Read my paper

[keermalec]

I wrote a short paper on using optimal launch orbits for manned interplanetaray travel and how these can slightly optimise the latest Mars Design Reference Mission-3.<br /><br />I would greatly appreciate it if you could give me your opinions and thoughts on this paper before it is made more public. I have included an Excel sheet with most of the calculations shown in the paper so they can be checked.<br /><br />The initial paper was removed. Following discussions on this thread, the updated paper is here and the updated Excel sheet is here .<br /><br />Many thanks for your thoughts. <br /><br /><br /> <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>

 

[kdavis007]

Very interesting.......

 

[billslugg]

In separating the thousands units, in the US and UK, commas are used, not apostrophes.<br />In the US/UK - commas separate thousands, periods separate whole numbers from fractions. xxx,xxx,xxx.xx<br />In the continental european tradition, periods separate thousands, commas separate whole from fraction. xxx.xxx.xxx,xx<br /><br />Boiloff - page 9 - should not be capitalized<br /><br />"centre" page 6 is in the British tradition, in US is spelled "center". Pick either one, but be consistent. <div class="Discussion_UserSignature"> <p> </p><p> </p> </div>

 

[spacester]

Hi Keermalec!<br /><br />Nice work. You're asking for a critic, right? I mean, as much as I'm very excited at what you appear to have come up with, I want to try to make sure it is valid before you 'publish'.<br /><br />I gotta tell you, it just doesn't ring true to me. I can't put my finger on it, but it just does not make sense that you reach an orbital altitude where the dV-to-Mars starts going up. You are higher in the gravity well, so it should take less energy to finish climbing out, but the energy to Mars should be the same no matter how you got there. So the sum should always be lower the higher you are.<br /><br />So do you mind if I grill you a bit here? In the interest of verifying your results. My experience is that you will not be taken seriously by the pros unless you make a VERY good case, and I think I can help you make a better case. If you're right, this is too important to rush, agreed?<br /><br />In particular, IMO you need to be able to provide an explanation as to why IGOs exists, independent of equations.<br /><br />Yet the only way I'm going to be convinced is with equations. If you can show me the math as being correct, I should be able to help you explain what is going on in words.<br /><br />My first question is about how you derived the second equation on page 6 of the pdf (section 2). I'm not convinced that it is valid - I'm thinking that Earth-relative velocities and Heliocentric velocities are being mixed here, but you need to convert from one to the other. I could be wrong, maybe this equation is fine. But that's where I'd like to start in verifying this idea.<br /><br />My other input is nit-picking - please show the units for all quantities on the spreadsheet, and note that cell D11 is incorrect on the units. Of no consequence, but still . . . <div class="Discussion_UserSignature"> </div>

 

[keermalec]

Thanks Billslug, for the corrections and Spacester, for these pertinent questions.<br /><br /><br />New corrected version of paper is here and the Excel sheet is here.<br /><br /><br />Equation 2 comes from equation 2.11 in Reppert 2006.<br /><br />However, just in case he was wrong, I worked it out for myself.<br /><br />Using the hyperbolic velocity equation:<br /><br />Vr = (2*u*((1/r)+(1/(2*a))))^0.5<br /><br />Where<br />Vr = hyperbolic velocity at radius r (km/s)<br />u = gravitational parameter for Earth = 398'600 km3/s2<br />r = orbital radius (km)<br />a = semi-major axis of hyperbola (km)<br /><br />I derived the semi-major axis of any escape hyperbola which has a velocity 2.94 km/s at the Earth SOI radius of 922'388 km.<br /><br />a = 1/(2*((2.94^2/(2*u))-(1/922388)))<br /><br />Having obtained the semi-major axis of my escape hyperbola, I was able to compute the necessary velocity at periapsis for any orbit around the Earth. IE: the velocity needed at that point, perpendicular to the gravity field, to attain 2.94 km/s at a distance of 922'388 km from the Earth.<br /><br />Subtracting orbital velocity from this escape velocity, I obtain the necessary delta-v to go from a circular orbit to a hyperbolic orbit with 2.94 km/s at SOI.<br /><br />Adding this to my Excel table, I was able to verify that there is indeed a low point at 93'300 km from the Earth. However, the true delta-v at MGO is not 2.08 but 1.98 km/s. The difference is due to the fact that when Reppert derived his equation, he assumed tha <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]

OK, thanks, I'm closer. You are good at this, kudos to you again.<br /><br />But here's what I <b>suspect</b> is the 'problem': the hyperbolic earth departure trajectory has not been shown to actually intersect with the circular orbit you are in when you apply the dV. Plus, it not only must intersect, but you either need to be tangent or do the vector math, which will show gravity losses.<br /><br />I apologize for not taking the time to get into this deeper, but that's life I guess. The best I can do ATM is point you to a link given on the Hohmann to Mars thread:<br />http://www.me.wustl.edu/courses/me301/me301orbpage.pdf<br /> <div class="Discussion_UserSignature"> </div>

 

[keermalec]

Spacester, the hyperbolic escape trajectory has a periapsis at 93'300 km, same as my Mars Gate Orbit. The (sm) and (ve) formulae below are derived from the Hyperbolic Orbit Velocity Formula.<br /><br />Note that for a given velocity, at a given distance from a center of given mass, all hyperbolas have the same semi-major axis. One of these has periapsis at 93'300 km.<br /><br />Here's the math (formulae are written to be copy-pasted into Excel): <br /><br /><br />Constants: <br />-------------------- <br />(u) Grav. Constant: 398'600.00 km3/s2<br />(v) Necessary delta-v at SOI: 2.94 km/s<br />(s) SOI radius: 922'388.00 km<br /><br /><br />Formulae: <br />-------------------- <br />(vo) orbital velocity = (u/r)^0.5 <br />(sm) escape hyperbola semi-major axis = 1/(2*(((v^2)/(2*u))-(1/s))) <br />(ve) escape hyperbola velocity at periapsis = (2*u*((1/r)+(1/(2*sm))))^0.5 <br />(dv) delta-v required at this orbit to go to Mars = ve - vo <br /> <br /> <br />Results: <br />-------------------- <br />For...<br />(r) orbital radius: 50'000.00 km<br />(vo) orbital velocity: 2.82 km/s<br />escape hyperbola periapsis: 50'000.00 km<br />escape hyperbola velocity at 922'388 km: 2.94 km/s<br />= /> (sm) escape hyperbola semi-major axis: 51'122.42 km<br />= /> (ve) escape hyperbola velocity at periapsis: 4.87 km/s<br />= /> (dv) delta-v required at this orbit to go to Mars: 2.05 km/s<br /><br />For...<br />(r) orbital radius: 93'300.00 km<br />(vo) orbital velocity: 2.07 km/s<br />escape hyperbola periapsis: 93'300.00 km<br />escape hyperbola velocity at 922'388 km: 2.94 km/s<br />= /> (sm) escape hyperbola semi-major axis: 51'122.42 km<br />= /> (ve) escape hyperbola velocity at periapsis: 4.04 km/s<br />= /> (dv) delta-v required at this orbit to go to Mars: 1.98 km/s<br /><br />For...<br />(r) orbital radius: 200'000.00 km<br />(vo) orbital ve <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]

Dang it.<br /><br />I really don't want to do this. But I have to tell you that you are wrong. I applaud all your efforts and was really hoping you were onto something here. But it just goes against everything I know about this subject. Now, I'm just some space dweeb on an internet message board, so I could be wrong, but math is math and I have found the error.<br /><br /><font color="yellow">Note that for a given velocity, at a given distance from a center of given mass, all hyperbolas have the same semi-major axis. One of these has periapsis at 93'300 km. </font><br /><br />No, sir. That is not true. It is not going to be easy to prove my position, and you've got a lot of energy invested in this project, but here goes.<br /><br />I can state my case succinctly, though I suppose it does constitute a proof.<br /><br />You are not recalculating the escape hyperbola semi-major axis but you need to do so, to make it match up with the circular orbit you are starting from. As you push the initial circular orbit out from Earth, the escape hyperbola changes.<br /><br />I created my own spreadsheet, (open office format) and MS Excel format) and my numbers match yours except that I don't hold the escape trajectory's semi-major axis constant. As suspected, the higher you go in the initial circular orbit, the less dV is required to go to Mars.<br /><br />There is an interesting thing going on there that I admit I don't completely understand just yet. I kinda do, but . . . not totally. So I'll leave it to you to ponder. No doubt you are going to try to save the concept of IGOs and frankly I hope you do. Maybe there's a clue in there. <br /><br />Note that semi-major axis is negative for hyperbolic orbits. I think the results shown are indicative of the difference between Earth considered alone in space, where a circular orbit of 461190 km is just a nudge away from a C3=0 parab <div class="Discussion_UserSignature"> </div>

 

[keermalec]

Hmm, we seem to be getting somewhere but there is one aspect of your effort I cannot duplicate. You say you work out the semi-major axis of each escape hyperbola for each orbital distance. How? the link to your Excel sheet does not work and the one forumla I have for Hyperbola semi-major axis does not reference periapsis:<br /><br />escape hyperbola semi-major axis = 1/(2*(((v^2)/(2*u))-(1/s))) <br /><br />derived from http://en.wikipedia.org/wiki/Hyperbolic_orbit<br /> <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>

 

[keermalec]

Having checked several references, I do believe I am right in saying:<br /><br />"...that for a given velocity, at a given distance from a center of given mass, all hyperbolas have the same semi-major axis. One of these has periapsis at 93'300 km. "<br /><br />This is supported on page 25 of this reference, on page 7 of this reference, where the author works out the hyperbola's semi-major axis BEFORE determining its periapsis, and makes sense if one considers the Hyperbolic velocity formula, which relates velocity, semi-major axis, and distance, but NOT periapsis.<br /><br />If semi-major axis is a constant for hyperbolas with a given velocity at a given distance from a center of given mass, as I believe they are, then there is an optimal launch orbit to attain a given hyperbolic velocity at a given distance, therefore IGOs exist.<br /><br /> <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>

 

[keermalec]

For a velocity at SOI of 2.94 km/s, a distance from Earth of 922'388 km, and a standard parameter of 398'600 km3/s2, let us consider the hyperbolic semi-major axis as being a constant -51'122 km.<br /><br />hyperbolic velocity at periapsis = 892(1/r - 1/-102244)^0.5 or = 892(1/r + 1/102244)^0.5<br /><br />circular orbital velocity = 631(1/r)^0.5<br /><br />Note that hyperbolic velocity has a larger coefficient than orbital velocity and an added constant within the square root of inverse radius. Each graph therefore has the characteristic S-shape of a y=square route of 1/x when X is plotted logarithmically. However, the hyperbolic velocity graph is flatter because as r becomes bigger, the effect of the 1/102244 constant becomes larger, finally overcoming the reducing effect of 1/r at r = 93'300. From then on, the two graphs diverge. Hence the minimum velocity difference at r = 93'300. <br /><br />Illustration here. <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]

Keermalec, I am very sorry for the delay in getting back to you here.<br /><br />But happily, I think you have me convinced! (There is still a gut-level 1% lingering doubt, but the math AND the geometry looks solid)<br /><br />In other words, I am thrilled to say that it appears I was wrong. <img src="/images/icons/laugh.gif" /><br /><br />Your latest paragraph and the Illustration did the trick. I can see it now, it makes sense. Orbital mechanics always has a trick up its sleeve, and it got me here. But much more than that, I do believe you have discovered something that is not in the literature anywhere. Something that lets us put more payload on Mars!<br /><br />Congrats!<br /><br />I would just add that I hope you take the utmost care in presenting this. You will face skepticism and any little error in presentation will be held against you. Not that the presentation now is bad, not at all. I'm just saying that IMO you will be well served by going over it even more carefully than you've already done. Getting rid of me as an obstacle will no doubt help you do that. <img src="/images/icons/wink.gif" /> <img src="/images/icons/laugh.gif" /><br /> <div class="Discussion_UserSignature"> </div>

 

[keermalec]

Thanks, Spacester, for these kind words.<br /><br />I also believe I should go over this more carefully. For one thing, I think I will follow Reppert's method of considering radius at infinity to calculate the semi-major axis of the hyperbola. I have used radius at SOI and in that manner shaved 0.1 km/s off the trip to Mars, but I believe using the patched conic approach here is dangerous and unecessary.<br /><br />I have been working on aerocapture delta-vs at Mars and it seems quite defendable to consider 0.9 km/s of fuelled delta-v for capture and insertion into a Low Mars Orbit. Considering 2.08 km/s to leave the Earth from the Mars Gate Orbit, total delta-v from Earth orbit to Mars orbit could just very well be under 3 km/s... <br /><br />That is of course considering circular orbits for Earth and Mars. As you have shown in your Hohmann Reference Thread, when Mars is at apoapsis, delta-v can be significantly reduced. Combining this with an MGO departure orbit should give us very very digestable delta-vs! :-D <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>

 

[keermalec]

So here's the final paper and here's the final Excel sheet. Any comments will be much appreciated.<br /><br /> <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]

Very cool.<br /><br />Into full nit-pick mode, then . . . <img src="/images/icons/laugh.gif" /> . . . all are suggestions only<br /><br />pdf:<br />pg. 4: Aw shucks, 'twern't nuthin. <img src="/images/icons/laugh.gif" /> Thanks for the nod. I prefer my username not be capitalized.<br /><br />pg. 5<br />para. 1: 'distance from center of mass' -- /> 'distance from orbited body' or 'distance from occupied node'<br />para. 2: 'solar orbital' -- /> 'heliocentric' . . . . delete 'with the sun as center of mass'.<br /><br />pg 6<br />equation 1: specify this is for a circular orbit<br />equation 2: This is the heart of your entire premise, I suggest showing a full derivation or otherwise establishing its validity beyond doubt.<br /><br />I highly recommend that you show a figure illustrating the hyperbolic trajectory, as referenced is these threads. (My doubts are creeping in again . . . )<br /><br />pg 7: Excellent. Maybe re-state the definitions of the three velocity variables for easy reference?<br /><br />pgs 8-11: Excellent. Perhaps a little discussion about the Van Allen belts, IIRC MGO is well above the belts, so mention it? Also, the numbers in Table 3 seem to come from nowhere, perhaps include an appendix or something to at least show the equations used?<br /><br />pg 12: 3rd para from bottom 'Mar' -- /> 'Mars' <img src="/images/icons/laugh.gif" /><br /><br />spreadsheet: Excellent. Maybe 'appendix' or 'supplement' instead of 'annex'? Other than that, I don't see anything to even nit-pick.<br /><br />Nice job, dude!<br /> <div class="Discussion_UserSignature"> </div>

 

[keermalec]

Thanks for all the comments, the final paper, with full mathematical proof is here. The Excel sheet is here.<br /><br />If anyone thinks there is a mistake, I will be happy to correct. <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>