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

Revision as of 11:43, 27 August 2014

Question: 9.81 m/s^2 in delta V calculations?

Isn't that figure the acceleration due to gravity on Earth at sea level?

<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)

Formula graphs might make the formulas easier to understand.

It might be a good idea to add graphs of the some of the formulas here, this way people could visually see what the pattern is, which makes estimating rocket builds a lot easier. Is there someone good with computer math software that wants to do this, or should I? (I can get Maxima set up and render the graphs in a few hours, however someone more experienced with computer math systems could clearly do a far better job in far less time.) --Ruedii (talk) 08:57, 20 February 2014 (CST)

Rocket equation correction for ascents

Note that the rocket equation ${\displaystyle \Delta v=I_{sp}\ln {\frac {m_{0}}{m_{1}}}}$ assumes no external forces, in particular abscence of gravity. That makes it sufficient for short maneuvers, but less well-suited for ascent, which is strongly affected by gravity. If gravity is included, the formula looks like this:

${\displaystyle \Delta v=I_{sp}\ln {\frac {m_{0}}{m_{1}}}+\int {\vec {g}}\,dt}$

where ${\displaystyle \int {\vec {g}}\,dt}$ is the integrated gravitational acceleration during the ascent (or any other maneuver). As it is hard to compute, we can conservatively estimate it by

${\displaystyle \left|\int {\vec {g}}\,dt\right|\leq {\frac {g_{0}+g_{1}}{2}}\Delta t}$

where ${\displaystyle \Delta t}$ is the duration of the burn and ${\displaystyle g_{0},g_{1}}$ are the gravity acceleration at its begin and end (this assumes that g(t) is concave, which is the case for a constant burn near a single gravity source). Exploiting that the burn duration is given by the mass of the expelled fuel over the fuel flow rate, this yields the nice formula

${\displaystyle \Delta v\geq I_{sp}\left(\ln {\frac {m_{0}}{m_{1}}}-{\frac {|{\overline {F_{G}}}|}{|F_{T}|}}\right)}$

for ascent, where ${\displaystyle {\overline {F_{G}}}={\tfrac {1}{2}}(g_{0}+g_{1})(m_{0}-m_{1})}$ is kind of an average fuel weight estimation and ${\displaystyle F_{T}}$ is the thrust of the engine. Now finally you can see why you need stronger engines in presence of stronger gravity. -- Cami (talk) 11:57, 26 August 2014 (CDT)

First of all a nice addition, and feel free to add it. But doesn't thrust-to-weight ratio already explain partially why you need stronger engines to lift off of a surface? And isn't it explained because when the engines are more powerful, the rockets accelerate faster and thus the ascent time is lower? And for low orbits you could approximate ${\displaystyle {\tfrac {1}{2}}(g_{0}+g_{1})\approx {\tfrac {1}{2}}(g_{1}+g_{1})={\tfrac {1}{2}}2g_{1}=g_{1}}$ as long as you are close to the body. For example it's about 90% for a 70 km orbit. And you should make avoid confusion with the conversion factor between I^sp in seconds and m/s. — xZise [talk] 06:43, 27 August 2014 (CDT)