Direction of Forces on an Euler Spiral Path | Intuitive Explanation

[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.

 

[A.T.]

But the forces along and perpendicular to the original velocity would still vary in magnitude from √2F (i.e. greater than F) to zero.

Where did you get that from?

 

[sophiecentaur]

Where did you get that from?

Sorry I didn't complete the sentence. I am referring to the centrally directed force (circular path) and, although it will take longer, the argument given by haruspex seems to imply that the two forces will always be less than F.

 

[A.T.]

Sorry I didn't complete the sentence. I am referring to the centrally directed force (circular path)

No matter what strategy you were talking about, perpendicular components can never be greater than the total magnitude. Isn't that obvious?

 

[sophiecentaur]

Yes, of course. But the magnitude that's involved with the 45° case (on either axis) is less than the highest magnitude in the centripetal case. Change in momentum (in either direction) depends upon the force and the time. To go from v to 0 and 0 to v requires the same momentum change, whatever path.
What I need is an explanation of how that two lines of reasoning actually applies when comparing the two cases. The actual wording seems to ignore the fact that, at some places on the path, the momentum change (or at least the force) will be greater in the circular case.
Look, I have accepted (the sums are trivial) that the 45° system does it faster. What I was questioning was the actual wording and I still doubt that the argument is actually valid, as it stands. All you are doing is to point out things I already know. Do you see (or have a glimmer of) what I am getting at?

 

[A.T.]

I still doubt that the argument is actually valid, as it stands.

I see no problem with the argument in post #43. It's a valid proof, as far I can see. And no, I can't make any sense of your objections to it.

The actual wording seems to ignore the fact that, at some places on the path, the momentum change (or at least the force) will be greater in the circular case.

What are you talking about? The force magnitude is F, all the time, in both cases.

 

[haruspex]

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

No, |Fx|=|Fy|=|F/√2|. In all paths, if the direction of the thrust at time t is $\theta(t)$ then $F_x(t)=F\sin(\theta(t))$ etc.

 

[Merlin3189]

... the total amount of energy available needs to be specified -...

If this is some sort of rocket thruster, do you mean energy or do you mean impulse?
If the path were circular, I'd have thought no energy is required, but to apply force does imply impulse.
Of course for a rocket to function at all requires energy input even if no useful energy is output.
Since this is a fixed thrust engine, both impulse and energy input are proportional to time.
So we come back to the original question, what is the shortest time we can operate the rocket to make the required change.

 

[sophiecentaur]

I see no problem with the argument in post #43. It's a valid proof, as far I can see. And no, I can't make any sense of your objections to it.What are you talking about? The force magnitude is F, all the time, in both cases.

It's interesting that haruspex, him,self is making no comment here. My difficulty is with this:

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.

where he talks about the net change in momentum in the 45° direction. The change in momentum in the original direction needs to be -P and the change in momentum at right angles needs to be P and the change in the 45° direction will have magnitude P√2, irrespective of the path taken or the direction of the thrusters at any time. How is it 'obvious' that the way that change is delivered is quickest if the thrust is always delivered along the 45° line? For other paths, the rate of momentum along the two axes will vary with time, being sometimes more in magnitude and sometimes less than P. Can you see at least something in my problem. I am not just arguing; I need some expansion of those few words in haruspex's assertion (not proof).

 

[haruspex]

It's interesting that haruspex, him,self is making no comment here. My difficulty is with this:

where he talks about the net change in momentum in the 45° direction. The change in momentum in the original direction needs to be -P and the change in momentum at right angles needs to be P and the change in the 45° direction will have magnitude P√2, irrespective of the path taken or the direction of the thrusters at any time. How is it 'obvious' that the way that change is delivered is quickest if the thrust is always delivered along the 45° line? For other paths, the rate of momentum along the two axes will vary with time, being sometimes more in magnitude and sometimes less than P. Can you see at least something in my problem. I am not just arguing; I need some expansion of those few words in haruspex's assertion (not proof).

P√2 = F t₄₅. Suppose we try some other strategy for time t₄₅. Let the 45 degree direction be the y direction. If for any part of the time we direct some of the available F in the x direction then during that time F_y < F. Thus the integral of F_y over time t₄₅ will be less than P√2, and we will not have achieved the desired result.

 

[A.T.]

I need some expansion of those few words in haruspex's assertion (not proof).

I think haruspex's explanation qualifies as proof. If you need expansion on it, you should pinpoint exactly which point of it you want expanded.

As an alternative consider the reference frame where the astronaut was initially at rest. Here he has to accelerate from 0 to some speed along the 45° line as quickly as possible. Still not obvious?