hubbazoot Posted May 2, 2012 Share Posted May 2, 2012 Does anyone know what type of function the relationship is between where the throttle is set, the fuel consumption, and the force output of the engine is? Link to comment Share on other sites More sharing options...
sal_vager Posted May 2, 2012 Share Posted May 2, 2012 I think closette, Kosmo-not and a few others figured out that it\'s a 1 to 1 ratio, no matter where you set the throttle you have the same fuel economy, the only real factor affecting this is atmospheric drag, so once in space there is nothing to be gained by throttling back. Link to comment Share on other sites More sharing options...
hubbazoot Posted May 2, 2012 Author Share Posted May 2, 2012 Alright, that raises my next question: What\'s the best way to optimize the burn while in the atmo? Link to comment Share on other sites More sharing options...
sal_vager Posted May 2, 2012 Share Posted May 2, 2012 What you need is to read up on the Goddard problem, unfortunately wikipedia is a bit light on details but there is a lot of info here.Basically you want to get up to speed as fast as possible then throttle back but keep accelerating, so atmospheric drag has the least affect on your ascent. Link to comment Share on other sites More sharing options...
The_Duck Posted May 2, 2012 Share Posted May 2, 2012 Alright, that raises my next question: What\'s the best way to optimize the burn while in the atmo?For a vertical ascent in atmosphere, the most fuel efficient strategy for a given rocket design is to try to keep your upward velocity equal to your ship\'s terminal velocity at the current altitude; that is, the velocity at which atmospheric drag equals one gee. Of course, if you\'re going for orbit you won\'t always be ascending vertically. But you\'ll probably find that most reasonable ships fall behind terminal velocity by the start of the pitchover (because terminal velocity increases rapidly with altitude as the air thins out). Once this happens you should probably always put the throttle to max. Link to comment Share on other sites More sharing options...
sal_vager Posted May 2, 2012 Share Posted May 2, 2012 At about 45km the air is too thin to do a lot, but it will depend a lot on your craft, something big and heavy just isn\'t going to accelerate fast enough to have to worry much about air drag, it\'s got enough problems just with it\'s own mass. Link to comment Share on other sites More sharing options...
hubbazoot Posted May 3, 2012 Author Share Posted May 3, 2012 For a vertical ascent in atmosphere, the most fuel efficient strategy for a given rocket design is to try to keep your upward velocity equal to your ship\'s terminal velocity at the current altitude; that is, the velocity at which atmospheric drag equals one gee. Of course, if you\'re going for orbit you won\'t always be ascending vertically. But you\'ll probably find that most reasonable ships fall behind terminal velocity by the start of the pitchover (because terminal velocity increases rapidly with altitude as the air thins out). Once this happens you should probably always put the throttle to max.Ah yeah, I knew about the terminal velocity bit. Is there an easy way to calculate the terminal velocity?I think I may have an idea. Link to comment Share on other sites More sharing options...
hubbazoot Posted May 3, 2012 Author Share Posted May 3, 2012 Ah yeah, I knew about the terminal velocity bit. Is there an easy way to calculate the terminal velocity?I think I may have an idea.Yup, my idea worked wonderfully. For a real practical way to calculate the '1g acceleration' point, just try to hover with your rocket on the pad. Link to comment Share on other sites More sharing options...
Kosmo-not Posted May 3, 2012 Share Posted May 3, 2012 The terminal velocity function is:Vterminal = sqrt(2*W/(m*?*CD))W = weight = mass * gravity? = density = .01*exp(-altitude/5000) altitude is in metersCD = drag coefficientHere is also a nice graph that can be used for all stock rockets (they have an overall drag coefficient very close to 0.2):Beyond about 10km, the terminal velocity increases too quickly to keep up with it (unless you have a super high thrust rocket). Link to comment Share on other sites More sharing options...
togfox Posted May 3, 2012 Share Posted May 3, 2012 If I could + rep the above post then I would. Thx. ;D Link to comment Share on other sites More sharing options...
Zephram Kerman Posted May 3, 2012 Share Posted May 3, 2012 The terminal velocity function is:Vterminal = sqrt(2*W/(m*?*CD))W = weight = mass * gravity? = density = .01*exp(-altitude/5000) altitude is in metersCD = drag coefficientOoh! Very, very handy graph.Question: does the W/m mean mass cancels out? (Not in real world, only in KSP.) That was noted in the Goddard Problem thread too. Link to comment Share on other sites More sharing options...
Recommended Posts