Reentry - An Orbital Simulator - Freedom 7 AAR


Apollo is there enough to do up to about Apollo 8, but there’s no manual as yet and you have to use the NASA original docs. Early access style! :slight_smile:

A Lunar Module in this, and the full Apollo 11 is the dream. So far I’m just transitioning to Gemini flights, and it’s a nice way to learn things bit by bit. It’s like the Hornet without a TPOD lol

In the last patch there seems to be Steam Workshop support too, so people are creating/adding missions.


So uh, I happen to have manuals and documentation for the Apollo Guidance Computer. Wow, never thought it would come in this handy!


Ha. just looked. There is a new Apollo draft manual. Oh man…

Yeah, you’re going to love it @Navynuke99


Yep, inevitably my wallet took “The Hit” last night…

I have been really short of time these last couple of days, or so, but managed to look at two Academy training missions. Loving this, hope I get a bit of time, soon.

It also caused me to have a brief look in on the “old favorite”, Orbiter, to see all was still working. That was the program that got me back into simming some time ago, incidentally, after a long break when I had sort of grown out of them. Still seems to be working, but a lot has been forgotten…

Looking forward to learning Reentry!


@fearlessfrog how are you doing with this? :slight_smile:

Still no time whatsoever to get really in depth, but I managed to finish all the Mercury Academy missions, at least (fortunately, they are pretty short and serve only for capsule familiarization). One of the most exciting things I am foreseeing here is the need to calculate your orbital elements mathematically, based on time and some observation only, as you have a very bobby-basic craft that does not give you all the parameters nicely laid out, like the MFDs in Orbiter.

For this reason I will be sticking with the Mercury for a while, with the Atlas LV, in free play. It does not even give you the altitude (or radius) or the mean orbital velocity, much less the inclination or any eccentricity. Then there are the de-orbit and re-entry parameters to contend with, if you are going to do anything that is not specifically the pre-programmed time for retro fire (ie; a specifically chosen splashdown point).

There will be more. This is excellent, not because of what there is, but because of what a faithfully recreated rendition of the craft lacks, and that would constitute a decision making resource toolbox for an “autonomously operating” astronaut.

Indeed, what a leap of faith it was for those who rode the originals, seeing what they had at hand in the cockpit, and contemplating it with a pilot’s perspective of the overpowering necessity to have information. It simply must have been overwhelmingly scary.


I’m at the same level I think. I saw a patch the other day fixed some mission issues I had as well.

I’m done with Mercury, apart from some longer Atlas things I want to try in terms of landings, and have put this on a laptop to get into Gemini when on vacation. Talking of which, getting shouted at for not actually packed as yet, so best get on with it!

See you all in a couple of weeks!


So, yeah. As we seem to have from @BeachAV8R the invitation to take one of these craft to Pago Pago, supposedly for the Christmas flight…

…I will not shy away from it. I do not know about @fearlessfrog’s participation in it (I believe he hinted he might do it, and my apologies for detracting from the exclusivity if so), but for me the opportunity is a bit of a God-send. IRL I am up to my eyes in an IOSA Certification for my admin department at work, and do not have the time to do any more lengthy aircraft flights for now. There is that, and the fact that on each of the previous two Christmas trips I have already “threatened” to use space craft. In fact, on the first one I took part in, I actually did a demo…

See, I am slightly nuts about space flight. And no, I have never used Kerbal Space Program. The inclusion of the little green men puts me off it. I am a bit of a follower of Dr. Martin Schweiger’s Orbiter. Anyway, let us cut to the chase.

I am establishing a couple of rules for myself;

First, I will be using the Mercury. I like the challenge of having next to no orbital information at all. It is an exercise in planning. I will not repeat all my impressions from what is stated in the posts I made above.

Second, I am making only one run at it. By that I mean, I will not be doing a series of deorbit tries to get as near as possible to Pago Pago, hit and miss fashion, and then publish the closest. Wherever I land on the first committed try will be from where I take up again the trip to Pago Pago (the closest airport, I guess, to splash down). I do not want to do any such cheating.

That said, I have already done the Academy missions, plus one free play flight into orbit for the sole purpose of establishing the accuracy of the program. I did use tried and trusted Orbiter as a yardstick to comparatively test my calculations, too. I was quite pleasantly surprised. For those at all interested,

Please read on...

No, I am not going to sit here writing a load of stuff about longitudes of ascending nodes and explaining all sorts of other basic orbital jargon. If you clicked to open this, you already know all of that. And if you know more, or a simpler way to do it, please shout. These are some formalae that I used in a planet simulation program of my own, some time ago.

What I am going to do is show some tables I generated in Excel. This approach seems the best for a simulation of a Mercury with rudimentary information. I am looking for good approximations for this mission, so what follows was created with some reasonably simple math.

Again, I say, I was looking for sim accuracy. There is some data generated on that table that will be useful again in the future. Here, the main goal was to see if orbital times were first, believable, and second, once establishing that, determine the mean altitude of the orbit, using the tables.

Now, in Orbiter you have a nice display that tells you when your rate of change N/S is zero, which would mean your inclination has taken you as far north or south as you will go. This is an ideal moment to start timing your orbit, until you are back at that point.

In Orbiter, I set up a 220 - 225 km orbit, this being as close to zero eccentricity as possible with all the perturbations turned on in the options, and timed it. Here is what I got…

Mark one…

Mark two…

16723 seconds - 11408 seconds = 5315 seconds = 88.6 mins

That compares quite favorably to the chart, for a 230 km altitude orbit. Also, the real time to game time elapsed was spot on, not surprisingly (see, I used the game clock and my phone stopwatch to time the orbit).

Timing the orbit for Reentry, with the Mercury, required me to use the orbital view, and I timed the orbit from descending node to the next descending node, across the equator. I got the following time data…

02:02:12 - 00:33:53 = 88.3 mins

By chance, I had come pretty close to the average altitude of the orbit with my preliminary estimate on Orbiter control test. I looked up some of Mercury’s orbit data from John Glenn, Scott Carpenter and Gordon Cooper’s flights, mainly for the perigee and apogee. Glenn’s average was 199 km, Carpenter’s 207 km and Cooper’s was 214 km. The periodicity for Glenn’s was 88.5 mins, Carpenter’s was 88.6 mins, and Cooper’s was 88.8 mins.

Again, the times were favorable to my table. Pleased with that, assuming the orbits that Mercury in the Reentry sim achieve are similar to the real ones.

So, that was training and preparation phase one over, with good results. I did notice one thing, however. The game clock for Reentry is a tiny bit fast. On the stop watch, I clocked the orbit at 86.9 mins, a full 1.4 minutes shorter in IRL time. This did not happen on Orbiter’s test, in which the game clock is well synchronized to real time passage. Still, the main thing is that the in game, panel mounted clock is accurate, and it is already great for new release in testing phase.

One thing I would have liked would have been better background stars representation, like Orbiter’s. I would have been able to use them very realistically instead of the “Orbital View” to ascertain when I was at zero N/S rate of change in Reentry…

Next time, I will look into an estimated drag model for the reentry process.

That is all for now.

Some notes mainly for myself on creating the tables, skipping opening this is advised:

First, to create the table, I went about calculating the degradation of gravity at various radii. For this, the formula…

Acceleration = Gravitaional Constant x ( Mass / radius^2 )

…does the trick. A brief example of that in action (mass in kilograms, radius in meters)…

A = 6.67E-11 x ( 5.98E+24 / 6.37E+6^2 )

A = 9.83 m/s^2

Looks familiar? It is approximately the gravitational acceleration caused while standing on the planet, at its average radius. Now, gravity degrades as you move farther away from a body, and it decays very predictably by the inverse square law (that is what the squared radius is all about in there). Any radiating energy decreases by this law, which is quite logical, as the area of a section of an expanding sphere is the square of what it was at half the distance, therefore that same energy is dispersed over the larger, squared area. Gravity is included.

This fact surprises some people, I have found, who think that as soon as you are out of the atmosphere, there is zero gravity. This is a myth. There is quite a bit of gravity still around for quite a way out from Earth. Zero gravity is, of course, an “illusion” caused by the fact that an orbit is a perpetual fall (in which everything floats around you), with a neat sidestep perpendicular to the surface at the same value of the downward velocity acheived by the gravitational acceleration. This “side step” is our orbital velocity, which not surprisingly also needs to “decay” accordingly the further away we are from the source of the gravity, for a nice stable circular orbit that is.

A modification of that formula will, not surprisingly, give us the value of the orbital velocity, as a velocity and not an acceleration. Such;

Velocity^2 / radius = Gravitational Constant x ( Mass / radius^2 )

…which boils down to…

Velocity = Square root of ( Gravitational Constant x ( Mass / radius ) )

Of course, we already have the radius. With that, we can easily get the circumference of the orbit. And armed with the now known orbital velocity and the circumference, we can get the time.

Finally (the math just keeps getting easier at this point) we can obtain the rate of orbital angular change, and the sidereal rate of angular change of the planet as it rotates. It is important to note that these calculations MUST take into account sidereal rotation of the planet, and not the solar. The orbit does not “twist” around with the planet while it orbits the sun, but stays aligned to a universal reference. All of this data is useful for navigation purposes.

I have already done a quick extrapolation of Mercury’s orbits, using my globe. I already know that I will probably miss Pago Pago. Not because of errors in calculations, all told, but because one orbit track is probably going to pass east of American Somoa, and the next one around will probably be west of it (unless I have made an error in the extrapolation). I have yet to calculate just how much of the available delta V can be used in the brief retro burn, slightly offset from the prograde path, to try and inch the closest one of the passes a bit closer to Pago Pago, and still have effected a deorbit. Then it is hope for the best. There are no aerodynamic controls to make any final, fine corrections.


Continuing the plotting and scheming. There have been no more flights, but…

More numbers...

And they all mean something. I have been having some fun with this, every now and then, and it has been a source of some great relaxation and distraction during some pressured times, lately.

In the last post, I had got as far as calculating the mean orbital velocities based on orbital radius. This is just scratching the surface. But before I start, there was one thing I did on that last Mercury flight which I omitted (because the post was getting long inside the summary).

I had timed the orbit at five points: Northernmost point (which appeared to be abeam Cape Hatteras, giving me the approximate inclination of the orbit at about 33.5º), crossing the equator descending, the southern most point, crossing the equator ascending, and finally again at the northernmost point. This gave me the interval between those points, which helped me to approximate the eccentricity of the orbit. Straight to the table I compiled for that data…


As can be seen, I got the intervals of 21.8, 21.6, 22.7 and 22.3 minutes. The longer the interval, as covered in the last post, the longer the radius of the orbit in that segment. I multiplied each of these by 4 (as they represent a quarter orbit) to obtain a period value for a total, circular orbit, then converted it to seconds for calculation purposes. I used the formula…

Radius = Cube Root of ( (Earth Mass kg * Grav Constant * Period Secs) / (4 * Pi^2) )

To obtain the approximate radius at each segment. I had to “dumb this down” for Excel, after looking up why it does not do cube roots. This did the trick, which is the same thing…

( (Earth Mass kg * Grav Constant * Period Secs) / (4 * Pi^2) ) ^(1 / 3)

Anyway. I was able to determine to a degree the eccentricity of the orbit from this information, and in which segments were the apogee and perigee. And as before, I deduced the velocity from the circumference and period. I did not need much more from this table, for now.

On to the next thing. The Mercury capsule and its retro rockets. These are three, which fire for ten seconds each, in overlapping sequence, and each produce 4.5 Kilo-Newtons. Now, I would be wrong to use the value of 4.5 kN x 3, because the nozzles are pointed outwards, and it is a component of 4.5 kN that is directed along the longitudinal axis. Of course, they are angled out like that for a reason; they point at the CoG of the capsule, so that their firing does not induce a rotation. I made an estimate that they point outwards about 24.2º. That is;

Arc Tangent ( 0.45 m / 1.0 m )

That angle gave me a longitudinal thrust component of…

_Cos ( 24.2º ) x 4.5 kN = 4.1 kN

…times three, for a total thrust vector of 12.3 Kilo-Newtons for the deorbit. But there is more. The retro-burn of the Mercury is done with the back end of the capsule pointed outside of the the orbit path, planar, at 34º (apparently). This further divides those 12.3 kN into a retro burn component, and a radial burn component, which tips the orbit. In effect, for this experiment, it is enough to say that it induces some “earthward” vertical velocity. As it is never going to complete the orbit, this assumption is fair enough. By how much is here…

Longitudinal component = Cos(34º) x 12.3 kN = 10.19 kN

Vertical component = Sin(34º) x 12.3 kN = 6.87 kN

By how much does all that affect the capsule? We need the capsule’s momentum, for that. The capsule’s mass is 1,360 kg, and in the first segment, its velocity is 7,823.73 m/s. Multiply to get the longitudinal path momentum…

1,360 x 7,823.73 = 10,640,272.8 N. (1.06E7)

…subtract the longitudinal retro burn force times the 10 seconds (and times 1,000, don’t forget, kN) from it, to get the new capsule momentum…

10,640,272.8 N - (10.19 N x 10 x 1,000) = 10,538,372.8 N

Divide that by the capsule mass to get the new velocity…

10,538,372.8 N / 1,360 kg = 7,748.8 m/s

For the radial burn component, things are easier. As the craft is moving along the path, we can assume zero momentum relatively, and the force acts only on the mass to change its vertical velocity…

(6,870 N x 10 secs) / 1,360 kg = 50.5 m/s.

And that is the vertical (or “perpendicular to path”) velocity induced by the burn. This was important, by the way. Scott Carpenter’s deorbit was a bit off angle, and resulted in a considerably longer deorbit and reentry path.

Well. Here is the table…


There are a lot more numbers already generated on other tables, but this is all I have had time for tonight. It strikes me, while I have been playing with this, that the whole object, eventually, is to identify just one particular second during the orbits when to fire the retro rockets. That must have been the world’s most expensive second, in terms of effort invested. LOL!

All interesting stuff! :slight_smile:

And a delightful little trick I stumbled upon while looking for the Excel cube root solution…


Yeah, not forgotten, there is tons more to put in here, just I have been short of time to organize the math procedures to make it as coherent as possible for the thread. I have had a lot of fun, with this…

However, there are a couple of things I want to mention.

  1. I do not want to hog @fearlessfrog’s the thread, and would like to put this into another, quiet, out of the way thread called something like “orbital scratch pad”, or the likes.

  2. I have been thinking… I do not want to usurp @fearlessfrog’s flight, if he was doing it, to Pago Pago. The math I have done will work with Orbiter just as well as with Reentry. So that said, if there is any ill feeling about me in my enthusiasm wanting to do it in the Mercury/Atlas, there are plenty of realistic spacecraft available for Orbiter with which I can do it. Just say the word, and I will switch to that. :smiley:

All the best.