Cheapest rocket possible

[spacester]

<font color="yellow">V = Ve ln(Mo/ Me ) <br /><br />where V is vehicle velocity, Ve is exhaust gas velocity, Mo is gross mass, Me is empty mass, and ln is the natural logarithm. The, direct relationship between vehicle velocity and exhaust gas velocity accounts for the interest in high energy propellants that yield high exhaust velocities. </font><br /><br />Very nice presentation otherwise but herein lies your error. You are not the first to do so; you will be best served by taking this opportunity to admit your error.<br /><br />V is NOT Vehicle Velocity<br /><br />V is dV = Delta V = CHANGE in velocity<br /><br />This small adjustment in your understanding will allow you to apply the change of reference knowledge just so eloquently provided to your best advantage. <div class="Discussion_UserSignature"> </div>

 

[MBA_UIU]

I thought I could make this argument, through the use of qualitative research, on my own. But at last I am not a rocket engineer, scientist, or theorist. Every post I have used here comes from the leading experts in the field of rocketry too include: NASA JPL, Northwestern University Department of Astrophysics, Allstar Network at FIU-NASA, M.I.T, the University of California at Berkeley and the Australian Military – Rocket Prolusion Laboratory (Note to Vogon13; if you cannot accept that these are the leading doctorates, scientist and engineers in the sciences of rocketry than maybe you need to go back to arguing with BB.). <br /><br />Each statement that I have represented on these pages comes directly from these sources (Just go back and try each link to see where it takes you). If you think that you have found an error then it is not my error but an error made by these doctorates, scientist and engineers that developed these systems. If each of you reading these posts would have just read the information provided by these links (and I understand that some of these pages have a lot of information that is not the easiest to understand) then I think this argument would have been over long ago without any attempts at personal attacks. <br /><br />With this in mind here is my last post on the subject. This is taken directly from NASA JPL and sourced in one of my earlier post. If you do not believe what it says then I guess you really should be working for NASA as they could sure use the help and your abundant knowledge that overrides everything they seem to know<br /><br />. Source: NASA JPL<br />specific impulse <br />The specific impulse (commonly abbreviated Isp) of a propulsion system is the impulse (change in momentum) per unit of propellant.<br />Depending on whether the amount of propellant is expressed in mass or in weight (by convention weight on the Earth) the dimension of specific impulse is that of speed or time, respectively, differing by a factor of g, the gravita <div class="Discussion_UserSignature"> <p><strong><font color="#0000ff"><br /><br /> <br /><img id="268587ce-7170-4b41-a87b-8cd443f9351a" src="http://sitelife.space.com/ver1.0/Content/images/store/6/8/268587ce-7170-4b41-a87b-8cd443f9351a.Large.jpg" alt="blog post photo" /><br /></font></strong></p> </div>

 

[MBA_UIU]

Read the post above what you just wrote then explain it to NASA-JPL (and ask for a job). As I stated above this information comes directly from their website so if it is incorrect then you must know way more then they do. <div class="Discussion_UserSignature"> <p><strong><font color="#0000ff"><br /><br /> <br /><img id="268587ce-7170-4b41-a87b-8cd443f9351a" src="http://sitelife.space.com/ver1.0/Content/images/store/6/8/268587ce-7170-4b41-a87b-8cd443f9351a.Large.jpg" alt="blog post photo" /><br /></font></strong></p> </div>

 

[tap_sa]

Information you posted from all those links is correct. None if it proves your claim. Re-study the rocket equation, dV = V_exhaustln(Mo/Me). Based on that, please explain why it would impossible for dV to be bigger than V_exhaust. Is it impossible to find a number whose natural logarithm is bigger than one?

 

[vogon13]

MBA_UIU, meet bigbrain<br /><br />bigbrain, meet MBA_UIU<br /><br /><br />Well, we've got the introductions out of the way.<br /><br /><br /><br /> <div class="Discussion_UserSignature"> <p><font color="#ff0000"><strong>TPTB went to Dallas and all I got was Plucked !!</strong></font></p><p><font color="#339966"><strong>So many people, so few recipes !!</strong></font></p><p><font color="#0000ff"><strong>Let's clean up this stinkhole !!</strong></font> </p> </div>

 

[tap_sa]

Another good example is the original Atlas. Almost no staging and quite abysmal Isp (effective exhaust velocity ~3000m/s) and yet it made into orbit.

 

[pu_aero]

Perhaps it would help to think of the transfer of momentum involved in the rocket thrust equation directly.<br />MBA_UIU, you are correct in the sense that propellant velocity is a limiting factor in the desing of a chemical rocket, as it is in all rockets, but that is only telling half the story. The other half is of course the "mdot" in the "Mdot*Ve" of the momentum thrust. This represents the mass flow rate of the propellant. Though Ve is limited by the energy contained within the fuel, in theory, if we can build a big enough rocket, we can carry as much mass as we want, allowing us to go as fast as we want (as long as we increase the propellant mass ratio. <br /><br />Let's illustrate this with a simplified example. Suppose we have a rocket with 1kg of structural mass, which carries 100kg of propellant. We then instantaneously propell this mass to 1m/s, what one considers a slow propellant speed. ("Instantaneously" just so we can skip over the calculus, which would force us to rederive the fundemental rocketry equation, DV = Ve ln(M0/Me), but essentially the point is the same)<br /><br />By conservation of momentum (the basis for momentum thrust in the rocket propulsion equation):<br /><br />m1*V1 =m2*V2 + m3*V3<br /><br />(1kg + 100kg) * V1 = 100kg * V2 + 1kg * V3<br /><br />V1 is the inital velocity of the rocket/propellant system, which we will assume was at rest (V1 = 0).<br /><br />V2 is the propellant velocity. If we define the forward direction of the rocket to be positive, and the propellant is ejected backward, then V2 = -1 m/s.<br /><br />We don't know V3, the velocity of the rocket after it expells the propellant, so lets solve for it:<br /><br />(101kg)*0m/s = 100kg * (-1m/s) + 1kg*V3<br /><br />0 kg*m/s = -100 kg*m/s + 1kg*V3<br /><br />100kg*m/s = 1kg*V3<br /><br />Thus V3 = 100 m/s, and our rocket is travelling 100 times faster than Ve!<br /><br />Note that if we remove the instantaneous bit of the equation and employ the fundamental rocketry equation, the answer becomes:

 

[nexium]

MBA may be correct that a rocket cannot exceed the average speed of air while in the atmosphere by more than the speed of the ejection mass. In the vacuum of space, we can more realisticaly compare the speed of the ejection mass to the speed of the rocket which means the rocket can accellerate to almost c, except no design has enough fuel and/or ejection mass. <br /> Remember speed is relative. Neil

 

[gunsandrockets]

"As a useful rule of thumb the delta-v that can be produced with a propellant mass of 63.2 % of the initial total mass is equal to the exhaust velocity (see Rocket equation.) "<br /><br />Ohh, so close and yet not there. Have you ever heard of mass ratio? The description quoted above is for a rocket with a mass ratio of about 3. With that high a mass ratio a rocket burnout speed will equal it's exhaust velocity. A rocket with a mass ratio less than 3 will have a burnout speed less than the exhaust velocity. And important point here, a rocket with a mass ratio greater than 3 will have a final speed greater than it's exaust velocity!<br /><br />http://en.wikipedia.org/wiki/Mass_ratio<br /><br />"This equation indicates that a delta-v (ɢv) of n times the exhaust velocity requires a mass ratio of en. For instance, for a vehicle to achieve a ɢv of 2.5 times its exhaust velocity would require a mass ratio of e2.5 (approximately 12.2)."<br /><br />That's why rockets use stages. Staging is an engineering trick to improve mass ratio. That's why it takes a three stage rocket to achieve Earth escape velocity using chemical rockets. Chemical rockets have such a low exaust velocity it takes a huge mass ratio for a rocket to reach a final speed greater than 25,000 mph.<br /><br />Don't take my word for it, just spend some time googling "rocket mass ratio" and see what you find.<br />

 

[gunsandrockets]

"But Andrews sees a better alternative in a particle beam boosted magsail (PBBM) that substitutes a neutral plasma beam for the laser and uses a magnetic sail instead of a lightsail."<br /><br />Pardon my pettiness, but how can a particle beam which is neutrally charged interact with a magnetic field so as to impart a propulsive force?

 

[MBA_UIU]

Someone else stated this a few post back, and I posted it probably 8 to 10 replies ago, the velocity of the exhaust is measured <b><font color="yellow"> relative to its x axis on the rocket’s body </font></b> <br /><br />If a person tosses a 90 mph fastball forward from a truck’s bed moving at 60 mph to the stationary observer that ball has a forward traveling speed of 150 mph. But to the driver of the truck that ball has a forward traveling speed of 90 mph, <font color="yellow">thus speed is relative to location of the observer and Newton’s third of law of motion is conserved.</font>Hence, a rocket cannot go faster then its exhaust when observed from the x axis of the rocket. I stated this a few post back also.<br /><br />Has the velocity of a rocket increases so to does it mass by means of momentum (or mass like energy). Nearing ½ c the energy requirements to change the delta velocity becomes so great, because of the mass like energy stored in the rocket, that a chemical rocket would need all the mass in the universe to make this change. <br /> <div class="Discussion_UserSignature"> <p><strong><font color="#0000ff"><br /><br /> <br /><img id="268587ce-7170-4b41-a87b-8cd443f9351a" src="http://sitelife.space.com/ver1.0/Content/images/store/6/8/268587ce-7170-4b41-a87b-8cd443f9351a.Large.jpg" alt="blog post photo" /><br /></font></strong></p> </div>

 

[chriscdc]

Look if physics works like you think it does then you would not be able to jump. When you jump the energy comes from you, right? Therefore the earth could be considered the exhaust particle. Somehow though you end up moving up faster than the earth moves down. <br />Look the maths works out. When you start learning physics you thick you can see these sorts of problems. It takes some time until you finally see the problem in your logic. At that point you will kick yourself, so do yourself a favour and stop insisting you are right as you will just appear more of a dork when you finally realise.<br /><br />On the neutrally charged particle beam. The ship or the laser emitter could charge the sail just before it reaches the ship. Perhaps the problem with a charged sail would be that it would interact with too much of the stuff in space eg solar and intersteller plasmas. Also you would only have to worry about gravity changing the course of the sail.<br /><br />This has gone slightly off topic, but I like a good discussion about intersteller transport as much as anyone else here.

 

[mrmorris]

I honestly thought you'd given up on this sillyness when there were no posts from you yesterday. Apparently not. OK -- let's use some information from the U Penn link that *you* gave back here<br /><br />UPenn Link<br /><br />Here's a quote from the later portion of that paper where they are referring to the velocity of the rocket as the mass fraction of the propellant increases:<br /><br /><i>"Now put in a sensible value for the exhaust velocity. Chemical rockets normally cannot generate exhaust speeds above 4000 m/s, so let's put in that value for uex and plot what we get...<br /><br />The logarithmic rise in speed predicted from the rocket equation shows that rocketry is very demanding technically. After about half the fuel is consumed, the rocket is only moving at uexln(2) or 70% of the exhaust velocity. Since x represents the original mass of fuel compared to the final mass at time t, a value of x = 100 means that only 1 percent of the original mass is left. At that time, the rocket is moving at about 18 km/s, or 40,000 miles/hour. This is a high speed, but not very practical since the payload of the rocket must be less than 1%of the original mass even neglecting gravity and air resistance. "</i><br /><br />From this, you can see that if 99% of the mass of the craft is propellant, the end speed is 18,000 m/s even though the exhaust velocity is 4,000 m/s. So your own source is demonstrating that you are not understanding what you're reading.<br /><br /><font color="yellow">"Nearing ½ c the energy requirements to change the delta velocity becomes so great, because of the mass like energy stored in the rocket, that a chemical rocket would need all the mass in the universe to make this change. "</font>

 

[tap_sa]

<font color="yellow">"I bet that if I were to take the mass of Jupiter, use it as a propellant at 4000 m/s exhaust velocity to accellerate a small mass -- I could break 1/2 c by the end"</font><br /><br />You're on!<br /><br />Let's make the case as optimal for you as possible. Initial mass would be m_Jupiter + m_mrmorris and final mass only mrmorris himself, the craft itself is mystically weightless giving best possible massratio. I'm assuming 200lbs/~90kg for mrmorris and 1.899x10²⁷kg for Jupiter.<br /><br />*fiddling with Mathcad*<br /><br />I'm sorry, you reach only speed of 233km/s. The logarithm in the rocket equation starts to bite really hard when the desired velocity exceed a few multiples of your exhaust speed. <br /><br />edit: Should mrmorris argue that what if the payload is something smaller than himself ... I calculated what massratio would be needed for 0.5c, only to notice that Mathcad maxes out at 2839km/s, yielding massratio of 1.74x10³⁰⁸. If you divide Jupiter's mass with that the payload is <i>quite</i> small <img src="/images/icons/smile.gif" />

 

[mrmorris]

<font color="yellow">"I'm sorry, you reach only speed of 233km/s."</font><br /><br />Dang!<br /><br />Oh well -- seeing as you have Mathcad handy, can you work the equation in reverse for those of us interested in learning something new (MBA, you should probably cover your eyes at this point) and determine how much mass *is* required? <br /><br />Edit: Nevermind -- you edited faster than I questioned.

 

[mrmorris]

<font color="yellow">"yielding massratio of 1.74x10308. If you divide Jupiter's mass with that the payload is quite small "</font><br /><br />OK -- help me out with this one. Trying to work out just how much reaction mass we really need. The mass of the sun is 1.989e30 kg. The mass of the Milky Way galaxy is estimated at 200 billion solar masses. What kind of payloads are we talking about here with these numbers. Do we really <b>need</b> the whole universe, or can we skimp by just using a galaxy or two? <img src="/images/icons/smile.gif" />

 

[tap_sa]

<font color="yellow">"how much reaction mass we really need."</font><br /><br />To get to 0.5c with 4000m/s? Uh ... the known universe is not enough. Calculating the massratio maxes at 2839km/s, that's the top speed that mathcad still gives answer (result within limits of ordinary IEEE double precision value), the 1.74x10³⁰⁸. See the value of exponent, <i>308</i>. Speeding up one kilogram to 'only' 2839km/s / ~0.01c would require 4.374x10²⁶⁹ galaxies. How many we know to exit, a trillion or so?<br /><br />For fun I tried the other way, how fast exhaust speed the imaginary weightless craft needs to speed you up to 0.5c using Jupiter's mass of propellant. Result 2573<i>km/s</i> meaning specific impulse of 262300 seconds. Currently the 462 seconds of RL-10B-2 is the best you can get.<br /><br />Lesson here, reaching relativistic speed will be difficult <img src="/images/icons/wink.gif" />

 

[mrmorris]

<font color="yellow">"How many we know to exit, a trillion or so? "</font><br /><br />The solution is obvious. We need to locate more galaxies! <img src="/images/icons/smile.gif" />

 

[MBA_UIU]

"From this, you can see that if 99% of the mass of the craft is propellant, the end speed is 18,000 m/s even though the exhaust velocity is 4,000 m/s."<br />Mr.Morris, Yes I am and as I stated a few post back, and I posted it probably 15 to 20 replies ago, <b>the velocity of the exhaust is measured relative to its x axis on the rocket’s body<b>. Do you not understand what I am saying here as it is exactly what you posted above!! As such, to an observer in the rocket, the speed of it exhaust will be x even though to an outside observer the speed of the rocket is a factor of y. <br /></b></b> <div class="Discussion_UserSignature"> <p><strong><font color="#0000ff"><br /><br /> <br /><img id="268587ce-7170-4b41-a87b-8cd443f9351a" src="http://sitelife.space.com/ver1.0/Content/images/store/6/8/268587ce-7170-4b41-a87b-8cd443f9351a.Large.jpg" alt="blog post photo" /><br /></font></strong></p> </div>

 

[tap_sa]

Yes, if rocket's exhaust speed is 4000m/s then it naturally follows that ... rocket's exhaust speed is 4000m/s. Wow <img src="/images/icons/laugh.gif" />

 

[nexium]

Should I conclude that 1/10c is not practical unless the reaction mass is explled from the nozzel at 1/2 c or more, as a prohibitave amount of reaction mass is otherwise needed? Neil

 

[tap_sa]

massratio of 20 and exhaust speed 0.033c / 10000km/s yields 0.1c dV

 

[mrmorris]

<font color="yellow">"Mr.Morris, Yes I am and as I stated a few post back, and I posted it probably 15 to 20 replies ago, the velocity of the exhaust is measured relative to its x axis on the rocket’s body. Do you not understand what I am saying here as it is exactly what you posted above!! As such, to an observer in the rocket, the speed of it exhaust will be x even though to an outside observer the speed of the rocket is a factor of y."</font><br /><br />Hmmm -- let's go through your posts. Your first on the subject was here where you stated: <i>"Even the most advance rocket engine cannot go faster then the fuel it burns."</i>. No mention of an observer there.<br /><br />You next posted here with the statement: <i>"To assert that a rocket can go faster then the escape velocity of its exhaust, which is where the engine¡¦s thrust is released, violates Newton¡¦s third law of motion. "</i>. Still no mention of an observer.<br /><br />Third post here: <i>"You really need to stop posting and start reading simple books on natural laws. "</i>. No observer -- still harping on breaking the laws of physics.<br /><br />Fourth post here. This presumably is where you are referring to your statement about 'observers'. The text then is:<br /><br /><i>"Typically, rocket engineers measure the speed relat</i>

 

[ortemus74]

You could use a scramjet engine coupled with a kerosene/oxygen rocket engine... I'm sure in time that will be the cheapest way to orbit.<br /><br />-Matt T.

 

[MBA_UIU]

I am not back tracking on myself, but will admit that I have a rather hard time explaining some concepts, please read on. <br /><br />First I knew it takes more mass then there is in the universe to move a chemical rocket to 1/2c. In my minds eye this is apparent when you see that the increase in a rocket’s velocity also increases it mass like energy. In other words the faster a rocket travels the more mass it takes on in the form of energy stored in the rockets structure (yeah I know that this is E=Mc^2). It then seems logical to me that the energy requirements needed to increase its velocity are expediential in relationship to its current state of mass. I don’t know if this is correct, but I believe that right now the shuttle uses something like 20 times it empty mass to reach 2.5 times its exhaust velocity. <br /><br />Without doing any math I believe comes out to something like 40 times its mass to reach 3 times its exhaust velocity and so on. To me this says that without some kind of mechanical advantage, where a rocket can exceed the velocity of its exhaust, there is no way for it to go faster then its exhaust velocity (and I should have added in relationship to the rocket itself) and overcome this problem. We do it all the time in autos, where the final drive is greater then the initial input, through the use of overdrives but a rocket engine has no such advantage. I am not BB or the other dude who believes in every conspiracy theory that comes down the pike. In fact I debated against BB (using the speed is relative argument) in his deep impact hoax thread when he argued that there is no way it could have reached the comet.<br /><br /> My bother-in-law, Brian Cariveau, is the chief software design engineer in charge of GPS satellite programming and communications at Collins Radio (formally Rockwell Collins) here in Cedar Rapids, Iowa. (You can find some of his work in this publication Cariveau, B. K., and K. L. Therklesen (1988). "Satellite data management in DoD <div class="Discussion_UserSignature"> <p><strong><font color="#0000ff"><br /><br /> <br /><img id="268587ce-7170-4b41-a87b-8cd443f9351a" src="http://sitelife.space.com/ver1.0/Content/images/store/6/8/268587ce-7170-4b41-a87b-8cd443f9351a.Large.jpg" alt="blog post photo" /><br /></font></strong></p> </div>