Delta v for Tylo?

[quasarrgames]

How much delta v do i need to land and take off from tylo? I made a single stage NERVA lander with about 4.5km/s. Is that enough? (with a significant margin for error)

THANKS Everyone for telling me! I made a new lander with 7km/s of delta v. Can't wait to try it!

Edited September 22, 2014 by quasarrgames

[Godot]

2270 according to this chart:

http://i.imgur.com/UUU8yCk.png

(for each ... so to land and takeoff again 4540 ... and this might rather be a best case scenario than a normal one)

If a single NERVA is enough to land is another question, which depends on your rockets weight (and therefore on its TWR).

You should use Kerbal Engineer Redux in VAB (with Tylo as Reference Body)

Edited September 21, 2014 by Godot

[EdFred]

And this says 3070 - so double that plus margin for error.

[cantab]

2300 does seem low for a Tylo landing. Yet another chart, http://i.imgur.com/skAIz90.png , quotes 2600.

I expect the figures will normally assume a decent TWR. A low TWR lander will incur greater gravity losses and need more delta-V to make up for it, though it's possible it'll still be lighter overall due to less engine mass. With a very low TWR you may need to take a different approach and you'll have very little room for error; if your lander cannot lift off on Kerbin fully fuelled you're probably into that zone.

[Frank_G]

You will need about 3300k of deltaV to land with a small margin of piloting error included and about 3100 deltaV to take off, as far as my previous tests came out.

If you are a good pilot, you can land and launch with about 6200 m/s of deltaV. But that will be a ride on a horse from hell for sure...

[Red Iron Crown]

The few times I've done the Tylo landing/ascent I've budgeted about 7km/s of dV. It can definitely be done with less, I'm not the most efficient at vacuum landings and I carried a reserve for rendezvous with the mothership after ascent.

[EdFred]

I've been planning for about 6500dV. Yeah, I cut it close. It's not the orbit that eats into the DV, it's the suicide braking. I should probably come up with a formula to determine when to start the suicide brake based on TWR, current velocity and body, and altitude.

[cantab]

The old trick is after you've done your deorbit burn, set up a manoeuvre node on the surface and drag it fully retrograde. Then start your braking burn when the predicted duration equals the time to the node.

[EdFred]

The old trick is after you've done your deorbit burn, set up a manoeuvre node on the surface and drag it fully retrograde. Then start your braking burn when the predicted duration equals the time to the node.

Yeah, I've seen that but, in theory that only works when your approach angle doesn't change, right? I kinda like to land vertical.

[cantab]

Well it's an approximation. You've got things like the non-constant TWR due to fuel usage, the rotational speed of the body, and errors in the map view's display of the terrain. But it's usually enough to get a good idea and you can add extra safety margin if you want. Incidentally, an "ideal" suicide burn would have you only tilting vertical at the very last moment after you cut engines and are a just a couple of metres off the ground.

[EdFred]

Well it's an approximation. You've got things like the non-constant TWR due to fuel usage, the rotational speed of the body, and errors in the map view's display of the terrain. But it's usually enough to get a good idea and you can add extra safety margin if you want. Incidentally, an "ideal" suicide burn would have you only tilting vertical at the very last moment after you cut engines and are a just a couple of metres off the ground.

All good points, but I've been trying to do exactly that (only going vertical in the last 100m)

[KerikBalm]

Tylo is basically just Laythe without an atmosphere, as far as dV needed to acheive orbit is concerned.

They have the same surface gravity, although Tylo does have a larger diameter... but then again, you have no atmosphere.

Orbital velocity for Low laythe orbit is about 1875 m/s.

For Tylo, I guess it would be similar... lets say

a=v^2/r ... assume a is the same (it will vary a little due to the different radius of the planets) (v_1)^2/r_1 = (v_2)^2/r_2

r_1= 6/5 *r_2 -> v_2 = (6/5)^0.5* v_1

If v_1 = 1875, then v_2 should be about 2055 m/s

So, the chart says 2270, that sounds about right if you take off with a very high TWR, and pitch over right away (its got no atmosphere).

That also means you could land like that with a suicide burn... but again, this requires a very high TWR.

While I have yet to do much exploring of the Jool system (still setting up my base at laythe, it includes quite a few OTEs worth of fuel in orbit), but from what I'm thinking, the hard part of Tylo will be the gravity drag.

Every 10 seconds you spend thrusting downward is ~80 m/s of lost delta V. If you burn retrograde (from very low orbit, for ideal efficiency) and then drop straight down, but spend 1 minute in powered descent controlling your speed, thats 480 m/s of lost delta V.

Add that to the 2270 m/s "ideal", and you've got 2750m/s.

On ascent, assuming a TWR of 2:1, you can pitch over 30 degrees without falling down... and you still get 86.6% of the horizontal component... so 2,500 m/s for ascent should be plenty if you have a 2:1 TWR

I plan to go there budgeting 3,000 m/s for the landing, and 2,500 m/s for the ascent.