Hohmann Transfer to Mars Reference Thread

[spacester]

See, once again I leave something unexplained and sure enough, you ask about it. That's great feedback for me, this way I <i>know</i> you're really looking hard at this stuff. Very cool.<br /><br />Mass fraction is the simplest parameter for the mass allocation of a rocketship. It's the fraction you most often see in the various versions of the Rocket Equation.<br /><br />It's one of those "Mf" variables that leads to so much confusion. In an effort to be consistent, I'll try to remember to always spell it out. IOW I'll try to never again write<br /><br />mf = e ^ (dV/Ve) [CONFUSING]<br /><br />and instead will write<br /><br />mass fraction = e ^ (dV/Ve) [LESS CONFUSING]<br /><br />and here is why:<br /><br />mass fraction = (original spaceship mass) divided by (final spaceship mass)<br /><br />IOW<br /><br />mass fraction = mo / mf<br /><br />So from now on 'mf' will always be 'final spaceship mass'<br /><br />[And mass flow rate will be 'm-dot']<br /><br />Mass fraction is also known as mass ratio . . .<br /><br />Note that 'original' and 'final' mean the mass of the entire vehicle just before and just after the firing of the rocket motors. When you get into staging, the final mass of one stage becomes the original mass for the next stage.<br /><br />EDIT:<br />Um, maybe I didn't answer "What is it used for?"<br />It's used to figure out how much fuel, payload and inert mass you will have based on the deltaV and the Isp. It's kinda hard to work with, which is why I use those equations with Wi, Wp, and WL. In fact, if you try to work a problem without those "W equations", you'll prolly realize what I was getting at with the 1/(1-pf) question. <div class="Discussion_UserSignature"> </div>

 

[arobie]

Thank you for differentiating between the three "mf's".<br /><br />How would I even start to solve a problem without the "W" equation?

 

[arobie]

<font color="yellow">How would I even start to solve a problem without the "W" equation?</font><br /><br />Nevermind, I'm scrutinizing the slides that you posted a link to in the Xcore Aerospace thread. I'm going to see if I can understand them.

 

[arobie]

Ok, the slides lost me. How do I find out how much propellant I need without using the "W" equation?

 

[spacester]

If dV and Isp are known, you can find mass fraction<br /><br />mass fraction = mo / mf<br /><br />original mass = final mass plus propellant mass<br /><br />mass fraction = (mf + propellant mass) / mf<br /><br />mass fraction = 1 + propellant mass / mf<br /><br />mf * (mass fraction – 1) = propellant mass<br /><br />This could be said to be the “classic” Rocket Equation. <br /><br />The equations on the slides allow you to separate the payload mass and inert mass. <div class="Discussion_UserSignature"> </div>

 

[arobie]

Ah hah, I see now. Sorry I was a bit thick, but I get it now. Thank you. <img src="/images/icons/smile.gif" /><br /><br />Ok to check my understanding, I'm gonna work though my first example mission using this method.<br /><br />propellant mass = mf * (mass fraction - 1)<br />mf = 65 tons<br /><br />mass fraction = e ^ (dV / Ve)<br /><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 />dV = 4.7655 km/s<br /><br />Ve = g * Isp<br />g = 9.807 m/s^2<br />Isp = 525 (Remember, my engine is best. <img src="/images/icons/tongue.gif" />)<br />Ve = 9.807 * 525<br />Ve = 5149 m/s<br />Ve = 5.149 km/s<br /><br />mass fraction = 2.718 ^ (4.7655/5.149)<br />mass fraction = 2.523<br /><br />propellant mass = 65 * (2.523 - 1)<br />propellant mass = 99.02 tons<br /><br />99.02 is pretty close to 99.14 considering that I rounded some of those decimals. <img src="/images/icons/smile.gif" /><br /><br />I have one question about this though.<br /><br /><font color="yellow">mass fraction = (mf + propellant mass) / mf <br /><br />mass fraction = 1 + propellant mass / mf</font><br /><br />How where you able to drop the parenthesis and substitute "1" for "mf"?<br /><br />Thank you Spacester for your help, patience, and time. I really do appreciate all of this. <img src="/images/icons/laugh.gif" />

 

[jcdenton]

<font color="yellow">How where you able to drop the parenthesis and substitute "1" for "mf"?</font><br /><br />Simple.<br /><br />mass fraction = (mf + propellant mass) / mf<br />= mf/mf + pm/mf<br />= 1 + pm/mf <div class="Discussion_UserSignature"> </div>

 

[arobie]

<i>*Smacks self on head*</i><br /><br />Doh! I knew that. Distributive property from basic algebra.<br /><br />Thank you jcdenton.

 

[arobie]

F = Isp / Mf <br /><br />In that equation, what is the mf?

 

[spacester]

Hey Arobie,<br /><br />I've been traveling this week on a business trip . . .<br /><br />That Mf is mass flow rate, typically only used to find the thrust given how fast you pump propellant into the engine.<br /><br />Y'all get the short version this time cuz I gotta run . . . <img src="/images/icons/laugh.gif" /> <div class="Discussion_UserSignature"> </div>

 

[arobie]

Isp = F / Mf - Shuttle_Guy's equation<br /><br />F = Isp / Mf - (one of) Spacester's equation(s)<br /><br />Mf = Mass flow rate<br /><br />This doesn't make sense to me.<br /><br />If I rearrange S_G's equation to solve for thrust, I get:<br /><br />F = Isp * Mf<br /><br />which is different from Spacester's equation of:<br /><br />F = Isp / Mf<br /><br />Thrust cannot equal to both "Isp * Mf" and "Isp / Mf. They give two completely different solutions. Am I missing something?

 

[arobie]

*Innocent Whistle*<br /><br />bump

 

[spacester]

Hey Arobie, sorry I didn't see the question earlier . . .<br /><br />S_G's equation is correct. I can only plead that I never have actually used that equation; if I had I would have discovered the error.<br /><br />There is a good way to figure this kind of question out (besides google). Put in the units and see if they make sense.<br /><br />Of course you would have to know that the units for Isp are seconds, but . . . <br /><br />Isp = F / Mf<br />where<br />Isp is specific impulse in seconds<br />F is force, or thrust, in kilograms<br />Mf is mass flow rate, which we agreed to call "m-dot", in kilograms per second<br /><br />So substitute the units into the equation:<br /><br />seconds = kilograms / (kilograms/second) = seconds<br /><br />They check out.<br /><br />However, note that this equation is technically incorrect as well. It mixes kilograms-mass and kilograms-force, this is a no-no. I won't get into the whole SI vs. Imperial discussion.<br /><br />This is the actual correct equation, as shown here:<br /><br />Isp = F / (m-dot * g)<br />where g = 9.807 m/s^2 always always always<br /><br /><img src="/images/icons/laugh.gif" /> <div class="Discussion_UserSignature"> </div>

 

[keermalec]

Spacester, how do you calculate gravity losses? In your iriginal post you indicate gravity losses from 0.12 to 0.18 km/s at each burn. How can there be gravity losses if you are accelerating horizontally wrt the gravitational field? And if there are, how do you work them out? <div class="Discussion_UserSignature"> <p><em>“An error does not become a mistake until you refuse to correct it.” John F. Kennedy</em></p> </div>

 

[gunsandrockets]

<How can there be gravity losses if you are accelerating horizontally wrt the gravitational field?><br /><br />Exactly. I've asked about this myself in the past and the consensus was there were not any significant gravity losses.<br /><br />understanding 'gravity losses'<br /><br />The only penalty I can think of off the top of my head for low-thrust acceleration which spirals out from a planet is, that unlike high thrust propulsion, low-thrust can't take practical advantage of accelerating during low periapsis.

 

[spacester]

The short answer is that gravity losses occur because not all the momentum you think you're adding to your spacecraft is 'converted' into orbital energy. Some of it is wasted on the gravity field itself, it slows down the planet instead of speeding up the craft.<br /><br />Here's the long answer:<br /><br />Nearly everything in classical Newtonian Physics is solved by Conservation of Energy and Conservation of Momentum. It's easy to lose sight of the latter, and that's where the answer lies here. (Conservation of Mass is the third Law in play here, but that's what the Rocket Equation is all about.)<br /><br />Gravity losses happen because the momentum imparted to the craft is not 100% transferred into orbital momentum.<br /><br />An orbiting body can be described in terms of orbital energy. The energy imparted to the craft is 'bound' to the gravity field. The Vis-Viva equation shows us that once you achieve orbit, to change orbits 2/3 of your energy input is Kinetic and 1/3 is Potential. IOW to raise my orbit I spend 3 units of energy to raise the altitude the amount represented by 1 unit of Potential Energy, the other 2 units go to Kinetic Energy.<br /><br />But for any energy you try to add to the craft, to bind all that energy to the orbit, you need 100% momentum transfer. If you fail to do that, you have gravity losses.<br /><br />Momentum is given by<br />H = r * v * cos (phi)<br />where<br />r = distance from center of the occupied apse (the planet)<br />v = velocity magnitude<br />phi = angle between r and v<br />(This is a vector cross-product)<br /><br />The reason a Hohmann Transfer is the most energy-efficient way to change orbits is that when you fire your engines, phi = 0. This is true because you fire at periapse and apoapse, and at those <b>instants</b> in time, phi = 0, cos(phi)=1 and all the added v represents added momentum and added orbital energy. (See my new favorite quick-reference page for the <div class="Discussion_UserSignature"> </div>

 

[spacester]

<font color="yellow">. . . consensus . . . </font><br /><br />ROTFLMAO <div class="Discussion_UserSignature"> </div>

 

[gunsandrockets]

Are you claiming 'gravity loses' ARE significant? If so, then exactly how significant are 'gravity loses' for a low thrust spiral trajectory?

 

[spacester]

Low thrust spiral trajectories are practically all about gravity losses. For the entire time the engines are firing, you are failing to couple all the momentum transfer to the orbital energy.<br /><br />You can throw energy out the back all you want, but you have to bind that energy to the gravitational field to avoid gravity losses. "Bound energy" is not some abstract concept with no practical application.<br /><br />Did you follow the math? Did you read the links? Do you have specific questions?<br /><br />I'm not "making claims" or trying to "establish consensus".<br /><br />This is friggin mathematics, not politics! <div class="Discussion_UserSignature"> </div>

 

[gunsandrockets]

<Low thrust spiral trajectories are practically all about gravity losses.><br /><br />That's odd. That's not what you said in the thread I linked to.<br /><br />http://uplink.space.com/showthreaded.php?Cat=&Board=missions&Number=517802&page=&view=&sb=&o=&vc=1<br /><br />< Do you have specific questions? /><br /><br />I did, and still do since you have not answered my question, exactly how significant are 'gravity loses' for a low thrust spiral trajectory? <br /><br />I don't know how I could have been any more direct. No matter.<br /><br />

 

[keermalec]

OK, I understand the equations for determining gravity losses are complex but it would help to have a look at them, in order to have a practical example of how this works. (at the moment it feels a bit like trying to understand a windows function without the source code ;-)).<br /><br />Forgetting about low-thrust spiral paths, if we do a straight patched conic approach fro an earth to mars mission with just start and arrival burns, leaving Earth at an angle to its orbit and reaching mars at a tangent to the transfer orbit's periapsis:<br /><br /> illustration <br /><br />In theory there will be gravity losses at Earth but not at Mars, right? But if, when we depart earth, we accelerate along our Earth orbit, ie perpendicular to Earth's gravity field, then there obviously are no Earth-related gravity losses. When we reach SOI we have the necessary speed to get to mars on the afore-mentionned elliptic orbit. There is therefore NO acceleration in the sun's sphere of influence, so why should there be gravity losses wrt to the sun?<br /><br /> Earth departure <br />Image shamelessly copied from Reppert's excellent report. <div class="Discussion_UserSignature"> <p><em>“An error does not become a mistake until you refuse to correct it.” John F. Kennedy</em></p> </div>

 

[spacester]

You want direct? I'll give you direct! I like you, G&R but you continue to really tick me off on this subject.<br /><br />So we're playing gotcha, eh?<br /><br />Fine, I was wrong. You got me. I shouldn't have concurred with Shuttle_Guy.<br /><br />Now go attack Shuttle_Guy and leave me alone until you start acting like the intelligent guy you are.<br /><br />He was the one I concurred with. He's the one who made the conflicting statements. Look at the flat thread. Look at the SG post right above mine which I concurred with. It is in conflict with the SG post which you quoted above that.<br /><br />But SG can't handle the idea that I might know something. I don't work for NASA or its contractors and I don't kiss his ass so I must be wrong. That's the way he comes across anyway. Or used to, he's gotten a lot better. And hopefully so have I. Care to join us in a higher level of discussion?<br /><br />You are the one bringing this controversy back up. You could avoid that by reading the friggin math and talking about the math.<br /><br />I was trying to keep the thing from escalating further. I was trying to be nice. I concurred when I should have argued. It's not like you were making an honest effort to understand. It's hard to explain these things to a hostile audience. I stand by my rant. You are more interested in playing gotcha than understanding the math. Talk math, not politics. <br /><br />Keermalec is here now so we can make some progress.<br /><br />'Exactly how significant' . . . <br /><br />17.2368 m/s<br /><br />There, is that a good answer? It's exact. <img src="/images/icons/wink.gif" /> Don't ask a qualitative question and expect a quantitative answer. That's not direct. It's a stupid question, you should know better.<br /><br />Oh, you want an explanation to go with the exactitude?<br /><br />Then read the linked paper. Read the charts. Make an effort. It's complicated. Give me some feedback here that tells me you grasp what has already been presented. I laid it out for you a <div class="Discussion_UserSignature"> </div>

 

[keermalec]

Well that doesn't answer my question though.<br /><br />I think I get the concept behind gravity losses.<br /><br />If you throw a rock in the air at an angle its path is parabolic, right?<br /><br />Wrong! its path is a segment of an ellipse, not a parabola, because gravity always points towards the center of the Earth and therefore changes angle ever so slightly as the rock moves horizontally. the rock's path is a segment of the elliptic orbit it would take if the earth were a point mass and didn't have a surface 6371 km up to stop it.<br /><br />I understand that the vertical component of the rock's acceleration must overcome the Earth's gravity or it won't go up. IE I need to use more muscle power to throw a rock in the air at speed v than if I was in a weightless environment. Stands to reason doesn't it?<br /><br />Now if the rock is actually a small rocket with a thruster and, as it is going up at an angle, the thruster is fired, that thruster must also fight gravitational attraction in the up direction (not in the horizontal direction). Therefore if the thruster is firing along the direction of flight, its up-acceleration is reduced by 9.81 ms-2.<br /><br />If my rocket is now in outer space, in an elliptical orbit and firing along the direction of flight, it will have to fight gravity at any point along its orbit except periapsis and apoapsis, where the direction of flight happens to be horizontal wrt the gravity field.<br /><br />To calculate gravity loss one therefore has to <br /><br />1. determine the rocket's acceleration before applying gravity loss, a<br />2. determine the acceleration due to gravity at that distance from the central body's center (b=GM/d^2).<br />3. determine the angle from horizontal that the rocket is firing along, c<br />4. determine the rocket's actual acceleration after gravity losses, d<br /><br />If "a" is the rocket's acceleration, "b" is the central body's acceleration due to graity at that distance, "c" is the orbital angle wrt the gravity fi <div class="Discussion_UserSignature"> <p><em>“An error does not become a mistake until you refuse to correct it.” John F. Kennedy</em></p> </div>

 

[spacester]

Sorry about the rant, Keermalec. As you can see, this goes back a ways here. It really touches a nerve of mine. Five+ years of meticulously typing in error free space math got me zero benefit of the doubt.<br /><br />Well asked questions. Thanks a ton for the illustrations!<br /><br />The equations aren't so complicated, it's the terminology. It's basically just law of cosines: vector arithmetic. But to explain the application of those simple formulas is a huge project. The hostile crowd here is not worth the trouble. You're pretty much the second guy in 6 years with a true interest in learning from this lowly layman.<br /><br /><font color="yellow">Forgetting about low-thrust spiral paths, if we do a straight patched conic approach fro an earth to mars mission with just start and arrival burns, leaving Earth at an angle to its orbit and reaching mars at a tangent to the transfer orbit's periapsis: </font>dit: apoapse<font color="yellow"><br /><br />illustration<br /><br />In theory there will be gravity losses at Earth but not at Mars, right? </font><br /><br />Correct. At Earth, you cannot simultaneously thrust along the flight path AND perpendicular to the radius vector AND depart Earth's orbit with the full benefit of earth's orbital velocity. (Unless there's a trick the experts know about, but I highly doubt it.)<br /><br /><font color="yellow">But if, when we depart earth, we accelerate along our Earth orbit, ie perpendicular to Earth's gravity field, then there obviously are no Earth-related gravity losses. When we reach SOI we have the necessary speed to get to mars on the afore-mentioned elliptic orbit. There is therefore NO acceleration in the sun's sphere of influence, so why should there be gravity losses wrt to the sun? </font><br /><br />Well, first let's assume the engine burn is instantaneous, which is impossible but removes that aspect from the discussion.<br /><br />Now if the instant burn is as shown, adding deltaVo to Vc, the init <div class="Discussion_UserSignature"> </div>

 

[spacester]

We were typing at the same time . . . <br /><br />I have to get some sleep. Sorry, more tomorrow. Looks like you've got it though. <div class="Discussion_UserSignature"> </div>