Direction of Forces on an Euler Spiral Path | Intuitive Explanation

[thinkagain]

Trying to figure out direction of forces of an object traveling on an Euler spiral path. As an example if you had an astronaut with a jetpack and he wanted to change his direction 90 degrees he could aim his thrusters outward from the center of a circle and he would turn at a constant rate with a constant radius. But if he wanted to change his direction as quickly as possible and shrink the radius he was traveling on as quickly as possible in which direction would he aim his thrusters? I don't have a higher math understanding so please keep things intuitive if possible. Thanks

 

[haruspex]

Trying to figure out direction of forces of an object traveling on an Euler spiral path. As an example if you had an astronaut with a jetpack and he wanted to change his direction 90 degrees he could aim his thrusters outward from the center of a circle and he would turn at a constant rate with a constant radius. But if he wanted to change his direction as quickly as possible and shrink the radius he was traveling on as quickly as possible in which direction would he aim his thrusters? I don't have a higher math understanding so please keep things intuitive if possible. Thanks

First, I need to be sure we have the problem properly defined. There is an astronaut traveling 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?

 

[thinkagain]

Thruster can be pointed in any direction and offers constant thrust. Speed can vary. Let's say she is starting out at 100 mph in a straight line and wants to change direction 90 degrees by changing her radius of travel as quickly as possible. I would assume the path taken would be a Euler spiral. The idea is to get through 90 degrees of the spiral as quickly as possible. Trying to figure out ideal direction for her to point the thruster in order to accomplish this.

 

[haruspex]

Thruster can be pointed in any direction and offers constant thrust. Speed can vary. Let's say she is starting out at 100 mph in a straight line and wants to change direction 90 degrees by changing her radius of travel as quickly as possible. I would assume the path taken would be a Euler spiral. The idea is to get through 90 degrees of the spiral as quickly as possible. Trying to figure out ideal direction for her to point the thruster in order to accomplish this.

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?

 

[thinkagain]

Thanks for your help. I"m actually not sure what you are asking, but let me try posing the question a different way with illustrations to see if that helps.

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?

 

[sophiecentaur]

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.
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?

 

[thinkagain]

Thanks for the input. We'll say it's Sandra:) The velocity doesn't have to be maintained. Not really concerned with what the final velocity is just the quickest way to get to 90 degrees on the second line. I'm more concerned with the "squirt". What angle would that need to be at? Would it change and how could it be calculated?

 

[mrspeedybob]

The circular path will only be the quickest if the thruster's maximum thrust is exactly what would be needed to make a turn with a radius equal to the distance between the lines. If the thrust is less then that, then the astronaut will still have forward velocity when they reach the 2'nd line. If the thruster is capable of higher thrust then a quarter circle turn would have to be followed by a straight path to reach the second line.

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.

 

[thinkagain]

Thanks, yeah I was not thinking the circular path would be the ideal solution, I just put it there as an example of a possible solution to explain the problem better. I put the 100 mph, 100 feet, 1 g constraints because I would think that would require deceleration prior to the circular path. Or I was thinking ideally combining the deceleration with the turning in the Euler spiral example. Not sure if you can tell by the illustrations, but the blue line in the 2nd illustration is the 1st 90 degrees of a Euler spiral.

 

[sophiecentaur]

The circular path will only be the quickest if the thruster's maximum thrust is exactly what would be needed to make a turn with a radius equal to the distance between the lines. If the thrust is less then that, then the astronaut will still have forward velocity when they reach the 2'nd line. If the thruster is capable of higher thrust then a quarter circle turn would have to be followed by a straight path to reach the second line.
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.

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.

I'm more concerned with the "squirt". What angle would that need to be at?

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.

 

[haruspex]

Thanks, yeah I was not thinking the circular path would be the ideal solution, I just put it there as an example of a possible solution to explain the problem better. I put the 100 mph, 100 feet, 1 g constraints because I would think that would require deceleration prior to the circular path. Or I was thinking ideally combining the deceleration with the turning in the Euler spiral example. Not sure if you can tell by the illustrations, but the blue line in the 2nd illustration is the 1st 90 degrees of a Euler spiral.

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.

 

[thinkagain]

What would be the simpler solution than circular with constant speed? I don't actually care about final speed, but am curious what you mean here.

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.

 

[sophiecentaur]

but the astronaut is assumed to keep going after the end of the path at whatever final speed they reach

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.
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.

 

[thinkagain]

Thanks, for calculation purposes I had put that the mass and thrust of the astronaut allows 1 g of constant acceleration. Starting speed of 100 mph. The lines are 100 feet apart. Is there something else we need?

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.

 

[haruspex]

What would be the simpler solution than circular with constant speed? I don't actually care about final speed, but am curious what you mean here.

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.

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.)
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.

 

[thinkagain]

Let's assume we can create thrust in any direction instantaneously so we don't have to worry about rotation. I would think the stopping and then right angle approach would be slower than even the circular method. I was thinking the combined turning/slowing of the euler spiral shaped path would be closer to ideal

 

[haruspex]

Let's assume we can create thrust in any direction instantaneously so we don't have to worry about rotation. I would think the stopping and then right angle approach would be slower than even the circular method. I was thinking the combined turning/slowing of the euler spiral shaped path would be closer to ideal

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.

 

[thinkagain]

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? Honestly I would have no clue where to even begin calculating these things.

 

[willem2]

Another idea is to work in a reference frame where the object would start out at rest. If you start out at rest and need to get to a speed (v_x,v_y,v_z), I think it should be obvious in which direction you need to thrust.

 

[haruspex]

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?

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.
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.

 

[thinkagain]

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?

 

[haruspex]

if we are starting at 100 mph

It will be more educational if we stick to symbolic values and see how the formulae turn out.

we need to go 100 feet

What 100 feet? This seems like a new constraint you are adding.

1 g

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

it would create a path at least similar to a Euler spiral

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.

 

[A.T.]

Okay, just to make sure I understand, bringing your speed along the original line to zero and ...

Since you don’t care about final speed, you can stop right here.

 

[A.T.]

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.

Firing the thruster at 45° to the original direction is faster.

 

[sophiecentaur]

Firing the thruster at 45° to the original direction is faster.

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.
Volunteers please. I have to go shopping right now.

 

[haruspex]

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.
Volunteers please. I have to go shopping right now.

Each method discussed takes a time which is some factor c multiplied by mv/F. If the final speed doesn't matter then we just have to come to a halt: c=1. For the 45 degree solution, c=√2. For the circular quadrant, $c=\frac{\pi}2$.
To see that the 45 degree must be the fastest way of finishing with the same speed, consider that the net change in momentum required is in that direction. With a thrust limited to F in magnitude, it must be quickest to keep that F in the desired direction throughout. If we direct the thrust in any other direction for a time, it reduces the momentum change in the desired direction.

 

[A.T.]

Each method discussed takes a time which is some factor c multiplied by mv/F. If the final speed doesn't matter then we just have to come to a halt: c=1. For the 45 degree solution, c=√2. For the circular quadrant, $c=\frac{\pi}2$.

Those time factors also apply if the final speed equals initial speed. The 45 degree solution also requires less space, with only 1/√2 of the circular arc radius.

 

[willem2]

The graph you get with 45 degree thrust is a rotated parabola

with an initial speed of v in the x direction and a desired speed of v in the y direction.

x = vt - (1/2sqrt(2))at^2, y = (1/2sqrt(2))at^2 with t from 0 to vsqrt(2)/a
if v = 1, and a =1 you get the following graph

we start at t=0 going to the right, and end up at x=y=t=(1/2)sqrt(2) going up.
thanks to http://www.wolframalpha.com/input/?i=x=t-(1/4)sqrt(2)t^2,y=(1/4)sqrt(2)t^2

 

[haruspex]

Those time factors also apply if the final speed equals initial speed.

Yes, that's what I said. If the final speed doesn't matter then the solution is c=1. Otherwise it is the 45 degree solution, and the circular arc takes longer.

 

[thinkagain]

I had put the 100 mph, 100 foot, 1 g constraints because I was looking for a situation where the astronaut would be decelerating up to the second line. The parabola answer looks like it would be correct for same speed at beginning and end as they would decelerate and then accelerate, but this would take less initial speed or the lines would need to be further apart. What does c=1 mean?

The reason I brought up the Euler in the first place was because intuitively it seems like the fastest way to do this would be to travel on a path of a constantly shrinking radius. This seems like it would maintain the highest speed possible and provide the shortest path possible, but I'm interested in finding the real answer whatever it may be. I'd be happy to find out if I'm way off base.

Sorry for all the questions and thanks for all your help.

 

[thinkagain]

Okay, did some more (amateur) calculations. With the 100 foot 1 g constraint for a constant arc the astronaut would need to go 32 mph to negotiate the quarter circle. That would mean they would need to take the time to go from 100 mph to 32 mph on the original line and then the time through the quarter circle. This would be my orginal "how not to do it" illustration. The 90 degree parabola example seems like it might be a tiny bit faster.

My proposed ideal solution seems like it would involve starting that 100 foot sideways portion early, maybe right away or maybe at some other point. Seems like that would be a shorter path and higher average speed so lower elapsed time. Is this making any sense?

 

[A.T.]

I had put the 100 mph, 100 foot, 1 g constraints

I think for these constraints, the quickest path would still be a parabola segment, just not a symmetrical one as for the same speed speed constraint.

 

[thinkagain]

Okay thanks, seems we are getting somewhere. Any ideas how we can go about figuring out the ideal parabola shape and direction of thrust?

 

[A.T.]

Okay thanks, seems we are getting somewhere. Any ideas how we can go about figuring out the ideal parabola shape and direction of thrust?

The acceleration magnitude defines the parabola shape, which you have to rotate to achieve 90° slope change over the desired vertical distance.

 

[thinkagain]

Okay, so then the ideal direction of force is always one angle in relation to the astronaut?

 

[willem2]

It's in the same direction as the difference of the initial velocity and the desired velocity.

 

[thinkagain]

I'm confused, so is the parabola shaped illustration I originally posted at least close to an optimal solution? Could someone at least give a ball park idea which way the astronaut should aim their thruster?

 

[willem2]

I'm confused, so is the parabola shaped illustration I originally posted at least close to an optimal solution? Could someone at least give a ball park idea which way the astronaut should aim their thruster?

The parabola is what you get if you fire your thrusters in the direction of the difference of the desired velocity and the starting velocity.
In this case the initial velocity is (1,0) and the desired velocity is (0,1). The difference between those is (-1,1) and this is the direction you should accelerate.

 

[thinkagain]

I don't actually have a desired final velocity on the y axis. Although I do want 0 mph on the X axis. Just wanted it to travel 100 feet to the next line. Actually I would think there would be a certain velocity you would attain by doing this in the optimal minimum time.

 

[sophiecentaur]

I don't actually have a desired final velocity on the y axis. Although I do want 0 mph on the X axis. Just wanted it to travel 100 feet to the next line. Actually I would think there would be a certain velocity you would attain by doing this in the optimal minimum time.

It has already been said that, if you don't care what the final speed is, then the shortest time taken is when you fire directly in line with your initial course and give a vanishingly small burst in the perpendicular direction. The problem is hardly of any interest if the speed perpendicular can be as small as you like as it's just a matter of motion against a force being reduced to zero velocity (simple school physics problem).

 

[sophiecentaur]

Each method discussed takes a time which is some factor c multiplied by mv/F. If the final speed doesn't matter then we just have to come to a halt: c=1. For the 45 degree solution, c=√2. For the circular quadrant, $c=\frac{\pi}2$.
To see that the 45 degree must be the fastest way of finishing with the same speed, consider that the net change in momentum required is in that direction. With a thrust limited to F in magnitude, it must be quickest to keep that F in the desired direction throughout. If we direct the thrust in any other direction for a time, it reduces the momentum change in the desired direction.

I agree with your two time values ($c=\frac{\pi}2$ and c=√2) but I can't see how your later argument about varying forces necessarily applies.Clearly the momentum changes must be the same for all paths (which is the requirement for equal speeds and perpendicular course change) but is your argument, that a changing force must be less effective, a satisfactory one? The final result is correct but could you convince me that argument is correct? It seems post hoc, to me.

 

[haruspex]

It has already been said that, if you don't care what the final speed is, then the shortest time taken is when you fire directly in line with your initial course and give a vanishingly small burst in the perpendicular direction. The problem is hardly of any interest if the speed perpendicular can be as small as you like as it's just a matter of motion against a force being reduced to zero velocity (simple school physics problem).

Thinkagain, having thought again, seems to have added the constraint that although no specific velocity is required in the new direction, a certain displacement is. This will certainly complicate matters.
It becomes a rather nasty calculus of variations problem. If this were a 'real' problem (textbook etc.) I might bother to attempt it, but not in the present context. What is clear is that the answer would consist of favouring a lateral thrust initially, then swinging around towards more of a braking thrust opposing the original motion. Note that this is the antithesis of an Euler spiral, of which thinkagain is so fond. Or maybe it's an Euler spiral in the reverse direction, i.e. an increasing radius?

 

[haruspex]

I agree with your two time values ($c=\frac{\pi}2$ and c=√2) but I can't see how your later argument about varying forces necessarily applies.Clearly the momentum changes must be the same for all paths (which is the requirement for equal speeds and perpendicular course change) but is your argument, that a changing force must be less effective, a satisfactory one? The final result is correct but could you convince me that argument is correct? It seems post hoc, to me.

Let the y direction be the 45 degree direction. At any instant, the components of thrust are Fx(t), Fy(t), Fx²+Fy²<=F². If the thrust is exerted for time T, the momentum change in the y direction is $\Delta p_y=\int^T F_y(t).dt$. Since Fy<=F at all times, this is clearly maximised when Fy=F at all times.

 

[thinkagain]

It has already been said that, if you don't care what the final speed is, then the shortest time taken is when you fire directly in line with your initial course and give a vanishingly small burst in the perpendicular direction. The problem is hardly of any interest if the speed perpendicular can be as small as you like as it's just a matter of motion against a force being reduced to zero velocity (simple school physics problem).

I think it's this "vanishingly small burst" I am interested in. How do I go about figuring that out. It seems pretty important at what point it's done as well. Not sure how you mean it would be a reverse euler spiral. If you are suggesting the path radius would increase I would think that would require a lot more than 100 feet between the two lines to need additional acceleration? I I'm only using the Euler spiral as an example because it has a constantly shrinking radius as the length increases. I'm not seeing how that is wrong. If someone could explain if the path would be a different shape then please tell me. Thanks

 

[haruspex]

I think it's this "vanishingly small burst" I am interested in. How do I go about figuring that out. It seems pretty important at what point it's done as well. Not sure how you mean it would be a reverse euler spiral. If you are suggesting the path radius would increase I would think that would require a lot more than 100 feet between the two lines to need additional acceleration? I I'm only using the Euler spiral as an example because it has a constantly shrinking radius as the length increases. I'm not seeing how that is wrong. If someone could explain if the path would be a different shape then please tell me. Thanks

If there is a distance that needs to be covered in the new direction and within the time period then it is no longer vanishingly small.
Using y for the initial direction and x for the new direction, your problem now consists of minimising T subject to:
$\ddot x^2+\ddot y^2 = a^2$, $x(0) = y(0) = \dot x(0)=0$, $\dot y(0) = v$, $x(T)=s$, $\dot y(T)=0$.
I.e., in the y direction we have to achieve a certain change in velocity, but in the x direction a certain change in position.
To get a rapid change in position, it is important to get up to a decent speed early. Thus, the initial acceleration should favour the x direction. This probably will look something like an Euler spiral with increasing radius.
But even simple strategies get quite messy to compute.
1. All x at first, then switching to all y:
T = v/a + √(2s/a)
2. Constant direction of thrust
$T^4-\left(\frac va\right)^2T^2-4\left(\frac sa\right)^2=0$
It is not immediately apparent that either of these is always better than the other.

 

[pbierre]

Trying to figure out direction of forces of an object traveling on an Euler spiral path. As an example if you had an astronaut with a jetpack and he wanted to change his direction 90 degrees he could aim his thrusters outward from the center of a circle and he would turn at a constant rate with a constant radius. But if he wanted to change his direction as quickly as possible and shrink the radius he was traveling on as quickly as possible in which direction would he aim his thrusters? I don't have a higher math understanding so please keep things intuitive if possible. Thanks

The thruster has to be able to deliver torque in order to change the direction the thruster jet is pointing in space. You wouldn't equip a jetpack with a single thruster jet for this very reason. So, let's make it a better jetpack design and allow pairs of torque-producing jets, two sets to control azimuth angle acceleration and deceleration. Then, a complete set of 4 more to control elevation angle accel. and decell.. Now you have the minimum degrees of rotational freedom to point the axis of your pair of opposing "translation" jets whose axis passes thru the astronaut's center of mass, one for accel. and one for decell. The operational concept is to use the torque thrusters to orient the translation jets to the desired motion vector, then move (accelerate-glide-decellerate). It's a 3D point-then-move.

So, now you can have full freedom to move anywhere you desire. The Euler spiral would achieve what goal? It ends up spinning the astronaut, and would cause severe vertigo sickness.

 

[haruspex]

The thruster has to be able to deliver torque in order to change the direction the thruster jet is pointing in space.

That is not true. An astronaut can reorient herself in space without recourse to a thruster. It works the same way a cat manages to land on its feet.

 

[sophiecentaur]

Let the y direction be the 45 degree direction. At any instant, the components of thrust are Fx(t), Fy(t), Fx²+Fy²<=F². If the thrust is exerted for time T, the momentum change in the y direction is $\Delta p_y=\int^T F_y(t).dt$. Since Fy<=F at all times, this is clearly maximised when Fy=F at all times.

You seem to be comparing the wrong two things, here. Surely |Fy| = |Fx| = |√2F|, in the 45°case, where F is the available thrust. The mean force (over time) (i.e. mean Momentum change) for another path will not necessarily be less than this but your argument is just assuming it is. I feel that any proof, based on momentum change would really need to do the integral and not just rely on an inequality, based on the wrong premise. What have I missed?

 

[A.T.]

Let the y direction be the 45 degree direction.

Surely |Fy| = |Fx| = |√2F|, in the 45°case, where F is the available thrust.

This is not how haruspex defined y.

 

[sophiecentaur]

This is not how haruspex defined y.

But the forces along and perpendicular to the original velocity would still vary in magnitude from √2F (i.e. greater than F) to zero. They are not =<F, as he seems to be saying. That's my problem.