Orbiter-Forum [Question] Wideawake to Wideawake half orbit calculation
 User Name Remember Me? Password
 Register Blogs Orbinauts List Social Groups FAQ Projects Mark Forums Read

 MFD Questions & Help Post your questions here for help with the Multi-Function Displays.

 06-13-2012, 12:23 AM #16 dgatsoulis Orbinaut 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). Last edited by dgatsoulis; 10-19-2012 at 08:08 PM.
 Thanked by:
 06-13-2012, 01:22 AM #17 Gr_Chris_pilot DGIV Captain dgatsoulis check my previous post i hope i got it right
 06-13-2012, 01:35 AM #18 dgatsoulis Orbinaut 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.
 06-13-2012, 01:45 AM #19 Gr_Chris_pilot DGIV Captain 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... Last edited by Gr_Chris_pilot; 06-13-2012 at 02:24 AM.
 06-13-2012, 02:05 AM #20 dgatsoulis Orbinaut 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.
 Thanked by:
 06-13-2012, 02:10 AM #21 Tommy Orbinaut 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.
 Thanked by:
 06-13-2012, 02:08 PM #22 Gr_Chris_pilot DGIV Captain 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!!!
 Thanked by:

 Posting Rules BB code is On Smilies are On [IMG] code is On HTML code is Off You may not post new threads You may not post replies You may not post attachments You may not edit your posts
 Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Orbiter-Forum.com     Announcements     Meets & Greets Orbiter Space Flight Simulator     Orbiter Web Forum         OFMM         Orbiter Forum Space Station III         Simpit Forum     General Questions & Help     MFD Questions & Help     Hardware & Software Help     Tutorials & Challenges     Orbiter SDK     Orbiter Visualization Project » Orbiter Project Orbiter Addons     OrbitHangar Addons & Comments     Addons     Addon Development     Addon Requests     Addon Support & Bugs         Addon Developer Forums Far Side of the Moon     Spaceflight News     Math & Physics     Astronomy & the Night Sky     Backyard Rocketry     Brighton Lounge     International Forum

All times are GMT. The time now is 07:20 AM.

 Quick Links Need Help?