I Why do rockets turn horizontally so soon after launch?

[DTM]

My 12 year old asked me this question. I have a MS in Mechanical Engineering, so I can usually answer his physics questions, but this one stumped me. When lifting off, why do most rockets turn close to horizontal almost immediately? Of course we know they need mostly horizontal speed to achieve orbit, but they also need altitude. If the rocket gained all it's altitude first, vertically, and then made a 90deg turn to the horizontal and then sped up horizontally, the majority of it's high speed flight would be outside of the Earth's atmosphere, greatly decreasing air resistance and wasted energy. Wouldn't this be much more energy efficient?

 

[russ_watters]

My 12 year old asked me this question. I have a MS in Mechanical Engineering, so I can usually answer his physics questions, but this one stumped me. When lifting off, why do most rockets turn close to horizontal almost immediately? Of course we know they need mostly horizontal speed to achieve orbit, but they also need altitude. If the rocket gained all it's altitude first, vertically, and then made a 90deg turn to the horizontal and then sped up horizontally, the majority of it's high speed flight would be outside of the Earth's atmosphere, greatly decreasing air resistance and wasted energy. Wouldn't this be much more energy efficient?

No, because quite a bit of the energy would be wasted just in holding the rocket up against gravity. I read somewhere that at liftoff the shuttle astronauts experienced 3g's of acceleration, which means that fully 1/3 of the energy expenditure is just to hold it up.

 

[A.T.]

When lifting off, why do most rockets turn close to horizontal almost immediately?

Watch a successful launch, not just the crashes.

 

[glappkaeft]

Rocket do travel mostly vertically in the parts of the atmosphere that are thick to minimize atmospheric drag. They then turn gradually sideways (usually in some variant of a gravity turn to avoid a large angle of attack that would tear the rocket apart) to minimize the effects of gravity drag/loss. The earlier and faster you go sideways the less you have to fight against gravity. It's a bit un-intuitive but the same effect is why after reaching orbital velocity gravity is the only thing needed to maintain orbit.

Your method is actually very inefficient, the entire energy used to achieve orbital height is wasted.

 

[DTM]

No, because quite a bit of the energy would be wasted just in holding the rocket up against gravity. I read somewhere that at liftoff the shuttle astronauts experienced 3g's of acceleration, which means that fully 1/3 of the energy expenditure is just to hold it up.

I'm not sure I understand that. 3g's of acceleration is their rate of change of speed. About 96 ft/s^2. The larger their acceleration the quicker they can get into orbit. Not sure how that equates to 1/3 of the energy expenditure is "just to hold it up"? I would suggest a rockets energy can be divided into 3 parts. 1. The energy needed to accelerate it up to orbital speed (~17,000 mph) (KE or Kinetic energy). 2. The potential energy it gains by getting to orbital altitude (about 150 miles for LEO). and 3. the energy it wastes by pushing the air out of the way on it's way up (Air resistance). I'm still not sure why they don't gain the potential energy first, which would be done at a slower speed and therefor much less air resistance which is proportional to speed^2 or even speed^3. Then turn horizontal to gain the KE. Yes this would have to be done at an angle of attached pointing away from the Earth to counteract gravity, but it still seems like a more efficient trajectory. Although I'm sure I must be wrong, or the experts would do it this way. Just trying to understand why I'm wrong.

 

[DTM]

[QUOTE="Your method is actually very inefficient, the entire energy used to achieve orbital height is wasted.[/QUOTE]

Wasted? You have to achieve the potential energy of your orbital altitude (say 150miles). You have to gain that PE sometime, why is it wasted if you do it right off rather than gradually?

 

[russ_watters]

I'm not sure I understand that. 3g's of acceleration is their rate of change of speed.

Acceleration felt includes the 1g you feel just by standing on Earth's surface.

About 96 ft/s^2. The larger their acceleration the quicker they can get into orbit. Not sure how that equates to 1/3 of the energy expenditure is "just to hold it up"?

At a force required for 1 g of acceleration, per f=ma, provided by the engines, the rocket would simply hover just above the launch pad. 3x more force provides provides 2g of speed increases in addition to that 1g of "just holding it up".

Thinking about it more, my answer was only half an answer though. That explains why fast acceleration is better than slow acceleration, but not necessarily why pitching-over sooner is better than later. The other half is that the atmosphere gets thin really fast. Even with the fast pitch-over, the shuttle would reach its maximum aerodynamic pressure in less than a minute, at 35,000 feet, and would experience almost no aerodynamic pressure after 2 minutes:
https://www.nasa.gov/pdf/466711main_AP_ST_ShuttleAscent.pdf

Total time to orbit is about 9 minutes; the rest is spent just accelerating.

As a simple estimate, you could calculate the angle that gets you to 150 miles altitude and 17,500 mph in 9 minutes assuming uniform acceleration and a linear trajectory pretty easily...

 

[mpresic3]

In order to reduce the effects of atmospheric drag, rockets guide their thrust to get above the densest portion of the atmosphere near the Earth surface is quickly as possible, (say around 80000 feet). They direct the thrust near vertical. At the same time, they need to achieve orbital velocity (which is quite high) so if they usually direct their thrust closer to horizontal they can use the big thrust form the first stages, rather than relying on the smaller thrust from the later stages. All stages will be necessary, but the more velocity gained by the rocket in the first stage, the less velocity will need to be gained from the smaller engines in the second and subsequent stages.

It turns out, the orbital velocity for the rocket at the Earth surface is not that much greater than the orbital velocity at a typical low Earth orbit about 150 miles altitude. If you (both) want you can verify this by taking a square root. Orbital velocity is square root of GM/R (see wikipedia). In one calculation,use R is the radius of the Earth. Then compare it to a similar calculation with R + 150 miles substituted for R. (Square roots were not taught in school for 12 year olds in my day, these days with calculators, I would not be surprised if kids understand this.)

The rocket climbs near vertical, not to be able to attain a smaller orbital velocity, but to get above the atmosphere to reduce drag and structural loads on the rocket. The rocket pitches over relatively quickly to use the big engines, and save the smaller ones till later.

It might also be hard to direct the thrust vertical until a height, and all of a sudden turn near 90 degrees all at once,

 

[sophiecentaur]

It’s worth pointing out that a circular orbit has equal amounts of Kinetic and Potential Energy so the vehicle has to be given both. Simply lifting it up by 150 miles and letting go will have it crashing into the ground.

 

[russ_watters]

It’s worth pointing out that a circular orbit has equal amounts of Kinetic and Potential Energy...

That's true only for one orbit, as potential energy increases but kinetic energy decreases with altitude.

 

[A.T.]

Wasted?

The vertical thrust component has to offset gravity, and only the remainder adds energy to the rocket, while the entire horizontal thrust component adds energy to the rocket. This has to be balanced against the losses due to drag, and the optimum is some gradual transition from vertical horizontal.

 

[mpresic3]

Concerning the last two posts. My Resnick / Halliday freshman/sophomore (college) physics text gives the gravitational potential energy at r to be - GMm/r
For a circular orbit, you can verify that (with the equation for orbital velocity in a circular orbit that I mentioned earlier) KE = 0.5* m v squared = GMm /(2r).

The kinetic energy for a circular orbit is (only) half as large as the potential energy. Moreover, they are not even the same sign. Potential energy is less than zero, and kinetic energy is greater than zero.

At the end of my posting, I (jocularly) was about to suggest the original poster, compute the orbital velocity (perhaps with his 12-year-old), for the two different altitudes over a bowl of Cherrios, next morning. Getting all these concepts straight is going to take several bowls of Cherrios.

 

[russ_watters]

Concerning the last two posts. My Resnick / Halliday freshman/sophomore (college) physics text gives the gravitational potential energy at r to be - GMm/r
For a circular orbit, you can verify that (with the equation for orbital velocity in a circular orbit that I mentioned earlier) KE = 0.5* m v squared = GMm /(2r).

The kinetic energy for a circular orbit is (only) half as large as the potential energy. Moreover, they are not even the same sign. Potential energy is less than zero, and kinetic energy is greater than zero.

A couple of things to note:
1. That equation is potential energy at a radius relative to a point mass. A rocket launched from the surface of the Earth.

2. I think your note about the sign convention adds unnecessary confusion: The rocket is doing positive work because it is applying force in the direction of motion.

 

[russ_watters]

I would suggest a rockets energy can be divided into 3 parts. 1. The energy needed to accelerate it up to orbital speed (~17,000 mph) (KE or Kinetic energy). 2. The potential energy it gains by getting to orbital altitude (about 150 miles for LEO). and 3. the energy it wastes by pushing the air out of the way on it's way up (Air resistance). I'm still not sure why they don't gain the potential energy first, which would be done at a slower speed and therefor much less air resistance which is proportional to speed^2 or even speed^3.

I can at least give quantitative answers to the first 2:

Per kg, a common low Earth orbit (150 miles/240km) requires:
2.4 MJ of potential gravitational energy
30.7 MJ of kinetic energy

So the kinetic energy is greater by nearly a factor of 13.

In terms of the angle, assuming a flat Earth and linear trajectory, the angle with respect to the ground would be 6.6 degrees. That's pretty low, but since Earth is curved, it would be even lower.

So going up first is really going in the wrong direction, so aerodynamic drag would need to be pretty significant to make up for the otherwise bad trajectory.

wikipedia has some good content here:

The gravity turn is commonly used with launch vehicles such as a rocket or the Space Shuttle that launch vertically. The rocket begins by flying straight up, gaining both vertical speed and altitude. During this portion of the launch, gravity acts directly against the thrust of the rocket, lowering its vertical acceleration. Losses associated with this slowing are known as gravity drag, and can be minimized by executing the next phase of the launch, the pitchover maneuver, as soon as possible. The pitchover should also be carried out while the vertical velocity is small to avoid large aerodynamic loads on the vehicle during the maneuver.[1]

The pitchover maneuver consists of the rocket gimbaling its engine slightly to direct some of its thrust to one side. This force creates a net torque on the ship, turning it so that it no longer points vertically. The pitchover angle varies with the launch vehicle and is included in the rocket's inertial guidance system.[1] For some vehicles it is only a few degrees, while other vehicles use relatively large angles (a few tens of degrees). After the pitchover is complete, the engines are reset to point straight down the axis of the rocket again. This small steering maneuver is the only time during an ideal gravity turn ascent that thrust must be used for purposes of steering. The pitchover maneuver serves two purposes. First, it turns the rocket slightly so that its flight path is no longer vertical, and second, it places the rocket on the correct heading for its ascent to orbit. After the pitchover, the rocket's angle of attack is adjusted to zero for the remainder of its climb to orbit. This zeroing of the angle of attack reduces lateral aerodynamic loads and produces negligible lift force during the ascent.[1]

https://en.m.wikipedia.org/wiki/Gravity_turn

Basically, you pitch over a little and then let it fall the rest of the way.

 

[ZapperZ]

My 12 year old asked me this question. I have a MS in Mechanical Engineering, so I can usually answer his physics questions, but this one stumped me. When lifting off, why do most rockets turn close to horizontal almost immediately? Of course we know they need mostly horizontal speed to achieve orbit, but they also need altitude. If the rocket gained all it's altitude first, vertically, and then made a 90deg turn to the horizontal and then sped up horizontally, the majority of it's high speed flight would be outside of the Earth's atmosphere, greatly decreasing air resistance and wasted energy. Wouldn't this be much more energy efficient?

I do not understand this, and I question the validity of the observation.

What does it mean that the "... rockets turn close to horizontal almost immediately... "? I've watched many rocket launches. They do NOT "turn close to horizontal almost immediately". In fact, they only change orientation after several stages of launch!

So please define "immediately" here.

Zz.

 

[jbriggs444]

If the rocket gained all it's altitude first, vertically, and then made a 90deg turn to the horizontal and then sped up horizontally

The Pythagorean theorem is a hint that making a 90 degree turn is not optimal. If you add a vertical delta-v to a horizontal delta-v the magnitude of the total is less than the sum of the magnitudes of the components.

The Oberth effect is a better clue. Maximum energy efficiency is obtained when thrusting parallel to the current velocity vector and doing so when the craft is at the highest available speed already. Waiting for the craft to come to a stop some hundreds of miles in the air before thrusting horizontally fails to make use of this.

Edit: Barring atmospheric effects and subterranean trajectories, I believe that the most efficient launch would be an impulsive horizontal burn into a Hohmann transfer orbit.

 

[mpresic3]

I can at least give quantitative answers to the first 2:

Per kg, a common low Earth orbit (150 miles/240km) requires:
2.4 MJ of potential gravitational energy
30.7 MJ of kinetic energy

I agree with your figures. The kinetic energy requirement is 13 times the potential energy requirement.

KE = 1/2 (1 kg) vorbital squared = 30.1 Million Joules (per kilogram) at r = ( 6370 km + 240 km ): location of orbit.

PE at orbit location is twice this and negative = - 60.2 Million Joules at r = ( 6370 km + 240 km ): location of orbit, and
PE surface of the Earth = - GMm / (R) = - 62.6 Million Joules at r = ( 6370 km ) : location of Earth Surface

At launch you have - 62.5 million Joules and you need - 60.2 Million Joules. You have to supply the difference: ( - 60.2 - - 62.6) = 2.4 million Joules.
It is like you have to supply energy to get less in the hole. With this interpretation, the total (potential + kinetic) energy for a bound (ellipse or circular) is always negative.

You never quoted it but the comment by Sophiecentaur that the gravitational potential energy is equal to the kinetic energy for a circular orbit is incorrect. (As the above calculation supports, the potential energy is twice as great and negative from the kinetic energy). Sophicentaur's point that without horizontal velocity, the rocket would eventually fall back is true.

The Wikipedia article you supplied has good detail.

Any comments by anyone on my note that the early stages with greater thrust contribute horizontal velocity imposing less "velocity to be gained" demand on the later stages with lesser thrust to achieve orbit

 

[russ_watters]

I do not understand this, and I question the validity of the observation.

What does it mean that the "... rockets turn close to horizontal almost immediately... "? I've watched many rocket launches. They do NOT "turn close to horizontal almost immediately". In fact, they only change orientation after several stages of launch!

So please define "immediately" here.

Zz.

The observation is wrong, but not that wrong: the "immediately" part is correct. The attitude adjustments start as soon as practicable after clearing the tower. Googling, I'm seeing T+15s for the shuttle (which I think is actually 11s after launch).

The "close to horizontal" is the most wrong; it isn't anywhere close to horizontal for a long time. It's a much more gradual arc.

[edit] Looking at some videos, the pitch-over starts immediately after a roll to get the orbiter to the right orbit inclination. It is noticeable, but very slow.

Edit2:
Here's a launch profile:
https://spaceflightnow.com/shuttle/sts124/fdf/124ascentdata.html

You can see from it that it travels vertically until about the time it finishes its roll (+16sec), then it starts pitching and traveling downrange. But it is quite slow. But after a minute, it is 6.8 miles up and 3.3 miles downrange; a 30 degree angle from vertical.

 

[ZapperZ]

The observation is wrong, but not that wrong: the "immediately" part is correct. The attitude adjustments start as soon as practicable after clearing the tower. Googling, I'm seeing T+15s for the shuttle (which I think is actually 11s after launch).

The "close to horizontal" is the most wrong; it isn't anywhere close to horizontal for a long time. It's a much more gradual arc.

[edit] Looking at some videos, the pitch-over starts immediately after a roll to get the orbiter to the right orbit inclination. It is noticeable, but very slow.

Edit2:
Here's a launch profile:
https://spaceflightnow.com/shuttle/sts124/fdf/124ascentdata.html

You can see from it that it travels vertically until about the time it finishes its roll (+16sec), then it starts pitching and traveling downrange. But it is quite slow. But after a minute, it is 6.8 miles up and 3.3 miles downrange; a 30 degree angle from vertical.

Please note that in the first post of this thread, the OP claim that the rocket "... turn close to horizontal almost immediately... ". "30 degree angle from vertical" is not what I consider to be "close to horizontal". This is what I am disputing.

Zz.

 

[russ_watters]

Please note that in the first post of this thread, the OP claim that the rocket "... turn close to horizontal almost immediately... ". "30 degree angle from vertical" is not what I consider to be "close to horizontal". This is what I am disputing.

Zz.

Fair enough; taken as one statement it is pretty wrong. I was responding to what appeared to you to be taking it as two statements. In particular I thought this was confusingly put:

In fact, they only change orientation after several stages of launch!

I may not be understanding what you mean by "several stages of launch", but I think it should be clear the orientation starts changing almost immediately after launch.

 

[OmCheeto]

My 12 year old asked me this question. I have a MS in Mechanical Engineering, so I can usually answer his physics questions, but this one stumped me. When lifting off, why do most rockets turn close to horizontal almost immediately?

Interesting question: "Why do rockets turn horizontally so soon after launch?", from a 12 year old.

Having never watched a rocket launch in real life, I digitized the data from the Falcon Heavy launch from 2 months ago, at 10 second intervals.

My guess as to why rocket scientists do it this way is because they want to travel the least distance through the sand.

"Sand, water, and air" being things you might relate to your 12 yo as being somewhat similar to air density.

 

[Janus]

My 12 year old asked me this question. I have a MS in Mechanical Engineering, so I can usually answer his physics questions, but this one stumped me. When lifting off, why do most rockets turn close to horizontal almost immediately? Of course we know they need mostly horizontal speed to achieve orbit, but they also need altitude. If the rocket gained all it's altitude first, vertically, and then made a 90deg turn to the horizontal and then sped up horizontally, the majority of it's high speed flight would be outside of the Earth's atmosphere, greatly decreasing air resistance and wasted energy. Wouldn't this be much more energy efficient?

Let's look at two different ways of putting a rocket into orbit.
First let's consider your idea. straight up until you reach orbital altitude and then horizontal to go into orbit.
We will assume an orbit at an altitude of 300 km.
First you will have to get up to enough speed to reach that height. This works out to leaving the Earth's surface at ~2.37 km/sec
Once you get to the orbital height, you have to get up to orbital speed, at 300 km, this is ~7.73 km/sec so your total velocity change cost for this is ~10.1 km.sec

Second plan, we lay the rocket on its side on a track and accelerate it. If we get it up to 7.99 km/sec it will have entered an orbit with a perigee at the Earth's surface and an apogee at 300 km. We now just wait for it to reach apogee. When it does it will be moving horizontally and still moving at 7.64 km/sec. This means that we just have to give it an additional 0.09 km/sec to enter a circular orbit. total velocity change cost 8.08 km/s or 2 km/sec less. (with rockets, this change of velocity or "delta v", is a big issue. With today's rockets, that 2 km/sec extra velocity represents 2/3's as much more fuel. )

Of course it isn't practical to launch the rocket sideways like that for a number of reasons. As you pointed out, air resistance would be a huge problem.

The solution is a compromise. We launch the rocket vertically so that it has enough upward velocity that it won't fall back to Earth, and then slowly lean it over. As it climbs it gains speed and gets closer to that elliptical orbit trajectory, but without having to get up to its full orbital speed while in the atmosphere.

This type of take-off works for an airless world too. In the vernacular, this trajectory is called a "gravity turn" and in reverse is even used to land on places like the Moon.

 

[DTM]

I do not understand this, and I question the validity of the observation.

What does it mean that the "... rockets turn close to horizontal almost immediately... "? I've watched many rocket launches. They do NOT "turn close to horizontal almost immediately". In fact, they only change orientation after several stages of launch!

So please define "immediately" here.

Zz.

I meant to say they START to turn horizontally almost immediately. Of course they don't get to be totally horizontal until they're in orbit. Within a few seconds after clearing the tower, the space shuttle began it's roll maneuver. As it rolls, it is already pitching onto it's back. Within 2 minutes I would say it's beyond 45 degrees (closer to horizontal than vertical). I'm not positive about those exact numbers, but it certainly appears that way. Watch a youtube shuttle launch.

 

[OmCheeto]

I meant to say they START to turn horizontally almost immediately. Of course they don't get to be totally horizontal until they're in orbit. Within a few seconds after clearing the tower, the space shuttle began it's roll maneuver. As it rolls, it is already pitching onto it's back. Within 2 minutes I would say it's beyond 45 degrees (closer to horizontal than vertical). I'm not positive about those exact numbers, but it certainly appears that way. Watch a youtube shuttle launch.
View attachment 224176

Thanks!
That is a most awesome picture. I think I can learn some maths from that.

"Launch of Space Shuttle Endeavour from Pad 39B on mission STS-97. This is a five minute exposure taken from the Causeway (about 7 miles away from the pad). Taken with a borrowed all-manual Nikomat camera, 24 mm lens, f/16, Kodak Royal Gold 100 film. When the exposure was stopped in this shot, the shuttle was approximately 229 statute miles downrange." --- Ben Wang [ref]​

 

[rootone]

Atmosphere drag is significant for the first 10km or so of altitude.
The rocket needs to loose that as fast as possible.
Maneuvers after that are a lot more fuel efficient.

 

[rcgldr]

Not mentioned yet is one of the reasons for a vertical launch on large rockets is due to structure issues. Smaller rockets are often launched at a angle, but an angled launch for a large rocket would require a heavy structure for the supporting structure, and a relatively heavy structure within the rocket itself.

 

[sophiecentaur]

. Watch a youtube shuttle launch.

Is this a fair test? So far the thread has implied that it is. The angle of view of an observer / TV camera and the distance involved make it very hard to judge the actual trajectory. Add to that the acceleration, which also affects the appreciation of distance and we really can't rely on what you think we are seeing. The OP quotes the observation of a 12 year old whose appreciation of what was happening would probably be naive (in the good sense) so why are we surprised that the reality may be a bit different?
In a similar vein, I guess we have all watched games of professional tennis on TV and the path of the served ball on the serve is always shown from high over the server's shoulder. That view of the serve is for visual impact and does it tell us about the subtleties of the trajectory?

 

[DTM]

Is this a fair test? So far the thread has implied that it is. The angle of view of an observer / TV camera and the distance involved make it very hard to judge the actual trajectory. Add to that the acceleration, which also affects the appreciation of distance and we really can't rely on what you think we are seeing. The OP quotes the observation of a 12 year old whose appreciation of what was happening would probably be naive (in the good sense) so why are we surprised that the reality may be a bit different?
In a similar vein, I guess we have all watched games of professional tennis on TV and the path of the served ball on the serve is always shown from high over the server's shoulder. That view of the serve is for visual impact and does it tell us about the subtleties of the trajectory?

You're right that angle of view can make such observations difficult. However, as my 12 year old pointed out, if you watch and LISTEN to some of the youtube videos of shuttle launches, the shuttles altitude and range downfield is often announced at regular intervals. I don't recall the exact numbers, but it is very clear from listening to the numbers, that the range downfield is growing MUCH quicker than the altitude. This is a much better data for us to use. My son also noted that when Max Q is announced, the shuttle is way beyond 45 degrees (closer to horizontal then vertical). So the observation that the shuttle is going "partially sideways" way before it's out of the relevant atmosphere is confirmed by that.

 

[sophiecentaur]

Haha. Your son is smarter than I gave him credit for.

 

[A.T.]

So the observation that the shuttle is going "partially sideways" way before it's out of the relevant atmosphere is confirmed by that.

Do you now understand why? If my simple explanation in post #11 was not clear, here is the extended version:

https://en.wikipedia.org/wiki/Gravity_drag

 

[sophiecentaur]

Do you now understand why? If my simple explanation in post #11 was not clear, here is the extended version:

https://en.wikipedia.org/wiki/Gravity_drag

Revisiting that earlier post, it makes me wonder why they don't use wings to aid takeoff. Up as far as several tens of km, a wing would provide some lift. More efficient than a rocket at low speeds. All the combinations must have been tried over the years - as has aircraft-assisted takeoff.

 

[Eric Bretschneider]

Consider high school textbooks on orbital velocity. They commonly show a cannon on a mountain. As their muzzle velocity increases the cannon shells fall further and further away until they reach their starting point (orbital velocity).

Sharp turns cost a lot of energy. If a spacecraft were launched straight up to an altitude of ~300 miles with zero horizontal motion, what is to keep it from falling as it accelerates horizontally? Far better to begin that turn early so when orbital height is reached you are essentially at orbital velocity.

 

[DTM]

Do you now understand why? If my simple explanation in post #11 was not clear, here is the extended version:

https://en.wikipedia.org/wiki/Gravity_drag

A.T. Thank you! I had never heard of gravity drag. It totally makes sense now! I'll have to fully read and comprehend that article, but just reading the first paragraph really does answer my initial question. Thanks!

 

[DaveC426913]

Maybe a primitive comparison to a mundane scenario:

When you finish fixing a flat on the loose gravel shoulder of a highway (let's assume it's empty), do you:
- pull straight out onto the road then make a right turn and begin accelerating? Or do you
- pull out at an angle so that an ever-increasing component of your movement is in the direction you ultimately want to go?

(Oh, and the road is tilted, so that you have to steer to the left, just to keep from sliding back into the gravel shoulder.)

 

[sophiecentaur]

(Oh, and the road is tilted,

That's your "Gravity Drag".

 

[russ_watters]

Revisiting that earlier post, it makes me wonder why they don't use wings to aid takeoff. Up as far as several tens of km, a wing would provide some lift. More efficient than a rocket at low speeds. All the combinations must have been tried over the years - as has aircraft-assisted takeoff.

They don't fly at low speed for long, and that's a lot of added aerodynamic stress to design for. The takeoff weight is about 20x the landing weight!

 

[sophiecentaur]

They don't fly at low speed for long,

Why not a high speed wing design then? Not much use at low altitude but a comparatively long time at supersonic speeds? I'm not arguing - just after a fuller picture.

 

[russ_watters]

Why not a high speed wing design then? Not much use at low altitude but a comparatively long time at supersonic speeds? I'm not arguing - just after a fuller picture.

I don't see how this changes the answer: massive aerodynamic stresses for little* benefit, for a short time. Except perhaps to say that in addition to going fast, the atmosphere gets thin fast too.

[edit]
*Given that it isn't done at all, it is probably better to say no net benefit. Any lift you add increases drag too, and that's a net harm, not benefit, to what you are trying to achieve.

 

[sophiecentaur]

Except perhaps to say that in addition to going fast, the atmosphere gets thin fast too.

I guess that is the answer I was after - plus the fact that they don't tend to do it (except in experimental systems (X15 etc)
The principle is obviously not favoured for good reason.

 

[DTM]

Maybe a primitive comparison to a mundane scenario:

When you finish fixing a flat on the loose gravel shoulder of a highway (let's assume it's empty), do you:
- pull straight out onto the road then make a right turn and begin accelerating? Or do you
- pull out at an angle so that an ever-increasing component of your movement is in the direction you ultimately want to go?

(Oh, and the road is tilted, so that you have to steer to the left, just to keep from sliding back into the gravel shoulder.)

View attachment 224211

Very good analogy. Thanks DaveC.

 

[cjl]

My son also noted that when Max Q is announced, the shuttle is way beyond 45 degrees (closer to horizontal then vertical). So the observation that the shuttle is going "partially sideways" way before it's out of the relevant atmosphere is confirmed by that.

It's worth noting that this is incorrect. At max q, it's flying at a flight path angle a bit less than 45 degrees off vertical (though it's pretty close to 45 at that point), as can be seen in the STS-116 ascent data here: https://spaceflightnow.com/shuttle/sts116/fdf/116ascentdata.html

As for why this is done? Rockets weigh a lot and have a lot of thrust. Thus, aerodynamic losses are actually fairly small. On the other hand, gravity losses (which occur due to the thrust fighting gravity while flying vertically) are much more significant. As you pitch away from the vertical, gravity losses shrink, since more of the thrust is accelerating the rocket and less is being wasted just holding it up. From what I can find, out of a total ΔV budget of a bit over 9km/s, the space shuttle lost about 107 m/s to drag loss, and about 1222 m/s to gravity losses, so this gives a good idea why you would want to start accelerating away from vertical quickly, in order to reduce the gravity loss even at the cost of some drag, since gravity loss is 10x more significant anyways.

 

[DaveC426913]

As you pitch away from the vertical, gravity losses shrink, since more of the thrust is accelerating the rocket and less is being wasted just holding it up.

I don't doubt you, I'm just having trouble wrapping my head around this.
Why would gravity losses shrink? I mean, doesn't it take just as much fuel to hold up against gravity regardless of the angle?
Taking that logic to the limit, wouldn't that mean loss due to gravity would approach zero as you approach horizontal?

 

[cjl]

It's basically just a function of trigonometry. If your rocket is thrusting with 1.4 times the force of the rocket's weight (pretty standard right after liftoff), and you're aimed straight up, your net acceleration is 0.4G, with 1G loss (most of the thrust is wasted just holding the rocket up). If you've got the same scenario but you're pitched over to 45 degrees, the vertical component of the thrust is equal to the rocket's weight, so you have zero vertical acceleration, and the horizontal component is 1G (each is equal to 1.4/sqrt(2)). Even though the vertical component still loses 1G, the rocket is only accelerating at a rate 0.4G less than what would occur in a zero-gravity environment, so it is gaining velocity more efficiently.

Also, it's true that gravity loss goes to zero as you approach horizontal, but your ability to do that is constrained by the requirement that the rocket gain altitude and not crash during the ascent.

 

[russ_watters]

From what I can find, out of a total ΔV budget of a bit over 9km/s, the space shuttle lost about 107 m/s to drag loss, and about 1222 m/s to gravity losses, so this gives a good idea why you would want to start accelerating away from vertical quickly, in order to reduce the gravity loss even at the cost of some drag, since gravity loss is 10x more significant anyways.

That's a great factoid, largely answering the 3rd part of the OP's energy budget question.

 

[russ_watters]

It's basically just a function of trigonometry. If your rocket is thrusting with 1.4 times the force of the rocket's weight (pretty standard right after liftoff), and you're aimed straight up, your net acceleration is 0.4G, with 1G loss (most of the thrust is wasted just holding the rocket up). If you've got the same scenario but you're pitched over to 45 degrees, the vertical component of the thrust is equal to the rocket's weight, so you have zero vertical acceleration, and the horizontal component is 1G (each is equal to 1.4/sqrt(2)). Even though the vertical component still loses 1G, the rocket is only accelerating at a rate 0.4G less than what would occur in a zero-gravity environment, so it is gaining velocity more efficiently.

Also, it's true that gravity loss goes to zero as you approach horizontal, but your ability to do that is constrained by the requirement that the rocket gain altitude and not crash during the ascent.

Well; note it also drops to zero as your speed rises to orbital speed. That's the whole reason you are gaining the speed!

The equation for centripetal acceleration is a square function of velocity, so for your 1.4g engine output example, at half orbital speed the angle for zero vertical acceleration is 32 degrees instead of 45.

This is part of the reason why the total gravity drag is so small (about 15% of final speed) despite starting so large (about 67%* of takeoff acceleration).

*Note: I'm uncertain of that number. The profile I posted lists 1.5g, which I initially assumed was speed change, but looking at the speed numbers it appears to be total (0.5g of speed change). That's lower than I thought and also doesn't make sense as they also say acceleration start at 0, instead of 1. But I think I need to trust the speed numbers.

 

[cjl]

Well; note it also drops to zero as your speed rises to orbital speed. That's the whole reason you are gaining the speed!

The equation for centripetal acceleration is a square function of velocity, so for your 1.4g engine output example, at half orbital speed the angle for zero vertical acceleration is 32 degrees instead of 45.

This is part of the reason why the total gravity drag is so small (about 15% of final speed) despite starting so large (about 67%* of takeoff acceleration).

*Note: I'm uncertain of that number. The profile I posted lists 1.5g, which I initially assumed was speed change, but looking at the speed numbers it appears to be total (0.5g of speed change). That's lower than I thought and also doesn't make sense as they also say acceleration start at 0, instead of 1. But I think I need to trust the speed numbers.

That's about right - most rockets lift off with a thrust to weight ratio of between 1.2 and 1.6 to 1, so they waste most of the thrust just fighting gravity. However, note that the vast majority of the rocket's weight is fuel, so the thrust to weight ratio improves dramatically as the fuel burns away. As a result, most rockets finish their burns at 4-6G of acceleration, in some cases even throttling down to reduce structural loading as the acceleration climbs. This also helps minimize gravity drag later in the burn, since 5G at a nearly horizontal flight path angle really has next to no gravity loss.

 

[Curiose]

So that the SRB's can splash down and not fall directly on the space center they just took off from 50km ago :D It's not a very elegant answer mathematically but it does account for why they would need to move horizontally so quickly.

 

[OmCheeto]

So that the SRB's can splash down and not fall directly on the space center they just took off from 50km ago :D It's not a very elegant answer mathematically but it does account for why they would need to move horizontally so quickly.

Funny, I was just reading about that while looking for something called "speed bonus".

Finally, it should be noted that there are additional reasons for launching the Shuttle to the east in addition to the speed bonus. The primary reason for launching to the east from Cape Canaveral is because of range safety constraints. A Shuttle will not launch due west from the Cape since it would be launching over populated land. The same rule applies for launches off the West coast. If it is necessary to launch a rocket to the west or into polar orbit, it must be done from the West coast, again because of range safety.
- answer by Joe Yoon, 16 March 2003​

[ref: aerospaceweb.org ]

 

[A.T.]

Why would gravity losses shrink?

https://en.wikipedia.org/wiki/Gravity_drag

 

[OmCheeto]

Now I know why they call it "Rocket Science".

Ehr... Mehr... Gerd...

ps. Can anyone tell me what "downrange" means in "Rocket Science"?

From the video:

10:29 video time audio; "30 seconds into the flight, Atlantis almost 2 miles in altitude, almost 6 miles downrange from the Kennedy Space Center already, traveling 500 mph."

and from the video snapshot:

the range was only 0.1 statute miles.

pps. As a reminder, I am not a rocket scientist.