19 Feb 16
So you have a space craft capable of constant acceleration, say one to two G's or so and you want to go from one point on Earth to another point directly opposite, 20,000 odd km away. One way would be to establish a path along an equilateral triangle where the journey would be in 4 parts, say a triangle with sides all equal to 12,800 km per leg. You could accelerate plus 1 g to half way through the first leg, then decelerate the second half of that leg then turn 60 degrees and do the same on the second segment of the triangle and end up at the destination with zero relative velocity so you can land safely.
What if you were constrained for some reason to be forced into a perfectly circular path. What kind of vector forces of acceleration and deceleration would you need to apply to start with zero relative velocity on on side of the Earth and end up on the other side of the Earth also with zero relative velocity? Can it be done at all?
22 Feb 16
2 edits
Are you taking the earth's motion into account?
Why the triangle in the first part? Why not simply go straight and accelerate half the way and decelerate the other half? What is the complication?
Are you taking the earth to be a perfect sphere?
And by the way, equilateral triangles on the surface of a sphere do not have 60 degree angles and equilateral triangles that take you to the opposite side of the earth, do not exist.
22 Feb 16
Originally posted by sonhouseJust to check, the destination is an exact antipode, can we assume that the earth is a perfect sphere with no atmosphere? What path do you want, a triangular one or a circle? I doubt that it is possible to obtain a circular path with constant acceleration.
So you have a space craft capable of constant acceleration, say one to two G's or so and you want to go from one point on Earth to another point directly opposite, 20,000 odd km away. One way would be to establish a path along an equilateral triangle where the journey would be in 4 parts, say a triangle with sides all equal to 12,800 km per leg. You could a ...[text shortened]... d end up on the other side of the Earth also with zero relative velocity? Can it be done at all?
22 Feb 16
1 edit
Why a perfectly circular path? and at what altitude? I assume you start from the surface of the planet and that you want to touch down to the surface at the destination without multiple orbits, don't you? You need ballistics to solve this problem, not geometry.
http://www.shooterscalculator.com/ballistic-trajectory-chart.php?t=f630255c
22 Feb 16
Originally posted by DeepThoughtWe can assume it's airless so the atmosphere wouldn't need to be taken into account. The triangle was just an effort to allow one to start out at zero velocity relative to the surface and end up with zero relative velocity so you can land on the other side of the globe.
Just to check, the destination is an exact antipode, can we assume that the earth is a perfect sphere with no atmosphere? What path do you want, a triangular one or a circle? I doubt that it is possible to obtain a circular path with constant acceleration.
I was thinking some kind of vector thrust that would allow you to follow a circular path maybe accelerating half way then decelerating half way so you can touch down and of course the gravity of the planet has to be taken into account especially at touchdown.
22 Feb 16
Originally posted by sonhouseWell SpaceX has done it many times.
We can assume it's airless so the atmosphere wouldn't need to be taken into account. The triangle was just an effort to allow one to start out at zero velocity relative to the surface and end up with zero relative velocity so you can land on the other side of the globe.
I was thinking some kind of vector thrust that would allow you to follow a circular pat ...[text shortened]... wn and of course the gravity of the planet has to be taken into account especially at touchdown.
What are you looking for, exact formulas? Yes or no?
You still haven't said anything about the rotation of the earth as that is a critical factor. If you don't start at the equator then the direction of the thrust required to maintain a geodesic orbit relative to the surface of the earth would be quite complicated. If you only need the spacecraft's orbit relative to the centre of the earth (and not the surface) to be a 'straight line' then I see no major problems. You would thrust some of the way and slow down some of the way, but which way and how much depends on your location and direction relative to the earth's rotation.
I am sure all the space companies have software to do all that for you. For real life calculations it is not trivial.
22 Feb 16
1 edit
Originally posted by twhiteheadFor now, just assume the planet is not spinning, that should simplify that part. I just want to see if there is any kind of vector thrust that can do the job with the copnstraint of having to maintain a circular flight path, say a given distance above the surface of the planet where you are starting say 100 km above the surface and ending 100 km above the surface with zero relative velocity so you don't have to shed delta V getting to the surface.
Well SpaceX has done it many times.
What are you looking for, exact formulas? Yes or no?
You still haven't said anything about the rotation of the earth as that is a critical factor. If you don't start at the equator then the direction of the thrust required to maintain a geodesic orbit relative to the surface of the earth would be quite complicated. ...[text shortened]... ce companies have software to do all that for you. For real life calculations it is not trivial.
23 Feb 16
Originally posted by sonhouseI am still not sure why you think there would be any problem. What constraints are we missing? Obviously you would thrust down a fixed amount (1g) all the way to maintain height, and back some of the way to go forward then forward some of the way to slow down.
For now, just assume the planet is not spinning, that should simplify that part. I just want to see if there is any kind of vector thrust that can do the job with the copnstraint of having to maintain a circular flight path, say a given distance above the surface of the planet where you are starting say 100 km above the surface and ending 100 km above the surface with zero relative velocity so you don't have to shed delta V getting to the surface.
To make it simple, point your nozzle 45 degrees to the vertical aimed behind you and thrust at root 2 g for the first half of the journey then tilt so that you are 45 degrees to the vertical aimed in front of you. I believe that will give you a thrust of 1g upwards at all times thus maintaining height and 1g to the side first accelerating then decelerating.
23 Feb 16
Originally posted by twhiteheadDon't know if that matters, the constraint is a circular orbit with some vector of thrust, obviously it would have to be some equivalent of my triangular path where there is a net positive thrust in the direction of travel and at some point, halfway?, the reverse takes place, a deceleration but sticking to the circular path.
I have assumed above that you are going too slow for your orbit to give you any lift.
23 Feb 16
Originally posted by sonhouseI still don't understand what your triangular thrust was. Are you saying you head north west then turn south west, or are you talking about vertical triangles?
Don't know if that matters, the constraint is a circular orbit with some vector of thrust, obviously it would have to be some equivalent of my triangular path where there is a net positive thrust in the direction of travel and at some point, halfway?, the reverse takes place, a deceleration but sticking to the circular path.
When you say 'vector of thrust' are you saying you must thrust in only one direction, and if so are we talking about one direction relative to the surface of the earth or relative to space?
23 Feb 16
If you want to maintain a constant hight, then you need the vertical thrust to be exactly 1g.
So merely factor that out then solve it as if you are a car on the surface of the earth ignoring friction and solve for any horizontal thrust you need and then add the vertical 1g back in at the end.
23 Feb 16
Originally posted by twhiteheadPicture the Earth, in this case no atmosphere to muck things up. You are on the surface or on top of a mountain, and you draw a representation of a triangle where one leg is where you are standing and on the opposite side of the globe is another mountain which is leg #3. Leg #2 is pointing out into space and as close to equilateral as possible so if you drew a line directly down from leg #2 it would bisect Earth and go directly to the center of Earth if it was drawn long enough.
I still don't understand what your triangular thrust was. Are you saying you head north west then turn south west, or are you talking about vertical triangles?
When you say 'vector of thrust' are you saying you must thrust in only one direction, and if so are we talking about one direction relative to the surface of the earth or relative to space?
I hope that is clear.
Then you take off following the path of leg 1 and accelerate halfway through leg one at one g or whatever, maybe it takes 2 g's to overcome gravity, but halfway, you switch the engines around 180 degrees and thrust negative to slow you down to get to point 2, the point furthest from Earth, then you turn around 120 degrees so the engine is pointed back into space, accelerate at the same rate on leg #2 going back to Earth but aiming at the opposite end of the globe. So you accelerate at your 2 g's or so halfway through THAT leg the do the same maneuver reversing the thrust to slow down at the same g rate so when you get to the top of the mountain you land with zero velocity there.
Does that clear up the triangle path? I said on top of a mountain to get the triangles out far enough not to touch ground early. If for instance, you made the triangle more obtuse say with 40 degrees on the outbound legs you would be cutting through Earth before you got up in space. So I thought 60 degrees would be the minimum angle to undertake such a journey. It could be 65 degrees, don't know for sure.
For instance, if you were not sure and wanted to make totally sure, you could just go tangent to a line linking the apex of the triangle to Earth's center, parallel to that line up then instead of a triangular trip you would basically travel on three edges of a box where you go each leg half way, decelerate the rest of that leg, now do only a 90 degree turn now going perpendicular to the line of the previous triangle apex to Earth, do the same thing, then turn 90 degrees again for the last leg, do the same thing, halfway accelerate, halfway decelerate and end up in the same spot with zero velocity. Does that make sense?
23 Feb 16
Originally posted by sonhouseYes. It is totally impossible as you would see if you draw a circle on a piece of paper then try to draw your triangle. Note that the tallest mountains on earth have no real impact on this. If you drew a circle on a piece of paper to represent the outline of the earth, a mountain would be an wiggle in the line so small you could not detect it.
Does that clear up the triangle path?
The three sided figure you mention later is possible but would not use constant thrust to achieve because of gravity.
It is still far from clear what you really want to know.
23 Feb 16
Originally posted by twhiteheadJust if there is a acceleration/deceleration pattern that will allow a circular path going around a planet. Even if you count the 'planet' as a big massless bubble to make things easier. So you are in space millions of Km from any mass and you want to turn, what thrust vector(s) would you need to maintain a circular path in the turn? Like starting with zero velocity and doing 180 degree turn of some radius and at the end of the turn to be again at zero relative velocity.
Yes. It is totally impossible as you would see if you draw a circle on a piece of paper then try to draw your triangle. Note that the tallest mountains on earth have no real impact on this. If you drew a circle on a piece of paper to represent the outline of the earth, a mountain would be an wiggle in the line so small you could not detect it.
The three ...[text shortened]... hrust to achieve because of gravity.
It is still far from clear what you really want to know.