[Question] Wideawake to Wideawake half orbit calculation

[Gr_Chris_pilot]

Hi there!
I want to know how i can calculate my launch heading from Wideawake International (or any other base) to make half orbit and then reentry.... All i want is to find the correct launch heading so i dont have to make any inclination corrections but only a retrograde burn back to the base i started....
i Have tried LaunchMFD SuborbitalMFD and BasesyncMFD...

You are really gonna help with my company ''Global space turism inc.''

[dgatsoulis]

Hey patrioti!

A WIN to WIN trajectory, isn't exactly half an orbit. It's a full orbit and then some.

Have a look at the PDF attached to this post

Once you can calculate how much your target will "move" by the time you get there, then you can launch at the correct heading.

[Gr_Chris_pilot]

Thanks a lot Dimitri!

I think its time to use my calculator after all

Yes i know after landing its one complete orbit (around surface of the earth) but this is what my virtual space turists pay for

[dgatsoulis]

Just for fun, try to find the launch azimuth that would take you from WIN to WIN after 10 complete 200x200km orbits. (you land on the eleventh).

[blixel]

Quote:

Originally Posted by dgatsoulis

 Have a look at the PDF attached to this post

Once you can calculate how much your target will "move" by the time you get there, then you can launch at the correct heading.

This is really interesting to me. I downloaded your PDF and am trying to follow the steps. I'm using Wideawake to KSC for my experiment.

MapMFD St=8171 km brg 303.0
Vc = 8.1km/s
Ta = 11

(3*11) + (((8171)-(3-2200))/8.1/60)= 54.3333333

54.33 * 0.25 = 13.5825 degrees east

target longitude = 80.67

80.67 + 13.5825 = 94.2525 (new target longitude)

If all the above is correct, the last part is to derive the launch heading. I don't understand how to do that.

In your other post, you said "Then I calculated the bearing for the "new" coordinates. This would be my launch azimuth. The result was 204.25 degrees."

What steps did you use for the "Then I calculated" part?

[dgatsoulis]

Quote:

Originally Posted by blixel

 This is really interesting to me. I downloaded your PDF and am trying to follow the steps. I'm using Wideawake to KSC for my experiment.

{image}
[INDENT]MapMFD St=8171 km brg 303.0
Vc = 8.1km/s
Ta = 11

(3*11) + (((8171)-(3-2200))/8.1/60)= 54.3333333

The result of that is 36.2325 not 54.3333

(3*11)+(((8171-(3*2200))/8.1/60)= 33+((8171-6600)/8.1/60)=
33+(1571/8.1/60)=33+3.2325=36.2325 minutes

36.2325 * 0.25 = 9.058 degrees east

target longitude = 80.67

80.67W + 9.058E = 71.585W (new target longitude)

In your example, you moved the target longitude west not east. KSC will "be" at 71.585W when you get there. (relative to the position you are now).

Quote:

Originally Posted by blixel

 What steps did you use for the "Then I calculated" part?

In the first post of that thread I've posted links to the formulae for finding bearing, distance and other cool stuff. The second link is particularly useful and also provides shortcuts.

In this case you want to find the bearing of the new target coordinates. If you want to do it yourself with a calculator the formula is:

θ=atan2(sin(Δlong)*cos(lat2),cos(lat1)*sin(lat 2) − sin(lat1)*cos(lat2)*cos(Δlong))

θ= bearing
lat1= launch site lattitude
long1= launch site longitude
lat2= target site lattitude
long2= target site longitude
Δlong= (long2-long1) Difference in longitude.

*example of Δlong. West longitudes get a - sine and East a positive one. If you launch from 90W to 30W you have (-90)-(-30)=-90+30=-60. The difference in longitude is 60 degrees.
If you launch from 90W to 30E it is (-90)-(+30)=-90-30=-120 The difference in longitude is 120 degrees.

[C3PO]

Remember you need to add longitudes if you cross the 0° meridian.

[blixel]

Quote:

Originally Posted by dgatsoulis

 The result of that is 36.2325 not 54.3333

I guess google calculator can't be trusted. It says the answer is 54.3333

Quote:

Originally Posted by dgatsoulis

 80.67W + 9.058E = 71.585W (new target longitude)

In your example, you moved the target longitude west not east.

KSC will "be" at 71.585W when you get there. (relative to the position you are now).

Ah... I see. Going east, so you have to subtract the longitude, not add.

Quote:

Originally Posted by dgatsoulis

 In this case you want to find the bearing of the new target coordinates. If you want to do it yourself with a calculator the formula is:

θ=atan2(sin(Δlong)*cos(lat2),cos(lat1)*sin(lat 2) − sin(lat1)*cos(lat2)*cos(Δlong))

θ= bearing
lat1= launch site lattitude
long1= launch site longitude
lat2= target site lattitude
long2= target site longitude
Δlong= (long2-long1)

*example of Δlong. West longitudes get a - sine and East a positive one. If you launch from 90W to 30W you have (-90)-(-30)=-90+30=-60.
If you launch from 90W to 30E it is (-90)-(+30)=-90-30=-120

The rest of this is over my head. Bummer.

[dgatsoulis]

you probably made an error in typing: ...(3-2200) instead of (3*2200)

It's not that difficult really, just takes a bit of practise.

Here is an example for WIN to KSC. We've already found the target's "new" coordinates so let's write down what we know.

lat1= -7.95
long1= -14.43
lat2= +28.5208
long2= -71.585
Δlong= long2-long1= (-71.585)-(-14.43 )=-71.585+14.43=-57.155

Here is a link for an online calculator with the atan2 function.

Copy the line bellow and paste it to the calculator. (one line)

atan2(sin(Δlong*(PI/180))*cos(lat2*(PI/180)),cos(lat1*(PI/180))*sin(lat2*(PI/180))-sin(lat1*(PI/180))*cos(lat2*(PI/180))*cos(Δlong*(PI/180)))*180/PI

Replace the lat1, lat2 etc with the values above

atan2(sin(-57.155*(PI/180))*cos(28.5208*(PI/180)),cos(-7.95*(PI/180))*sin(28.5208*(PI/180))-sin(-7.95*(PI/180))*cos(28.5208*(PI/180))*cos(-57.155*(PI/180)))*180/PI

When you press enter the result is -53.874 degrees

When you get a result with (-) in front of it, you need to subtract it from 360 degrees.
Since you obviously will be making a trip with a direction south-east to north-west, your launch heading must be 360-53.874=306.126

[Tommy]

The problem here is that your calculation will give you a heading for the shortest route between the two points. Since our two points are the same place (just at different times) the "shortest route" will cover the distance WIA has moved - not a full orbit +/- the distance moved.

Would simply reversing the heading given result in the correct heading for a "long way around" course?

[blixel]

Quote:

Originally Posted by dgatsoulis

 you probably made an error in typing: ...(3-2200) instead of (3*2200)

You're right. When I fixed the typo, google calc gave the right answer.

Quote:

Originally Posted by dgatsoulis

 Since you obviously will be making a trip with a direction south-east to north-west, your launch heading must be 360-53.874=306.126

This may be a dumb question, but is there a way to incorporate Align Plane MFD so you can type in the Inclination and LAN before take off? That way you can use the RInc and Rate +0.xxx/-0.xxx to help guide you?

Or do you just take off, turn to 306.126 and hold that heading the entire way there?

[dgatsoulis]

Quote:

Originally Posted by Tommy

 The problem here is that your calculation will give you a heading for the shortest route between the two points. Since our two points are the same place (just at different times) the "shortest route" will cover the distance WIA has moved - not a full orbit +/- the distance moved.

For a WIN to WIN flight, you are exactly right Tommy. But if you calculate the TOF accurately, then the launch heading will get you to an orbit that passes over WIN when you get there. In this case the distance from the launch site to the landing site is equal to Earth's circumference.

---------- Post added at 11:42 PM ---------- Previous post was at 11:32 PM ----------

Quote:

Originally Posted by Tommy

 Would simply reversing the heading given result in the correct heading for a "long way around" course?

Not exactly, because the TOF will be longer thus the target will move more.

Quote:

Originally Posted by blixel

 Or do you just take off, turn to 306.126 and hold that heading the entire way there?

Take off to that heading and hold the direction. The heading will change as you travel along the path of the Great circle that passes over your launch site and your landing site.

This method that I've developed to estimate the Time of Flight is only based on experience and many many fun hours on our favorite simulator. It's not 100% and it works best for larger suborbital flights (distance>10000km). But it does work, and AerobrakeMFD usually helps with the small differences you get during the reentry part of the journey.

[Tommy]

Quote:

Originally Posted by dgatsoulis

 For a WIN to WIN flight, you are exactly right Tommy. But if you calculate the TOF accurately, then the launch heading will get you to an orbit that passes over WIN when you get there. In this case the distance from the launch site to the landing site is equal to Earth's circumference.

Ah, yes. You are correct. There would be a small error due to precession (if you have non-spherical gravity sources enabled), but that should be small enough to ignore.

This assumes that you want a prograde (eastward) trajectory (meaning the flight would be just over a full orbit), but why wouldn't you? Even though the distance is a bit longer, the dV is about 900 m/s lower! Even though you are suborbital, a trip that goes a full orbit (or close to it) still requires almost orbital velocities so you still benefit from the "head start" provided by the Earth's rotation.

[dgatsoulis]

Quote:

Originally Posted by Tommy

 This assumes that you want a prograde (eastward) trajectory (meaning the flight would be just over a full orbit), but why wouldn't you? Even though the distance is a bit longer, the dV is about 900 m/s lower! Even though you are suborbital, a trip that goes a full orbit (or close to it) still requires almost orbital velocities so you still benefit from the "head start" provided by the Earth's rotation.

I'm not really sure, because I have not fully tested this yet, but before reading your post, I thought they canceled out. Launching Prograde gives you the advantage of the Earth's spin, but you are chasing a target that is moving away from you.
Doing the opposite means that you are moving towards a target that is moving towards you.
I prefer to use Groundspeed in the equation, because it reflects the rate at which you are approaching your target.

[Gr_Chris_pilot]

Quote:

Originally Posted by blixel

 I guess google calculator can't be trusted.

I use this one....
http://web2.0calc.com/

I think it works for this formula too
θ=atan2(sin(Δlong)*cos(lat2),cos(lat1)*sin(lat 2) − sin(lat1)*cos(lat2)*cos(Δlong)) fill the gaps and copy paste

For a WIN to WIN i have:

Distance is 33600km (7.95 S 14.43 W opposite side of earth 7.95 N 165.57 E Dist~16.800km x2)

TOF: (3*9) + (((33600)-(3*2000))/7.4/60)~89.1min (9 minutes and 2000km because i'm using the DGIV with MarkV engine)
So 89.1 x 0.25= 22.275

WIN after 89min 7.95 S 7.845 E

atan2(sin(22.275*(PI/180))*cos(-7.95*(PI/180)),cos(-7.95*(PI/180))*sin(-7.95*(PI/180))-sin(-7.95*(PI/180))*cos(-7.95*(PI/180))*cos(22.275*(PI/180)))*180/PI= 91.559

Launch heading 91.55 it that correct?


Ps: dgatsoulis YOU ARE A PRO!!!