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I was doing some back-of-the-envelopish calculations on gravity turns. Now the basics are clear to me (I think), but this detail escapes me:

After liftoff we perform pitchover manouvre at time T+x, and start gaining downrange speed. After this the thrust vectoring is reset to point along the axis, and off we go, with zero angle of attack... except the angle of attack is not strictly zero. We need to reorient ourselves along the velocity vector, but what is the strict formulation here?

Do we hold the predetermined pitchover angle for some predetrmined amount of seconds, and then thrust vector to zero angle of attack?

Or do we hold ourselves at a solid pitchover angle until the velocity vector coincides, and then start following it?

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  • $\begingroup$ Great question. I've wondered about this. On one hand, the gravity turn is done to minimize aerodynamic stresses, so you want to get right to it without delay. On the other hand, if you thrust vector to zero AoA right away, you pitch the rocket up (as the velocity vector is lagging behind you from the pitchover maneuver). I've had to guess this before, and my choice was to wait until the velocity vector comes into alignment with the rocket's longitudinal (x) axis, at which point gravity turn is triggered. This seems to match the pitch plots I've seen better than hurrying on to gravity turn. $\endgroup$
    – user39728
    Commented Mar 15, 2021 at 0:55

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This depends on how stable your rocket is. If your rocket is aerodynamically stable, meaning its center of pressure is behind its center of mass, the rocket will likely be turned to its velocity vector (zero angle of attack) by aerodynamics alone.

A gravity turn is optimized for the least manual maneuvering possible. Any launch trajectory besides a perfect gravity turn uses up some energy (thruster fuel or drag from fins) in forcibly changing the rocket’s velocity vector by adding to the angle of attack. Directly after launch, there is an initial small maneuver to slightly off-vertical in the direction of the turn. The acceleration due to gravity turns the rocket’s velocity vector over time, and ideally, this results in horizontal attitude at your intended orbit’s perigee. There is usually some maneuvering required to compensate for wind, turbulence, and other disturbances. The free variables involved here are the starting maneuver final attitude, thrust curves on the rocket, aero properties on the rocket, etc.

I don’t know the exact math for determining rotation rates for a particular gravity turn, but I bet it involves getting the unit direction of the earth-centered-inertial frame total acceleration of the rocket, projecting that onto the rocket’s body-yz (body-x is forward) plane, and doing a cosine for an angular rate.

If the rocket is aerodynamically unstable, with a CoP forward of the CoM, or marginally stable, with a CoP very close to the CoM, active control is required to maintain the gravity turn (usually computer guidance). This requires more energy from thrusters or fins to correct the spontaneous disturbances from the unstable aerodynamics. More unstable means more energy.

If the rocket is overstable, as described here: https://www.rocketryforum.com/threads/open-rocket-stability-number.122399/, there may be even more energy needed for course corrections due to the “weathercock” effect, the tendency to turn into the wind. Think about a dart with large fins suddenly being hit in flight with a crosswind, and how that would affect its flight path.

Excerpt from the rocket stability forum post:

I usually aim for a stability of 1.0, a stability of 1 is the Center of Gravity (CG) is ONE caliber (body tube diameter) forward of the Center of Pressure (CP). Anything less than one is considered to be marginally stable, and anything over 1.0 is considered to be over stable (iirc). Overstable rockets usually want to weather cock (turn into the wind) to varying degrees, marginally stable rockets may do everything but fly straight.

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    $\begingroup$ Are any actual orbital rockets not aerodynamically unstable? Much of this discussion seems more applicable to model rockets, which do not perform gravity turns. $\endgroup$
    – Organic Marble
    Commented Dec 29, 2019 at 2:52
  • $\begingroup$ Thanks! After some digging it seems that achieving circular orbit from underneath an atmosphere is not a straightforward operation. On airless planet one would turn so that vertical thrust just cancels gravity drag minus angular acceleration. When velocity vector is tangential, orbit is circular, and thrust can be cut. Gravity turn on the other hand does not seem to lead into a circular orbit by itself. Or I'm missing something. $\endgroup$
    – Elmore
    Commented Dec 29, 2019 at 23:26
  • $\begingroup$ @Elmore There is usually some deviation from a gravity turn on a normal Earth orbit to account for spending less time at low altitude (high drag), and miscellaneous vehicle performance and safety requirements. The need for a “gravity turn” comes from a need to minimize drag by minimizing angle of attack. On an airless world like the moon, one can boost upwards for a few seconds to clear nearby terrain, then immediately turn to the most efficient attitude for increasing orbit height: horizontal. $\endgroup$
    – CourageousPotato
    Commented Dec 30, 2019 at 0:34
  • $\begingroup$ @OrganicMarble I don’t know how many rockets are aerodynamically unstable. I don’t think there are any duplicates of “What orbital launch vehicles are aerodynamically stable in their launch configuration?”. You could post that question if you want. $\endgroup$
    – CourageousPotato
    Commented Dec 30, 2019 at 2:43
  • $\begingroup$ I'm reasonably sure that the answer is "none". $\endgroup$
    – Organic Marble
    Commented Dec 30, 2019 at 4:18

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