A kerosene-fueled X-33 as a single stage to orbit vehicle.
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From: Robert Clark on 15 Jun 2010 15:51 On Jun 14, 12:42 pm, Robert Clark <rgregorycl...(a)yahoo.com> wrote: > On Jun 13, 11:36 am, Robert Clark <rgregorycl...(a)yahoo.com> wrote: > > > > > ...I what I want to investigate is > > if it is possible for a rocket that does not have wings or lifting > > surfaces to travel at such a straight-line trajectory at an angle from > > lift-off so that the achieved velocity will be in this range (but > > keeping in mind this might not be the same as the equivalent "delta-V" > > that actually has to be put out by the engines.) > > But I have question: if you angle the rocket launch from the start > > with the thrust vector along the center line with the trajectory angle > > such that the vertical component of the thrust equals the rocket > > weight could you have the rocket travel at a straight-line all the way > > to orbit? I'm inclined to say no because the gravity is operating at > > the center of gravity of the rocket not at the tail where the thrust > > is operating. This would certainly work if you had a point particle, > > but I'm not sure if it would work when your body has some linear > > extent. > > This method for traveling at a straight-line at an angle for some or > > all of the trip would make my calculation easier. However, I'll show > > in a following post there is another way to do it even if this first > > method doesn't work. > > The second method though would require some modification to the usual > > design of rockets and is more computationally complicated. > > > Bob Clark > > The question can be boiled down to this: imagine you have a long > cylindrical object, could be a pencil, could be broom stick. You can > give it an initial thrust at the bottom and push it away at an angle. > It will then follow a parabolic trajectory with its center of mass > following a parabolic arc, disregarding air drag. > What I'm asking is will it work to supply a continual push at the > bottom with the force maintained at the bottom at a fixed angle to the > horizontal so that the vertical component of this force is the > cylindrical body's weight? > Will the body maintain a continual straight-line trajectory at this > set fixed angle? This is really actually a question in continuum mechanics sometimes called solid mechanics. In physics we sometimes idealize a body subject to forces as a point particle. Idealized this was, the thrust force applied to the rocket would add as a vector to the force of gravity so it would cancel gravity no matter what the angle of the trajectory. But in continuum mechanics you have to consider the physical extent of the body and where on the body the forces are applied. I imagine this is a common type of problem addressed in mechanical engineering and civil engineering. A force applied at one position on the body won't have the same effect as when it is applied to another point for instance in regards to the torque produced. Torque is measuring the turning "force" on the body. But it's defined as the cross product of the force vector times the radial vector to the center of rotation. Intuitively, what we have to worry about is rotation of the rocket with the thrust applied only at the tail. But if the rocket were to rotate it would be about the center of gravity. However, if we make it so the thrust is always along the center line, the radial vector to the cg and the force vector are parallel, resulting in a 0 torque vector. That would mean there would be no rotation around the center of gravity so we should be able to maintain our straight-line trajectory. This would be valid if the center of gravity were fixed. But the cg is accelerating as the thrust is applied. So I'm not sure if this argument applies in that case. Bob Clark
From: Alain Fournier on 15 Jun 2010 21:30 Robert Clark wrote: > On Jun 13, 11:36 am, Robert Clark <rgregorycl...(a)yahoo.com> wrote: > >>...I what I want to investigate is >>if it is possible for a rocket that does not have wings or lifting >>surfaces to travel at such a straight-line trajectory at an angle from >>lift-off so that the achieved velocity will be in this range (but >>keeping in mind this might not be the same as the equivalent "delta-V" >>that actually has to be put out by the engines.) >>But I have question: if you angle the rocket launch from the start >>with the thrust vector along the center line with the trajectory angle >>such that the vertical component of the thrust equals the rocket >>weight could you have the rocket travel at a straight-line all the way >>to orbit? I'm inclined to say no because the gravity is operating at >>the center of gravity of the rocket not at the tail where the thrust >>is operating. This would certainly work if you had a point particle, >>but I'm not sure if it would work when your body has some linear >>extent. >>This method for traveling at a straight-line at an angle for some or >>all of the trip would make my calculation easier. However, I'll show >>in a following post there is another way to do it even if this first >>method doesn't work. >>The second method though would require some modification to the usual >>design of rockets and is more computationally complicated. >> >> Bob Clark > > > The question can be boiled down to this: imagine you have a long > cylindrical object, could be a pencil, could be broom stick. You can > give it an initial thrust at the bottom and push it away at an angle. > It will then follow a parabolic trajectory with its center of mass > following a parabolic arc, disregarding air drag. > What I'm asking is will it work to supply a continual push at the > bottom with the force maintained at the bottom at a fixed angle to the > horizontal so that the vertical component of this force is the > cylindrical body's weight? > Will the body maintain a continual straight-line trajectory at this > set fixed angle? For low speeds that would be a good approximation. As you get faster you get nearer to weightlessness. Therefore the force required to fight gravity gets smaller. At low speeds the effect is small, at higher the change in weight per change in speed gets bigger. So if you have a large delta-V or if you start at high speeds, the approximation is no longer very good. I'm not sure why you are asking the question. If what you had in mind is that you could by this way reach orbit without needing to have the rocket turn and therefore not needing to have directional control on the rocket then I would say that you are both misguided (you and your rocket). Even if you your flight plan doesn't require to turn the rocket you need directional control. It is like holding a stick vertically on your finger, if you place it perfectly vertical and you don't move at all and there is absolutely no air movement, the stick will hold steady. But in practice, you didn't put the stick perfectly vertical, you do move a little, there is some airflow and you need some control to keep it from falling. You could spin stabilize it, but good luck doing so for a rocket giving the kind of acceleration needed to reach orbit. Alain Fournier
From: Jeroen Belleman on 16 Jun 2010 04:07 Robert Clark wrote: > On Jun 14, 12:42 pm, Robert Clark <rgregorycl...(a)yahoo.com> wrote: >> On Jun 13, 11:36 am, Robert Clark <rgregorycl...(a)yahoo.com> wrote: >> >>> ...I what I want to investigate is >>> if it is possible for a rocket that does not have wings or lifting >>> surfaces to travel at such a straight-line trajectory at an angle from >>> lift-off [...] >> Will the body maintain a continual straight-line trajectory at this >> set fixed angle? The system is dynamically unstable. In control system terminology, the orientation as a function of thrust contains a double integration. The slightest disturbance or error will make it diverge ever faster. So, no, it won't maintain a straight-line trajectory. Jeroen Belleman
From: Robert Clark on 16 Jun 2010 12:50 On Jun 15, 9:30 pm, Alain Fournier <alain...(a)sympatico.ca> wrote: > ... > > > The question can be boiled down to this: imagine you have a long > > cylindrical object, could be a pencil, could be broom stick. You can > > give it an initial thrust at the bottom and push it away at an angle. > > It will then follow a parabolic trajectory with its center of mass > > following a parabolic arc, disregarding air drag. > > What I'm asking is will it work to supply a continual push at the > > bottom with the force maintained at the bottom at a fixed angle to the > > horizontal so that the vertical component of this force is the > > cylindrical body's weight? > > Will the body maintain a continual straight-line trajectory at this > > set fixed angle? > > For low speeds that would be a good approximation. As you get faster > you get nearer to weightlessness. Therefore the force required to > fight gravity gets smaller. At low speeds the effect is small, at > higher the change in weight per change in speed gets bigger. So if > you have a large delta-V or if you start at high speeds, the > approximation is no longer very good. > > I'm not sure why you are asking the question. If what you had in mind > is that you could by this way reach orbit without needing to have > the rocket turn and therefore not needing to have directional control > on the rocket then I would say that you are both misguided (you and > your rocket). Even if you your flight plan doesn't require to turn > the rocket you need directional control. It is like holding a stick > vertically on your finger, if you place it perfectly vertical and you > don't move at all and there is absolutely no air movement, the stick > will hold steady. But in practice, you didn't put the stick perfectly > vertical, you do move a little, there is some airflow and you need > some control to keep it from falling. You could spin stabilize it, but > good luck doing so for a rocket giving the kind of acceleration needed > to reach orbit. > > Alain Fournier The idea behind it is straight-line is the shortest distance between two points. This works also for velocities. That is if you want a body to have two specified speeds in the x and y directions, then the *magnitude* of the velocity you need to achieve can be minimized by traveling at an angle so that x and y components have the speeds you want. For correction of variations in the trajectory, the usual idea of gimbaling the engines would work in this scenario. Bob Clark
From: Alain Fournier on 16 Jun 2010 20:19
Robert Clark wrote: > On Jun 15, 9:30 pm, Alain Fournier <alain...(a)sympatico.ca> wrote: >> >>> The question can be boiled down to this: imagine you have a long >>>cylindrical object, could be a pencil, could be broom stick. You can >>>give it an initial thrust at the bottom and push it away at an angle. >>>It will then follow a parabolic trajectory with its center of mass >>>following a parabolic arc, disregarding air drag. >>> What I'm asking is will it work to supply a continual push at the >>>bottom with the force maintained at the bottom at a fixed angle to the >>>horizontal so that the vertical component of this force is the >>>cylindrical body's weight? >>> Will the body maintain a continual straight-line trajectory at this >>>set fixed angle? >> >>For low speeds that would be a good approximation. As you get faster >>you get nearer to weightlessness. Therefore the force required to >>fight gravity gets smaller. At low speeds the effect is small, at >>higher the change in weight per change in speed gets bigger. So if >>you have a large delta-V or if you start at high speeds, the >>approximation is no longer very good. >> >>I'm not sure why you are asking the question. >> >>Alain Fournier > > The idea behind it is straight-line is the shortest distance between > two points. This works also for velocities. That is if you want a body > to have two specified speeds in the x and y directions, then the > *magnitude* of the velocity you need to achieve can be minimized by > traveling at an angle so that x and y components have the speeds you > want. For reaching orbit in the absence of an atmosphere, the most energy efficient path is to keep the vertical component of the force just enough to keep you from going higher or lower until your horizontal component is equal to orbital speed. So this somewhat what you were saying up there. But do keep in mind that the vertical component of the force needed goes down as your orbital velocity approaches orbital speed, and as the mass of the rocket goes down due to expenditure of fuel. So the vertical component of the force goes down until it reaches zero at orbital velocity. Alain Fournier |