Height at which pilot must pull out of dive before hitting 5g

[negation]

Oops. I still would use a 4 here instead of a 5. The difference is reserved to deal with the omnipresent small red arrow pointing straightupwards. In particular at point E in the last moment before letting go of the stick. To explain this, I took quite some time to compose #45. Might have made mistakes (again -- if so, please point them out), but it's better than post #25.

The 5 is applicable in Iewando's scenario number 3. But the Earth is damn nearby.

I sure wish I could come up with an animated version of the picture in #44. Things like this make me want to learn a new software tool to do so efficiently. Any tips?

Regardless the 4g or 5g, how did you arrived at the 4g to the right? I would really appreciate the explanation

 

[BvU]

To the right of what ? I thought we had at last had agreed upon a coordinate system. This by now is a long thread.

Let me make a wild but educated guess. (btw It's the guessing what the other guy means exactly, that makes this whole thing so time consuming, almost exasperating. For all parties involved.).

I thought we had at last agreed upon a coordinate system. I use it. I posted a bunch of pictures with letters and arrows. I refer to picture in post 44 and my educated guess is you are referring to point B, point (0,0). There is an eight line explanation with a cliffhanger: to be justified later. Under point E explanation the veil is lifted: the happy end is "hence the 4g at B".

So the way I think about this thing is:
1. Don't complicate it more than necessary: assume constant speed.
2. You suffer 1g behind your desktop. If you pass out at 5 g, you can stand 4g extra from acrobatics like looping and such -- in a worst case where Earth and acrobatics pull in the same direction.
3. If all you want is to get out of a nosedive, a quarter circle is what you need and have sufficient knowledge to do some calculations on. Iewando's complications are for SMH (what on Earth or in heaven is that?) or post-BS.
4. Combine 2 and 3: a quarter circle with at max 4g normal force in the vertical upward direction to deliver the centripetal force needed to describe a circular motion is what the exercise wants us to work on.
5. We know a constant acceleration is needed for circular motion. Acceleration is a vector. It points to the center of the circle. Acceleration from Earth gravity is also a vector. It points downward (or rather: towards the center of gravity of the Earth - don't ask). From a simple sketch it becomes clear that the worst case (i.e. Earth and acrobatics pulling in the same direction, so that the vector addition comes down to a simple magnitude addition is at point E, at the very end of the quarter circle.
6. Combine 1 and 5: 4g is available for the quarter circle.

So in response to your question (as I guess it was meant): 4g is the max available. It the trajectory is to be a quarter circle and the speed constant, the centripetal acceleration has to be constant in magnitude and has to point to the center of that circle. So initially, at point B, after a jerk on the stick, it has to point in the positive x-direction in the coordinate system we agreed upon. And its magnitude is 4g. Big red arrow.

Now, what makes this exercise so difficult for you: There is a difference between what the pilot feels and the net acceleration. Behind a desktop (on earth, please), you feel 1g. But you don't fall because the chair holds you in place by exercising exactly 1g upwards. The normal force in school books. In this exercise the pilot feels an acceleration that is equal to the opposite of the purple arrows (purple is unique so far, so I don't specify that it's in the picture of post #45). The net acceleration is ZERO until point B (hence a straight line, constant speed. Newton!), then from B to E the net acceleration is a vector, magnitude 4g and pointing to (r,0), the centre of the circle (hence the circle, and hence a radius v²/4g), and after point E the net acceleration is ZERO again, hence a straight line, grazing the surface of the Earth at constant speed.

Unhealthy, but we were asked to calculate the minimum height, regardless of tree heights ;-)



(Why does my W7 IE11 all of a sudden wants to capitalize each occurrence of the word Arrow, but not the word arrows? Some copyright or trademark that is suddenly a part of the english language ?)

 

[negation]

To the right of what ? I thought we had at last had agreed upon a coordinate system. This by now is a long thread.

Let me make a wild but educated guess. (btw It's the guessing what the other guy means exactly, that makes this whole thing so time consuming, almost exasperating. For all parties involved.).

I thought we had at last agreed upon a coordinate system. I use it. I posted a bunch of pictures with letters and arrows. I refer to picture in post 44 and my educated guess is you are referring to point B, point (0,0). There is an eight line explanation with a cliffhanger: to be justified later. Under point E explanation the veil is lifted: the happy end is "hence the 4g at B".

So the way I think about this thing is:
1. Don't complicate it more than necessary: assume constant speed.
2. You suffer 1g behind your desktop. If you pass out at 5 g, you can stand 4g extra from acrobatics like looping and such -- in a worst case where Earth and acrobatics pull in the same direction.
3. If all you want is to get out of a nosedive, a quarter circle is what you need and have sufficient knowledge to do some calculations on. Iewando's complications are for SMH (what on Earth or in heaven is that?) or post-BS.
4. Combine 2 and 3: a quarter circle with at max 4g normal force in the vertical upward direction to deliver the centripetal force needed to describe a circular motion is what the exercise wants us to work on.
5. We know a constant acceleration is needed for circular motion. Acceleration is a vector. It points to the center of the circle. Acceleration from Earth gravity is also a vector. It points downward (or rather: towards the center of gravity of the Earth - don't ask). From a simple sketch it becomes clear that the worst case (i.e. Earth and acrobatics pulling in the same direction, so that the vector addition comes down to a simple magnitude addition is at point E, at the very end of the quarter circle.
6. Combine 1 and 5: 4g is available for the quarter circle.

So in response to your question (as I guess it was meant): 4g is the max available. It the trajectory is to be a quarter circle and the speed constant, the centripetal acceleration has to be constant in magnitude and has to point to the center of that circle. So initially, at point B, after a jerk on the stick, it has to point in the positive x-direction in the coordinate system we agreed upon. And its magnitude is 4g. Big red arrow.

Sorry mate, I should have been more specific. I was only curious about how the value of 4g (towards the right, in direction of the positive x-axis) was arrived at.
You meant SHM? I think if the entity kept to a constant predictable loop, indefinitely, a simple harmonic motion model could be drawn out-a sine phase.

Now, what makes this exercise so difficult for you: There is a difference between what the pilot feels and the net acceleration. Behind a desktop (on earth, please), you feel 1g. But you don't fall because the chair holds you in place by exercising exactly 1g upwards. The normal force in school books. In this exercise the pilot feels an acceleration that is equal to the opposite of the purple arrows (purple is unique so far, so I don't specify that it's in the picture of post #45). The net acceleration is ZERO until point B (hence a straight line, constant speed. Newton!), then from B to E the net acceleration is a vector, magnitude 4g and pointing to (r,0), the centre of the circle (hence the circle, and hence a radius v²/4g), and after point E the net acceleration is ZERO again, hence a straight line, grazing the surface of the Earth at constant speed.

Unhealthy, but we were asked to calculate the minimum height, regardless of tree heights ;-)

Yes, there was an initial confusion arising from the conflation of what the pilot experience and the opposing force indicated schematically. But, in time, I realize only the opposing force to what the pilot experience matters.

 

[BvU]

Ah, I guessed sophisticated human modelling, so there you see what damage an abbrev can cause!

You Australian, mate?

And no, there is nothing pointing to SHM here. I can imagine a sine pops up if you start decomposing, but a sine is not Always going to SHM. (Always is capitalized automatically too!?)
SHM tell-tales: force opposite wrt motion and proportional to deviation from equilibrium, periodicity, etc. Proportional being a nice word for linear, which is often a first approximation. You'll get plenty exercise with SHM, don't worry. All the way through to QFT (an abbreviation!)

This confusing g scale pilots and entourage use is a boon for teachers looking for exercises. Same thing about pressure measurements in Barg (or worse, psi without mentioning the g) instead of N/m². So: beware.

But real world physics is also physics, so we have to deal with the jargon

If you study physics a bit longer, everything becomes relative. I go on and on about this because you need a good basis further on, no matter what you do or study.

 

[negation]

Ah, I guessed sophisticated human modelling, so there you see what damage an abbrev can cause!

You Australian, mate?

And no, there is nothing pointing to SHM here. I can imagine a sine pops up if you start decomposing, but a sine is not Always going to SHM. (Always is capitalized automatically too!?)
SHM tell-tales: force opposite wrt motion and proportional to deviation from equilibrium, periodicity, etc. Proportional being a nice word for linear, which is often a first approximation. You'll get plenty exercise with SHM, don't worry. All the way through to QFT (an abbreviation!)

This confusing g scale pilots and entourage use is a boon for teachers looking for exercises. Same thing about pressure measurements in Barg (or worse, psi without mentioning the g) instead of N/m². So: beware.

But real world physics is also physics, so we have to deal with the jargon

If you study physics a bit longer, everything becomes relative. I go on and on about this because you need a good basis further on, no matter what you do or study.

I'm not Aussie but been here for a while. Yes, I understand the importance of relativity. And yes I will be doing Physic. I'm going into my 2nd year next month- first year was foundation. I'm reading Astrophysics and Math as majors. Been doing lots of self-study during this break for preparation.
The reading materials aren't sufficient to give deep analysis and most lecturers, in my experience, don't pile a deep underlying foundation either.

 

[BvU]

Good. I'm in NL, so GMT+1. Use your time efficiently: go to the next exercise.

 

[homemade]

What if a pilot were submersed in a liquid container in flight, wouldn't that significantly reduce g force? The liquid you can breathe would be even better.