- #1

- 19

- 0

- Thread starter thinkagain
- Start date

- #1

- 19

- 0

- #2

- 35,263

- 6,319

First, I need to be sure we have the problem properly defined. There is an astronaut travelling with velocity v. She has a thruster that can be pointed in any direction and delivers constant thrust. She wishes to travel at the same speed but at right angles to v, and do so in the least time. Right?

What is the overall momentum change? What is the relationship between force, momentum and time?

- #3

- 19

- 0

- #4

- 35,263

- 6,319

I don't know why you think the answer should be an Euler spiral. The application of those to transport is to avoid too rapid a change in acceleration. It sounds like that is not a constraint here.

I repeat my original questions. If the mass is m, the initial velocity is ##\vec v_i## and the final velocity is ##\vec v_f## :

What is the overall momentum change?

What is the relationship between force, momentum and time?

- #5

- 19

- 0

Let's say the astronaut is traveling along a line at 100 mph. They need to reach a parallel line as quickly as possible, but they need to have stopped all their velocity along the initial axis by the point they hit the second line. We can say the lines are 100 feet apart if that helps. We can also say their combined mass and thrust capabilities allowed 1 g of acceleration. One way of doing this would be to slow down to whatever speed was needed and then take a circular path with constant radius and velocity by pointing their thruster in the opposite direction to the center point of a circle they traveled on.

Alternatively I would think it would be faster if they used their thruster to slow down and turn at the same time. I would think that would give them a Euler spiral shaped path, but please let me know if I am wrong. I'm primarily wondering where the center of force they aim their thruster opposite of would be for this path. Would it move?

- #6

sophiecentaur

Science Advisor

Gold Member

2020 Award

- 25,386

- 4,959

F = mv

I can't think of a path that would achieve a 90° change of direction, any quicker, with the same final speed.

There are variations around this; for instance, when she/he doesn't insist on the speed being maintained. The quickest way to just get a 90° course change would be to fire them backwards and leave just enough 'squirt' to give a finite velocity at right angles to the original.

PS Are we discussing Sandra Bullock or George Clooney here?

- #7

- 19

- 0

- #8

- 867

- 61

This is actually a very interesting problem. I'm not sure what the solution is yet, but its only a quarter circle in 1 very special case.

- #9

- 19

- 0

- #10

sophiecentaur

Science Advisor

Gold Member

2020 Award

- 25,386

- 4,959

I was assuming that the turn is to be as quick as possible and with no change of speed (and that gives you the radius). If you have more thrust, then the radius (and the time taken) can both be less.

This is actually a very interesting problem. I'm not sure what the solution is yet, but its only a quarter circle in 1 very special case.

How long is a piece of string? If you are interested in a 'vestigial' final speed - just in the wanted direction - the direction would be in the reverse direction, with a squirt at right angles at the very end. The two could be combined into just one constant direction, of course and if the reverse thrust is much larger than the lateral thrust, you could simplify the situation into x and y co ordinates with the thrust being directed in a global sense, rather than relative to the craft. You would then just have motion under constant acceleration in both x and y directions, so the resultant would be the combination of two parabolas. I expect that curve will have been given a name by someone but it would be easy enough to plot it out with the help of Excel or equivalent. I don't think that would give an optimal course, though - that is if the circular path would be optimal.I'm more concerned with the "squirt". What angle would that need to be at?

- #11

- 35,263

- 6,319

If you only care that the final speed equals the initial speed then there is a quicker and simpler solution than a circular path. The questions I asked you lead to this.

If you don't even care about the final speed then the question ceases to make sense. You can drop the final speed to zero, so that there is not even a direction at the end. The problem then reduces to stopping as quickly as possible.

- #12

- 19

- 0

Not sure what you mean by question not making sense if I don't care about final speed. Maybe people are misunderstanding the problem. The colored paths just show the curved part in between the two lines, but the astronaut is assumed to keep going after the end of the path at whatever final speed they reach. If their speed needed to drop to zero that seems like it would take longer than if they were able to turn 90 degrees and cross the line while they were still going 50 mph or whatever speed is attainable on that radius at the end of the spiral.

- #13

sophiecentaur

Science Advisor

Gold Member

2020 Award

- 25,386

- 4,959

That doesn't make sense to me. "whatever final speed" implies it can be arbitrarily low, as haruspex states, and the transition is accomplished in the shortest possible time.but the astronaut is assumed to keep going after the end of the path at whatever final speed they reach

There is another point to consider and that is the total amount of energy available needs to be specified - otherwise, the final speed could be as high as you like.

I think it is necessary to restate the OP more precisely if we are to progress this further.

- #14

- 19

- 0

I would be interested in how to calculate the speed they are at as they reach the second line, but mainly interested in which direction their thruster would be pointed to accomplish this movement in the minimum time possible.. Intuitively I would think there is a certain amount of force needed to stop all movement along the original axis and then a certain amount needed to cross the 100 feet at the same time. I would think the thruster would want to be aimed primarily opposite their original direction of travel, but aimed away from the second line at some shallow angle. Maybe this angle would ideally change as they approached the second line. Hopefully this is making sense. Sorry, I don't have much formal physics/math training beyond high school.

- #15

- 35,263

- 6,319

I mean there is a simpler solution in regards to what the astronaut has to do. (Remember, you have provided no means by which the astronaut will rotate on her own axis, so we can assume she faces in a constant direction. Arranging to move in a circular path could be tricky.)

Not sure what you mean by question not making sense if I don't care about final speed. Maybe people are misunderstanding the problem. The colored paths just show the curved part in between the two lines, but the astronaut is assumed to keep going after the end of the path at whatever final speed they reach. If their speed needed to drop to zero that seems like it would take longer than if they were able to turn 90 degrees and cross the line while they were still going 50 mph or whatever speed is attainable on that radius at the end of the spiral.

If she does move in a circular path under constant thrust, the speed will be constant. You can deduce the radius from F=mv^2/r. How long will it take that way?

If you do not care about final speed, she could just use the thruster to come to a stop, then with one tiny blast at right angles move off in the desired direction. How long will that take? Which is quicker?

The interesting case is where the final speed is required to equal the original speed but you don't require it to be constant during the transition. For this, think in terms of the overall change in momentum required, and how that would be most obviously achieved.

- #16

- 19

- 0

- #17

- 35,263

- 6,319

You may think those things, but you would be wrong.

The reason you think stopping and then moving off at a right angle would be slow is, I suggest, that intuitively you are thinking in terms of then accelerating to regain the original speed. But you have said that is not required, so it is only the time taken to stop that counts.

You seem reluctant to do the calculations. They are not difficult. If you show some attempt I'm happy to assist.

- #18

- 19

- 0

- #19

- 1,988

- 269

- #20

- 35,263

- 6,319

Only if you really do not care what your new speed is. Applying the thrust for a tiny fraction of a second in the new direction is enough to get you moving, so does not effectively contribute to the total time.Okay, just to make sure I understand, bringing your speed along the original line to zero and then turning the thruster 90 degrees and accelerating directly toward the second line is the fastest way to do it?

How long does it take to come to a stop, starting at speed v and decelerating under a constant force F? Since the force is constant (in magnitude and direction) the acceleration is constant, F/m. Are you familiar with the SUVAT equations? If not, use your favourite search engine.

- #21

- 19

- 0

Okay so if we are starting at 100 mph then it would take 4.56 seconds to reach 0 mph if I'm doing that right. Then we need to go 100 feet so not sure how to calculate exactly how long that would take at 1 g, but I am estimating around 2 seconds if we can do 32 feet per second squared. That would be around 6.56 seconds total.

edit: I think I underestimated the time it would take to travel the 100 feet by a little bit.

Correct me if I'm wrong, but intuitively that doesn't seem like the fastest way because if we switched the order and did the 100 feet first we would continue traveling at approx. 64 fps past the line as we slowed down to 0 mph on the initial axis of travel and would be 291 feet past the second line by the time we finally stopped moving along our initial axis.

If we started our sideways thrust first and wanted to time it so that we hit the line at the same time we reached 0 on initial axis it seems we would need to create an initial sideways velocity of 22 feet per second which would take around 2/3 of a second of thrust. Then we turn our thrusters and slow down to 0 on the initial axis which would again take 4.56 seconds. Total time of a little over 5 seconds. Except for the initial little sideways thrust this seems like it would create a path at least similar to a Euler spiral. Would this be the quickest way? Or would combining the 2 different thrust periods into one single event at some angle in the middle be even faster than this?

edit: I think I underestimated the time it would take to travel the 100 feet by a little bit.

Correct me if I'm wrong, but intuitively that doesn't seem like the fastest way because if we switched the order and did the 100 feet first we would continue traveling at approx. 64 fps past the line as we slowed down to 0 mph on the initial axis of travel and would be 291 feet past the second line by the time we finally stopped moving along our initial axis.

If we started our sideways thrust first and wanted to time it so that we hit the line at the same time we reached 0 on initial axis it seems we would need to create an initial sideways velocity of 22 feet per second which would take around 2/3 of a second of thrust. Then we turn our thrusters and slow down to 0 on the initial axis which would again take 4.56 seconds. Total time of a little over 5 seconds. Except for the initial little sideways thrust this seems like it would create a path at least similar to a Euler spiral. Would this be the quickest way? Or would combining the 2 different thrust periods into one single event at some angle in the middle be even faster than this?

Last edited:

- #22

- 35,263

- 6,319

It will be more educational if we stick to symbolic values and see how the formulae turn out.if we are starting at 100 mph

What 100 feet? This seems like a new constraint you are adding.we need to go 100 feet

1g? Is that the available thrust? Again, let's stick to symbols: Initial speed u, mass m, available thrust F, time t.1 g

I don't know why you're so keen on Euler spirals in this context. As I mentioned, the benefit of Euler spirals is in minimising 'jerk'. There will definitely be solutions that take less time by not worrying about jerk.it would create a path at least similar to a Euler spiral

- #23

A.T.

Science Advisor

- 10,909

- 2,431

Since you don’t care about final speed, you can stop right here.Okay, just to make sure I understand, bringing your speed along the original line to zero and ...

- #24

A.T.

Science Advisor

- 10,909

- 2,431

Firing the thruster at 45° to the original direction is faster.If the thrusters deliver a fixed thrust, the best direction to fire them must (surely?) be towards the centre of a circular path with a radius, set by the equation

F = mv2/r

I can't think of a path that would achieve a 90° change of direction, any quicker, with the same final speed.

- #25

sophiecentaur

Science Advisor

Gold Member

2020 Award

- 25,386

- 4,959

I guess someone needs to do the actual sums to prove or disprove that. It would be a matter of comparing the time to zero velocity with acceleration of -a√2 and the time for a quarter turn with a centrepetal acceleration a.Firing the thruster at 45° to the original direction is faster.

Volunteers please. I have to go shopping right now.

- Last Post

- Replies
- 4

- Views
- 5K

- Last Post

- Replies
- 2

- Views
- 4K

- Replies
- 3

- Views
- 1K

- Replies
- 18

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 3K

- Last Post

- Replies
- 3

- Views
- 5K

- Last Post

- Replies
- 0

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 5

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 5K