23 Feb 16
Originally posted by sonhouseYou thrust to one side to turn. To achieve a circle you would need to vary your sideways thrust depending on your velocity. The faster you are going the more sideways thrust you will need.
Just 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 ...[text shortened]... 180 degree turn of some radius and at the end of the turn to be again at zero relative velocity.
If you want to do a circle with constant total thrust, your thrust would start directly behind you and slowly rotate around you ending up thrusting directly in front of you.
I am not certain whether you would rotate it at constant angular speed or not.
But it now has no real relation to the initial question.
23 Feb 16
Originally posted by twhiteheadIt's a simplified version of the original, say you are on the top of a very high platform, lets say there are two space elevators and you are at the top of one where that would be something like 30K above the surface. Now you want to go to the other elevator also at 30Km above the surface but for some reason you have to maintain a circular path. Leaving station one at zero relative velocity and you land on the other at zero relative velocity like taking off from an airport, you start with zero velocity and end with zero velocity.
You thrust to one side to turn. To achieve a circle you would need to vary your sideways thrust depending on your velocity. The faster you are going the more sideways thrust you will need.
If you want to do a circle with constant total thrust, your thrust would start directly behind you and slowly rotate around you ending up thrusting directly in front o ...[text shortened]... e it at constant angular speed or not.
But it now has no real relation to the initial question.
Does that make more sense. You still have to deal with a gravity field but not as steep as being on Earth's surface. Not sure what the gravity would be like at the geo stationary point if you were on top of a 30k high platform. Probably roughly 1/4th surface gravity. Say 2 m/s accel at that point.
23 Feb 16
Originally posted by sonhouseVery very much simplified. Gravity is one of the largest components in the original.
It's a simplified version of the original,
say you are on the top of a very high platform, lets say there are two space elevators and you are at the top of one where that would be something like 30K above the surface.
30km is nothing relative to the diameter of the earth.
Does that make more sense. You still have to deal with a gravity field but not as steep as being on Earth's surface. Not sure what the gravity would be like at the geo stationary point if you were on top of a 30k high platform.
You have your units all wrong. Geostationary orbit is at just over 35,000km.
Probably roughly 1/4th surface gravity.
Well if you are geostationary, then g is essentially zero because it is counteracted by your obit. To get to the other space elevator you just rise a little bit which causes you to slow down relative to earth then wait for the elevator to come to you.
Orbital mechanics and gravity are so critical that you can't be wishy washy about whether or not to include them.
24 Feb 16
Originally posted by twhiteheadI meant 36,000 Km, sorry. Geo sync orbit on the platform of a space elevator. So here you want to go to the other platform also at 36,000 Km up and constrained to a circular path, starting at zero velocity relative to tower and ending the same on the other platform exactly opposite Earth. Looks like a circular path of around 260,000 Km, half circle.
Very very much simplified. Gravity is one of the largest components in the original.
[b]say you are on the top of a very high platform, lets say there are two space elevators and you are at the top of one where that would be something like 30K above the surface.
30km is nothing relative to the diameter of the earth.
Does that make more sense ...[text shortened]... and gravity are so critical that you can't be wishy washy about whether or not to include them.
- Joined
- 10 Dec 06
- Moves
- 8528
24 Feb 16
1 edit
From what I skim read of the discussion, I think you are just looking for centripetal acceleration in circular motion.
If this is the case and you want to move in a perfect circular arc while maintaining constant tangential velocity. Your Thrust would have to always maintain a "magnitude" of:
F_t =M_craft*a_c = M_craft*(v^2/R)
It would need to be perpendicular at all times to the vector that initially described your linear motion. So I think in practice lets say your out in free space(no external forces on the spacecraft) moving (coasting) at a constant velocity "v". You want to move in a circle with radius "R". You need to fire your thrusters such that the magnitude of the thrust force is:
F_t = M_craft*(v^2/R)
AND the direction is perpendicular your linear (previous straight line) direction of motion.
I think that is right, hope that helps.
24 Feb 16
1 edit
Originally posted by joe shmoDon't forget you start with zero relative velocity and end the same, you don't want to crash into the tower at 10,000 Km/sec or whatever🙂 So that requires an accel phase and a decel phase while transcribing a circular path.
From what I skim read of the discussion, I think you are just looking for centripetal acceleration in circular motion.
If this is the case and you want to move in a perfect circular arc while maintaining constant tangential velocity. Your Thrust would have to always maintain a "magnitude" of:
F_t =M_craft*a_c = M_craft*(v^2/R)
It would need to be ...[text shortened]... linear (previous straight line) direction of motion.
I think that is right, hope that helps.
- Joined
- 10 Dec 06
- Moves
- 8528
24 Feb 16
5 edits
Originally posted by sonhouseOk, so you want to muck things up a bit. You want to change your tangential velocity while traveling in the same circular path of radius "R". The very thing I stated needed to be constant for the above relationship to hold...
Don't forget you start with zero relative velocity and end the same, you don't want to crash into the tower at 10,000 Km/sec or whatever🙂 So that requires an accel phase and a decel phase while transcribing a circular path.
I'm at work now. I'll get back to you with a solution (at least I'll try), unless someone else comes up with it for you in the mean time. 😉
- Joined
- 10 Dec 06
- Moves
- 8528
25 Feb 16
3 edits
Originally posted by joe shmoSo in circular motion (n-t : normal and tangential) coordinates The magnitude of the acceleration is given by:
Ok, so you want to muck things up a bit. You want to change your tangential velocity while traveling in the same circular path of radius "R". The very thing I stated needed to be constant for the above relationship to hold...
I'm at work now. I'll get back to you with a solution (at least I'll try), unless someone else comes up with it for you in the mean time. 😉
|a| = √( (a_t)² + (a_n)² ) ...Eq(1)
The tangential velocity
Assumptions: constant tangential acceleration "a_t" ; Initial tangential velocity = 0
v = (a_t)*t Eq(2)
a_n = v²/R Eq(3) -> Sub Eq(2) -> Eq(3) ==> ((a_t)*t)²/R ...Eq(4)
Sub Eq(4) --> Eq(1) ==> Eq(5)
|a| = √( (a_t)² + (((a_t)*t)²/R)² ) --> √( (a_t)² + ( (a_t)^4 * t^4 / R² ) ) --> Eq(5)
You want to "accelerate" half way between point A and B on the circumference, then "decelerate" to a stop on that same circumference.
Let the angle subtended by arc AB be represented as "α"
Let the distance from A-B be represented as S:
S = R*α ...Eq(6)
Let half the circumfrential distance be represented by:
S' = S/2 = R*α/2 ...Eq(7)
with a_t constant:
S = 1/2*(a_t)*t² ...Eq(8)
Now eliminate "t" in Eq(5) from Sub Eq(7) --> Eq(8) and solve for "t"
t² = α*R/(a_t) --> t^4 = (α*R/(a_t))² ...Eq(9)
Sub Eq(9) --> Eq(5) for t^4
|a| = (a_t)*√( (1 + α² )
And the cos of the angle "ß" it makes with respect to the normal axis is given by:
cos( ß ) = α/(√( (1 + α² ))
So the Magnitude of the Thrust would be:
M_craft*|a| ==> M_craft*(a_t)*√( (1 + α² )
and the direction relative to normal:
ß = arccos( α/√ (1 + α² ) )
Hopefully someone will confirm this result.