Launch into angle of target, orbit calculations

[Clear Air Turbulence]

Hi there

I have been looking through the tutorial threads without finding a clear answer to these questions

I am in sandbox, need to rescue 2 Kerbals adrift in orbit around K

(1) How do I launch into angle of target without Mechjeb? I find Mechjeb makes my rocket tumble end over end. I can get into orbit without it OK (and without somersaults) but I am always 6-7 degrees (and therefore an expensive corrective burn) out of alignment with my target

(2) How do I calculate the Delta-V needed for getting into various orbits? I see there is a simple orbital calculator mod, but I can't get it to work & there's no answer on the thread in question (http://forum.kerbalspaceprogram.com/threads/115990-1-0-4-SOC-Simple-Orbit-Calculator-v1-4-0-(6-23-2015)

thanks!

[mhoram]

1) If the target inclination is not 0°, then it gets tricky.

As an alternative you can adjust your inclination after reaching orbit to match the one of your target. Adjusting a 6-7° inclination difference does not take that much dV.

2) For the transfer between two orbits you usually use a Hohmann-Transfer consisting of two burns.

The dV needed consists of the following

- velocity difference between your current orbit and the transfer orbit at the location of the first burn

- inclination change

- velocity difference between the transfer orbit and the target orbit at the location of the second burn

Calculating the velocities of your ship in the three orbits is done via the Vis-Viva equation.

For calculating the dV for the inclination change you can find the relevant formulas here: http://forum.kerbalspaceprogram.com/threads/69036

[nekogod]

FYI For larger inclination changes it can be cheaper delta V wise to go into a higher slower orbit change inclination then drop back into the lower orbit.

[SpeedDaemon]

You can pretty easily launch into an inclined orbit for an intercept. It's hard to get it right on the money, but getting close from the very beginning means less dV expended later.

- Time your launch so that you're launching at the ascending or descending node (when KSC is at the point where your target's orbit crosses the equator).

- Instead of going straight east (90 deg heading), offset it by your target's inclination (if your target's orbit is 15 deg, and you launch at the ascending node, your heading should be 15 deg north of east (75 deg))

- Once you launch and can see the AN/DN markers, you can tweak yourself north or south to get closer to the right orbit.

- Once you make orbit, you can make finer adjustments.

Unfortunately, I don't know that you can get the exact inclination # anywhere for those rescue missions, so you kind of have to eyeball it and hope for the best.

This also applies for satellite contracts (which you can get the exact inclination for).

[KSK]

What SpeedDaemon said. Although a 6-7 degree inclination change is about what you need to align with Minmus from an equatorial orbit and if I remember rightly that only requires about 250m/s of delta-V. Not too crazy.

Although, if you want to try launching into the right orbit then go for it! I don't do it very often but I always feel BadS when I do.

[SpeedDaemon]

What SpeedDaemon said. Although a 6-7 degree inclination change is about what you need to align with Minmus from an equatorial orbit and if I remember rightly that only requires about 250m/s of delta-V. Not too crazy.

Aligning with Minmus should generally be done outside of the Mun's orbit (about halfway out to Minmus, and ignoring the AN/DN), if you're starting with an equatorial LKO orbit, which limits dV to mostly negligible amounts. My strategy for Minmus is usually to get into LKO, burn prograde when Minmus is on the horizon until my AP is about 44Mm, then throw down a node halfway out to adjust for the intercept (usually no more than ~5-20 m/s).

[OhioBob]

Clear Air Turbulence said:

(1) How do I launch into angle of target without Mechjeb? I find Mechjeb makes my rocket tumble end over end. I can get into orbit without it OK (and without somersaults) but I am always 6-7 degrees (and therefore an expensive corrective burn) out of alignment with my target

When you say "angle of target" I assume you mean inclination. Inclination is the angle between Kerbin's equatorial plane and the plane of the orbit. What you need to do is go to the map view and watch for the moment when the launch site rotates into the orbital plane of the target. This happens twice per day. I do this by (1) first clicking on Kerbin to center it in the map view, (2) moving the map view around until I see Mun's orbit edge on (Mun's orbit is the same as Kerbin's equatorial plane), and (3) rotating the view left or right until I see the orbit of the target edge on. Once you see both orbits edge on, they appear as two intersecting lines. You want to warp ahead until the launch site moves to a point where it lies along the line of intersection of the two planes. There are only two places on opposite sides of Kerbin where this happens. Once you're at one of those two points, you want to launch in either a slightly northerly or southerly direction so that you insert your spacecraft into the same orbital plane as the target. By looking at the map view you can see whether you need to launch north or south. If the target's orbit crosses the launch site moving north to south, then you want to launch in a southerly direction, and if the opposite is true, then you launch in a northerly direction. How much north or south depends on the inclination of the target. The angle of your launch vehicle's flight path in relation to the equator should be roughly equal to the inclination of the target. For instance, if the target's inclination is 10 degrees, then you want to rotate your launcher as soon as it leaves the launch pad 10 degrees, either north or south as required, and then pitch over (actually yaw over) so that you're flying at an angle of 10 degrees in relation to the equator. This should put you into an orbital plane that matches that of the target. You'll probably still have to make a small plane change adjustment, but with practice you can get very close.

I said that your rocket's flight path "should be roughly equal to the inclination of the target" because you actually have to account for the fact that you have an initial due east velocity of about 175 m/s due to Kerbin's rotation. You therefore have to over compensate a bit and fly at an actual heading a little greater than the target's inclination. This correction is small for low inclinations and can be ignored in most cases. However, for high inclinations, such as a polar orbit, the correction can be several degrees. For example, to launch into an orbit the flies directly over the north pole, it is necessary to launch along a heading that is 4 or 5 degrees west of north.

Clear Air Turbulence said:

(2) How do I calculate the Delta-V needed for getting into various orbits?

The delta-v depends a lot of the design of your launch vehicle and payload, so there is no simple answer. However, if you just want to know how much extra delta-v it takes to launch into an orbit other than a non-inclined prograde orbit, that's easy. A zero inclination prograde orbit is what you get when you launch due east from Kerbal Space Center. You get a free ride of about 175 m/s when you do this because of Kerbin's rotation. If you launch due north or due south, you don't get any free ride from the planet's rotation, so you have to make up that missing 175 m/s with your launch vehicle. Thus the delta-v required to reach orbit goes up 175 m/s. If you launch due west, you are now fighting Kerbin's rotation. Not only have you lost the 175 m/s free ride, but you are now moving 175 m/s in the wrong direction. The delta-v required to reach orbit is now 350 m/s greater than a due east launch. Computing the free ride that you get from Kerbin's rotation involves vector mathematics, but it is approximately equal to 175*sin(A), where A is the azimuth of your flight path, measured eastward from north.

Edited April 22, 2019 by OhioBob

[Clear Air Turbulence]

Thanks, that's very useful. Much appreciated!

[bigcalm]

You're welcome to use my google docs spreadsheet, though the landing / take-off numbers should be taken with a large pinch of salt as they really need calculus to do properly.

[Sacs]

If you have Kerbal Alarm Clock, while sitting at the KSC waiting to launch, you can set your target to the ship you need to rescue, then set an alarm for the next An/Dn. The alarm will go off as the targets orbit is passing over the KSC then you can launch into the target inclination as others have mentioned.

[OhioBob]

I just derived an equation to allow for a more accurate determination of the heading one must fly to insert into a target inclination. The equation is,

Heading = atan[ (V_orb*sin(i)) / (V_orb*cos(i)-174.6) ]

where V_orb is the orbital velocity, i is the orbital inclination, and heading is measured from the equator.

For example, let's say you want to insert into a 75 km orbit with an inclination of 10 degrees. For a 75 km orbit, V_orb = 2287.4 m/s, and, of course, i = 10^o. We therefore have,

Heading = atan[ (2287.4*sin(10)) / (2287.4*cos(10)-174.6) ] = 10.82^o

If we are launching at the target orbit's ascending node, then the heading is northerly; and if we are launching at the target orbit's descending node, then the heading is southerly.

[utopein]

On 9/24/2015 at 11:53 PM, OhioBob said:

I just derived an equation to allow for a more accurate determination of the heading one must fly to insert into a target inclination. The equation is,

Heading = atan[ (V_orb*sin(i)) / (V_orb*cos(i)-174.6) ]

where V_orb is the orbital velocity, i is the orbital inclination, and heading is measured from the equator.

For example, let's say you want to insert into a 75 km orbit with an inclination of 10 degrees. For a 75 km orbit, V_orb = 2287.4 m/s, and, of course, i = 10^o. We therefore have,

Heading = atan[ (2287.4*sin(10)) / (2287.4*cos(10)-174.6) ] = 10.82^o

If we are launching at the target orbit's ascending node, then the heading is northerly; and if we are launching at the target orbit's descending node, then the heading is southerly.

hello i must be missing thing here but the heading give a number not a vectorxyz direction how do know the direction ? 

[bewing]

A compass heading is not a vector. Just an angle.

[utopein]

11 minutes ago, bewing said:

A compass heading is not a vector. Just an angle.

allright  but relative to what ? at lease you still need latitude longitude  relative to the compass

 

degree left or  right 

 

degree up or down ? 

[OhioBob]

@utopein, I explain it a couple posts earlier.  The heading that my formula gives is the angle relative to due east.  So you would launch in a direction due east +/- the heading angle.

[utopein]

18 hours ago, OhioBob said:

@utopein, I explain it a couple posts earlier.  The heading that my formula gives is the angle relative to due east.  So you would launch in a direction due east +/- the heading angle.

 you don t have by any chance a formula that work on all direction ?

[Zhetaan]

4 hours ago, utopein said:

 you don t have by any chance a formula that work on all direction ?

Technically, it already does.  The problem is that Kerbin has a surface velocity and that velocity is defined to be in the eastward direction.  Any formula that accounts for that surface velocity is going to need to account for its direction, as well.  Since @OhioBob's formula essentially calculates and applies a correction factor to account for that velocity during launch, it is both mathematically and reasonably appropriate to consider the correction relative to the velocity that needs correcting.

Of course, it's not really a problem:  if Kerbin didn't have rotation and the attendant surface velocity, then it would never be possible to launch into orbits whose ascending or descending nodes were not directly over the KSC.

The plus-minus part of the formula is similarly unavoidable, and exists because of the difficulty of applying a two-dimensional correction factor to a three-dimensional, rotating frame of reference--rotating, that is, relative to the launch point, provided that we hold the launch point stationary and equatorial, and always launch to a heading from that point, which, being the KSC, is always the case.  If we were to apply it to launches from various landing sites, then it gets quite a lot more tricky.

In those cases, you need to account both for the fact that the surface velocity changes (it diminishes to zero as you travel from the equator to the poles) and for the fact that you cannot launch into an equatorial orbit from non-equatorial locations, which means that you need a correction relative to the inclination you'd start with, but the range of possibilities is restricted because there are certain inclinations that are no longer available.  To wit, you do not need a surface velocity correction if you launch from the exact north pole.  But you also cannot launch into anything but a polar orbit from there.

[pincushionman]

4 hours ago, utopein said:

 you don t have by any chance a formula that work on all direction ?

Given that the compass heading for north is 0° and east is 90°, then your heading h = (90 - a), where a is your desired angle from@OhioBob‘s equation, and is a positive number for heading northwards and a negative number heading south.

Do note, however, once your vessel gets more than a few degrees north or south of the equator (that is, your position is), don’t bother trying to match the navball heading anymore and focus on getting to orbit, as your actual heading throughout a great circle route (which all orbits are) changes continuously. You should be close and any orbital correction later on should be cheap.

You’ll need to convert from -180<h<180 to 0<h<360 as appropriate, but when you need to do that it should be obvious.

[OhioBob]

7 hours ago, utopein said:

 you don t have by any chance a formula that work on all direction ?

I don't know what you mean exactly.  The formula is for launches from Kerbal Space Center, no place else.  If you want an orbit with zero inclination, then you launch due east*.  If you want to launch into an orbit that has an inclination i, then you have to launch along a heading that is north or south of due east by an amount equal to i + c, where c is a correction to account for Kerbin's rotation.  The formula computes the value of i + c.  It doesn't make much sense to do it any other way.

* This is exactly correct because KSC isn't exactly on the equator.  But it's close enough not to split hairs over.
 

Edited April 23, 2019 by OhioBob