Kerbal Space Program

Kerbal Space Program

Calculate Gravity Turns?
I'm not enjoying the tutorials, videos and Scott Manely vids. I'm not very fond of trial and error so I ask (after googling in vain): Is there any way to calculate a gravity turn for the space craft? Like taking into weight of craft, thrust, drag and gravity into account.
Last edited by Hole config question; Jul 4, 2017 @ 5:12pm
< >
Showing 1-7 of 7 comments
internetrules Jul 4, 2017 @ 5:26pm 
if you are not into trial and error in KSP than get ready for a future career in NASA cuz you are gonna have to learn alot of math n physics n ♥♥♥♥ (unless you are gonna use KER and MechJeb
Last edited by internetrules; Jul 4, 2017 @ 5:26pm
I start turning to the 45° when I'm entering into the 3rd atmospheric layer (or whatever that last dark blue layer on the gauge is called). Once I get past the atmosphere, I will start calculating my final maneuvers to get me into a solid orbit.

I don't think it matters too much as long as you stay below the sound barrier in the first and second atmospheric layers.
JanaSzar Jul 4, 2017 @ 5:44pm 
https://en.wikipedia.org/wiki/Gravity_turn#Mathematical_description

Though, what I usually do is just hit 100~200mps then start to slightly tilt about 5*.
This won't answer your question, but it gives an idea.

https://www.nasa.gov/pdf/315950main_Microgravity_Around_the_World.pdf

The example of a cannonball shot off a mountain explains the required velocity. In KSP it's about 2200m/s velocity, at 70,000m altitude. Then you have to take into account that the planet is rotating below you and when you hit about 50km altitude it shows you orbital velocity rather than ascension. How to calculate this would most of all require compensating for wind resistence at various altitudes.

You could try to calculate this in a simplified way for a planet with no atmosphere, such that you know if you were driving a car on the surface what speed it needs to go to attain orbit (or just skip along near orbit).

Good luck. When you finish, please post a link to your paper with all the calculations so we can all review it before you apply for that job at NASA!
flying diomedea Jul 4, 2017 @ 6:10pm 
For a gravity turn you need to solve a system with these equations:
- v(t) = ∫t (Thrust(ρ)/Mass(t) - Drag(v,ρ) + g(z) cos(pitch)) + vi
- pitch = atan(g(z)*t/v(t))
where:
- Thrust is function of air density (ρ) (or of external pressure, ρ = pressure/unitary volume);
- Mass is function of time (t) = Initial Mass - fuelflow * t (note that fuelflow depends on throttle)
- Drag is function of speed (v) and air density (ρ) (also, of Reynolds which also depends from v)
- gravity (g) is function of altitude (g(z))
- pitch is always oriented in the prograde direction (necessary condition for a gravity turn)
- air density (ρ) also is function of altitude
- also, beware about initial velocity (vi). It starts with the tangential velocity from the rotation of the planet at the launch site, though it won't be noticed until after the frame of reference is switched from surface to orbital.

Some of those terms are pretty complex in themselves. That's why I compute gravity turns by numerical integration, rather then with differential equations.
Originally posted by flying diomedea:
For a gravity turn you need to solve a system with these equations:
- v(t) = ∫t (Thrust(ρ)/Mass(t) - Drag(v,ρ) + g(z) cos(pitch)) + vi
- pitch = atan(g(z)*t/v(t))
where:
- Thrust is function of air density (ρ) (or of external pressure, ρ = pressure/unitary volume);
- Mass is function of time (t) = Initial Mass - fuelflow * t (note that fuelflow depends on throttle)
- Drag is function of speed (v) and air density (ρ) (also, of Reynolds which also depends from v)
- gravity (g) is function of altitude (g(z))
- pitch is always oriented in the prograde direction (necessary condition for a gravity turn)
- air density (ρ) also is function of altitude
- also, beware about initial velocity (vi). It starts with the tangential velocity from the rotation of the planet at the launch site, though it won't be noticed until after the frame of reference is switched from surface to orbital.

Some of those terms are pretty complex in themselves. That's why I compute gravity turns by numerical integration, rather then with differential equations.
Thanks!.
maj.solo Jul 5, 2017 @ 4:59am 
you sum up the force vectors. Calculating drag I have not figured out yet. Is that data available? Or is it going to be guesswork?

People are overthinking this. Just launch and see what happens.
Last edited by maj.solo; Jul 5, 2017 @ 5:01am
< >
Showing 1-7 of 7 comments
Per page: 1530 50

Date Posted: Jul 4, 2017 @ 5:09pm
Posts: 7