Challenge KSC to Wideawake - "The Classic" Speed Run

[SolarLiner]

Nope. Off once again to Jupiter and Beyond !

 

[thepenguin]

What is the proper launch heading (compass degrees) from KSC to Wideawake? This will be a great test for an autopilot I have been doing some work on!

 

[vsfx]

It is intense,I keep skipping out. Dous landing gear damage disabled count as a cheat,it's so sensitive. :blush:

Nope. Off once again to Jupiter and Beyond !

I'm glad I'm not the only one struggling! If I get one part right I will screw up another, or I'll push the temps too far.

Damage simulation should definitely be on because diving down and crashing onto the runway as your "landing" would unfairly save you a lot of time.

My first video; There were a couple of mistakes but overall it was a pretty good run. Timed at 18 minutes 16.8 seconds from "throttle up" to "wheels stop". (XR2 MET counter: 18 min 07 seconds). Max velocity (GS) was 14.62 km/s

I added a lua script to display some info of the run and a keypress logger to show the keys. Flight was made without a joystick.

I will update if I get a better score.

Added. That was nicely flown dgat. Very smooth movements, I would of had no idea you were using a keyboard. :tiphat:

 

[thepenguin]

What is the proper launch heading (compass degrees) from KSC to Wideawake? This will be a great test for an autopilot I have been doing some work on!

nvm, found it in a video. 110 degrees.

 

[blixel]

nvm, found it in a video. 110 degrees.

108 degrees is a bit better.

 

[thepenguin]

108 degrees is a bit better.

Thanks. It's not exactly reassuring when his next words in the video were "the heading doesn't matter too much for the time being".

do you have numbers that are more accurate (after the decimal place)? (I wouldn't expect it if you do things by sight, but I can hope)

 

[SolarLiner]

:woohoo:
Finally managed to do a proper run !
No, I didn't recorded it, and I actually resumed from one quicksave. Here's the proof or now:

I have chosen 11.6 km/s GS for MECO at 70 km (my super cruise altitude)
I think I got a bit better than blixel on the ascent, but the thing that knocked of was the reentry. I got the AoA so bad that I bounced off to space, making my final reentry from a 123 km with almost no ground speed, to 10km, just over the runway. Since I cannot land like that, I did a little circle in the hope to reduce my speed, with the airbrakes. And my landing was creepy - as always with a numpad.

Anyway, VSFX, you can add my score to the board if you want. The pic says 26:42 but the real Wheel Stop was around 26:37 (the time to deep breathe, switch to the upper panel and take a screenshot, before letting the joy fill my feels (We all know that feel, bros )) with GS of 11 600 m/s.

 

[jthill]

What is the proper launch heading (compass degrees) from KSC to Wideawake? This will be a great test for an autopilot I have been doing some work on!

The stock surface HUD will show you this, set the base as your target on the Map MFD and look at the heading tape.


Edit:

Watching that 16m35 playback makes me think agentgonzo may not have chosen his own nickname.

edit2:

... and nobody who was there will ever say "Wideawake" quite the same way again either. Did the reentry boom from 3km straight up shatter every window on the base?

 

[dgatsoulis]

do you have numbers that are more accurate (after the decimal place)? (I wouldn't expect it if you do things by sight, but I can hope)

Estimate the TOF (Time Of Flight) and calculate how many degrees the Earth will rotate by the time you get there. Add the result to the target's longitude and calculate the bearing for the new target coordinates.

Example:
The average flight time for this challenge is ~20 minutes. In that time Earth rotates by [math] \frac{20 \cdot 60}{86164.1} \cdot 360^\circ \; = 5.014^\circ [/math]
WIN's longitude is -14.43°. Add to that the 5.014° of the rotation.
-14.43° + 5.014° = -9.416°

Cape Canaveral
long1 = -80.675 = -1.408 rad
lat1 = +28.5208 = 0.4978 rad

WIN(after rotation)
long2 = -9.416° = -0.1643 rad
lat2 = -7.95° = -0.1388 rad

Now we have everything we need to calculate the bearing, which will be our launch azimuth.

Δlong = long2 - long1 = (-0.1643) - (-1.408) = 1.2437

θ = atan2(sin(Δlong)*cos(lat2), cos(lat1)*sin(lat2) - sin(lat1)*cos(lat2)*cos(Δlong))

= atan2(sin(1.2437)*cos(-0.1388), cos(0.4978)*sin(-0.1388) - sin(0.4978)*cos(-0.1388)*cos(1.2437)) = 1.8545492758098

Turn the result back to degrees: 1.8546 * 180/π = 106.26°

The more accurately you know your TOF the more accurate the calculation. If you managed to get there 2 minutes earlier, the azimuth should have been different, because the Earth would have rotated ~0.5° less.
If you redo the calculation for a flight time of 18 minutes, you get a launch heading of 106.51°

 

[thepenguin]

Estimate the TOF (Time Of Flight) and calculate how many degrees the Earth will rotate by the time you get there. Add the result to the target's longitude and calculate the bearing for the new target coordinates.

Example:
The average flight time for this challenge is ~20 minutes. In that time Earth rotates by [math] \frac{20 \cdot 60}{86164.1} \cdot 360^\circ \; = 5.014^\circ [/math]
Turn the result back to degrees: 1.8546 * 180/π = 106.26°

The more accurately you know your TOF the more accurate the calculation. If you managed to get there 2 minutes earlier, the azimuth should have been different, because the Earth would have rotated ~0.5° less.
If you redo the calculation for a flight time of 18 minutes, you get a launch heading of 106.51°

Ok, this looks great! where exactly does that 86164.1 come from, though?
I think it has something to do with the seconds in a day (86400), but I'm not sure.

 

[C3PO]

Estimate the TOF (Time Of Flight) and calculate how many degrees the Earth will rotate by the time you get there. Add the result to the target's longitude and calculate the bearing for the new target coordinates.

Would it be cheating to fly the other way? The Earth's rotation would bring KSC closer during the flight, but the TAS would be higher.

And remember that the bearing has to be continuously updated. The heading depends on your actual position and not on your starting point.

---------- Post added at 11:25 AM ---------- Previous post was at 11:17 AM ----------

Ok, this looks great! where exactly does that 86164.1 come from, though?
I think it has something to do with the seconds in a day (86400), but I'm not sure.

That's the time Earth takes for one rotation. 86400 is the time between noons (the time the Sun is highest over the horizon). They are different because Earth orbits the Sun.

 

[MattBaker]

Ok, this looks great! where exactly does that 86164.1 come from, though?
I think it has something to do with the seconds in a day (86400), but I'm not sure.

The calculation [math]\frac{20 \cdot 60}{86164.1} \cdot 360^\circ \; = 5.014^\circ[/math] basically was "flight time in seconds (20 minutes * 60)" divded by "rotation period of Earth (roughly four minutes short of 24 hours)" so you'd get the fraction of a day the flight would take.
And then when you have the fraction of a day you can just multiply by 360° since one day=360° rotation so 0.1 days are 36° etc.

I hope that was understandable.

 

[thepenguin]

That's the time Earth takes for one rotation. 86400 is the time between noons (the time the Sun is highest over the horizon). They are different because Earth orbits the Sun.

Right, forgot about that giant ball of gas out there. I get it now.

 

[dgatsoulis]

And remember that the bearing has to be continuously updated. The heading depends on your actual position and not on your starting point.

No, not continuously. The whole point of taking Earth's rotation into account is to avoid that. You take off, point to that heading and level out.

Yes there is a little bit of adjustment, because the trip isn't with constant speed all the way, but you get fairly good results.

---EDIT--
Ah, unless you mean it the other way. That you hold the original heading/direction and don't try to follow the calculation's azimuth all the way to WIN.

 

[vsfx]

What is the proper launch heading (compass degrees) from KSC to Wideawake? This will be a great test for an autopilot I have been doing some work on!

The Google Earth ruler gives 108.03 deg (and a distance of 8,159 km), but dgatsoulis seems to have things a bit more accurate!

:woohoo:
Anyway, VSFX, you can add my score to the board if you want. The pic says 26:42 but the real Wheel Stop was around 26:37 (the time to deep breathe, switch to the upper panel and take a screenshot, before letting the joy fill my feels (We all know that feel, bros )) with GS of 11 600 m/s.

Added. Well done SolarLiner! :thumbup:

 

[SolarLiner]

I recorded my third attempt (first one burnt in atmosphere while ascending, second with 26:37). Video soon

 

[C3PO]

No, not continuously. The whole point of taking Earth's rotation into account is to avoid that. You take off, point to that heading and level out.

Yes there is a little bit of adjustment, because the trip isn't with constant speed all the way, but you get fairly good results.

---EDIT--
Ah, unless you mean it the other way. That you hold the original heading/direction and don't try to follow the calculation's azimuth all the way to WIN.

When you travel on the surface along a great circle arc, your heading changes continuously. If you're in a plane you don't correct for Earth's rotation because the frame of reference rotates with the Earth.

The problem is that the heading on the HUD uses a frame of reference that rotates with the Earth. You cannot travel on a constant heading between two points on the surface on a sphere, unless they have the same longitude or are on the equator. If you follow a constant heading (like when you trace the course on a map) you are actually traveling in an arc over the spherical surface. If your trip is long enough it can mean that you much longer than the shortest distance.
In this particular scenario the difference isn't that large, because the initial azimuth is rather close to 90°.

---------- Post added at 05:36 PM ---------- Previous post was at 05:07 PM ----------

Google is your friend :lol:

Try measuring the distance from Oslo Norway to New York in Google Earth. That's a great circle track of ~3200 NM. Now open Google Map and place a ruler between both places. You will notice that GE misses Newfoundland to the north but GM misses it to the south. The 'straight' line in GM is over 3400 NM long.

 

[dgatsoulis]

{snip}..You cannot travel on a constant heading between two points on the surface on a sphere, unless they have the same longitude or are on the equator. If you follow a constant heading (like when you trace the course on a map) you are actually traveling in an arc over the spherical surface...{/snip}

Yes, I know.

Anyway, the thread has been jacked for too long with this launch azimuth and heading stuff.

Time for another run. First one under 18 minutes. XR2 MET: 17:53
Playback attached.*

*The counter in the playback starts at throttle up, 12 seconds earlier than
the real flight.

 

[jthill]

So the apparent heading changes are pure artifact, then -- it's a great circle, viewing the path as an orbit there are no plane changes.

Yes, of course: viewing your path from directly above its (orbital) equatorial crossing, every longitudinal lines is curved differently.

And now I'm going to try to get on that board.

 

[vsfx]

Time for another run. First one under 18 minutes. XR2 MET: 17:53
Playback attached.*

*The counter in the playback starts at throttle up, 12 seconds earlier than
the real flight.

Updated your time. Epic flying once again!