A kerosene-fueled X-33 as a single stage to orbit vehicle.
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From: hallerb on 13 Jun 2010 20:00 On Jun 13, 7:03�pm, "Androcles" <Headmas...(a)Hogwarts.physics_z> wrote: > <hall...(a)aol.com> wrote in message > > news:e183261a-8f6a-4b59-b411-f4416c22b558(a)20g2000vbi.googlegroups.com... > On Jun 13, 11:36 am, Robert Clark <rgregorycl...(a)yahoo.com> wrote: > > > > > > > It is known that you can reduce the fuel requirements to orbit by > > using a lifting trajectory. For instance it was explored for use on a > > reusable orbital craft in the 80's: > > > Fire in the sky: the Air Launched Sortie Vehicle of the early 1980s > > (part 1) > > by Dwayne Day > > Monday, February 22, 2010http://www.thespacereview.com/article/1569/1 > > > In the first post in this thread I suggested this fact be used for a > > lifting-body SSTO. Here is the argument I made: > > > ------------------------------------------------------------------ > > "Then the gravity losses could be further reduced by flying a lifting > > trajectory, which would also increase the payload capability by a > > small percentage. > > "The trajectory I'll use to illustrate this will first be straight- > > line at an angle up to some high altitude that still allows > > aerodynamic lift to operate. At the end of this portion the vehicle > > will have some horizontal and vertical component to its velocity. > > "We'll have the vertical component be sufficient to allow the vehicle > > to reach 100 km, altitude. The usual way to estimate this vertical > > velocity is by using the relation between kinetic energy and potential > > energy. It gives the speed of v = sqrt(2gh) to reach an altitude of h > > meters. At 100,000 m, v is 1,400 m/s. > > "Now to have orbital velocity you need 7,800 m/s tangential, i.e., > > horizontal velocity. If you were able to fly at a straight-line at a > > constant angle to reach 7,800 m/s horizontal velocity and 1,400 m/s > > vertical velocity and such that the air drag was kept at the usual low > > 100 to 150 m/s then you would only need sqrt(7800^2 + 1400^2) = 7,925 > > m/s additional delta-v to reach orbit. Then the total delta-v to orbit > > might only be in the 8,100 m/s range. Note this is significantly less > > than the 9,200 m/s delta-v typically needed for orbit, including > > gravity and air drag. > > "The problem is with usual rocket propulsion to orbit not using lift > > the thrust vector has to be more or less along the center-line of the > > rocket otherwise the rocket would tumble. You can gimbal the engines > > only for a short time to change the rocket's attitude but the engines > > have to be then re-directed along the center line. However, the center > > line has to be more or less pointing into the airstream, i.e., > > pointing in the same direction as the velocity vector, to reduce > > aerodynamic stress and drag on the vehicle. But the rocket thrust > > having to counter act gravity means a large portion of the thrust has > > to be in the vertical component which means the thrust vector has to > > be nearly vertical at least for the early part of the trip when the > > gross mass is high. Then the thrust vector couldn't be along the > > center line of a nearly horizontally traveling rocket at least during > > the early part of the trip." > > ------------------------------------------------------------------ > > > The key fact I want to focus on is that sqrt(7800^2 + 1400^2) = 7,925 > > m/s number. Actually if you add on the ca. 460 m/s velocity you get > > for free from the Earth's rotation you might be able to reduce this to > > sqrt(7400^2 + 1400^2) = 7,531 m/s. So 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 math is over my head, but if the vehicle can refuel along the way, > with air to air refueling done in the military all the time the > payload can be a lot more > > ==================================================== > Altitude of International Space Station = 200 miles. > Altitude of Mount Everest = 6 miles. > Speed of International Space Station in orbit = 17,000 mph > Speed of Military Tanker with fuel = 500 mph > Let's refuel at 30,000 feet, 500 mph. > Only 194 miles and 16,500 mph to go. > What is your idea of a "lot" more? > Are we there yet? > Are we there yet? > Are we there yet?- Hide quoted text - > > - Show quoted text - refuel at 50,000 feet, acceleration from that high should be easier the vehicle out of the worst of the gravity well and air nice and thin..... less aerodynamic slowing. ideally probably a 2 stage vehicle.... the ground based portion optimized for air breathing engines. the 2nd stage to orbit a small shuttle like vehicle to take people to orbit big dumb boosters for heavy cargo
From: Androcles on 13 Jun 2010 20:29 <hallerb(a)aol.com> wrote in message news:8a62361f-b1eb-4eaf-82ad-b5b6e73d2d8e(a)u7g2000yqm.googlegroups.com... On Jun 13, 7:03�pm, "Androcles" <Headmas...(a)Hogwarts.physics_z> wrote: > <hall...(a)aol.com> wrote in message > the math is over my head, but if the vehicle can refuel along the way, > with air to air refueling done in the military all the time the > payload can be a lot more > > ==================================================== > Altitude of International Space Station = 200 miles. > Altitude of Mount Everest = 6 miles. > Speed of International Space Station in orbit = 17,000 mph > Speed of Military Tanker with fuel = 500 mph > Let's refuel at 30,000 feet, 500 mph. > Only 194 miles and 16,500 mph to go. > What is your idea of a "lot" more? > Are we there yet? > Are we there yet? > Are we there yet?- Hide quoted text - > > - Show quoted text - refuel at 50,000 feet, ======================================== Bwahahahahahahahahahahahahaha! Tankers can't fly to 50,000 feet, that's U2 spy plane altitude. http://www.inquisitr.com/wp-content/u2.jpg Big wings in thin air, light payload, it carries a pilot and a camera. Only 194 miles and 16,500 mph to go. The math is over your head. Are we there yet? Are we there yet? Are we there yet? What is your idea of a "lot" more?
From: Pat Flannery on 14 Jun 2010 03:20 On 6/13/2010 4:29 PM, Androcles wrote: > > Tankers can't fly to 50,000 feet, that's U2 spy plane altitude. KC-135 tankers can reach 50,000 feet; U-2's operate in the 70,000 - 75,000 foot area. However, the point is valid...other than reduced air drag as the spacecraft accelerates to orbital altitude and speed, the air-refueled option isn't really a game changer, and adds a lot of complexity to the whole launch plan. In fact, when you consider that the spacecraft to be refueled must be flying horizontally and slow enough to link up with the tanker aircraft, it may make more sense just to launch it vertically and put up with the air drag. Pat
From: Robert Clark on 14 Jun 2010 12:42 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? Bob Clark
From: Jeff Findley on 14 Jun 2010 16:32
In article <2b63305f-94cd-4fe0-bdc8-63b724ffcb01 @z26g2000vbk.googlegroups.com>, rgregoryclark(a)yahoo.com says... > > 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? You seem very confused about the fundamentals of rocket stability in the atmosphere (which are obviously different than in vacuum). You need to read up on model rocket design and pay close attention to the parts which talk about stability, center of gravity, and center of pressure. There are some very good books out there, but the ones I vaguely remember are from the 70's and 80's. Fortunately, the fundamentals have not changed over the years. I'm sure there is something on the web about this, but I'm not going to do your searching for you. Jeff -- The only decision you'll have to make is Who goes in after the snake in the morning? |