[Question] Wideawake to Wideawake half orbit calculation

[dgatsoulis]

Let's try to answer the OP's question. How would you calculate the launch azimuth for a WIN to WIN flight.

Since the Earth is rotating at a rate of 360 degrees/day = 0.25 degrees/minute, we know that we are trying to hit a moving target. We must aim ahead of it's path, in order to hit it when we get there.

The steps to take are:
1.Estimate how long it will take us to get there.
2.Estimate how many degrees to the East our target will move during that time.
3.Add the result in our target's longitude.
4.Find the bearing (launch azimuth) for our target's "new" coordinates.
5.Refine the result.

The first thing we need to know is the ship that will be making the flight. Specifically, the time it takes to reach orbital speed and how many km downrange the ship travels during that time.

I understand that Chris' favourite ship is the DGIV, so I'll use that in this example.
A typical ascent to orbit with the DGIV takes about 11 minutes and the distance traveled is about 2100 km downrange.

The next thing we need to know is the distance between our launch site and our landing site. In this case they are the same site, which we'll get to after one orbit, so the distance is equal to the circumference of the Earth ~40000km

Now we can use this equation to estimate the Time Of Flight (in minutes):

TOF=(3*Ta)+(((St)-(3*Sa))/Vc/60)

TOF= Time Of Flight (in minutes)
Ta= Time of ascent (in minutes) - In this case =11
St= Distance to target (in km) - In this case 40000
Sa= Distance travelled downrange during Ta (in km) - in this case 2100 km
Vc= Ground speed when we reach orbital speed (in km/s) - in this case I cannot predict beforehand, what the GroundSpeed of my orbital path will be, so I arbitrarily choose a 7.8 km/s "polar" orbit GS.

(I explain the reasoning behind this equation, why I chose a Vc of 7.8 km/s and some data on different ships on the PDF attached to this post.)

Step1. Calculate the TOF

TOF=(3*Ta)+(((St)-(3*Sa))/Vc/60) = (3*11)+(((40000)-(3*2100))/7.8/60)=105.01 minutes

Step2. During the TOF our target will move 105.01*0.25=26.2525 degrees to the East.

Step3. WIN's "new" coordinates will be:
lat= -7.95
long= -14.43+26.2525=+11.8225

Step4. Find the bearing for the "new" coordinates.

The formula is:

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

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

lat1= -7.95
long1= -14.43
lat2= -7.95
long2= +11.8225
Δlong= long2-long1= (+11.8225)-(-14.43)=11.8225+14.43=26.2525

(I prefer to use this online calculator, with the atan2 function. Degrees have to be converted to radians and then the result is converted back to degrees, that's why you see all those 180/PI and then the result multiplied by PI/180 to get an angle in degrees. It might seem a bit much at first, but all it is, is the equation for the bearing, written in the notation that the calculator uses).

If you are going to use the same calculator, you'll find this next line very useful.

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 and Δlong with the values above:

atan2(sin(26.2525*(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(26.2525*(PI/180)))*180/PI

Hit the enter button and you'll get a result of 91.847 degrees.

Step5.Obviously this is an Eastward launch azimuth that will give us a GroundSpeed of ~7.3km/s when we reach orbital speed. So let's redo the calculation with this speed to refine it.

Step1. Calculate the TOF

TOF=(3*Ta)+(((St)-(3*Sa))/Vc/60) = (3*11)+(((40000)-(3*2100))/7.3/60)=109.941 minutes

Step2. During the TOF our target will move 109.941*0.25=27.48525 degrees to the East.

Step3. WIN's "new" coordinates will be:
lat= -7.95
long= -14.43+27.48525 =+13.05525

Step4. Find the bearing for the "new" coordinates.

lat1= -7.95
long1= -14.43
lat2= -7.95
long2= +13.05525
Δlong= long2-long1= (+13.05525)-(-14.43)=13.05525+14.43=27.48525


atan2(sin(27.48525*(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(27.48525*(PI/180)))*180/PI

= 91,937 degrees.

It's up to you to see if you want to take step No5 or if the difference is too small to even bother with it. (Aerobrake MFD can easily cope with this).

The answer to Chris' question for what would be the launch azimuth for a WIN to WIN journey is 91.937 degrees (In a DGIV).

[Gr_Chris_pilot]

dgatsoulis check my previous post i hope i got it right

[dgatsoulis]

The way you did the calculation is "correct". (Meaning that it's the same calculation I use. I'm not sure if it's correct, but it does provide good results).

The only thing I spotted as incorrect is the distance. You used 33600 km when the Earth's cirmcumference is about 40000 km and the actuall distance you will travel will be slightly higher than that.

Other than that - which will result in a slight change in the heading- everything is correct.

[Gr_Chris_pilot]

Quote:

Originally Posted by dgatsoulis

 The way you did the calculation is "correct". (Meaning that it's the same calculation I use. I'm not sure if it's correct, but it does provide good results).

The only thing I spotted as incorrect is the distance. You used 33600 km when the Earth's cirmcumference is about 40000 km and the actuall distance you will travel will be slightly higher than that.

Other than that - which will result in a slight change in the heading- everything is correct.

I calculated the distance using the earth opposite coordinates... the WIN is not on the equator so a 90deg heading will result in a ''great circle'' over the earth...

The shortest distance between any two points on the surface of a sphere is the arc of a Great Circle
http://en.wikipedia.org/wiki/Great-circle_distance
But i am not sure if this still apply here...

[dgatsoulis]

Quote:

Originally Posted by Gr_Chris_pilot

 I calculated the distance using the earth oposite coordinates... the WIN is not on the equator so a 90deg heading will result in a ''great circle'' over the earth...

The shortest distance between any two points on the surface of a sphere is the arc of a Great Circle
http://en.wikipedia.org/wiki/Great-circle_distance
But i am not sure if this still apply here...

It applies, but what happens when the points overlap? Isn't the distance equal to going once around the Earth?

Still, I cannot find anything wrong with your reasoning for finding the distance.

EDIT:

I think I got it. You didn't take into account the "movement" of the opposite coordinates due to Earth's rotation.

[Tommy]

Quote:

Originally Posted by dgatsoulis

 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.

I guess it depends on whether you are trying for a ballistic trajectory, or plan for a multiple skip re-entry. If you plan to go ballistic, you would need a much higher groundspeed to attain the ballistic trajectory - around 8.4 km/s. If you'll use aerodynamics (ie, a Sanger Re-Entry), I suspect they more or less cancel out.

In my experience, the retrograde option has always cost me significantly more fuel.

[Gr_Chris_pilot]

Quote:

Originally Posted by dgatsoulis

 It applies, but what happens when the points overlap? Isn't the distance equal to going once around the Earth?

Still, I cannot find anything wrong with your reasoning for finding the distance.

EDIT:

I think I got it. You didn't take into account the "movement" of the opposite coordinates due to Earth's rotation.

I think you are right! but i reallized even with errors like this the actual change of launch heading is less than 0.5deg probably 100k to correct at reentry soooo the passengers will never know

---------- Post added at 14:08 ---------- Previous post was at 02:13 ----------

After a few tests my distance calculation method put me only 5km north of my target (BasesyncMFD) I used LanchMFD for my flight set for 8.10deg inclination (91.55 southern heading) Thanks a lot for the help!

Happy orbiting!!!