How to calculate a rendezvous??

[Der Anfang]

So, exactly how do you catch up with something in orbit? Let's say I am in Earth orbit, 500 km above sea level (semi major axis of 6878.1 km, circular orbit), and I am trying to catch up with a satellite in a 750 km circular orbit. Both I and the satellite are moving prograde, and are coplanar. My position relative to the satellite is 4 o clock.

What do I need to know to solve for this? How do I know how long the rendezvous will take me. Also, how do you know how much delta-v you will need to perform the maneouvre?

Edited October 9, 2015 by Der Anfang

[Bill Phil]

The Dv is pretty much the same for a Hohmann transfer. I'd add a good margin though.

You use the period of the transfer to figure out when you need to burn onto the transfer orbit.

This probably hasn't helped much, sorry.

[YNM]

Mainly Keppler's 3rd Law and Vis-Viva equation. The idea for rendezvous is that you get into an orbit in resonance with the object you want to rendezvous to (like, say you're 8 min ahead, then get to an orbit with a period 8 min longer/slower, or if you're 8 min behind, get into an orbit 8 min shorter/faster). For direct launch -> rendezvous, that'd either require empirical data or some advanced simulation...

[GoSlash27]

Der Anfang,

What YNM said.

You know what the angular velocity is for both bodies in circular orbit and you can easily calculate the transfer using the Vis-viva.

The tricky part is knowing the precise angular separation in KSP.

For problems like that, I use periodic orbits so that the longitude of a body is a function of time, but as a practical matter I'd simply use the maneuver node. Dial in the prograde to match the target apoapsis, then drag it around my orbit until I have an intercept.

Is there some reason that's not an option?

Best,

-Slashy

[Meithan]

It's the same problem as a transfer to another planet; it's all just a matter of phase angles.

The math is not difficult if the orbits are circular, coplanar and prograde. Let's also assume that you're planning to intercept the target using a Hohmann transfer - which is a good idea if you're interested in miniziming delta-v.

Here's is a diagram of the maneuver (which is simply a Hohmann transfer):

You start at point A, your target is somewhere along its orbit, let's say at point B, and since you're using a Hohmann transfer your intercept point is at C. Your initial orbit has radius r₁ = (6378.1 + 500) km = 6878.1 km, while the target's has radius r₂ = (6378.1 + 700) km = 7078.1 km.

The time required by the transfer is simply that of the standard Hohmann transfer:

where μ is the gravitational parameter of the body you're orbiting. For Earth, this is μ = 3.986e5 km³/s² (notice I'm using km for all distances; one must be careful that the units match), so the transfer time is:

We also know that the transfer completes half an ellipse, which means that when you reach the target orbit you will have traveled 180° around, and you'll be on the opposite side of the primary.

Now, the crucial thing for the rendezvous to work is to start the transfer at the right time so that when you reach point C the target is also there. We know that in time t_H you travel 180°, but how much (in terms of angle) does the target travel? Let's call this angle θ. Since the target's orbit is circular, this is easy to compute:

where ̉ۡ_target is the angular speed of the target, which for any circular orbit is given by , with r the radius of the orbit. Substituting and simplifying:

Computing this, we get:

The correct phase angle for the rendezvous, , is then the difference between the angle you travel, 180°, and the angle the target travels:

In this case, we get:

This means that the target must be 3.8 degrees ahead of you when you start the transfer if you are to intercept it at C.

Finally, the delta-v requirement is simply that of the Hohmann transfer:

for the initial burn at A, and

for the circularization burn at C, which in this case is the burn to match speeds with the target. The total delta-v is thus about 100.8 m/s.

If initially the target is not +3.8 degrees ahead of you, then you'll have to wait for the phase angle between you two to reach that value. If you're at 4 o'clock and the target is at 12 o'clock (and moving counter-clockwise), that means your current phase angle is +120° (target 120° ahead). Since you're in a lower orbit, you're moving faster and are catching up, and thus the phase angle is reducing with time. If you wait enough, it'll eventually come down to +3.8°, at which time you should start the transfer.

We can figure out how long that will take by computing the phase angle's rate of change, which is simply the difference between your target's angular speed and yours:

Thus, for the phase angle to reduce from +120° to +3.8°, you'll need to wait about 12.1 hours (which is about 7.5 orbits).

Edited October 10, 2015 by Meithan

[Der Anfang]

It's the same problem as a transfer to another planet; it's all just a matter of phase angles.

The math is not difficult if the orbits are circular, coplanar and prograde. Let's also assume that you're planning to intercept the target using a Hohmann transfer - which is a good idea if you're interested in miniziming delta-v.

Here's is a diagram of the maneuver (which is simply a Hohmann transfer):

http://i.imgur.com/0sCGe6p.png

You start at point A, your target is somewhere along its orbit at point B and since you're using a Hohmann transfer your intercept point is at C. Your initial orbit has radius r1 = (6378.1 + 500) km = 6878.1 km, while the target's has radius r2 = (6378.1 + 700) km = 7078.1 km.

The time required by the transfer is simply that of the standard Hohmann transfer:

tH = pi * sqrt( (r1+r2)^3/(8*mu) )

where mu is the gravitational parameter of the primary celestial body you're orbiting. For Earth, for which mu = 3.986e5 km^3/s^2 (notice I'm using km for all distances; one must be careful that the units match), we get:

tH = 2900.6 s

= 48 m 20.6 s

We also know that the transfer completes half an ellipse, which means that when you reach the target orbit you will have traveled 180° around, and you'll be on the opposite side of the primary.

Now, the crucial part for the rendezvous to work is to start the transfer at the right time so that when you reach point C the target is also there. We know that in time tH you travel 180°, but how much (in terms of angle) does the target travel? Since the orbit is circular, this is easy to compute:

theta_target = omega_target * tH

where omega_target is the angular speed of the target, which for a circular orbit is given by omega = sqrt(mu/r^3), with r the radius of the orbit. Thus:

theta = sqrt(mu/r2^3) * tH

= sqrt(mu/r2^3) * pi * sqrt( (r1+r2)^3/(8*mu) )

= pi * sqrt( (r1+r2)^3/(8*r2^3)

= pi * sqrt( (1+r1/r2)^3 / 8 )

Computing this, we get:

theta = 3.07544 rad

= 176.20 deg

The correct phase angle for the rendezvous, phi, is then the difference between the angle you travel, 180°, and the angle the target travels:

phi = pi - theta

= pi - omega_target*tH

= pi - pi * sqrt( (1+r1/r2)^3 / 8 )

= pi * (1 - sqrt((1+r1/r2)^3 / 8) )

In this case, we get:

phi = 0.066293 rad

= 3.8 deg

This means that the target must be 3.8 degrees ahead of you when you start the transfer if you are to intercept it at C.

Finally, the delta-v requirement is simply that of the Hohmann transfer:

dv1 = sqrt(mu/r1) * (sqrt( 2*r2/(r1+r2) ) - 1)

= 0.054353 km/s

= 54.35 m/s

for the initial burn at A, and

dv2 = sqrt(mu/r2) * (1 - sqrt( 2*r1/(r1+r2) ))

= 0.053965 km/s

= 53.96 m/s

for the circularization burn at C, which in this case is the burn to match speeds with the target. The total delta-v is thus about 100.8 m/s.

If initially the target is not +3.8 degrees ahead of you, then you'll have to wait for the phase angle between you two to reach that value. If you're at 4 o'clock and the target is at 12 o'clock (and moving counter-clockwise), that means your current phase angle is +120° (target 120° ahead). Since you're in a lower orbit, you're moving faster and are catching up, and thus the phase angle is reducing with time. If you wait enough, it'll come down to +3.8° at which point you should start the transfer.

We can figure out how long that will take by computing the phase angle's rate of change, which is simply the difference between your target's angular speed and yours:

d phi / dt = omega_target - omega_you

= sqrt(mu/r2^3) - sqrt(mu/r1^3)

= -4.6578e-05 rad/s

= -9.6073 deg/h

Thus, for the phase angle to reduce from +120° to +3.8°, you'll need to wait about 12.1 hours (which is about 7.5 orbits).

Very helpful, I think. But what is tH? Also, can we use metres in this equations or does it HAVE to be in km? Also, is rad "radians?"

[Meithan]

Very helpful, I think. But what is tH?

The time the Hohmann transfer takes (going from point A to point C). I should have written it with a subscript H, like this: t_H, but I was a bit lazy .

Also, can we use metres in this equations or does it HAVE to be in km?

You can use any unit of length you want, as long as you're consistent. If you use metres for the radii r₁ and r₂, then you also need to write all other quantities that involve length in metres. For instance, you should write μ in m³/s².

Also, is rad "radians?"

Yup, sorry for not clarifying. When you compute angle stuff, you usually get the answers in radians. To convert radians to degrees, you multiply by 180/À.

[Der Anfang]

The time the Hohmann transfer takes (going from point A to point C). I should have written it with a subscript H, like this: t_H, but I was a bit lazy .

You can use any unit of length you want, as long as you're consistent. If you use metres for the radii r₁ and r₂, then you also need to write all other quantities that involve length in metres. For instance, you should write μ in m³/s².

Yup, sorry for not clarifying. When you compute angle stuff, you usually get the answers in radians. To convert radians to degrees, you multiply by 180/À.

Okay, thank you. So, that makes sense and the equations are doable for me. However, what changes if we add a little bit of eccentricity to our target?

[PB666]

gee, a couple of simplifyers

1. period is a function of semi major axis, if you do a transfer between two ellipses whose pe and ap are approximately the same theta from a reference line, then you will need exactly a half period of the transfer orbit to transfer. In general ap will be close to targets alt, if you start transfering close to either its ap or pe theta then you will rendevous at its pe or apo, respectively.

2a = pe + ap + cb diameter. cb = central body, but in a binary system wher you are outside the binaries orbit, this is the systems center of mass. pe and ap are variables in the transfer orbit.

https://upload.wikimedia.org/math/5/0/f/50f9cad0c149bd3fe30446a17b834884.png

2. Since you know the time it takes for ship one to reach the rendevous point then all you have to know

is keplers sweep law, an object in orbit sweeps the same area per unit of time. Since you have the time its simply a matter of taking the area of the target orbit (pi a finding its period, this tells the area sweep rate per time.

3. Almost there, multiply the transfer time by sweep-area/time and you have target's sweep area during the transfer.

- shortcut take the average radius within the swept area and calculate dTheta. Works great if the target in a circular orbit. r= alt + r[cb]. dTheta = swept area in transfer period/ r

-longcut create a integral starting at the rendevous and working backwards in reasonable angles until the integral consumes the area. works better for intercepting eccentric orbits

4. This then gives you the angular distance (dtheta) and time that thevtarget needs to be from from rendevous to coincide with the transfer ship when it arrives. if we recall that one takes a half period then its 180 minus dTheta, this means that traget ship leads transfer ship by theta degrees if in higher orbit or lags the transfer ship by -theta degrees if in a lower orbit.

This is not perfect but its generally good enough, minor adjustments in transfer ships path can be made when 3pi/4 radians (focal centered; 135') have been crossed. Such adjustments generally have to be made anyway unless the burn is perfect and all attitude controls are silenced. It works in KSP, null and void if you work for NASA, ;^}.

Edited October 10, 2015 by PB666

[Meithan]

Okay, thank you. So, that makes sense and the equations are doable for me. However, what changes if we add a little bit of eccentricity to our target?

Ah, eccentricity (of either orbit) makes things considerably more complicated, since, for instance, angular speed is no longer constant (in an elliptical orbit, you move slower near apoapsis than near periapsis). The overall logic would still be the same, though: you do a Hohmann transfer to "touch" the target orbit at some point, and you time the start of the transfer so that the target is there when you reach apoapsis of your transfer half-orbit.

Btw, I just changed all the equations in the original post for nicely typeset ones (thanks to codecogs!), so everything is easier to read now. Whew. This forum really needs LaTeX support.

[FancyMouse]

And this is the math behind arbitrary eccentricity and inclination - and this is the math behind AlexMoon's page/TransferWindowPlanner.

[Meithan]

And this is the math behind arbitrary eccentricity and inclination - and this is the math behind AlexMoon's page/TransferWindowPlanner.

Quite right. Lambert's Problem is the more general version of the rendezvous problem: if I'm at arbitrary position r₁ (in 3D) at some time t₁, what trajectory gets me to some other position r₂ exactly at time t₂?

[InsaneDruid]

CheersKevin has some nice videos where he writes various KOS scripts that cope with this problem:

The video #10 - Space Rescue! is the one about the rendezvous, but some of the scripts used (like to calculate orbit times etc) are developed in previous videos. IIRC he solves it like Meithan described vith the variation that he first burns for a transfer orbit, THEN burns for a higher "wait/sync" orbit with a period of <period of target>+phase time. So a bit less elegant and less efficient but this works for launch to rendezvous without waiting on the ground and you are guaranteed to have a sync orbit outside the atmo (for its always has higher APo than the target ones), making it a nice KOS base script.

[lajoswinkler]

So, exactly how do you catch up with something in orbit? Let's say I am in Earth orbit, 500 km above sea level (semi major axis of 6878.1 km, circular orbit), and I am trying to catch up with a satellite in a 750 km circular orbit. Both I and the satellite are moving prograde, and are coplanar. My position relative to the satellite is 4 o clock.

What do I need to know to solve for this? How do I know how long the rendezvous will take me. Also, how do you know how much delta-v you will need to perform the maneouvre?

You need a radio contact with a bunch of slide rule monkeys in the mission control.

[Der Anfang]

Quite right. Lambert's Problem is the more general version of the rendezvous problem: if I'm at arbitrary position r₁ (in 3D) at some time t₁, what trajectory gets me to some other position r₂ exactly at time t₂?

So, what exactly is the math you use to rendezvous with something on it's way out of the solar system, but it's coming close to the Earth? I'm going to assume the math is similar. Is the angular speed (rads/s) always constant or no?

I'm sorry, I am still fairly new to this stuff, so trying to understand it all. I managed to figure out and understand(physically worked it all on paper) rendezvous for a circular orbit scenario, though!

Also, thank you again so much for your help. You are a godsend, lol.

[PB666]

CheersKevin has some nice videos where he writes various KOS scripts that cope with this problem:

The video #10 - Space Rescue! is the one about the rendezvous, but some of the scripts used (like to calculate orbit times etc) are developed in previous videos. IIRC he solves it like Meithan described vith the variation that he first burns for a transfer orbit, THEN burns for a higher "wait/sync" orbit with a period of

+phase time. So a bit less elegant and less efficient but this works for launch to rendezvous without waiting on the ground and you are guaranteed to have a sync orbit outside the atmo (for its always has higher APo than the target ones), making it a nice KOS base script.

There are less math solutions to the problem. Given: radius central body = 600k, a rescue target at 100k alt on equitorial orbit (90'), atm psuedo limit at 70k, launch site on equator, surface gravity = 9.8

Method 1. Direct intercept. Also known as "watch your target zip by on close approach as you are struggling to get your 1g*mass thruster to throw out 1000 dV in desired 2 seconds."

0. Nu is the sqrt of (mu). Nu is our cheat, I keep it close to my heart. mu = 3.5316E12 (obtain by multiplying accel(=9.81)*r(=0.6M)^2: Nu = sqrt(9.81) * r(=0.6M) = 1.88M

1. Calculate target orbit

a. nu/sqrt® = velocity = 2246 m/s. The circumferance is what pi*rad(=700k)*2 = 4398229m. The period circ/vel = 1959 seconds.

b. angular velocity(omega) = 2*pi()/period = 0.0033 radians/sec - done

2. Emperically a vessel takes 5 minutes to reach near orbital velocity. During the first minute of flight its horizontal velocity is near zero, but climbs from near zero to orbital over 4 minutes.

a. thus it is traveling about 2/5ths orbital velocity (discounting ground movement) for 5 minutes, and orbital velocity thereafter.

b. this means if a target is directly overhead it need to warp 3 minutes forward to catch up after reaching target orbit. But since we know angular velocity we can calculate what? lets see

c. 3min(=60sec)*0.0032 = 0.57 radians = 33 degrees.

3. So if you know the longditudinal coordinates of the launch site (in mech jeb this will be given as angle to prograde) and you know the angle to prograde of the target, just wait until the launch site preceeds the anlge to prograde of the target by 33, launch and set the target altitude at 100. The only trick here is to make the orbit realtively flat over 42k. With reference to the targets humor, if you are not experienced in making your gravity turn you might have to thrust in the -radial direction to get orbital velocity were it will intercept. If you are not, then allow your self a few extra degrees. When you reach the target your horizontal velocity relative to the orbital velocity (target velocity also)/2*accel(=thrustMax/mass) is the lead time you need over the target at your apo. If you are one of those who likes to do most of the circularization at target altitude, then you need to factor this time into your lead over the target at launch. For example, if you are 1000 m/s slower and your thrust is 1 g, then it takes 100 seconds, you will need half this time. so 50 seconds = 0.16rad (=10 degrees).

Oh, and how much dV do you need, its the same as orbital insertion from launch. IOW practice making orbit at 75k and then use that method and extend your apo from 75k to 100k at an altitude of 60k or so.

In any case if you know your insertion methods (wipe that grin off your face) and you know how to guestimate targets position relative to launch site even goofing this method and sill having a 100k equitorial then you are going to get much closer to your target than you might imagine, 5k is ideal, more than that and you might have to increase or decrease your orbit a tad to close the gap.

Method 2.

Overkill phasing. This method is less risky than the first, but really overkills on fuel

1. So we know what our Nu is, is that cute, so how do we get an orbit going half the velocity. r (=600K+alt) needs to be 4 fold. So to get an orbit half as fast we need an orbit 4 times the radius or 2800K, that means we need an alt of 2200K. But then our target is 1/4th the radius, so its radians per second are v/r and ours is (v/2)/4r = 1/8th its making 8 oribt for our 1 and basically we have a window every 37 minutes. Thats not to clever, but on the bright side we are long 400km from a KSO.

2. So we are going to have to close the gap, but do we actually need to think about this too much just match the planes. Actually this does not need much math either. Just burn retro until the orbit is nearly the targets orbit alt or even the targets orbit. It is recommended that you do this at your targets theta(=apo) this way your intercept is at the targets pa. Now we might get lucky and get close to the target, but actually chances are no and don't waste the effort trying. When the ship hits apo, just retro, as you retro your ships ellipse will intercept the target maybe six times as you close your orbit, you are interested in the smallest circle. Each of these you can calculate with relative ease if you know the period. Mech jeb tells you the periods of your orbit and your target will give its period. so basically you can use this to estimate the intercept if you know the angle to prograde.

Method 3

Smart phasing. So basically we have done the direct, and we have overkilled on the phasing. How about a better phasing. Once again target is at 100 k

1. This time you want to be about 90 degrees ahead of your target.

2. Launch to about 125 to 150 km alt.

3. Once at target apo, correct the pe to 100 km alt, (125 alt apo = ~ 0.0031 rad/sec, 150 alt apo = ~0.00305 rad per second)

4. At pe squeeze the apo in just as in method 2

5. time the intercept.

Note this can be done at any launch position, but who actually wants to wait

So the question is what makes method 1 and 3 work. The short answer is we are not lingering around waiting for a transfer window, in method 1 we are essentially allowing our non-inertial reference frame (=surface of Kerbin see Einstien's theory of relativity) to phase us with a target in its inertial reference frame. We only need to wait 32 minutes at most to launch, and we are trying to get within a 5 km window where our docking manuevers are more important than transfer dynamics.

In method 3 we know a higher orbit has a slower angular velocity, so we allow ourselves to get about 60 degrees ahead of the target, and then use that lead to close our orbit with the target successively such that when we are ready to make the final closure we are on top of the targets. In method three we retro at pe which is targets alt, so no fuel is wasted in making multiple retros at the same kick position. The better you are at approximating your launch with target intercept, the lower altitude that is convienient (avoiding wasted time in orbit).

And the lesson we learn is that if we had to rescue a naut in crisis its almost always best to plan a launch window that will intercept a target in a more or less direct manner. Simple math on the ground is alot cheaper on the brain than complex math in space.

[Meithan]

So, what exactly is the math you use to rendezvous with something on it's way out of the solar system, but it's coming close to the Earth? I'm going to assume the math is similar. Is the angular speed (rads/s) always constant or no?

It requires solving the Lambert problem mentioned above. In general, unless something is in a circular orbit its angular speed is not constant and figuring out its position along its orbit as a function of time ("where will it be at some specific time t?") is complicated. Circular orbits are the easy case.

Just so you get an idea of how complicated it gets, here's one approach to solving Lambert's problem, with an Earth-Venus transfer as worked example: Lambert’s Problem - Prof. Jeffrey S. Parker - UC Boulder

Edited October 10, 2015 by Meithan

[PB666]

yes but you know its velocity at pe and ap. And if it were a circle at pe (an oxymoron), then you know how much velocity you need to transfer, you just dont know when the target will coincide. Since you know you speed at pe or ap and you know how much would be required if it was a circle then you how much more or less you would need to transfer. The problem is the target, but we can fix this.

The most efficient places to begin the transfer out is at pe and to transfer in is at apo. So we can use these to our advantage, infact kicking ion drives at pe essentially does this. So the planning is in kicking out enough that at the next cycle through pe, if we kick so that our apo ~ crosses the target orbit at coincidence the target. The kicking out also allows one to match planes at the optimal lowest velocity. This startegy is advantageous under almost all circumstances except if time is a constraint. The 67p mission is essentially doing this because they lacked a dV budget to reach their target, they basically are using various oberth boosts to get to their target.

To do this you have to know exactly when the target passes the orbiters theta at apo. this is simple becuase of the sweep rule, you ca determine the time to that theta, and you know the period thus its the series t(toTheta) + period*n(=0,1,2,3; easy to set up in excel, or the free linux equiv). Then, how much time to pe you can calculate and subtract from each item in the series. Then its a matter of fideling with 2 periods one generated by the intermediate apo and on by the final apo to get a two orbit cumulative that coincides at the second apo with any element in that series target. its usually going to be n= 0 or 1 and the orbiter may need additional orbits to hit 0.

The problem with this stategy, the only one that is functional and not subjective (i.e having to do with mission needs such as food or battery life) is that when the two orbits apo and pe do not coincide, there are two inefficiencies, one is non optimal dV at vector matching, this is most pronounced when the targets apo is at the orbiters pe, such as in making a transit to moho starting the mission with a kerbin theta at moho's pe. In this case the orbiter may want to transfer his pe into an apo and would plan for the

some plane matching to be approximate to this point, if possible. The other problem is that some radial dVs will be required if Pe or apo are not close in angle.

Edited October 10, 2015 by PB666

[Der Anfang]

It requires solving the Lambert problem mentioned above. In general, unless something is in a circular orbit its angular speed is not constant and figuring out its position along its orbit as a function of time ("where will it be at some specific time t?") is complicated. Circular orbits are the easy case.

Just so you get an idea of how complicated it gets, here's one approach to solving Lambert's problem, with an Earth-Venus transfer as worked example: Lambert’s Problem - Prof. Jeffrey S. Parker - UC Boulder

Phew! That looks like a mess! Lol, I'll have to spend some time on that one. Thank you for the reference, though. That is exactly what I was looking for! I am almost done here; if you don't mind, I'd like to burden you with one more question.

So, now that I know how to calculate the intercept angle and such as well as the time and delta-v required... let's say I wanted to intercept my target in only half of the time of the optimal intrcept. Let's say we have enough delta-v and don't really care about efficiency. How do you calculate an angle of intecept based on the desired travel time? I hope that question makes sense. Thank you in advance!

[Meithan]

Phew! That looks like a mess! Lol, I'll have to spend some time on that one. Thank you for the reference, though. That is exactly what I was looking for! I am almost done here; if you don't mind, I'd like to burden you with one more question.

So, now that I know how to calculate the intercept angle and such as well as the time and delta-v required... let's say I wanted to intercept my target in only half of the time of the optimal intrcept. Let's say we have enough delta-v and don't really care about efficiency. How do you calculate an angle of intecept based on the desired travel time? I hope that question makes sense. Thank you in advance!

The question makes total sense. If you're willing to spend more delta-v than a Hohmann transfer, you can get a faster intercept by doing a one-tangent burn transfer. The basic idea is that the transfer orbit is a higher-energy orbit (which can be elliptical or even hyperbolic) that crosses your target's orbit instead of touching it tangentially (as the Hohmann transfer does):

At intercept, you'll be moving considerably faster relative to your target (and, more importantly, along a different direction), so the speed matching burn will be costly. Robert Braeunig's excellent website (from which I stole the above picture) has the mathematical formulae needed for this kind of transfer.

In particular, look at the Orbit Maneuvers section of his Orbital Mechanics page. Due to the way these expressions are written, you'd have to first find the size of the transfer orbit (i.e. its semimajor axis, a_tx in Robert's notation) by trial and error, repeatedly choosing a value of a_tx and computing the corresponding time of flight (equation 4.71) until you get the desired transfer time.

Then, you can use the other equations to compute the delta-v of both transfer burns as well as the angular position of the intercept (the "true anomaly at second burn"). The phase angle for the start of the transfer can then be deduced from this quantity and the time of flight (in a way similar to what we did for the Hohmann transfer).

Edited October 10, 2015 by Meithan

[Rune]

Hey, I have a much, much easier KSP-related solution: throw a maneuver node and see what happens.

Seriously, I do this several times a day. These days, I don't even bother waiting for a target to get to the "if I launch now, I'll get close" position, too much variability on account of TWR, and honestly who cares if your mission takes an ingame day instead of two hours. It's a really easy thing to do, in four easy steps:

1-Get into a coplanar orbit with the target, either over or under it, but consistently so.

2-Drop a maneuver node that gets your orbit to intercept the target's, pretty much anywhere. If you are in a higher orbit, this a retrograde burn to bring your Pe to the target's altitude, otherwise it is a prograde vector to bring you apoapsis to it. Note that doing this from a lower orbit is slightly more efficient, if you are coming form the ground. A "close approach" will show up when you do this, don't worry what it shows, just that it shows a single encounter at the right height, ideally over an ascending/descending node if you were not 100% accurate matching inclination.

3-Right-click on the node, and play with doing the node one or more orbits later. You will notice this changes the separation at the close approach, the more orbits you let pass before doing it. If the separation is going down, continue to add orbits until just before you catch up to it. If the separation is going up... do the same thing, only it will take longer because you have to go the long way around. This encounter won't be anything acceptable, but it will put you in the position of getting one that is, on the next orbit after it.

4-Once you have actually done that, drop another node just after the close approach. It will start showing the next close approach after that one, and placing a maneuver that brings your orbit closer to the targets will get you a perfect encounter on the next orbit, 100% of the time, guaranteed. Just match orbits, but not quite, until the new encounter distance reads under 1km.

This works because if you are in an orbit with a different semi-major axis, your orbital period is different (you take different time to complete a revolution), so you will "gain" (lower orbit) or "lose" ground (higher orbit) with respect to your target on each revolution. Since how much you "win" or "lose" depends only on the difference of your periods, when you make your orbit more like the target's, that "gain" can be tailored to any amount you might want. One and a half times the orbital period in one of them? You will gain or lose half an orbit each revolution. 1/100th difference? You will gain or lose a few degrees (specifically, 360º/100=3.6º) on each orbit. So just get things close while maintaining an intersection point (the close approach, never maneuver so it changes your altitude at that point), and you can adjust things on the last-before-overshooting pass so the next one is just the right difference.

Oh, and done this way, dV is the bare minimum to get this done, BTW. After all, you are doing something that very closely matches an ideal Hohmann, only you break the final circularization burn into two in order to fine-tune the timing on the last orbit. And when you get good at it, you can disregard inclinations and just make the stuff happen at the Ascending/Descending nodes, leaving the inclination change for the last velocity-matching burn (that you should do, incidentally, when you are frighteningly close to your close approach, by pointing retrograde with the velocity indicator on "target" mode).

Rune. Basically, if I can't do the math in my had playing KSP, I don't. I still manage gravity assists and the like, just because I know how the math works, and the game has awesome predicting software incorporated.

Edited October 10, 2015 by Rune

[Der Anfang]

Hey, I have a much, much easier KSP-related solution: throw a maneuver node and see what happens.

Seriously, I do this several times a day. These days, I don't even bother waiting for a target to get to the "if I launch now, I'll get close" position, too much variability on account of TWR, and honestly who cares if your mission takes an ingame day instead of two hours. It's a really easy thing to do, in four easy steps:

1-Get into a coplanar orbit with the target, either over or under it, but consistently so.

2-Drop a maneuver node that gets your orbit to intercept the target's, pretty much anywhere. If you are in a higher orbit, this a retrograde burn to bring your Pe to the target's altitude, otherwise it is a prograde vector to bring you apoapsis to it. Note that doing this from a lower orbit is slightly more efficient, if you are coming form the ground. A "close approach" will show up when you do this, don't worry what it shows, just that it shows a single encounter at the right height, ideally over an ascending/descending node if you were not 100% accurate matching inclination.

3-Right-click on the node, and play with doing the node one or more orbits later. You will notice this changes the separation at the close approach, the more orbits you let pass before doing it. If the separation is going down, continue to add orbits until just before you catch up to it. If the separation is going up... do the same thing, only it will take longer because you have to go the long way around. This encounter won't be anything acceptable, but it will put you in the position of getting one that is, on the next orbit after it.

4-Once you have actually done that, drop another node just after the close approach. It will start showing the next close approach after that one, and placing a maneuver that brings your orbit closer to the targets will get you a perfect encounter on the next orbit, 100% of the time, guaranteed. Just match orbits, but not quite, until the new encounter distance reads under 1km.

This works because if you are in an orbit with a different semi-major axis, your orbital period is different (you take different time to complete a revolution), so you will "gain" (lower orbit) or "lose" ground (higher orbit) with respect to your target on each revolution. Since how much you "win" or "lose" depends only on the difference of your periods, when you make your orbit more like the target's, that "gain" can be tailored to any amount you might want. One and a half times the orbital period in one of them? You will gain or lose half an orbit each revolution. 1/100th difference? You will gain or lose a few degrees (specifically, 360º/100=3.6º) on each orbit. So just get things close while maintaining an intersection point (the close approach, never maneuver so it changes your altitude at that point), and you can adjust things on the last-before-overshooting pass so the next one is just the right difference.

Oh, and done this way, dV is the bare minimum to get this done, BTW. After all, you are doing something that very closely matches an ideal Hohmann, only you break the final circularization burn into two in order to fine-tune the timing on the last orbit. And when you get good at it, you can disregard inclinations and just make the stuff happen at the Ascending/Descending nodes, leaving the inclination change for the last velocity-matching burn (that you should do, incidentally, when you are frighteningly close to your close approach, by pointing retrograde with the velocity indicator on "target" mode).

Rune. Basically, if I can't do the math in my had playing KSP, I don't. I still manage gravity assists and the like, just because I know how the math works, and the game has awesome predicting software incorporated.

Lol, yes. I love KSP for that type of convenience, and I use whatever I can in-game, but because I am writing a sci-fi space exploration story, I'd like to know some of the actual math behind these things, since ksp isn't going to tell me how those things are actually calculated.

[Meithan]

Hey, I have a much, much easier KSP-related solution: throw a maneuver node and see what happens.

[...]

Rune. Basically, if I can't do the math in my had playing KSP, I don't. I still manage gravity assists and the like, just because I know how the math works, and the game has awesome predicting software incorporated.

I agree with you Rune: in practice I don't do the calculations either, I just throw a maneuver node and play with it too. As you said, what's important is understanding the fundamental principles behind it (the orbital mechanics).

For rendezvous, I actually don't use a lower or higher phasing orbit. What I do is first match my target's orbit (close to zero relative inclination and close-ish Ap/Pe), trying to get as close to my target but behind it. Then, I do an elliptical phasing orbit: I raise my Ap a bit so that my orbital period increases. By toying with the maneuver node I can get the extra period to be exactly the time I'm lagging behind the target. The result is a very close intercept on the next orbit once I get back to Pe. The relative speed will also be quite low (a few 10's of m/s usually), so the transfer is close to optimal.

[PB666]

Hey, I have a much, much easier KSP-related solution: throw a maneuver node and see what happens.

Seriously, I do this several times a day. These days, I don't even bother waiting for a target to get to the "if I launch now, I'll get close" position, too much variability on account of TWR, and honestly who cares if your mission takes an ingame day instead of two hours. It's a really easy thing to do, in four easy steps:

I didn't plan launch windows either until career mode hard. Two things that are costly in career mode

1. Not double checking you sat. equipment required

2. Misplanned launch windows

Now its true you can get a sat up with extra fuel to correct, but you can do other things with those sats.

Launch windows, ok, so why

The target orbit is polar. Its silly not to use launch windows. The alternative is your orbit coukd be oof by 180 degrees and requires the user to create a highly elliptical orbit which is the matched at apo and shrunk, cost is around 3500 dV.

Minmos flights. Save a few dV on the plane matching.

That sat you saved fuel with you can put in Kerbol orbit near Kerbin. By using the flight planner you can roughly test to see when the best launch times. So when is this important moho! Its not a day, it could be years of capital wasted to get a good dV. The best transfer is when you launch window is closest to mohos apo and the transfer orbit pe coincides with moho pe. Obviously this is hard to do, but moho travels around 5 times per each kerbin orbit so at worst you are going 36' from apo. So in this instance you may have to wait half a year, and waiting a up to a year and a half you could get within 18 degrees. Once you get the launch window you can plan a launch which optimizes plane mtaching. So this is an intance where planning luanch windows can save 1000s of dV.

Edited October 11, 2015 by PB666
fix an annoying typo

[Rune]

Lol, yes. I love KSP for that type of convenience, and I use whatever I can in-game, but because I am writing a sci-fi space exploration story, I'd like to know some of the actual math behind these things, since ksp isn't going to tell me how those things are actually calculated.

Oh, ok, you want the nitty-gritty detail. Then, I refer to the mathematical formulae that are already posted here by all these nice folks, with a caveat: once your orbit touches the target's, preferably at a descending/ascending node, it's all a matter of getting the orbital periods to phase the right way, and orbital period is directly proportional to your semi-major axis (semi-major axis=(Ap+Pe)/2). With that in mind, the math makes much more sense, and the problem isn't even hard to solve. Say your target is X degrees in front of you, then you want an orbit with a period X/360 times lower than your target's. So, for example, for a target 90º before you, you want your orbital period to be 90º/360º=3/4ths of the target's (or in other words, a semimajor axis 3/4ths of the target's), so you gain 1/4 of an orbit, or 90º, on each revolution. Or, you know, gain 1/8th on each orbit if you have an orbital period 7/8th of your target's, and you'll get there in two orbits with less dV expenditure. Or 1/16th, to get there in four, or 1/12th to get there in three. Or you could of course go higher, say 9/8ths of its period, and then you will get there the long way around in... 270º/45º=6 orbits.

See how that makes a lot of intuitive sense, if you look at it form an orbital period perspective?

I agree with you Rune: in practice I don't do the calculations either, I just throw a maneuver node and play with it too. As you said, what's important is understanding the fundamental principles behind it (the orbital mechanics).

For rendezvous, I actually don't use a lower or higher phasing orbit. What I do is first match my target's orbit (close to zero relative inclination and close-ish Ap/Pe), trying to get as close to my target but behind it. Then, I do an elliptical phasing orbit: I raise my Ap a bit so that my orbital period increases. By toying with the maneuver node I can get the extra period to be exactly the time I'm lagging behind the target. The result is a very close intercept on the next orbit once I get back to Pe. The relative speed will also be quite low (a few 10's of m/s usually), so the transfer is close to optimal.

Myself, I launch into temp. ~70x70 orbit, then see where the target is at, relative to me. If it's far away, I run the clock for a while until I'm closer. Once I'm less than a quarter orbit away, I raise my Ap to the target's orbit in any one node, and go from there, always approaching from behind so I don't waste dV going higher than my target. An if I'm short on dV or want to be extra-efficient I try to set a very tiny burn and explore what will happen a lot of orbits from now by dropping a second node, and then moving that forward in time to see my subsequent close approaches. I feel so professional when I set up a rendezvous something like eight orbits form now... nice trick to know!

I didn't plan launch windows either until career mode hard. Too things that are costly in career mode

1. Not double checking you sat. equipment required

2. Misplanned launch windows

Now its true you can get a sat up with extra fuel to correct, but you can do other things with those sats.

Launch windows, ok, so why

The target orbit is polar. Its silly not to use launch windows. The alternative is your orbit coukd be oof by 180 degrees and requires the user to create a highly elliptical orbit which is the matched at apo and shrunk, cost is around 3500 dV.

Minmos flights. Save a few dV on the plane matching.

That sat you saved fuel with you can put in Kerbol orbit near Kerbin. By using the flight planner you can roughly test to see when the best launch times. So when is this important moho! Its not a day, it could be years of capital wasted to get a good dV. The best transfer is when you launch window is closest to mohos apo and the transfer orbit pe coincides with moho pe. Obviously this is hard to do, but moho travels around 5 times per each kerbin orbit so at worst you are going 36' from apo. So in this instance you may have to wait half a year, and waiting a up to a year and a half you could get within 18 degrees. Once you get the launch window you can plan a launch which optimizes plane mtaching. So this is an intance where planning luanch windows can save 1000s of dV.

Oh, this is for launching stuff to rendezvous around the same planet you launched from. Interplanetary transfers are another thing entirely, and then it really pays of to figure out the right time and angle to eject. I usually just find out the window with KAC, and experiment until I get the ejection angle right.

Rune. Rocket science is actually a very easy science, at least as far as the maths go. Either they equations are unsolvable (n-body problem), or they are dirt easy.