Euler spiral forces?

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  • #51
A.T.
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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?
 
  • #52
sophiecentaur
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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.
 
  • #53
A.T.
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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?
 
  • #54
sophiecentaur
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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?
 
  • #55
A.T.
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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.
 
  • #56
haruspex
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|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.
 
  • #57
Merlin3189
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... 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.
 
  • #58
sophiecentaur
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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).
 
  • #59
haruspex
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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 Fy < F. Thus the integral of Fy over time t45 will be less than P√2, and we will not have achieved the desired result.
 
  • #60
A.T.
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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?
 
  • #61
sophiecentaur
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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 Fy < F. Thus the integral of Fy over time t45 will be less than P√2, and we will not have achieved the desired result.
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?
Thanks for persevering, chaps. I think I have it now.
 
  • #62
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So is the 45 degree angle the optimum direction of thrust then? Thanks!
 
  • #63
A.T.
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So is the 45 degree angle the optimum direction of thrust then?
For the equal speed constraint, not for the fixed turn space constraint .
 
  • #64
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I see, so we still don't have an answer for that one then?
 
  • #65
haruspex
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I see, so we still don't have an answer for that one then?
As I posted, that would make quite a hard problem. In all likelihood, there is no analytical solution. Certainly out of the scope of a homework forum.
 
  • #66
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I just figure after all that, if you can't make the turn quick enough you all stop thrust until you can make the turn by gradually redirecting to the y so you just barely have no x just in time and all y thrust at the boundary. If you have more than enough thrust you start all y direction until you have just enough time to all x up to the boundary and shift 90 degrees at the exact moment you cross, which in my simple mind... the latter resembles an inverse Euler spiral.
 

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