Talk:Specific impulse

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...— Preceding unsigned comment added by Sirplentifus (talk • contribs) 11:14, 19 April 2014‎ (UTC)

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 ? — Preceding unsigned comment added by Jarod E. McRolle (talk • contribs) 10:59, 30 July 2014‎ (UTC)

Comparing Engine

I did just a little work for the german translation of this page and i'm not the greatest mathematician, but is the equation $W={\frac {I_{sp}^{2}}{F}}\left(m_{f}+(m_{p}+m_{e})\log \left({\frac {m_{p}+m_{e}}{m_{p}+m_{e}+m_{f}}}\right)\right)$ really right? Because it is said based on this formula that 4 LV-N are better than a poodle for a craft with 1 ton payload and 10 tons of fuel. Well if i do a dV calculation for such a craft, the result says that the poodle will deliver about 170m/s more with the discribed values. Also i don't get this discribed 1.9 value of "betterness" if i use the equation in a excel-sheet. I mean it should look like this in excel wouldn't it? =(Isp^2/F)*(m fuel+(m payload+m engine)*log((m payload+m engine)/(m fuel+m payload+m engine))). Maybe i sound just really dumb here, but in my opinion something isn't right there.--Amöbius (talk) 14:24, 20 July 2015 (UTC)