I Why do rockets turn horizontally so soon after launch?

[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.

 

[rootone]

Simply means the amount of distance the rocket has traveled horizontally with regard to the Earths surface.
The distance between the place where it is now directly above the surface compared to the launch site.
I'm not sure why the audio mentions a different figure to what the data display is showing.
There is also a discrepancy with the velocity which has been achieved at this point.

 

[sophiecentaur]

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

Don't worry; it's not brain surgery.

 

[russ_watters]

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.

I don't think you need to be a rocket scientist to tell that you can't get 6 miles in any direction by accelerating from zero to 500mph in 30 sec...

 

[OmCheeto]

I don't think you need to be a rocket scientist to tell that you can't get 6 miles in any direction by accelerating from zero to 500mph in 30 sec...

That's what I thought, until I watched another shuttle launch.
I assumed that the person in the first video(STS-129) didn't see a decimal point or something.
But then...

STS-127

video time: 1:03

liftoff​

video time: 1:33

t+ 30 seconds

"400 mph
1 mile in altitude and
7 miles downrange"​

video time: 2:13

t+ 70 seconds

"1100 mph
10 miles in altitude and
10 miles downrange"​

So it went sideways really fast, and then it went upways really fast?
??

Do they add in the 915 mph the moment spacecraft leave the ground?
I'm grasping for rocket science straws here.

 

[jbriggs444]

video time: 1:33

t+ 30 seconds

"400 mph
1 mile in altitude and
7 miles downrange"​

video time: 2:13

t+ 70 seconds

"1100 mph
10 miles in altitude and
10 miles downrange"​

Sounds like non-zero starting offset from the tracking station.

 

[russ_watters]

Sounds like non-zero starting offset from the tracking station.

Yeah, I was thinking the same thing. @OmCheeto , coming from multiple sources it probably is on purpose. My guess is they are using a non-rotating reference frame since ultimately that's what your orbital speed is calculated with respect to. The numbers seem to work; it's 840 mph, which sounds about right (dont feel like trying to calculate what it should be).

 

[jbriggs444]

I found some data that looks reliable. It is, at least, detailed.

https://spaceflightnow.com/shuttle/sts124/fdf/124ascentdata.html

Edit: *blush*, old news. Sorry @russ_watters

 

[russ_watters]

I found some data that looks reliable. It is, at least, detailed.

https://spaceflightnow.com/shuttle/sts124/fdf/124ascentdata.html

Did you find it on page 1 of the thread?

That data is nice because it includes the speed with respect to a stationary Earth in a separate column from rocket velocity. And it doesn't include that speed in its range calculation, which makes better sense to me. Since I hadn't looked at that column, I didn't note the speed: 914mph. That's 7.6 miles in 30 seconds; in the same ballpark as @OmCheeto 's numbers.

 

[jbriggs444]

Did you find it on page 1 of the thread?

That data is nice because it includes the speed with respect to a stationary Earth in a separate column from rocket velocity. And it doesn't include that speed in its range calculation, which makes better sense to me. Since I hadn't looked at that column, I didn't note the speed: 914mph. That's 7.6 miles in 30 seconds; in the same ballpark as @OmCheeto 's numbers.

The problem with that theory is the video's claim of 10 miles at T+70. That's another 3 miles in 40 seconds. The speed is decreasing?!

But yes, the posted speed numbers in the table seem to be based on an Earth-centered inertial frame.

 

[russ_watters]

The problem with that theory is the video's claim of 10 miles at T+70. That's another 3 miles in 40 seconds. The speed is decreasing?!

Well, I had a thought about how the orbit's inclination angle affects how the the downrange speed and distance change (the Earth isn't quite spinning in the right direction for that 914mph to be completely helpful. I haven't tried to put any effort into developing that thought...