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Space Forum / Astronomy / July 2008Short Mars travel times at high speed.
The distance from Earth to Mars is about 60,000,000 km at closest approach. If we have a 30 km/sec initial velocity to Mars, which might be achievable with airbreathing(scramjet) or nuclear propulsion then the travel time might be 23 days if you make a simplifying assumption of a straight-line trip. However, the time required to make the journey might be made significantly better than this 23 days. The key fact is that the Earth itself has a 30 km/s velocity around the Sun that can be used to give us an extra velocity boost toward the orbit of Mars. In this new estimate I'll simplify the analysis by assuming that at this high velocity and at the short travel time achieved, the path will be essentially straight, rather than the actual ellipse. The famous Hohmann transfer orbit gives a minimal delta-v and energy solution for traveling from one orbit to another but this is known for its long travel times, 6 to 7 months for a Earth to Mars trip. We want to shorten that for a manned trip to reduce the exposure to radiation and to reduce the effects of long periods in zero-g. I'll take the Earth orbit radius to be 150 million km and the Mars orbit radius to be a little more than its distance at perihelion 210 million km. If we went in a tangential direction to Earth's orbit we would have a total velocity toward the orbit of Mars of 60 km/s. The problem here is that we would also have a longer straight-line travel distance. This would result in the travel time being longer than moving radially at 30 km/s. So the idea is to move at an optimal angle that can use the Earth's orbital velocity while at the same time not making the travel distance too long. See the image here for the diagrams to illustrate the addition of velocities at an angle θ (theta) and the travel distance at the angle θ calculations: http://www.advancedphysics.org/forum/attachment.php?attachmentid=282 (may need to do a free registration at www.advancedphysics.org to access the image.) In the diagrams v is the total velocity, r is Earth's orbital radius, R is Mars orbital radius and d is the straight-line travel distance. Applying the law of cosines for the velocities gives for the total velocity: v^2 = 30^2 + 30^2 -2(30)(30)cos(180-θ) = 2(30^2)(1 + cos(θ)) So v = 30*sqrt(2(1 + cos(θ)) Applying the law of cosines for the travel distance gives the equation: R^2 = r^2 + d^2 -2(r)(d)cos(90 + θ) = r^2 + d^2 + 2(r)(d)sin(θ) Solving for the travel distance d using the quadratic formula gives: d = -rsin(θ) + sqrt(R^2 -(rcos(θ))^2) I created a table using various angles θ in fractions of π (pi) radians to find the shortest trip time: θ (radians)| time (days) ----------------------------- 0 | 28.36 π/2 | 16.4 π/3 | 11.4 π/4 | 12.9 π/5 | 14.8 π/6 | 16.4 π/7 | 17.7 π/8 | 18.7 π/9 | 19.6 We see the shortest time at π/3 or 60 degrees is a surprising 11.4 days. Quite a significant advantage than taking a 6 month long Hohmann transfer orbit. I'll assume like Robert Zubrin, author of the "Mars Direct" plan, that aerobraking can be used to stop at Mars on arrival even at such high speeds. How reasonable is the assumption that at such high speeds and the resulting short travel times the straight-line approximation is accurate? Bob Clark The distance from Earth to Mars is about 60,000,000 km at closest approach. If we have a 30 km/sec initial velocity to Mars, which might be achievable with airbreathing(scramjet) or nuclear propulsion then the travel time might be 23 days if you make a simplifying assumption of a straight-line trip. However, the time required to make the journey might be made significantly better than this 23 days. The key fact is that the Earth itself has a 30 km/s velocity around the Sun that can be used to give us an extra velocity boost toward the orbit of Mars. In this new estimate I'll simplify the analysis by assuming that at this high velocity and at the short travel time achieved, the path will be essentially straight, rather than the actual ellipse. The famous Hohmann transfer orbit gives a minimal delta-v and energy solution for traveling from one orbit to another but this is known for its long travel times, 6 to 7 months for a Earth to Mars trip. We want to shorten that for a manned trip to reduce the exposure to radiation and to reduce the effects of long periods in zero-g. I'll take the Earth orbit radius to be 150 million km and the Mars orbit radius to be a little more than its distance at perihelion 210 million km. If we went in a tangential direction to Earth's orbit we would have a total velocity toward the orbit of Mars of 60 km/s. The problem here is that we would also have a longer straight-line travel distance. This would result in the travel time being longer than moving radially at 30 km/s. So the idea is to move at an optimal angle that can use the Earth's orbital velocity while at the same time not making the travel distance too long. See the image here for the diagrams to illustrate the addition of velocities at an angle è (theta) and the travel distance at the angle è calculations: http://www.advancedphysics.org/forum/attachment.php?attachmentid=282 (may need to do a free registration at www.advancedphysics.org to access the image.) In the diagrams v is the total velocity, r is Earth's orbital radius, R is Mars orbital radius and d is the straight-line travel distance. Applying the law of cosines for the velocities gives for the total velocity: v^2 = 30^2 + 30^2 -2(30)(30)cos(180-è) = 2(30^2)(1 + cos(è)) So v = 30*sqrt(2(1 + cos(è)) Applying the law of cosines for the travel distance gives the equation: R^2 = r^2 + d^2 -2(r)(d)cos(90 + è) = r^2 + d^2 + 2(r)(d)sin(è) Solving for the travel distance d using the quadratic formula gives: d = -rsin(è) + sqrt(R^2 -(rcos(è))^2) I created a table using various angles è in fractions of ð (pi) radians to find the shortest trip time: è (radians)| time (days) ----------------------------- 0 | 28.36 ð/2 | 16.4 ð/3 | 11.4 ð/4 | 12.9 ð/5 | 14.8 ð/6 | 16.4 ð/7 | 17.7 ð/8 | 18.7 ð/9 | 19.6 We see the shortest time at ð/3 or 60 degrees is a surprising 11.4 days. Quite a significant advantage than taking a 6 month long Hohmann transfer orbit. I'll assume like Robert Zubrin, author of the "Mars Direct" plan, that aerobraking can be used to stop at Mars on arrival even at such high speeds. How reasonable is the assumption that at such high speeds and the resulting short travel times the straight-line approximation is accurate? Bob Clark ============================================== Incredibly naive... Look at a shuttle and the size of the fuel tank needed to lift that to Earth orbit so that it travels at 27,755 km per HOUR to match speed with the ISS. http://spaceflight.nasa.gov/realdata/tracking/ Now you want to travel at more than 3,600 times that, 27,755 km per SECOND. That means you have to lift into Earth orbit a rocket so huge it just doesn't bear thinking about and THEN fire THAT up to travel to Mars. But that's only half the story, you have to carry as much fuel mass at the other end to stop it again. Aero braking? Columbia broke up using aero braking, and your ship is going 3,600 times faster than that. And for the return trip... What you need is a sleigh as fast as Santa's that can deliver all the prezzies to all the kiddies all in one night. In other words, magic. Do your sums again. ... > Incredibly naive... > Look at a shuttle and the size of the fuel tank needed to lift that [quoted text clipped - 12 lines] > 3,600 times faster than that. > And for the return trip... You misread the speed. Orbital velocity for low Earth orbit which the shuttle reaches is about 7500 m/s or 7.5 km/sec. I'm suggesting a speed of 4 times that 30 km/sec. Every shuttle on return uses aerobraking under its normal aerodynamic configuration to land. It was because a wing suffered severe damage that caused Columbia to lose its normal configuration for landing that caused it to break up. Bob Clark | ... | > Incredibly naive... [quoted text clipped - 15 lines] | | You misread the speed. I gave the web page above. Here it is again. http://spaceflight.nasa.gov/realdata/tracking/ Tell NASA they don't know what they are talking about, I'm sure they'd enjoy the laugh. | Orbital velocity for low Earth orbit which the | shuttle reaches is about 7500 m/s or 7.5 km/sec. I'm suggesting a | speed of 4 times that 30 km/sec. Ok. 7710.05 mps over Japan as I look right now. E = 1/2 mv^2 Four times the velocity, 16 times the energy. You want to lift 16 FULL fuel tanks into orbit in preparation for the Mars trip, outgoing, 16 more for stopping, and 32 for the return trip. Total 64 times as much. How much fuel will you need to lift the fuel? Oh wait... you are going to develop a nuclear powered drive by 2020, right? NASA currently has Discovery, Atlantis, Endeavour. Challenger and Columbia are no more. Let's build a bigger fleet, a hundred should be enough. And the reason for this is to make a nice, comfortable short trip to Mars for the crew so that the poor darlings don't get bored on a long trip and have to ride an exercise bicycle to keep fit as they do on the ISS. | Every shuttle on return uses aerobraking under its normal aerodynamic | configuration to land. Yes. | It was because a wing suffered severe damage | that caused Columbia to lose its normal configuration for landing that | caused it to break up. | | Bob Clark Yes... and you plan on travelling 4 times faster and burn up 16 times the kinetic energy in heating up the atmosphere after you've spent 64 times as much money as a slow boat to Mars will cost, all for creature comfort. How about if the oh-so-terrible government raises your taxes 64 times to pay for this trip to a dust bowl in a desert? At least you'll be able to plant a flag there, right? Hey, I'm no greenie, I'm not a tree-hugger and I'm all for exploring the planets, but hell man... have SOME common sense. Neither of you got it quite right - or said it quite clearly enough.. To get an idea of what's going on you need an elementary knowledge of the rocket equation - how rockets build speed - and an elementary knowledge of orbital mechanics. The speed and transit times for minimum energy orbits are calculated - and once that is understood, we can then proceed to see what the benefits and costs of adding speed are; ROCKET EQUATION The velocity of a rocket propelled projectile is given by the Tsiolkovsky Equation; Vf = Ve*LN(1/(1-u)) Where Vf = final velocity Ve= exhaust velocity LN( ..) = natural logarithm function u = propellant fraction. So, if a rocket is 50% by weight propellant and its exhaust speed is 4 km/sec we can compute a final velocity for the rocket of; Vf = 4.0 x LN(1/(1-0.5)) = 4.0 x LN(2) = 4.0 x 0.693147 = 2.772 km/ sec If a rocket is 80% by weight propellant and its exhaust speed is 4.5 km/sec we can compute a final velocity for the rocket of; Vf = 4.5 x LN(1/(1-0.8)) = 4.5 x LN(5) = 4.5 x 1.609438 = 7.242 km/ sec Temperatures and pressures achievable with modern materials limit the exhaust speeds of rocket engines. Strength to weight of materials limit the amount of propellant a tank can carry. These are the design parameters we have to work within. There are several types of rocket engines that have been developed over the years, and many practical systems proposed that are capable of both high thrust and high performance. 1) solid propellant rockets - Ve = 2.5 km/sec 2) hypergolic liquid propellant rockets - Ve=3.2 km/sec 4) cryogenic liquid propellant rockets - Ve=4.5 km/sec 5) solid core nuclear thermal rocket with cryogenic propellant - Ve = 9.0 km/sec 6) gas core nuclear thermal rocket with cryogenic propellant - Ve = 15.0 km/sec 7) nuclear pulse rockets- Ve = 25.0 km/sec In recent years, after the advent of SDI, some have proposed replacing the nuclear heat source in the last 3 rocket types with laser energy beamed efficiently to the rocket - providing an increase of thrust to weight. ORBITAL MECHANICS Orbital velocity on Earth surface is 6.5 km/sec to 8.2 km/sec. Minimum Moon trajectory 10.9 km/sec (4 days) Escape velocity on Earth surface is 11.2 km/sec Minimum Mars Trajectory 11.8 km/sec (9 months) This does not count air drag losses or gravity drag losses during ascent. These add 1.2 km/sec to 2.0 km/sec depending. The ideal final velocity for the Space Shuttle is 9.2 km/sec. So, any vehicle with that delta vee capacity -in other words that Vf- can fly the same flight envelope from Cape Canaveral Florida. We can re-arrange the rocket equation to solve for propellant fraction needed to attain these velocities u = 1 - 1/EXP(Vf/Ve) Orbital velocity Vf = 6.5 km sec Ve = 2.5 km/sec ---> u = 0.9257 = 3.2 u = 0.8688 = 4.5 u = 0.7641 = 9.0 u = 0.5143 =15.0 u = 0.3517 =25.0 u = 0.2290 This shows that when the desired speed of a rocket exceeds the exhaust velocity, its best to achieve that velocity in stages to reduce overall mass. A gas core or nuclear pulse rocket are practical ways to achieve very high velocities - from the surface of the Earth. Upper stages need only achieve 0.6 km/sec or more - to attain interplanetary speeds if already on an escape trajectory put there by an existing rocket. Since no existing rocket is large enough to send a manned payload to escape velocities, when considering manned travel to Mars, we are talking about multiple launches of existing rockets, and assembly on orbit, OR, the construction of brand new rocket systems much larger than the ones we use at present. Both paths are extremely expensive - not as expensive as our invasion of Afghanistan - but expensive nevertheless. Interplanetary flight occurs along minimum energy trajectories - called hohmann transfer orbits. http://en.wikipedia.org/wiki/Hohmann_transfer Which gives you minimum energy transfer delta vee - 'budgets' http://en.wikipedia.org/wiki/Delta-v_budget#Interplanetary_budget Basically add - 0.6 km/sec to the escape velocity from Earth's surface, and you can get to mars. You asked about transfer times. For that you need an introduction to orbital mechanics. http://www.braeunig.us/space/orbmech.htm Which gets you to Androcles post - though I didn't check his math... so, I can't warrant that. However, Keplers third law of motion can be helpful here to understand transfer times. The squares of the orbital periods are equal to the cubes of the semi- major axes. So, the semi-major axis (radius) of Earth's orbit is 1 au, and its period it 1 year. 1 x 1 = 1 x 1 x 1 The semi-major axis of Mars' orbit is 1.523679 AU and its period is 1.8808 years 1.523679 x 1.523679 x 1.523679 = 1.8808 x 1.8808 So, a hohmann transfer orbit has a perihelion of 1.0 AU and an apohelion of 1.523679 AU add those up to get a major axis of 2.523679 - divide by 2 to get 1.261845 - cube that - (multiply by itself 3 times) to get 2.00091 and take the square root - to get 1.417454 years - this is the time it takes to undergo a complete circuit - divide by two to get the transfer time - 0.70872 years - multiply by 12 to get 8.504 months. To understand all those sines and cosines understand what we're doing - we're taking segments of the orbit, and calculating the transit time over those segments - once we know the orbit http://en.wikipedia.org/wiki/Image:Kepler_laws_diagram.svg Generally speaking, if you accelerate a vehicle continuously along a trajectory at accelerations that are large compared to the sun's local gravity - at Earth and at Mars - then you can use straight line approximations. If you are kicking the payload with a rocket blast at the beginning of a journey - and the speed change due to the accumulation of solar gravity influence is small during transit - then you can use straight line approximations again. The acceleration at the surface of the sun is 28.02 gees (274 m/s/s) and the solar radius is 0.00452 AU (679,000 km) at 1 AU solar gravity is reduced by a factor of 1/r^2 - or 1/48,800 th the gravity at the surface. That's 574 micro gees (5.636 mm/s/s). At 1.52 AU that's reduced to 248 micro gees.(2.435 mm/s/s) Velocity is equal to acceleration times time. So, for our 8.5 months = 22.35 million seconds 5 mm/s/s x 22.35 mega seconds = 126 km/sec Over the course of the transit along a minimum energy orbit - you have 100 km/sec delta vee due to solar influence. A delta vee of substantially higher velocity will look like a straight line and lose little of its velocity. Also, a transit at 1 gee - to maintain gee forces aboard the spacecraft - will also look like a straight line - since solar influence will be nil at 1/2 milligee. So, here's an interesting calculation; 60 million km = 60 billion meters 1/2 this value is 30 billion meters D = Vf^2 / 2*9.82 rearranging terms Vf = SQRT(2* 30e9 * 9.82) = 767.59 km/sec This takes a sort of ship that we haven't seriously considered yet - one that uses say anti-matter to efficiently detonate fusion reactions with exhaust velocities at 1,000 km/sec or more. Where Th = time of the transfer > The distance from Earth to Mars is about 60,000,000 km at closest > approach. If we have a 30 km/sec initial velocity to Mars, which might [quoted text clipped - 70 lines] > resulting short travel times the straight-line approximation is > accurate? Straight lines are always faster trips at the same speeds, as long as you do actually go in the straight line. Why not reduce the speed in half and double the time and we still get there in about 23 days and don't have to worry about as much aerobraking. :)  SignatureJames M Driscoll Jr Spaceman On Jul 5, 11:30 am, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > The distance from Earth to Mars is about 60,000,000 km at closest > > approach. If we have a 30 km/sec initial velocity to Mars, which might [quoted text clipped - 81 lines] > James M Driscoll Jr > Spaceman You need the high speed to reduce the effect on the *overall* shape of the trajectory by Earth's gravity, so the straight-line approximation is accurate. At slower speeds the actual curved elliptical path becomes dominant and you have to consider the effect of that curved path on the travel time. Think of throwing a ball horizontally. At a slow speed the curved path is obvious. At a high speed the path becomes straighter and it is able to reach a longer distance. My guess is that at 15 km/sec this is so close to the escape velocity of 11 km/sec that the curved elliptical path would become dominant and the trip time would be only a little better than the curved Hohmann transfer orbit time of 6 to 7 months. Bob Clark > My guess is that at 15 km/sec this is so close to the escape velocity > of 11 km/sec that the curved elliptical path would become dominant and > the trip time would be only a little better than the curved Hohmann > transfer orbit time of 6 to 7 months. > > Bob Clark Guess? Do the calculations! > > My guess is that at 15 km/sec this is so close to the escape velocity > > of 11 km/sec that the curved elliptical path would become dominant and [quoted text clipped - 4 lines] > > Guess? Do the calculations! It's a two body problem in *two* dimensions, not one. Doable, but not trivial. The closest I've seen to it on the net is this presentation: Basic of Space Flight: Orbital Mechanics:Orbit Altitude Changes. http://www.braeunig.us/space/orbmech.htm#maneuver But this takes the angle of departure as only tangential to the initial orbit so you don't find the optimal angle to minimize the trip time. Bob Clark >>> My guess is that at 15 km/sec this is so close to the escape velocity >>> of 11 km/sec that the curved elliptical path would become dominant and [quoted text clipped - 9 lines] > Basic of Space Flight: Orbital Mechanics:Orbit Altitude Changes. > http://www.braeunig.us/space/orbmech.htm#maneuver Thanks for the online reference... > But this takes the angle of departure as only tangential to the > initial orbit so you don't find the optimal angle to minimize the trip > time. With proper timing, the optimal angle IS tangential, making the most of the orbital velocity. > Bob Clark ... > > The closest I've seen to it on the net is this presentation: > [quoted text clipped - 11 lines] > > > Bob Clark No. Just draw two concentric circles to see why it isn't. The shortest distance from one to the other is radial, but then you a get a smaller velocity boost from the Earths speed. If you travel tangentially you get the biggest boost from the Earth's speed but then the distance is longer. For shortening the *time* an intermediate angle between these two extremes is optimal, assuming the initial speed is high enough that the straight-line approximation is accurate. Bob Clark > ... > [quoted text clipped - 26 lines] > > - Show quoted text - They're all conic sections - if you can choose your departure time, the hyperbolic trajectory from Earth will be tangential to Earth's orbit - an excess energy orbit will not be tangential to Mars' orbit upon arrival however. Start with a Hohmann transfer ellipse - and you'll see you've got 8.5 months transit time - for minimum energy. As you add energy to the orbit - you get straighter and straighter hyperbolic conic sections connecting the two worlds - and the Sun's influence on the trajectory is less and less. It is greatest at the minimum energy orbit - where it contributes 120 km/sec to the velocity of the payload over 8.5 months - and only 0.6 km/sec is added to the escape velocity - adding more than 120 km/sec - gets you a pretty straight line. A constant 1 gee acceleration also gets you straight lines - and gives you a delta vee for the rocket of over 700 km/sec. Even 1/3 gee - gives you a pretty straight line - though everything is taken into account with precise nagivation. Also, everything occurs in 3 dimensions - with a matrix giving the components needed for plane changes in transit at any particular time of departure and arrival. Though approximations can be made by assuming planar transfers. Given the precision with which we know things in the solar system these days, it should be possible to create a GPS enabled interplanetary navigation chipset - that takes GPS signals (which include precise time as well as location) and determine minimum delta vee - and boost trajectories from any point on Earth's surface to arrive at any point on Mars surface. The problem remains, propulsion. When delta vee requirements exceed exhaust speeds, propellant fractions grow to unmanageable levels. This is the case for chemical and even nuclear thermal type rocket systems. Nuclear pulse, anti-matter catalyzed fusion, laser detonation thrusters - provide adequate propulsive capabilities to create an interplanetary cruiser. > > > My guess is that at 15 km/sec this is so close to the escape velocity > > > of 11 km/sec that the curved elliptical path would become dominant and [quoted text clipped - 16 lines] > > Bob Clark If you actually want to take this trip its a calculation in 3 dimensions of all the significant bodies of the solar system gravitation influence including the sun upon the vehicle your travelling in. Then theres the new materials you're going to need to protect you and your travellers. Then again you will need a real engine to take you there. Not a chemical rocket. Not even a nuclear plasma rocket provides the required force times time to get you there. (not intended as an equation, but more of a 'reminder' as to lack of real space engines we now have). Its fun enough to calculate and dream. My personal favorite is the solar sail. Its not only great because the sun already provides the high speed atomic particles and stripped nuclei for free, but also if it is a strong enough influence upon your vehicle (read a bigger sail is better), then you no longer have to travel along orbital lines between planets but now may enjoy the privilege of simply 'driving' there like an airplane or a car. (the math for this is even worse because the solar wind varies) Slowing down is tricky with a solar sail, you probably cant 'tack into the wind like an earthbound sailing ship, but there are other possibilities presented by solar wind not enjoyed by earth's winds on the water. It is possible, but not as easy. For instance the sail could double as a lens ahead or even a mirror focusing the solarwind behind to a small point inside your vehicles engine "combustion" chamber to superheat material as a plasma rocket. Or it may be discovered that a material exists similiar to beryillium with fast neutrons and radiation that reflects the solar wind. Maybe even beryillium itself can do it, its light, but not very durable except as a much heavier alloy. Of course once you get to your destination you can use the solar sail just like an earthbound parachute to slow your orbital travel in relation to the sun to simply "fall" out of orbit back to the earth, and at high relative speed to the earth too. Just don't miss !!!! :) Its these kind of science based premise notions that form the very best of science fiction. Something we can almost do. Sorta. > > > > My guess is that at 15 km/sec this is so close to the escape velocity > > > > of 11 km/sec that the curved elliptical path would become dominant and [quoted text clipped - 54 lines] > > Sorta. A one G speed up and a one G slow down should more than do the trick without hardly a hitch, at least according to our resident nuclear fly- by-rocket wizard William Mook. Got a spare trillion $ to blow? - Brad Guth Brad_Guth Brad.Guth BradGuth > On Jul 5, 11:30 am, "Spaceman" <space...@yourclockmalfunctioned.duh> > wrote: [quoted text clipped - 97 lines] > the trip time would be only a little better than the curved Hohmann > transfer orbit time of 6 to 7 months. Then it comes down to plotting a straight line once as far out of the problem of escape velocity curving needed. We should not be shooting just a ball, we should be shooting a ball with controllable rockets on it. :) Once we can, we change course to "as close to straight line as possible". :) It all does come down to timing also. There is no doubt that straight lines at same speeds will shorten trips. So, finding the straight line when we can is the best method and should shorten the trip simply. Just as the shortest distance is a straight line. and the shortest distance would also be a "Straight" path. :) 60,000,000 kilometers.in a straight line at 15 km/sec = 4000000 seconds about 1112 hrs 46 days if you actually travel a straight line at that speed, and curves in the path of course will lengthen that time. and increase in speed on the straight line shortens the trip. Why do we take the long way when we really don't have to if the "shooting" is timed correctly from the correct shooting platform. :) SignatureJames M Driscoll Jr Spaceman > Why do we take the long way when we really don't have to > if the "shooting" is timed correctly from the correct shooting > platform. With enough fuel almost any desired path and transit time (within reason) is possible. Unfortunately, the energy cost is just too high to make it practical. There is a time versus fuel tradeoff (or more technically correct, a time versus energy tradeoff). The Hohmann transfer orbit provides the least energy means of getting from one orbit to another, barring tricks like gravity assist manouvers where momentum is "stolen" from other bodies. Just getting fuel into space is expensive (where here "expensive" means the cost in fuel mass). Look how much fuel the shuttle burns just to end up in low Earth orbit with practically empty tanks and having discarded its solid rocket boosters. >> Why do we take the long way when we really don't have to >> if the "shooting" is timed correctly from the correct shooting [quoted text clipped - 9 lines] > gravity assist manouvers where momentum is "stolen" > from other bodies. So, all we really need to do is "trick" our ships to steal Earths gravitational momentum to get the speed we really want and time such to be "as close to straight line" we also "really" want to shorten the trip. > Just getting fuel into space is expensive (where here > "expensive" means the cost in fuel mass). Look how > much fuel the shuttle burns just to end up in low > Earth orbit with practically empty tanks and having > discarded its solid rocket boosters. And more silly is traveling around the globe to get to the spot that is three feet away from you. :) Straight lines should use less total fuel in the final trip. And.. tada... they always do. :) An energy trade off to not go straight is just stupid in the end. How does taking a longer trip, use less fuel in the end? It would not, is the real answer. :) >>> Why do we take the long way when we really don't have to >>> if the "shooting" is timed correctly from the correct shooting [quoted text clipped - 15 lines] > to be "as close to straight line" we also "really" want to shorten > the trip. That would be nice if it were possible. But it's not possible to steal enough momentum to approach a straight line path in the solar system. The gravity of the Sun makes sure that reasonable velocity trajectories are all curved (conic sections). Momentum is "stolen" by doing gravity assist trajectory flybys. Problem is, each such fly by adds some velocity to the craft, but you need to swing around and come back for another pass to do it again. Each such pass uses an unpowered orbit, so each successive pass would take longer. Eventually the craft will reach Earth escape velocity (less than 11 km/sec) and not come back at all. >> Just getting fuel into space is expensive (where here >> "expensive" means the cost in fuel mass). Look how [quoted text clipped - 7 lines] > Straight lines should use less total fuel in the final trip. > And.. tada... they always do. Not in space though, where unpowered motion is the most fuel efficient way to go -- just travel inertially, free-falling from one place to another after an initial shove. These paths are all conic sections. > :) > An energy trade off to not go straight is just stupid > in the end. Such is life in the real world. In space, near large masses where things orbit, straight line paths are amongst the most expensive ways to go! > How does taking a longer trip, use less fuel in the end? Because you don't burn any for most of the trip; you just glide along on inertia. > It would not, is the real answer. Wrong. >>>> Why do we take the long way when we really don't have to >>>> if the "shooting" is timed correctly from the correct shooting [quoted text clipped - 21 lines] > of the Sun makes sure that reasonable velocity > trajectories are all curved (conic sections). Fighting such a small gravitational pull of the Sun, on such a small object as a spaceship is not as hard as one wants to make it. :) > Momentum is "stolen" by doing gravity assist trajectory > flybys. Problem is, each such fly by adds some velocity [quoted text clipped - 3 lines] > longer. Eventually the craft will reach Earth escape > velocity (less than 11 km/sec) and not come back at all. And if such speed is achieved the straight path becomes more possible to even fight the Suns pull also. Let me ask you this, If a curved path is better, Why do we try to basically shoot "striaght" upward first at all? According to the curve sillyness, we should be shooting sideways first. :) > Not in space though, where unpowered motion is the > most fuel efficient way to go -- just travel > inertially, free-falling from one place to another > after an initial shove. These paths are all conic > sections. Enough speed will allow the straight path with an unpowered flight that will be just enough to negate the Sun's pull. > Such is life in the real world. In space, near large > masses where things orbit, straight line paths are > amongst the most expensive ways to go! No, Try thinking of a trip to the top of a mountian. Would you burn more fuel by taking a curved path to the top, or a striaght line towards the top? Do you think an airplane will burn more fuel making a curved line to the top than it would by traveling a straight shot towards it? > Because you don't burn any for most of the trip; you > just glide along on inertia. the inertia is the savings. If the amount of inertia was used in a straight line. It would be faster to get there and burn less fuel in the end. Again would you shoot a bullet to the side to get higher altitude from the "inertia" or stright up? >> It would not, is the real answer. > > Wrong. No, And cars an airplanes prove you are wrong every single day. oh ya.. and so don't rockets. :) SignatureJames M Driscoll Jr Spaceman >>>>> Why do we take the long way when we really don't have to >>>>> if the "shooting" is timed correctly from the correct shooting [quoted text clipped - 25 lines] > on such a small object as a spaceship is not as hard > as one wants to make it. It is what it is. The acceleration due to the Sun's gravity depends on position (distance from the Sun), not velocity. >> Momentum is "stolen" by doing gravity assist trajectory >> flybys. Problem is, each such fly by adds some velocity [quoted text clipped - 7 lines] > becomes more possible to even fight the Suns pull > also. Again, the acceleration due to the Sun's gravity depends on position, not velocity. > Let me ask you this, > If a curved path is better, > Why do we try to basically shoot "striaght" upward first at all? > According to the curve sillyness, we should be shooting sideways > first. Horizontal takeoff would be ideal if it weren't for the atmosphere. The atmophere causes drag, so you want to get through it as quickly as possible. If you look at the actual paths of rockets taking off, you'll see that they start off going straight up, then bank over as they get higher, eventually travelling parallel to the Earth's surface directly below them as they reach orbital height. > :) > [quoted text clipped - 7 lines] > unpowered flight that will be just enough to negate > the Sun's pull. No. You can approach stright line, but never achieve it. The fastest thing around is light, and even it's path is made curved by the Sun's gravity. >> Such is life in the real world. In space, near large >> masses where things orbit, straight line paths are [quoted text clipped - 6 lines] > Do you think an airplane will burn more fuel making a curved > line to the top than it would by traveling a straight shot towards it? The devil is in the details. Running a non 100% efficient engine for longer time against the vagaries of friction on the path will ultimately use more energy for a longer path. The ideal, least energy path to the top of the mountain would be to be fired from a cannon. Expend the requisite energy in one efficient burst to impart the kinetic energy corresponding to the potential energy difference between the bottom and top of the mountain. In space there is (essentially) no friction. So unpowered motion is free. >> Because you don't burn any for most of the trip; you >> just glide along on inertia. > > the inertia is the savings. The fuel is the savings. Intertia is, well, inertia. ' > If the amount of inertia was used in a straight line. Inertia isn't 'used'. Inertia is the tendancy of a body to resist change in momentum. > It would be faster to get there and burn less fuel in the end. Burning no fuel is always cheapest. Burning just the needed fuel as efficiently as possible is the next cheapest. > Again > would you shoot a bullet to the side to get higher altitude > from the "inertia" or stright up? Staight up, of course, if highest altitude was your goal. Air friction determines the ideal initial angle. If there were no air friction, and the bullet could pass freely through the Earth (obviously a hypothetical situation) any angle would do as the bullet would go into an orbit with a major axis fixed by the initial position and velocity. It would reach the same height above the "surface" eventually (apogee). Our notions of efficient paths on Earth are misleading when it comes to space travel, since down here we tend to take into account the "difficulty" of terrain and air friction and other losses. In space the cheapest way to travel is to just coast after making the most efficient expenditure of fuel that we can (usually meaning a short efficient rocket firing). In space you don't have to worry about air friction. > It is what it is. The acceleration due to the Sun's > gravity depends on position (distance from the Sun), > not velocity. I never said its pull depends on the objects velocity, I said the velocity if high enough negates that pull. > Horizontal takeoff would be ideal if it weren't > for the atmosphere. The atmophere causes drag, [quoted text clipped - 5 lines] > the Earth's surface directly below them as they > reach orbital height. If you want to get there as quickly as possible, You travel a straight line. wind or not, To get to a higher altitude, a straight line is the shorter path always. Ans a shorter path will burn less fuel in the end. again. Traveling to a mountain top proves it. Why don't you think about such? >> Enough speed will allow the straight path with an >> unpowered flight that will be just enough to negate [quoted text clipped - 3 lines] > it. The fastest thing around is light, and even it's > path is made curved by the Sun's gravity. Because light doesn ot have built in rockets. So..it can't fight gravity, Spaceships can. Again you seem to like to ignore that the shortest distance is a striaght line and the least fuel burned comes from a straight line also. > The devil is in the details. Running a non 100% > efficient engine for longer time against the vagaries of > friction on the path will ultimately use more energy > for a longer path. Correct, longer path = more fuel use. always. Now think about that for once. :) > The ideal, least energy path to the top of the mountain > would be to be fired from a cannon. Expend the requisite > energy in one efficient burst to impart the kinetic energy > corresponding to the potential energy difference between > the bottom and top of the mountain. The ideal path would be a straight line. Again, you seem to ignore the fact that a straight line uses the least fuel. > In space there is (essentially) no friction. So unpowered > motion is free. And if enough fuel that brings you to that curve is used in a straight path instead, you will be getting there faster for the same amount of fuel. :) > The fuel is the savings. Intertia is, well, inertia. Fuel is always saved by traveling a straight line. Do you actually think taking a longer path uses less fuel? Again, I invite you to prove such to a trip the the mountain top. :) > Burning no fuel is always cheapest. Burning just the > needed fuel as efficiently as possible is the next > cheapest. Again, shorter paths will burn less fuel. You seem to want to ignore that fact. > Staight up, of course, if highest altitude was your goal. > Air friction determines the ideal initial angle. If there [quoted text clipped - 4 lines] > would reach the same height above the "surface" eventually > (apogee). > Our notions of efficient paths on Earth are misleading > when it comes to space travel, since down here we tend [quoted text clipped - 5 lines] > > In space you don't have to worry about air friction. In reality the shortest path is the least fuel burned. and not once is there 0 gravity towards Mars. so the straight path will still win for least fuel burned once speed is achieved. :) Again, If you don't want to think about taking the trip to the top of the mountain and the trip back will be "basically free" then you don't get the fact about shortest distance will use least fuel all the time. :) the quickest path between two places is a slalom (brachistochrone or tautachrone per Liebniz' calculus. I always thought that the efficient "burn" of fuel, required burningot throughout the trip, halfway in acceleration & deceleration. it seems that such speeds on the approach to Mars, might strip its atmosphere, before you stopped (or, the aerobrakeing'd take longer than the trip, or cover a significant portion of Mars with the 'chute !-) > If you don't want to think about taking the trip to > the top of the mountain and the trip back will > be "basically free" then you don't get the fact about > shortest distance will use least fuel all the time. thus: I should add that these may be really important features in the bugset: your cones are not generally quadric surfaces, and Joe's facets are not generally convex. just because you guys tried to hide that, or merely averted your eyes in the patent-pending or paper-pretending, doesn't mean it isn't "all good." like, I'd call yours,"post-whacked dickonoids, " or "flying funny shingles" or some thing; seriously ... after a *lot* more of math, or at least systematic experiment (that is to say, with a good write-up of observables, not just a virtual tour of the latter-day shack). >http://uspto.gov/web/patents/patog/week26/OG/html/1331-4/US07389612-20080624.html --Seargent Barracks Soros McCheeny Pepper, "Give jihad a chance in The Sudan, Rhodesia, and other former colonial moments -- Yahoo!TM; you're going to feel my computerized draft, boys'n'girls: NO AMERICAN MIDDLESCHOOLER LEFT BEHIND; NO RHODESIA SCHOLARS IN HARM'S WAY!" http://larouchepub.com/lar/2008/3526lar_soros_pamph.html http://larouchepub.com/other/2008/3526save_nations_parasites.html http://larouchepub.com/other/2008/3526zim_brit_op.html > the quickest path between two places is a slalom > (brachistochrone or tautachrone per Liebniz' calculus. The why do downhill skiers make it past the finish line first and faster. :) The quickest path at a set speed is a straight line. The least fuel burned at the same set speed is also a straight path. Why are such simple facts ignored so well to come up with this curved path crap? Such ignorance of reality is amazing. LOL SignatureJames M Driscoll Jr Spaceman >> the quickest path between two places is a slalom >> (brachistochrone or tautachrone per Liebniz' calculus. [quoted text clipped - 7 lines] > up with this curved path crap? > Such ignorance of reality is amazing. You're ignoring the effect of gravity and asssuming that the ship has a road to run on. Skiiers must follow the mountain terrain and are not free to choose their own 3D trajectory. If they were, and could ignore the ski gates, they would follow a curved path sa "spudnik" wrote. It's a classic 2nd year calculus problem. >>> the quickest path between two places is a slalom >>> (brachistochrone or tautachrone per Liebniz' calculus. [quoted text clipped - 10 lines] > You're ignoring the effect of gravity and asssuming > that the ship has a road to run on. No I am not. I am ignoring your ignorance of a longer path needed more fuel than a shorter path would. And sadly, you are ignoring that fact completely to come up with your "curved path nonsense". :) Hint Greg: There is no spot between here and Mars where there is no gravity. so even when you are "freefalling" you are taking a longer path and will need to make it up with fuel again to once again fight the gravity taking the longer trip. > Skiiers must > follow the mountain terrain and are not free to > choose their own 3D trajectory. If they were, and > could ignore the ski gates, they would follow a > curved path sa "spudnik" wrote. It's a classic > 2nd year calculus problem. It's a joke and you don't seem to know how stupid it really is in reality. The shortest distance is the straight line, and the straght line (even when fighting gravity) will burn less fuel. SignatureJames M Driscoll Jr Spaceman >>>> the quickest path between two places is a slalom >>>> (brachistochrone or tautachrone per Liebniz' calculus. [quoted text clipped - 23 lines] > path and will need to make it up with fuel again to > once again fight the gravity taking the longer trip. No. You give the ship the necessary impulse at the beginning of the trip and allow it to coast all the way to the vicinity of Mars' orbit. Then apply another short burst to circularize the orbit to coincide with that of Mars. >> Skiiers must >> follow the mountain terrain and are not free to [quoted text clipped - 5 lines] > It's a joke and you don't seem to know how stupid it really > is in reality. Reality is not stupid. Ignoring it by trying to apply your backyard experience is. > The shortest distance is the straight line, and the straght > line (even when fighting gravity) will burn less fuel. No. The shortest distance may be a straight line, but it's not always the most energy efficient. Remember, you don't need to keep your engine running the whole way. > No. The shortest distance may be a straight line, but > it's not always the most energy efficient. Remember, > you don't need to keep your engine running the whole > way. Let me correct a goof up I have said, I used the "fuel" word where I should have used the energy word. The shortest distance being a straight line will always take the least amount of energy, and the longer the distance will take more energy always. and A slingshot effect is simply grabbing that extra energy needed for the longer trip. Once you have used up the slingshot, you are back to square one. The shortest distance is best again, unless you can get another slingshot. but sadly curving slowly outward away from the sun, by taking the "curved path to Mars" is not "gathering energy" and hence the longer path doing such is stupid and will not use less energy and in fact would need more energy coming from the spaceship to do such. So, I admit my use of the word fuel was wrong, but the energy will always be more used for the longer path. :) Now show me a circular path away from gravity that will use less energy than the straight path would Greg. :) >> No. The shortest distance may be a straight line, but >> it's not always the most energy efficient. Remember, [quoted text clipped - 25 lines] > but the energy will always be more used for the longer > path. In space travel, fuel is intimately associated with energy. That's why I didn't take you to task for the usage in the first place. It's common to equate fuel with energy expenditures. Further, spacecraft manouvers are usually calculated in terms of "delta v", that is, the required velocity changes, which in turn require a certain energy and certain amount of fuel for a given craft. So, for example, departing Earth orbit to begin a trip to Mars may require a delta V of 4km/sec. At the Mars end, the ship may need to make a delta V of 3 km/sec to circularize into Mars' orbit. > Now show me a circular path away from gravity that > will use less energy than the straight path would Greg. Any elliptical orbit section that gets you where you want to go will take less energy than a straight path would, simply because you don't need to be burning energy contantly just to maintain a straight path (note that much of the energy required to maintain a straight path will *not* be getting you to your destination quicker... it will be at an angle to your direction of motion, in the direction opposing the gravitational acceleration). > In space travel, fuel is intimately associated with > energy. That's why I didn't take you to task for [quoted text clipped - 20 lines] > motion, in the direction opposing the gravitational > acceleration). You truly like to ignore the energy given by the slingshot huh? That is pretty sad ignoring all that energy you used to take the longer path. You did nto have to store it true, but you still needed it. Again, once the slingshot is done. The uphill trip to Mars is still there. You better have enough energy stored now. I know a straight path up the hill will need less energy than the spiral path you wish to take. :) It is pretty sad you won't admit that fact at all. Apparently you do have some form of violation of the conservation of energy occuring in your magical uphill spiral path. :) SignatureJames M Driscoll Jr Spaceman >> In space travel, fuel is intimately associated with >> energy. That's why I didn't take you to task for [quoted text clipped - 23 lines] > You truly like to ignore the energy given by the > slingshot huh? What energy is that? In order to gain energy by slingshot manouver around the Earth, you first need to spend several months travelling away from the Earth and picking up speed to lap the Earth in its orbit (thus coming up on it from behind). If you're lucky and you do the manouver just right, you can almost double your velocity (just like the elastic collisions we wer talking about in another thread). You still need to burn quite a bit of fuel (and time!) to get into position to make the manouver. Granted the available boost is significant and free after that, but it's another example of a fuel versus time tradeoff in space travel. That velocity boost *still* won't allow you to travel in a straight line to Mars. It'll just give you a segment of a larger ellipse to follow on your way out. You'll need to burn fuel and aerobrake at the other end to insert into Mars orbit. > That is pretty sad ignoring all that energy you > used to take the longer path. > You did nto have to store it true, but you still > needed it. In space travel, anything you don't have to carry is free. The only currencly that matters is available Delta-V = fuel = energy. You can't make a potential versus kinetic energy trade in space without burning *some* fuel. > Again, > once the slingshot is done. > The uphill trip to Mars is still there. > You better have enough energy stored now. If you did the manouver correctly, you have it stored as kinetic energy of motion. > I know a straight path up the hill will need less > energy than the spiral path you wish to take. No, then you'll be burning energy (fuel) to try to follow a straight path, when nature will let you travel the elliptical one for free. > It is pretty sad you won't admit that fact at all. I won't because it's not true. > Apparently you do have some form of violation > of the conservation of energy occuring in your > magical uphill spiral path. Apparently you can't come to terms with the different environment for travel that space provides as opposed to travel on the surface of the Earth where we can always grab onto the surface to manouver, break for free, and so forth. In space, any manouver at all that departs from an inertial path costs fuel. And fuel is more precious than gold in space. > What energy is that? In order to gain energy by > slingshot manouver around the Earth, you first need [quoted text clipped - 16 lines] > your way out. You'll need to burn fuel and aerobrake > at the other end to insert into Mars orbit. WRONG! If you have enough energy to make it up a hill spirally to point B then you had enough energy to travel straight up that hill to point B instead. You truly have no clue about facts like such. And I can see you will never admit you are wrong about it. Try doing an experiement with marbles or something Greg, the spiral path up will not take you as far up the hill as the straight path will. go ahead I dare you to try such. You might learn something from a "Real" experiment you actually do yourself for once. :) <snipped rest of ignorance til Greg rolls the freakin marble up the hill and admits he is wrong.> :) It's easy to set up too Greg, One ramp for downhill inertia, and 2 setups of paths up the hill, one a striahgt line from the bottom of your inertial ramp and one with a curved ramp around the hill. both ramps will end at point B on the hill. Se which one gets there first and at the highest speed when it gets there. IF your curved ramp can even do as high as the straight ramp will! LOL SignatureJames M Driscoll Jr Spaceman >> What energy is that? In order to gain energy by >> slingshot manouver around the Earth, you first need [quoted text clipped - 24 lines] > And I can see you will never admit you are wrong > about it. You keep changing the scenario from space to hills, which have some differences. In space there's no surface to follow to keep you from falling in the direction of gravity. On a hillside you use the hills surface to provide that support. In space, the only way to travel a straight line in free-fall is to travel radially in the gravitational field. In the earth to Mars scenario you would have to travel outward along the line that passes through the Sun and Earth, at least for the start of your journey when the Earth's gravity is a significant factor. Of course, the Earth is in orbit and won't stay aligned with you for long. If you choose to travel this radial path, then you must discard the "free" velocity that you have already got by simply travelling along with the Earth in its orbit. That's throwing away about 30 km/sec of speed. Not only that, but you can't just apply breaks to shed that velocity. You have to perform an engine burn to do it, a Delta-V of 30 km/sec. That's before you even get started on your way to Mars. After that you need to expend the energy to reach the higher potential energy of Mars' orbit. After *that* you need to expend the energy to match Mars' speed in its orbit. That's another 24 km/sec Delta-V. So I figure you've burnt about 60 Delta-V units in fuel to drive your straight line. Compare this with a Delta-V of about 5 or 6 for a Hohmann transfer. Rocket scientists aren't dummies. > Try doing an experiement with marbles or something > Greg, > the spiral path up will not take you as > far up the hill as the straight path will. If there's no friction, it certainly will. You're just trading kinetic energy for potential energy. >>> What energy is that? In order to gain energy by >>> slingshot manouver around the Earth, you first need [quoted text clipped - 27 lines] > You keep changing the scenario from space to hills, > which have some differences. Not really, The Sun is the hill. <snipped rest of babble based upon not thinking the Sun's gravity will cause "drag" just as a hill would.>. SignatureJames M Driscoll Jr Spaceman : "Spaceman" <spaceman@yourclockmalfunctioned.duh> : <snipped rest of babble based upon not thinking : the Sun's gravity will cause "drag" just as a hill would.>. Whether you claim it's "babble" or not, it's still true. Gravity causes no drag. Friction causes drag. Wayne Throop throopw@sheol.org http://sheol.org/throopw >> "Spaceman" <spaceman@yourclockmalfunctioned.duh> >> <snipped rest of babble based upon not thinking >> the Sun's gravity will cause "drag" just as a hill would.>. > > Whether you claim it's "babble" or not, it's still true. > Gravity causes no drag. Friction causes drag. So What does gravity cause that stops us from jumping off Earth? LOL Wayne, If the same amount of energy is needed to get to the same spot on the hill no matter the path. Then guess what? The straight path to Mars will be faster. and not use anymore energy once the slingshot gives the needed velocity. Sheesh. This is funny, First.. there was no way the short path could work at all and now both paths use the same energy. You both should make up your mind. LOL If you take the long way. I will beat you there and use the same amount of energy in the end, even according to both of you now. Simple as that. LOL :) SignatureJames M Driscoll Jr Spaceman : "Spaceman" <spaceman@yourclockmalfunctioned.duh> : If the same amount of energy is needed to get to the same spot on the : hill no matter the path. Then guess what? The straight path to Mars : will be faster. You guessed wrong. The fastest freely falling trajectory to mars is going to be curved. In fact, it works better if you don't guess, and actually try calculating or simulating the cases. : First.. there was no way the short path could work at all You of course ahve a reference to where I said a short path "will not work at all". Please refresh my memory, cupplying a cite, since near as I can tell, I never said that, first, last, or in between. A "short path" or "stright path" can, in theory, be made to work. With vast excess and impractical expenditures of delta-v. Trajectories starting at rest wrt earth and ending at rest wrt mars which take least delta-v are curved. If the delta-v is high enough, it can be made to look fairly straight, but it will be curved. Wayne Throop throopw@sheol.org http://sheol.org/throopw >> "Spaceman" <spaceman@yourclockmalfunctioned.duh> >> If the same amount of energy is needed to get to the same spot on the [quoted text clipped - 4 lines] > is going to be curved. In fact, it works better if you don't guess, > and actually try calculating or simulating the cases. sheesh man, you don't "free fall" to mars. You could "freefall" to Venus though. You have to go "uphill" to mars.. The sun is laughing at you and your "freefall" to Mars. It is sorta like freefalling up from Earth. :) So Lets try again. What uses more energy to get to the top of the hill, a curved path around and up the hill or a straight path up the hill? And remove friction if you want to. A:) Both the same B:) The short path C) The long path. > So > Lets try again. [quoted text clipped - 6 lines] > B:) The short path > C) The long path. Absent friction, both take the same amount of energy. >> So >> Lets try again. [quoted text clipped - 8 lines] > > Absent friction, both take the same amount of energy. Yes, If an curved path to mars has enough energy to make it to mars, a straight path would also have enough. The difference of energy needed will only come from the change in course to achieve orbit and of course slowing down to do such.. SignatureJames M Driscoll Jr Spaceman > >> So > >> Lets try again. [quoted text clipped - 12 lines] > If an curved path to mars has enough energy to make it > to mars, a straight path would also have enough. The only possible straight-line motion between earth orbit and mars orbit that does not require continual acceleration just to keep on that path would have to move effectively anti-parallel to the sun's attraction. And this sort of orbit would itself waste HUGE amounts of energy, first when leaving earth to kill the built in earth orbital velocity around the sun and then when reaching Mars orbit to gain mars orbital velocity. >>>> So >>>> Lets try again. [quoted text clipped - 17 lines] > that path would have to move effectively anti-parallel to the sun's > attraction. What? If you have enough energy to make the curved path to Mars you have enough to make the straight path also. The hill is the Sun, the paths are the paths taken. These paths are picked after a slingshot from Earths gravity to give them enough energy at all. If I am wrong about such, you are wrong about the hill question above. :) and if the curve path uses less energy than the straight path would just to hit mars.. (ignoring orbital maneuvers) You have found a violation of the conservation of energy. :) SignatureJames M Driscoll Jr Spaceman : "Spaceman" <spaceman@yourclockmalfunctioned.duh> : and if the curve path uses less energy than the straight path would : just to hit mars.. (ignoring orbital maneuvers) : You have found a violation of the conservation of energy. No, just found yet another thing you don't understand. Not surprising by now. Wayne Throop throopw@sheol.org http://sheol.org/throopw > > The only possible straight-line motion between earth orbit and mars > > orbit that does not require continual acceleration just to keep on [quoted text clipped - 4 lines] > If you have enough energy to make the curved path to Mars > you have enough to make the straight path also. I was not disputing your statement, but only commenting that any straight line free-fall orbit must be parallel (or anti-parallel) to the sun's attraction. > The hill is the Sun, the paths are the paths taken. > > These paths are picked after a slingshot from Earths gravity > to give them enough energy at all. Slingshotting from earth produces maximal velocity when departing earth in the direction of earths motion, but for the straight line orbit to mars, that is exactly the wrong slingshot. >>> The only possible straight-line motion between earth orbit and mars >>> orbit that does not require continual acceleration just to keep on [quoted text clipped - 17 lines] > earth in the direction of earths motion, but for the straight line > orbit to mars, that is exactly the wrong slingshot. Actually As long as the direction is not backwards the final speed "energy" will be the same. The spin speed of Earth and the Orbital speed of Earth both are included in any slingshot angle that is not backwards to the direction of travel. Again, try such in Euclidian 3D space :) The final energy (I will call it speed) will be the same as long as the direction is not backwards from the orbital path. Think about this one: A train is moving at 100 mph. A tennis ball shooter that shoots at 100 mph is mounted on the train (we will neglect air resistance for this of course.) When you shoot the ball upward or forward the speed of the ball will still be 200 mph in it's path speed whether it is shot upward or forward and it only does not work like such when you shoot it against the forward motion. In fact if you shoot it upward the angle it leaves will be 45 degrees and the speed will still be 200 mph on that angle. The forward speed will still only be 100 and the upward speed will still only be 100 but the path speed of the ball itself will be 200. SignatureJames M Driscoll Jr Spaceman >>>> The only possible straight-line motion between earth orbit and >>>> mars orbit that does not require continual acceleration just to [quoted text clipped - 20 lines] > Actually As long as the direction is not backwards > the final speed "energy" will be the same. No. It works the way elastic collisions do (it is in fact a form of elastic collision). The way to "steal" momentum from the larger object, in this case the Earth, is to run into it and "bounce" off of it in its forward direction. For maximum effect that means approaching from a forward direction, looping around using the Earth's gravity to hold you, and coming back out again in a forward direction. The more directly you approach from the frontward direction and the more frontward your departure angle, the greater will be the speed you pick up. > The spin speed of Earth and the Orbital speed of Earth > both are included in any slingshot angle that is not backwards > to the direction of travel. The Earth's spin doesn't enter into it. The spacecraft never touches the Earth during its flyby. > Again, try such in Euclidian 3D space > :) > The final energy (I will call it speed) will be the same > as long as the direction is not backwards from the orbital > path. No, as explained above, it will depend critically on the angle of approach and departure. > Think about this one: > A train is moving at 100 mph. [quoted text clipped - 4 lines] > upward or forward and it only does not work like such > when you shoot it against the forward motion. Velocities don't add that way, even in Newtonian Euclidean reckoning. The net speed of the ball for the upward hit will be sqrt(100^2 + 100^2) = 141.2 mph. > In fact if you shoot it upward the angle it leaves will be > 45 degrees and the speed will still be 200 mph on that angle. Nope. That's like saying that the diagonal of a 100 by 100 square is 200. > The forward speed will still only be 100 and the upward > speed will still only be 100 but the path speed of the ball itself > will be 200. Nope. > Velocities don't add that way, even in Newtonian Euclidean > reckoning. The net speed of the ball for the upward hit > will be sqrt(100^2 + 100^2) = 141.2 mph. Oops sorry, thanks for the wakeup on that one.. got stuck in my linear addition sillyness. Typing to fast and thinking of linear only for some silly reason. I wonder why I did that. Probably because of so many telling me the linear addition is wrong.. but anyways.. Again, I admit.. I am wrong. This is a bad thread for me with word replacements and wrong condition use of math. But I will still be traveling at the orbital speed of Earth at a lesser elipse than you. :) So How fast is that Greg? That is not fast enough to get to Mars at that speed and a shorter path than yours? SignatureJames M Driscoll Jr Spaceman On Sun, 6 Jul 2008 17:07:28 -0400, in a place far, far away, "Spaceman" <spaceman@yourclockmalfunctioned.duh> made the phosphor on my monitor glow in such a way as to indicate that: >> Velocities don't add that way, even in Newtonian Euclidean >> reckoning. The net speed of the ball for the upward hit [quoted text clipped - 5 lines] >Typing to fast and thinking of linear only for some silly reason. >I wonder why I did that. I'm quite confident that it's because you don't know WTF you're talking about. You're ignorant of *both* physics and mathematics (not surprising, of course). >> Velocities don't add that way, even in Newtonian Euclidean >> reckoning. The net speed of the ball for the upward hit [quoted text clipped - 17 lines] > That is not fast enough to get to Mars at that speed > and a shorter path than yours? Once again, the devil is in the details. The trajectory you end up on depends upon the specifics of the slingshot encounter, and that depends upon the prior setup. Getting the spacecraft into the right approach path plus having the planet in the right place at the right time is not a simple exercise. Spacecraft with little Delta-V capability that need to reach the outer solar system have used some pretty elaborate maneuvers to make the requisite slingshot boosts, like looping past Venus for a second pass at Earth after the first slingshot. Very time consuming, but necessary when the fuel budget isn't there. With enough time and planning, you can certainly make it to Mars with a gravity assist maneuver around Earth, and yes, you can arrange things so that the inter-orbit travel time is less than the minimal energy transfer. This discounts the possibly months long diddling about in the inner solar system to get the optimum encounter lined up, of course. Any trajectory you can attain with gravity assist maneuvers could also be done with sufficient available fuel, in much short time. It's the usual time versus fuel tradeoff that is a constant feature of space travel. > >>> The only possible straight-line motion between earth orbit and mars > >>> orbit that does not require continual acceleration just to keep on [quoted text clipped - 23 lines] > both are included in any slingshot angle that is not backwards > to the direction of travel. To end with a velocity vector perpendicular to Earth's orbit, the exit direction from any sling shot operation starting from EARTH will have to have a considerable component opposite to the direction of travel of earth. > Again, try such in Euclidian 3D space > :) > The final energy (I will call it speed) Energy is not the same as speed, and only a fool would call it so. It still remains a fact that the the only straight line orbits, such as "Spaceman" recommends, due only to the sun's gravitational field, are those in directions parallel to that field, i.e., radial. And any such orbit as a passage from Earth to Mars will be horribly wasteful of energy. >>> So >>> Lets try again. [quoted text clipped - 15 lines] > from the change in course to achieve orbit and of course > slowing down to do such.. And said changes in course will be punnishingly expensive in terms of fuel, because you're throwing away the velocity you start with, then adding most of it back at the other end, all with fuel expenditure. : "Greg Neill" <gneillREM@OVEsympatico.ca> : And said changes in course will be punnishingly expensive : in terms of fuel, because you're throwing away the : velocity you start with, then adding most of it back : at the other end, all with fuel expenditure. Note another way to look at it: even if you assume lithobreaking at the far end, your radial velocity will be far less than the delta-v your ballistic launch is capable of; it's a simple matter of vector addition. Given a specific launch delta-v, your speed (ie, absolute value of velocity) wrt earth will start out the same, but in getting to a radial-from-the-sun trajectory (which is the only straight trajectory),you'll end with less speed wrt the sun than on a curved trajectory. And therefore the time before you lithobrake on mars will be longer. Wayne Throop throopw@sheol.org http://sheol.org/throopw >> "Greg Neill" <gneillREM@OVEsympatico.ca> >> And said changes in course will be punnishingly expensive [quoted text clipped - 11 lines] > trajectory. And therefore the time before you lithobrake on mars will > be longer. Wow, If an object is thrown radially from a moving orbit, It does not actually travel perfectally radial. The orbit speed is still there and a look from another point or from the Sun itself will not show a radial path. It seems both you and Greg think I want to somehow cancel out the Earths orbital velocity? LOL That is funny, You both are making the same mistake now. The radial "shot" does not cancel the Earth orbital velocity just as running and throwing a tennes ball striaght up in the air would not cancel the forward motion. Sheesh. You both need a refresher course in basic physics. WOW. SignatureJames M Driscoll Jr Spaceman >>> "Greg Neill" <gneillREM@OVEsympatico.ca> >>> And said changes in course will be punnishingly expensive [quoted text clipped - 19 lines] > It seems both you and Greg think I want to somehow cancel > out the Earths orbital velocity? Yes, you said that you want your spacecraft to take a straight radial path outward from the Sun (thus the shortest distance between orbits). Are you now shifting from that position? > LOL > That is funny, > You both are making the same mistake now. > The radial "shot" does not cancel the Earth orbital velocity > just as running and throwing a tennes ball striaght up in the air > would not cancel the forward motion. Then it's not a radial shot any more, but just another curved, elliptical trajectory. > Sheesh. > You both need a refresher course in basic physics. > WOW. : "Spaceman" <spaceman@yourclockmalfunctioned.duh> : If an object is thrown radially from a moving orbit, : It does not actually travel perfectally radial. Duh. That's what everybody's been trying to tell you for many tens of posts now. Took you a while, but you finally got there. The *path* is curved wrt the sun. You said the straight path was the best. You now tacitly agree you were wrong about that. Progress, of a sort. : The orbit speed is still there and a look from another point or from : the Sun itself will not show a radial path. Or, for that matter, from the earth, since the earth is accelerated during the trip. It's accelerated towards the sun just like everything else in the solar system. : It seems both you and Greg think I want to somehow cancel : out the Earths orbital velocity? That's largely because you said y ou wanted a straight path. : You both are making the same mistake now. You said you wanted a straight path. We tried to point out to you why that's a bad idea. You now agree it's a bad idea. Again, progress of a sort. : LOL It's good that you are easily amused. Perhaps you could play with some car keys, or a brightly colored mobile. Wayne Throop throopw@sheol.org http://sheol.org/throopw >> "Spaceman" <spaceman@yourclockmalfunctioned.duh> >> If an object is thrown radially from a moving orbit, [quoted text clipped - 4 lines] > The *path* is curved wrt the sun. You said the straight path was > the best. You now tacitly agree you were wrong about that. I said the stright shot was best, If I goofed and typed path instead of shot when I stated such, then I admit my fault with such a statement. but yet again, I can admit faults unlike you. So we know you are so perfect you are aslo perfectally flawed. :) SignatureJames M Driscoll Jr Spaceman > >> "Greg Neill" <gneillREM@OVEsympatico.ca> > >> And said changes in course will be punnishingly expensive [quoted text clipped - 15 lines] > If an object is thrown radially from a moving orbit, > It does not actually travel perfectally radial. But Spaceman's supposed orbit IS perfectly radial, so he has to spend energy to kill tall tranverse velocity to get onto that orbit. > The orbit speed is still there Not for a straight line orbit, relative to an inertial coordinate system. and a look from another point > or from the Sun itself will not show a radial path. > It seems both you and Greg think I want to somehow cancel > out the Earths orbital velocity? Since you want a straight line free fall orbit, you must kill all transverse velocity relative to the sun, to get it. > LOL > That is funny, > You both are making the same mistake now. > The radial "shot" does not cancel the Earth orbital velocity Then it does not produce a straight line orbit. > just as running and throwing a tennes ball striaght up in the air > would not cancel the forward motion. > Sheesh. > You both need a refresher course in basic physics. > WOW. Actually, I would flunk Spaceman on physics. > But Spaceman's supposed orbit IS perfectly radial, so he has to spend > energy to kill tall tranverse velocity to get onto that orbit. You are now resorting to the typicle twisting of the relativist. That is sad. Next thing you know, you will be saying that I said blue nosed nymphs power the spaceship. LOL > Since you want a straight line free fall orbit, you must kill all > transverse velocity relative to the sun, to get it. I again will say it. If I said radial path instead of radial Shot I made that mistake. but again I make mistakes because I am human and apparently you must not be and must be a perfect human replica that never makes mistakes. So perfect, that you are also perfectally flawed. :) SignatureJames M Driscoll Jr Spaceman |
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