Difference between revisions of "Talk:Tutorial:Advanced Rocket Design"
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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? --[[User:Hovissimo|Hovissimo]] ([[User talk:Hovissimo|talk]]) 19:45, 25 December 2012 (UTC)<P><P> | 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? --[[User:Hovissimo|Hovissimo]] ([[User talk:Hovissimo|talk]]) 19:45, 25 December 2012 (UTC)<P><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". [[User:Kahlzun|Kahlzun]] ([[User talk:Kahlzun|talk]]) 07:37, 2 January 2013 (UTC) | 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". [[User:Kahlzun|Kahlzun]] ([[User talk:Kahlzun|talk]]) 07:37, 2 January 2013 (UTC) | ||
| − | ::Isn't that kind of arbitrary, though? Wouldn't Δv = Isp · ln(m<sub>0</sub>/m<sub>1</sub>) make more sense? | + | ::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) |
Revision as of 19:48, 23 March 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)
