Reentry speed coming back to Kerbin

[Warzouz]

I did a Jool test few days ago. That's not the point. As I played in creative and want to end the mission before unreasonable bed time, I ignore Alexmoon correct transfer window and got a dirty transfer back.

That wasn't seemed to be catastophic. The return nearly touched Eve orbit and encounter Kerbin 30° later. 

The issue is that I hit Kerbin Pe around 6300m/s. Even the fuel I had left (1600m/s) didn't allow my ship to survive the reentry. I used all the fuel before hitting the atmosphere, but, even my ship had a big heat shield :

• Hitting a 35km wasn't enough to slow down to a capture speed
• Hitting a 25km turn ship and crew to (mostly) gazeous form

Sure, my transfer was awful. I usually plan much better and always try to get then encounter at 180° of the starting position hitting the target orbit with an tangent approach.

My 2 questions

• Is there evaluations of return speed from all bodies when hitting Kerbin atmo (or PE) ?
• Was there a way to handle with my dreadful encounter at 6300-1600m/s slow down = 4700 to 5000m/s

[Snark]

2 hours ago, Warzouz said:

 

• Is there evaluations of return speed from all bodies when hitting Kerbin atmo (or PE) ?
• Was there a way to handle with my dreadful encounter at 6300-1600m/s slow down = 4700 to 5000m/s

 

So there's going to be a certain minimum return speed  from any body, even if you have an ideal Hohmann transfer.  You can get all calculatory or spreadsheety if you like, but the easy way to calculate it is to do this:

1. got to http://ksp.olex.biz
2. set up a transfer from Kerbin parking-orbit-at-70-km to whatever-planet-it-is (Yes, I know, we're going to Kerbin, not from Kerbin, this is deliberately doing it backwards so we get an arrival speed rather than departure speed)
3. Look at what dV it tells you
4. Add that to Kerbin orbital velocity at 70 km (about 2200 m/s)
5. That's the speed you'll arrive at.

That's the minimum.  If you have a crappy launch window and arrive in some other way than tangentially, then it'll be higher.  How much higher will completely depend on your trajectory, so there's no way to tell the answer without knowing the trajectory.

As for how to handle super-high-speed encounters:  Well, for any given ship design, there's a certain maximum limit that it can handle.  Sounds like you were well in excess of yours-- i.e. for that particular arrival speed for that particular ship, your situation was hopeless.  However, you can design a ship to withstand higher speeds.  The main way to boost the maximum survivable reentry speed is to try to reduce tons of ship mass per square meter of heatshield.

Ships that will have difficulty are long skinny lawn-darts with small heatshields on the front.  Ships that will do well and handle high speeds are pancakes.

This means:

1. use the biggest-diameter heatshield possible (even if you don't need one that wide to cover your ship
2. reduce your ship mass
3. if #1 and #2 are insufficient, you can add some "outriggers" sticking out to the side that are basically just mount points for additional forward-facing heatshields.  They're not actually shielding anything important (just the outriggers themselves, whose only purpose is to hold the heatshield)-- they're just there to generate additional drag and slow the ship down.

If you do add outriggers as described in #3, put them at the back of the ship, so they'll act like airbrakes and help to keep you pointed forward.

Edited January 7, 2016 by Snark

[rodentgun]

32 minutes ago, Snark said:

 

1. got to http://ksp.olex.biz
2. set up a transfer from Kerbin parking-orbit-at-70-km to whatever-planet-it-is (Yes, I know, we're going to Kerbin, not from Kerbin, this is deliberately doing it backwards so we get an arrival speed rather than departure speed)
3. Look at what dV it tells you
4. Add that to Kerbin orbital velocity at 70 km (about 2200 m/s)
5. That's the speed you'll arrive at.

That's the minimum. 

So, if I perform this calculation for Eeloo, I get the Ejection burn dV at 2071. Add 2200 from Kerbin low orbit velocity and I should expect to return to Kerbin with a speed of around 4270 m/s?

[Warzouz]

OK, I never thought of "Escape velocity" from Olex tool. So that gives (rounded, starting from 30km "orbit")

• Moho : 4100 m/s (varies due to eccentricity)
• Eve : 3500 m/s
• Mun : 3300 m/s
• Minums : 3200 m/s
• Duna : 3500 m/s
• Dres : 4000 m/s (varies due to eccentricity ?)
• Jool : 4200 m/s
• Eeloo : 4500 m/s (varies due to eccentricity)

If tangent is not coplanar, speed would be higher (1+sinus of plane angle ?). which can be quite high coming from Moho, Dres or Eeloo without plane change.

 

 

Yes, my ship was mostly long : A big heatshield, a empty tank adapter to MK3 passenger and a double entpy tank, then again, a empty rockamax adapter tank and one battery, one probe core, and one reaction wheel. (I dumped the 3 LVN engines)

I think that on my second try, the ship suffered from body lifting and quit prograde trajectory. The heatshield wasn't upfront and the ship exploded.

 

 

[Snark]

35 minutes ago, pistolhamster said:

So, if I perform this calculation for Eeloo, I get the Ejection burn dV at 2071. Add 2200 from Kerbin low orbit velocity and I should expect to return to Kerbin with a speed of around 4270 m/s?

Yes, exactly.  That's the speed you'd be at when you hit atmosphere at 70 km-- you'll be going faster before you slow when you get farther down in the atmosphere.  Thanks to @Warzouz for the calculations for various planets.

 

26 minutes ago, Warzouz said:

If tangent is not coplanar, speed would be higher (1+sinus of plane angle ?). which can be quite high coming from Moho, Dres or Eeloo without plane change.

If you're not coplanar, then yes, there will be more reentry speed.  It won't be as simple as a 1-plus-sine multiplier, you'd have to do some vector math.  It's not super complicated, but I suspect more tedious than it's worth, since it would mean you can't use the orbit calculator at olex.biz; you're back to working out all the numbers manually.

Basically, you'd work out the vector difference in speed between the ship and Kerbin when the ship arrives at Kerbin's SoI boundary, i.e. the X and Y components thereof.  Kerbin's X component will be its orbital velocity, Y component is zero.  For the ship, you'd do the math to work out its Sol-orbital velocity at the position of Kerbin's orbit, then its X component will be velocity times cosine of inclination, Y component will be velocity times sine of inclination.  The X component of dV will be Kerbin's X velocity minus ship's X velocity; similar for Y component.  Then the total Kerbin-relative velocity will be sqrt(dV_x² + dV_y²).  That gives the speed of the ship at the SoI boundary, then you need to do the math for how much faster it'll be going when it arrives down at Kerbin.  Roughly speaking this will be sqrt(v_o² + v_e²), where the two velocities involved are the speed-at-SoI boundary (which you just calculated) and Kerbin's escape velocity.  That's just an approximation which assumes that Kerbin has no atmosphere and you're falling all the way to the surface.  To do better than that would require more math.  Not super hard, but yet more tedium.

So overall, I'd say ignore the inclination, do the simple calculation with the orbit tool, and just add some padding if there's a significant inclination difference.

Edited January 7, 2016 by Snark

[Streetwind]

Word of advice: rate Moho higher than that - overall it's the highest dV destination (from Kerbin) in the game. There might be certain rare alignments that may let you get away with slightly less than Eeloo takes, but your average window costs more. Additionally, Eeloo is super forgiving, while Moho is a diva. Make a small mistake on your Kerbin ejection burn, and you just bought yourself an 50 m/s mid-course correction and a final capture burn 200 m/s higher. Mistime the moment of burn on the way back to Kerbin, and the same could happen - and suddenly you can no longer aerocapture safely.

TL;DR: It's a trap! Don't underestimate Moho.