Difference between revisions of "Talk:Tutorial:Advanced Rocket Design"
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::Isn't that kind of arbitrary, though? Wouldn't Δv = Isp · ln(m<sub>0</sub>/m<sub>1</sub>) make more sense? [[User:Ninenineninefour|Ninenineninefour]] ([[User talk:Ninenineninefour|talk]]) 14:48, 23 March 2013 (CDT) | ::Isn't that kind of arbitrary, though? Wouldn't Δv = Isp · ln(m<sub>0</sub>/m<sub>1</sub>) make more sense? [[User:Ninenineninefour|Ninenineninefour]] ([[User talk:Ninenineninefour|talk]]) 14:48, 23 March 2013 (CDT) | ||
:::Wait, I just answered my own question: It's not the equation that is arbitrary, it's the definition of specific impulse itself, which is the exhaust velocity divided by standard gravity. What would make more sense, in fact, is Δv = V<sub>e</sub> · ln(m<sub>0</sub>/m<sub>1</sub>). The reason why Isp is used instead of exhaust velocity is so that the efficiency of engines can be easily compared, even when they are measured with the imperial system instead of the (much better) metric system. | :::Wait, I just answered my own question: It's not the equation that is arbitrary, it's the definition of specific impulse itself, which is the exhaust velocity divided by standard gravity. What would make more sense, in fact, is Δv = V<sub>e</sub> · ln(m<sub>0</sub>/m<sub>1</sub>). The reason why Isp is used instead of exhaust velocity is so that the efficiency of engines can be easily compared, even when they are measured with the imperial system instead of the (much better) metric system. | ||
| + | ::::Oh I didn't saw this post when I [http://wiki.kerbalspaceprogram.com/w/index.php?title=Tutorial:Advanced_Rocket_Design&diff=14240&oldid=14231 undid] this change. At first I also find this very strange until Scott Manley uploaded a video an explained it. After this it “meh” of course ;). Actually sometimes the I<sub>sp</sub> is given in <math>\frac{Ns}{kg}</math> which is equal to <math>\frac{m}{s} = kg \cdot \frac{m}{s^2} \cdot \frac{s}{kg}</math> (with <math>N = kg \cdot \frac{m}{s^2}</math>). That is also why I added extra explanation everywhere I<sub>sp</sub> is used. — [[User:XZise|xZise]] <small valign="middle">[[[User talk:XZise|talk]]]</small> 11:38, 29 April 2013 (CDT) | ||
Revision as of 16:38, 29 April 2013
Question: 9.81 m/s^2 in delta V calculations?
Isn't that figure the acceleration due to gravity on Earth at sea level?
I'm assuming that the devs used an earth-like parameter for Kerbin, and obviously this math is working for you; but do you use a different acceleration parameter when calculating the delta-V of your Munar return stage? --Hovissimo (talk) 19:45, 25 December 2012 (UTC)<P> No, the 'g' is a standard unit of measurement. Changing it would be like changing the length of a metre because the Mun "has less of them". Kahlzun (talk) 07:37, 2 January 2013 (UTC)
- Isn't that kind of arbitrary, though? Wouldn't Δv = Isp · ln(m0/m1) make more sense? Ninenineninefour (talk) 14:48, 23 March 2013 (CDT)
- Wait, I just answered my own question: It's not the equation that is arbitrary, it's the definition of specific impulse itself, which is the exhaust velocity divided by standard gravity. What would make more sense, in fact, is Δv = Ve · ln(m0/m1). The reason why Isp is used instead of exhaust velocity is so that the efficiency of engines can be easily compared, even when they are measured with the imperial system instead of the (much better) metric system.
- Oh I didn't saw this post when I undid this change. At first I also find this very strange until Scott Manley uploaded a video an explained it. After this it “meh” of course ;). Actually sometimes the Isp is given in which is equal to (with ). That is also why I added extra explanation everywhere Isp is used. — xZise [talk] 11:38, 29 April 2013 (CDT)
- Wait, I just answered my own question: It's not the equation that is arbitrary, it's the definition of specific impulse itself, which is the exhaust velocity divided by standard gravity. What would make more sense, in fact, is Δv = Ve · ln(m0/m1). The reason why Isp is used instead of exhaust velocity is so that the efficiency of engines can be easily compared, even when they are measured with the imperial system instead of the (much better) metric system.
- Isn't that kind of arbitrary, though? Wouldn't Δv = Isp · ln(m0/m1) make more sense? Ninenineninefour (talk) 14:48, 23 March 2013 (CDT)
