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Let's say we're at T+0, just as the rocket starts to move, what keeps it upright as it clears the tower and gains speed?

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You can recreate the problem by placing a pencil point-first on your finger. Try to keep the pencil/rocket upright by moving your hand back and forth. If you managed it for more than a few seconds, congratulations! You're doing better than Proton 535-43 did.

In the very early stages of flight (before aerodynamics has any major effect) the rocket can be described as an inverted pendulum, just like the pencil. In order to keep itself stable the rocket must ensure that the thrust vector from its engines passes directly through its center of gravity.

enter image description here From http://exploration.grc.nasa.gov/education/rocket/gimbaled.html

Most modern rockets gimbal their engines to direct the thrust, but it's not the only way to achieve thrust vectoring. Here are a few more: enter image description here enter image description here

See Source for more info. (Note: Parts of the site seem to be identical to Rocket Propulsion Elements, George P. Sutton, Oscar Biblarz.)

One of the best ways of solving the inverted pendulum problem is with a proportional-integral-derivative controller (PID controller). I'll let Wikipedia explain:

A proportional-integral-derivative controller (PID controller) is a control loop feedback mechanism (controller) commonly used in industrial control systems. A PID controller continuously calculates an "error value" as the difference between a measured process variable and a desired setpoint. The controller attempts to minimize the error over time by adjustment of a control variable, such as the position of a control valve, a damper, or the power supplied to a heating element, to a new value determined by a weighted sum:

$$u(t) = K_pe(t) + K_i\int^t_0e(\tau)d\tau+K_d\frac{de}{dt}$$

where $K_p$, $K_i$, and $K_d$, all non-negative, denote the coefficients for the proportional, integral, and derivative terms, respectively (sometimes denoted $P$, $I$, and $D$).

Control Solutions, Inc. has a very good (and quite easy to follow) explanation on their website of the bare basics of a PID controller.

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    $\begingroup$ I am quite terrified by that video. The RSO there either just didn't do the job at all, or waited what seems like far too long to say "this is not working" and blow it up. Any more details on that? $\endgroup$ – Nathan Tuggy Sep 25 '15 at 21:44
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    $\begingroup$ @NathanTuggy IIRC the Russians don't use explosive FTS. To abort they just shut down the engines, which apparently happened at t+17. $\endgroup$ – ForgeMonkey Sep 25 '15 at 21:47
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    $\begingroup$ Fair enough; I saw a lot of dispute about exactly when the abort happened (4 seconds, 17 seconds, wasn't possible until 45 seconds...). $\endgroup$ – Nathan Tuggy Sep 25 '15 at 21:50
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    $\begingroup$ Seems they cut off one of the six engines in order to safely (hmm) steer it away from the launch pad, just as planned for such an eventuality. It was designed as a super-ICBM 50 years ago so maybe some of its safety philosophy is a bit different than for modern rocketry. A couple of gyroscopes had been installed upside down so I think the launch looked pretty stable considering that. $\endgroup$ – LocalFluff Sep 26 '15 at 2:36
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    $\begingroup$ "One of the best ways of solving the inverted pendulum problem is with a proportional-integral-derivative controller (PID controller)" I doubt that. PID is an easy-to-use way of feedback control, but it is bare-bones. I don't know if it is even sufficient for this task, but it is definitely not one of the best ways. $\endgroup$ – Peter Mortensen Sep 26 '15 at 9:21
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The other answers here are correct: gimbaling or other active correction measures are used.

While most launchers do try and maintain a vertical flight off the pad, the Antares rocket is known for the intentional "Baumgartner Maneuver" it does on takeoff, deliberately gimbaling the engine to maneuver away from the tower in the first seconds of flight, as you can see here.

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  • $\begingroup$ Staying "vertical" off the pad can be more exciting with multiple engines. If you lose one engine that results in an asymmetric thrust pattern from the remaining engines, but still have enough thrust to accelerate upwards, you have a choice. You can tilt the rocket's attitude and keep the center of mass going up vertically, or you can hold the attitude but the now asymmetrical thrust is going to push the vehicle sideways somewhat. This kind of thing is even more exciting to consider when a pad has a big tower on the side of it! $\endgroup$ – Matt Jessick Feb 4 '16 at 2:58
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    $\begingroup$ Most launchers have a hold-down system where the rocket isn't released until it's confirmed the first-stage engines are performing properly. Single-engine failures in the narrow window between hold-down release and clearing the towers must be vanishingly small, no? $\endgroup$ – Russell Borogove Feb 4 '16 at 4:01
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Most rockets gimbal their engines actively to maintain stability. Shifting the axis of thrust slightly works just fine to keep it upright.

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Comparing a rocket to an pencil on your finger is the pendulum rocket fallacy. When you balance the pencil, the direction of the normal force depends on the direction of gravity, the pencil's attitude, and your finger's position, but is generally up. As the pencil tips, it's unstable because the disturbing torque increases. The torque from a gimballed rocket engine does not depend on the attitude or gravity, since the angle of the thrust is in relation to rocket itself, not the ground. The only thing that matters is the actuation angle of the gimbal, so there is no runaway. It is no more difficult to stabilize than if the engine were at the top—if anything it's easier since the actuation is further from the center of mass, which is closer to the nose. Robert Goddard famously made this misjudgment when he pioneered rocketry with early rocket prototypes that placed the motor at the top. It had no effect on stability and with no active control or aerodynamic stability by design it careened into the ground. It's not an inverted pendulum because gravity acts on all parts of it equally and there is no constrained pivot. It's just a mass in free fall with a input torque that can be imagined to act at all points simultaneously, if you disregard flex in the structure.

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    $\begingroup$ Welcome to space! The structure of stack exchange is as question and answer, and you have posted a what appears to be a comment on another answer as a stand alone response to the question of mechanically how rockets are stabilised during launch. Suggest either deleting the answer or converting it into something focusing on the question as asked. $\endgroup$ – GremlinWranger Mar 27 at 7:01
  • $\begingroup$ Once you reach 50 reputation points you'll be able to post comments (like this) under other people's posts. Answer posts here need to attempt to answer "How is a rocket stabilized during the initial, slow speed, portion of launch?" I've just asked What is the “pendulum rocket fallacy” as it relates to analogizing a pencil balanced on a finger to maintaining attitude of a hovering rocket?. If you think some of this applies there, you can consider adding an answer there instead. Thanks, and Welcome to Stack Exchange! $\endgroup$ – uhoh Mar 27 at 12:18

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