Hohmann Transfer to Mars Reference Thread

[arobie]

Periapse altitude=Low point in orbit<br />Apoapse altitude=High point in orbit<br /><br />Is that correct? I'm sorry, I know this is the basics, but I have not learned it. I would like to understand it though.

 

[najab]

Yup. Though I think it's periapsis and apoapsis.

 

[spacester]

Periapsis and Apoapsis are the generic terms for any satellite, no matter what body it is orbiting.<br /><br />For the Earth, the terms become perigee and apogee.<br /><br />For the Moon, they are perilune and apolune. Or periselene and aposelene.<br /><br />For Jupiter, perijove and apojove.<br /><br />My turn for a question: on cybersix's post, what do you suppose CFM stands for? Computational Fluid Mechanics? Somehow I think it means something else. <img src="/images/icons/laugh.gif" /> <div class="Discussion_UserSignature"> </div>

 

[arobie]

To all,<br /><br />Could someone explain specific impulse to me. So a rocket has Isp of 375 seconds...What does that mean?<br /><br />Halman,<br /><br /><font color="yellow">Even if we still had the Saturn 5 around, we would be limited to a payload to Mars equivalant to the Apollo Command Module, Service Module, and Lunar Excursion Module, if we used Hohmann orbits.</font><br /><br />Why would the payload be limited using Hohmann orbits?

 

[arobie]

Ok, so Isp is a measure of efficiency. Is there an ideal Isp? Also, which is better, a higher Isp or a lower one?<br /><br />Thank you for clearing up my misunderstanding about the limitations of going to Mars. I see now.

 

[JonClarke]

Back in the dark ages halman wrote: <br /><br />"Even if we still had the Saturn 5 around, we would be limited to a payload to Mars equivalant to the Apollo Command Module, Service Module, and Lunar Excursion Module, if we used Hohmann orbits."<br /><br />Apollo 15 CSM+LM massed almost 47 tonnes. Say 40 tonnes as an indication of what a Saturn V could send to Mars. There was a lot of growth potential to the Saturn V that was never ralised either.<br /><br />Forty tonnes is a very useful payload and using 2-3 launches are quite adequate for a useful Mars mission with ~50 tonnes on the surface of Mars.<br /><br />Jon <div class="Discussion_UserSignature"> <p><em>Whether we become a multi-planet species with unlimited horizons, or are forever confined to Earth will be decided in the twenty-first century amid the vast plains, rugged canyons and lofty mountains of Mars</em>  Arthur Clarke</p> </div>

 

[spacester]

Well that’s one of the few questions on this subject that’s easily answered. All else being equal, the higher the Isp the better. Period, end of story.<br /><br />But d’ya think I’m gonna go with such a short post when a long one seems in order? <img src="/images/icons/laugh.gif" /><br /><br />The rest of the story is that with chemical propulsion, there is a <i>maximum possible</i> Isp. The laws of chemistry tell us that we’ll get the highest possible Isp when we “burn” LH2 (Liquid Hydrogen) with LOX (Liquid Oxygen). IIRC, the theoretical maximum is around 525 seconds but in the real world, the SSME is prolly as good as it’s ever going to get at um IIRC 475 sec. (There’s several rocket engine experts around here . . . I’m not one of them)<br /><br />Now a lot of people hear that fact and conclude we need to develop some new, revolutionary propulsion technology cuz chemical is limited and we’ve already done the best we can do. Many of us reject that simplistic answer and point out that the correct performance criterion to apply to getting to orbit is economic and not fuel efficiency. BDBs and all that: no one has ever designed a launcher with strict economic criteria. It’s a subject for another thread, but I can’t write the previous paragraph without writing this one.<br /><br />Good for you on pursuing this subject, Arobie! You’ve learned about orbits, now you need to learn the rocket equation. That’s where you see why Isp is so important. The rocket equation is where the mission requirements (deltaV) and the spaceship performance (Isp) come together so that you can design a rocketship to do the job.<br /><br />(Oh, cool, the other subjects I wanted to talk about have been addressed by s_g and Jon)<br /><br />This is the part where most people’s eyes glaze over. The see an equation coming and yikes they are outta here! It’s not that hard! And no, there is no other way to figger this thing out, so please, gentle reader, try to stick with it.<br /><br />pf = e ^ (dV / Ve) (EDIT: THI <div class="Discussion_UserSignature"> </div>

 

[najab]

><i>Ok, so Isp is a measure of efficiency.</i><p>Isp is called specific impulse for a reason. It is a measure of the acceleration (impulse) that you get from 1 mass unit of propellant (1 unit hence specific). Remember that for all practical purposes, Isp is the inverse of thrust - as Isp goes up, thrust goes down. That's what shuttle_guy was alluding to with his reference to Ion thrusters. They have a very high Isp (around 1000s), but very low thrust, as compared to the chemical rocket engines which have lower Isp but much higher thrust.<p>spacester explains the maths behind it way better than I can, so take a look at his equations - they're not that hard, sometimes even I understand them! <img src="/images/icons/smile.gif" /></p></p>

 

[arobie]

Thank you S_G, Spacester, and najaB. All of the help and information great! I really appreciate all of this. <img src="/images/icons/laugh.gif" /><br /><br />I have been sitting here for almost an hour trying to comprehend all of the information that has been put out(I have enjoyed this hour), and I think I am beginning to understand. I understand the concepts, and I mostly understand the equations given out. I understand how they impact each other. <br /><br />But of course, I have questions.<br /><br />Does e always equal 2.718?<br /><br />What is dV, and is it interchangeable with Vf? <br />(When you figured the propellant requirements for getting the water to Mars orbit, you used the formula: <br /> <font color="yellow">"pf = 1 - 1 / (e ^ (<b>Vf</b> / Ve))"</font><br /><br />Also why is the 1 - 1 in the formula?<br /><br />When you were discussing "Gravity Losses", you said <font color="yellow">"mf = mass of rocket with fuel ("wetmass")"</font> When shuttle_guy was explaining Isp he said <font color="yellow">"Propellant flow rate (Mf)"</font> Which does Mf stand for, or does it stand for both in different situations?<br /><br />"[(mf)(Isp)(g)] / F" is the formula for burn time, correct?<br /><br />That's all the questions that I have for now. While looking through all of this info, I will probably come up with more.

 

[arobie]

Another question:<br /><br />So when you find the pf, you multiply it by the mass of your ship to find out how much propellant you will need. Is that correct?

 

[najab]

><i>Does e always equal 2.718?</i><p>Yes. That is the mathmatical constant e (the base of natural logs).<p>><i>Also why is the 1 - 1 in the formula?</i><p>It's not 1 - 1 since that would make the result of the equation zero! It's just that HTML isn't the best way to write formulae - it should be written pf = 1 - [1 / (e ^{<small>(V_f/V_e</small>})]</p></p></p>

 

[spacester]

Ooops, that’s what I get for “working without a net”. Sorry for the confusion, Arobie.<br /><br />I wanted to show the simplest possible form of the rocket equation and was concentrating on the right-hand side.<br /><br />Thank you for the clarification najaB, I was too lazy to use superscripts.<br /><br />pf = e ^ ^{(dV / Ve)} is WRONG<br /><br />najaB’s version is CORRECT<br /><br />mass fraction = e ^ ^{(dV / Ve)} is CORRECT<br /><br />mass fraction is yet a THIRD variable that could be written as mf !! There’s mass flow rate, final spaceship mass, and mass fraction !! Tres confusing, yes?<br /><br />Mf as used in shuttle_guy’s thrust equation is mass flow rate. Back in the day of pencil and paper, this would be called “m-dot”, written as an “m” with a dot on top of it. The dot-on-top is universally recognized as a rate-of-change-with-respect-to-time variable. For example, if v is velocity then “v-dot” is the rate of change of velocity, which is the definition of acceleration. I don’t think there’s a way to write it in html.<br /><br />So m-dot is the rate of change of mass over time, which is simply the mass of propellant being consumed per second.<br /><br /><font color="yellow">"[(mf)(Isp)(g)] / F" is the formula for burn time, correct?</font><br /><br />Yes. Here, mf is the mass of the spaceship. Two points:<br />1. This is just our old friend F = m * a , essentially it is the definition of Isp<br />2. Note that the mass of the spaceship is changing during the burn time because you’re throwing stuff out the back, so mf is not constant, so this equation is an approximation based on assuming the mass of the ship is much greater than the mass of propellant, which of course is NOT true. So this equation is just an estimate. But if you use the average of the starting spaceship mass and the ending spaceship mass, it’s pretty close.<br /><br /><font color="yellow">So when you find the pf, you multiply it by the mass of your ship to find out how much propellant you</font> <div class="Discussion_UserSignature"> </div>

 

[arobie]

Don't worry about the mistake, it's not that big. Now it's "mf=e^(dV/Ve)" instead of the pf there. (Hey! I recalled that equation off the top of my head.<img src="/images/icons/smile.gif" />) I didn't get too confused. No problem.<br /><br />Now if pf is not the mass fraction, what is it?<br /><br /><font color="yellow">"Wp = [(Wi * pf) + (WL * pf)] / (1 - pf )"</font><br /><br />What is "Wp", "Wi", and "WL"? (EDIT: Nevermind, I figured it out. WL is the payload, Wi is everything else (except propellant) and Wp is the propellent.)<br /><br />"F = m * a"<br />Thrust = mass * acceleration<br />Do I have it correct?<br /><br />Ok, It's all starting to sink in. I've gotten out my calculator and started to throw in numbers into the formulae. As soon as I understand all of it I will put it all together and actually create an example mission. I'm also determined to find why pf - 1 is important. I've written down all of the formulas that have been put out. I'm going to go think on it and try to piece it all together.

 

[spacester]

<img src="/images/icons/cool.gif" /> <img src="/images/icons/cool.gif" /> <img src="/images/icons/cool.gif" /> <br /><br />I’m afraid I’ve sent you off on a bit of a wild goose chase. I don’t want to distract you if you’re planning a mission, that’s much more cool.<br /><br />The question should have been<br />“Why is the number<br />(1 / pf) – 1<br />important?”<br /><br />Make sure you refer to the thread with my 'Hey kids, want to design a rocketship?' post.<br /><br />Hint: try working a problem without that cool yellow equation and instead start with<br /><br />Pf = Wp / GLOW<br /> <br />(edit: fixed the link) <div class="Discussion_UserSignature"> </div>

 

[bobvanx]

This is a great thread.<br /><br /><img src="/images/icons/cool.gif" />

 

[arobie]

You're not distracting me from planning a mission. My goal is only to understand all of this in this thread. I only plan on eventually planning a mission because you said:<br /><font color="yellow">"I encourage you to create your own example mission and work thru the numbers."</font><br /><br />What you're explaining to me in this thread is what I want to learn and understand. I would like to continue the train of thought that we have going.<br /><br />I still don't understand what pf is. Could you explain to me? I got confused after the mistake in the formula. <br /><br />PS: You link doesn't work. Could you fix it?

 

[spacester]

This is so cool. I appreciate your interest and diligence.<br /><br />Try the link again, I referred to it because I explain pf there. But don't hesitate to keep asking after you've read that. It's the, um, 38th post in that thread. <div class="Discussion_UserSignature"> </div>

 

[arobie]

Thank you, and thank you also for fixing the link.<br /><br />So pf is the propellant fraction of the craft. It is, the propellant mass over the entire ship mass.<br />Wp/GLOW<br /><br />Hey that's easy. <img src="/images/icons/smile.gif" />

 

[spacester]

<img src="/images/icons/laugh.gif" /> <img src="/images/icons/cool.gif" /> <img src="/images/icons/laugh.gif" /> <div class="Discussion_UserSignature"> </div>

 

[arobie]

I decided to plan a mission because I was looking at all of the information in this thread and the other thead. As I was reading though it, nothing was confusing me, but I knew that I wouldn't know whether I understood or not untill I tried it, so I decided to try.<br /><br />Ok, this is my experiment example mission. I'm going to send supplies to Mars in preparation for human visitors. I'm going to send 50 tons worth of supplies to the surface. The tanks and engines will be about 15 tons. <br /><br />WL = 50 tons (Pulled out of thin air)<br />Wi = 15 tons (ditto)<br />Wp = [(Wi * pf) + (WL * pf)] / (1 - pf)<br />1st burn, dV, corrected = 3.8077 km/s --- 2020-JUL-26<br />2nd burn, dV, corrected = 0.9578 km/s --- 2021-MAY-02<br />Total dV = 4.7655 km/s<br />Isp = 525 sec (Theoretical highest Isp using chemical rockets. My rocket's best. <img src="/images/icons/tongue.gif" />)<br />pf = 1 - 1 / (e ^ (dv / Ve))<br />dV = 4.7655 km/s<br />Ve = 9.807 * 525 = 5149 m/s = 5.149 km/s<br />pf = 0.604<br />Wp = [(15 * 0.604) + (50 * 0.604)] / (1 - 0.604) = 99.14 tons <br />99.14 tons of propellant to arrive at Martian Orbit starting from LEO<br /><br />99.14 tons of propellant just seems too high to me. I've checked my math and I haven't found any errors. I think I might have made a mistake using the formula's or might have made a mistake in my understanding. It just doesn't look right to me. (Nevermind)

 

[arobie]

Actually 99 tons might not be too high. For the water buffalo in Mars orbit scenario, it took 609 tons of propellant to get 448 tons to Mars. So nevermind, I take back my statement that 99 tons of propellant looks to high. <br /><br />I would still appreciate any critiqueing on my calculations. If there are any errors, let me know. I will also review my work in the morning. It's after midnight and I'm losing concentration, so I'm going to get some sleep.

 

[spacester]

Nice work, looks 100% correct to me! <img src="/images/icons/cool.gif" /><br /><br />Yeah, it's late . . . <br /> <div class="Discussion_UserSignature"> </div>

 

[arobie]

<font color="yellow">“Why is the number <br />(1 / pf) – 1 <br />important?”</font><br /><br />I don't have any idea. Any hints? <br /><br />Very nice "Hey Kids, want to design a rocketship? post. Alot of useful information in it like always. <br /><br />I do have a question about an equation in it though.<br /><br /><font color="yellow">Vf = Ve * LN (1/(1-pf))</font><br /><br />What is "LN"?

 

[spacester]

"LN" is the thing that freaks everybody out. <img src="/images/icons/laugh.gif" /><br /><br />It's the *OMG* Natural Logarithm Function<br /><br />Run away! Run away!<br /><br />Wolfram and wikipedia and Univ of BC all came up on a Google search, but chances are those will just confuse you.<br /><br />Here's all you really need to know for now:<br />The Natural Logarithm is the "opposite" (the inverse, actually) of putting something as the exponent of the number e=2.718 . . . <br /><br />IOW, if you "take the natural log" of e^x then you get x<br /><br />So <br />LN(e^x) = x<br /><br />and<br /><br />LN(e^(dV/Ve)) = dV/Ve<br /><br />and<br /><br />e^LN(dV/Ve) = dV/Ve<br /><br />Knowing this, you can now compare<br /><br />dV = Ve * LN (1/(1-pf)) <br /><br />with<br /><br />pf = 1 - 1 / (e ^ (dV / Ve)) <br /><br />and show yourself how they are the same thing, just rearranged using basic algebra. <div class="Discussion_UserSignature"> </div>

 

[arobie]

Ok, thank you for explaining natural logarithms to me. I don't quite understand it yet, but I found that it is in my Algebra II textbook.<br /><br />What is mass fraction? I have the equation: mf = e ^ (dV/Ve), but what is mf a measure of, and what is it used for?