Difference between revisions of "Talk:Tutorial:Advanced Rocket Design"

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(dammit, why won't the math work??)
(well then, fuck you too, TEX.)
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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 <math>\Delta v = I_{sp}\cdot \ln\left(\frac{m_1}{m_2}\right)</math> 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?

Revision as of 23:14, 22 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?