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

Revision as of 02:12, 31 December 2013 (view source) Lancastrian (talk | contribs) (→‎Cleaning Explanation of Delta V calculations?) ← Older edit		Revision as of 13:25, 31 December 2013 (view source) XZise (talk | contribs) (→‎Cleaning Explanation of Delta V calculations?) Newer edit →
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	* <math>r_1</math>= The radius of our first orbit from the center of the orbited body (in this case 700km, or our orbital altitude of 100 km plus Kerbin's radius of 600km).		* <math>r_1</math>= The radius of our first orbit from the center of the orbited body (in this case 700km, or our orbital altitude of 100 km plus Kerbin's radius of 600km).
	--[[User:lancastrian|lancastrian]] ([[User talk:lancastrian|talk]]) 22:16 CST, 29 December 2013 (CST)		--[[User:lancastrian|lancastrian]] ([[User talk:lancastrian|talk]]) 22:16 CST, 29 December 2013 (CST)
		+	:This is of course possible, but I was thinking as everything in KSP is showing the distance relative to the surface it should be made clear that the distance to the center is important. But as you can simply redefine what r<sub>1</sub> you can also say that it is the distance from the center. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 07:25, 31 December 2013 (CST)

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Revision as of 13:25, 31 December 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 = V_e · 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 I_sp is given in ${\displaystyle {\frac {Ns}{kg}}}$ which is equal to ${\displaystyle {\frac {m}{s}}=kg\cdot {\frac {m}{s^{2}}}\cdot {\frac {s}{kg}}}$ (with ${\displaystyle N=kg\cdot {\frac {m}{s^{2}}}}$). That is also why I added extra explanation everywhere I_sp is used. — xZise [talk] 11:38, 29 April 2013 (CDT)

Cleaning Explanation of Delta V calculations?

In the "Calculating Transfer Maneuvers" section could we make it more clear that R1 and R2 are the radius from the center of the planet and so you need to add 600km to the altitude to get correct numbers. Also, it might be useful to include that the solutions to the two delta-V calculations are 73.65 m/s for burn #1 and 71.23 m/s for burn #2 for a total delta-V of 144.88 m/s.--lancastrian (talk) 22:16 CST, 29 December 2013 (CST)

Good point, I added the radius into the formula. I don't know if the ${\displaystyle r_{1}+R+r_{2}+R}$ should be simplified to ${\displaystyle r_{1}+r_{2}+2R}$ because I try to make in the first clear that you need the distance to the center not the surface. I've also validated and added the calculated Δv values. — xZise [talk] 05:44, 30 December 2013 (CST)

I wonder if we could keep the equation the same as it was before but specify that R1 and R2 are the radius from the center of the planet. So, the explanation for R1 would read:

• ${\displaystyle r_{1}}$= The radius of our first orbit from the center of the orbited body (in this case 700km, or our orbital altitude of 100 km plus Kerbin's radius of 600km).

--lancastrian (talk) 22:16 CST, 29 December 2013 (CST)

This is of course possible, but I was thinking as everything in KSP is showing the distance relative to the surface it should be made clear that the distance to the center is important. But as you can simply redefine what r₁ you can also say that it is the distance from the center. — xZise [talk] 07:25, 31 December 2013 (CST)