Difference between revisions of "Talk:Specific impulse"
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:Though I'm not a rocket scientist, so this might not be correct. — [[User:XZise|xZise]] <small>[[[User talk:XZise|talk]]]</small> 11:26, 18 April 2014 (CDT) | :Though I'm not a rocket scientist, so this might not be correct. — [[User:XZise|xZise]] <small>[[[User talk:XZise|talk]]]</small> 11:26, 18 April 2014 (CDT) | ||
I don't know much about that, but if you manipulate the first formula for I_sp in the article, you get that it is equal to the exhaust speed in Tsiolkovsky's equation. I_sp = (m*dv/dt)/(dm/dt) = (m*dv)/(dm)=dv*(m/dm) <=> dv = I_sp*(dm/m). Integrating you get: delta_v = I_sp*ln(m_final/m_initial). This is the Rocket Equation with I_sp instead of v_e, which would suggest that, at least in KSP, I_sp (without being multiplied by g_0) is equal to the exhaust speed... | I don't know much about that, but if you manipulate the first formula for I_sp in the article, you get that it is equal to the exhaust speed in Tsiolkovsky's equation. I_sp = (m*dv/dt)/(dm/dt) = (m*dv)/(dm)=dv*(m/dm) <=> dv = I_sp*(dm/m). Integrating you get: delta_v = I_sp*ln(m_final/m_initial). This is the Rocket Equation with I_sp instead of v_e, which would suggest that, at least in KSP, I_sp (without being multiplied by g_0) is equal to the exhaust speed... | ||
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| + | == Fuel Consumption == | ||
| + | |||
| + | It is written : "In KSP the fuel consumption on most engines depend on the atmospheric pressure with the lowest consumption in vacuum".<br /> | ||
| + | Is there an equation to determine the real fuel consumption according to the pressure ? | ||
Revision as of 10:59, 30 July 2014
Sorry if this is a dumb question, but,
What is the font used in the images? Mozziedoo (talk) 02:51, 6 February 2014 (CST)
- Are you talking about the formulas, because they are rendered automatically by the MediaWiki software. Unfortunately I can't really give you a better explanation, but you might want to check Help:Displaying a formula on the Wikipedia. — xZise [talk] 10:09, 6 February 2014 (CST)
Suggestion
I did some math, and I_sp before being divided by g_0 is equal to the exhaust speed (v_e) used in Tsiolkovsky rocket equation. The same is said in wikipedia. Is it still true in KSP? Can I edit this information into the article? — Preceding unsigned comment added by Sirplentifus (talk • contribs) 14:17, 18 April 2014 (UTC)
| “ | Although the unit of the specific impulse is a velocity it is lower than the exhaust speed usually, because some of the fuel consumed isn't used for propelling directly, but runs the turbopumps to fuel the engine. | ” |
- Though I'm not a rocket scientist, so this might not be correct. — xZise [talk] 11:26, 18 April 2014 (CDT)
I don't know much about that, but if you manipulate the first formula for I_sp in the article, you get that it is equal to the exhaust speed in Tsiolkovsky's equation. I_sp = (m*dv/dt)/(dm/dt) = (m*dv)/(dm)=dv*(m/dm) <=> dv = I_sp*(dm/m). Integrating you get: delta_v = I_sp*ln(m_final/m_initial). This is the Rocket Equation with I_sp instead of v_e, which would suggest that, at least in KSP, I_sp (without being multiplied by g_0) is equal to the exhaust speed...
Fuel Consumption
It is written : "In KSP the fuel consumption on most engines depend on the atmospheric pressure with the lowest consumption in vacuum".
Is there an equation to determine the real fuel consumption according to the pressure ?
