey there! Send me a message! 02:13, 22 March 2025 (UTC)[reply]

The acceleration of the rocket equals its mass divided by the thrust. The combined thrust is constant (assuming no engine runs out of fuel); it equals the sum of the thrusts of all engines. Let ${\displaystyle T_{\mathsf {E}}}$denote the thrust of engine E. The total thrust is then given by the sum of all individual thrusts: ${\displaystyle T=T_{1}+T_{2}+...\,.}$The difficulty of the problem is that the acceleration is not constant. The mass decreases as rocket fuel is consumed, which means that the acceleration increases. To solve this requires to use some integral calculus. We need to know the total mass of the system at the start; let's call it ${\displaystyle M}$. How does it decrease as fuel is burned? Let ${\displaystyle I_{\mathsf {E}}}$denote the the specific impulse of engine E. The contribution of this engine to the rate of fuel mass consumption equals ${\displaystyle Q_{\mathsf {E}}=T_{\mathsf {E}}/I_{\mathsf {E}}.}$Sum this quantity over all engines to get the total mass flow rate ${\displaystyle Q=Q_{1}+Q_{2}+...}$. After a time ${\displaystyle t}$has elapsed, the mass of the whole system will have decreased to ${\displaystyle M-Qt.}$So the acceleration at time ${\displaystyle t}$after the start equals ${\displaystyle T/(M-Qt).}$To get ${\displaystyle \Delta v}$after a burn time ${\displaystyle \Delta t}$we need to integrate this:

${\displaystyle \Delta v=\int _{0}^{\Delta t}{\frac {T}{M-Qt}}\,{\text{d}}t={\frac {T}{Q}}\log \left({\frac {M}{M-Q\Delta t}}\right).}$

The ${\displaystyle \log }$function in this formula is the natural logarithm, so, for example, ${\displaystyle \log(1.35)\approx 0.30\,.}$ ​‑‑Lambiam 11:49, 22 March 2025 (UTC)[reply]

Specific pulse is the ration of the trust to the mass flow of the propellant. You can do the simple calculation knowing thrusts and specific pulses of all engines. Ruslik_Zero 20:02, 22 March 2025 (UTC)[reply]

This only works if the mass of the fuel consumed during the burn (${\displaystyle Q\Delta t}$in the formula above) is insignificant, compared to the total mass. Writing ${\displaystyle q={\frac {Q\Delta t}{M}},}$we can expand ${\displaystyle {\frac {M}{T}}{\frac {\Delta v}{\Delta t}}}$as a Taylor series:

${\displaystyle {\frac {M}{T}}{\frac {\Delta v}{\Delta t}}=1+{\frac {q}{2}}+{\frac {q^{2}}{3}}+\cdots \;.}$

 ​‑‑Lambiam 07:48, 23 March 2025 (UTC)[reply]

It will work in any case:

${\displaystyle sfc={\frac {\sum thrust_{i}}{\sum {\frac {thrust_{i}}{sfc_{i}}}}}}$.

All these engine parameters (thrusts and specific fuel consumptions) are just constants. Ruslik_Zero 19:05, 23 March 2025 (UTC)[reply]

Actually in your formula ${\displaystyle T/Q=sfc}$. Ruslik_Zero 19:10, 23 March 2025 (UTC)[reply]

How do you get from ${\displaystyle {\mathit {sfc}}\,}$to ${\displaystyle \Delta v}$?  ​‑‑Lambiam 09:19, 24 March 2025 (UTC)[reply]

By tsiolkovsky rocket equation as usual. Ruslik_Zero 19:52, 24 March 2025 (UTC)[reply]

Tsiolkovsky's formula is the answer I gave; ${\displaystyle M-Q\Delta t}$is the remaining mass after a burn time ${\displaystyle \Delta t.}$Tsiolkovsky's formula assumes that all fuel is consumed. In the case of multiple engines with different characteristics, I don't think we can assume all burn out at the same time, and then the formula no longer works. You need to calculate ${\displaystyle \Delta v}$to the first burn-out, then add the ${\displaystyle \Delta v}$from that point till the next burn-out, and so on.  ​‑‑Lambiam 23:23, 24 March 2025 (UTC)[reply]

Is it AI-generation prompt or AI-generated prompt Trade (talk) 11:43, 22 March 2025 (UTC)[reply]

Neither of the two sounds natural to me. If it is clear that the context is generative AI, just "prompt" will generally do just fine. If, in the context, you need some attribute to distinguish it from other kinds of prompts, you can use "GenAI prompt"^[1] (or possibly, depending on the audience, the longer term "generative-AI prompt").  ​‑‑Lambiam 12:00, 22 March 2025 (UTC)[reply]

"AI-generated prompt" is the least correct of the options the OP and Lambiam have proposed. It means that the prompt is generated by AI, not that the prompt is used to have AI generate something. DMacks (talk) 12:18, 22 March 2025 (UTC)[reply]

"AI prompt" is used in documentation for ChatGPT, DeepSeek, and DALL-E. The word "generative" has special meaning in AI. You can't just toss it around willy-nilly. 12.116.29.106 (talk) 17:34, 25 March 2025 (UTC)[reply]