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

[negation]

Homework Statement

A jet is diving vertically downwards at 1200kmh^-1. If the pilot can withstand a max acceleration of 5g before losing consciousness, at what height must the plane start a quarter turn to pull out of the dive.
Assume the speed remains constant.

The Attempt at a Solution

I do not understand the question fully. Could someone give me a leg-up?

 

[rcgldr]

I would assume the problem is asking what is the lowest height so that the pilot can pull ouf of the dive without hitting the ground or exceeding 5 g's. I would assume the 5 g limit needs to take the effect of gravity into account.

 

[BvU]

Oh boy, a lot of parallel questions. What do you think is meant when you read "quarter" turn ?

 

[negation]

Oh boy, a lot of parallel questions. What do you think is meant when you read "quarter" turn ?

That the pilot makes a pi/2 turn?

 

[negation]

I drew a circle for this and take the radius (the -y component) as the height along which the pilot travels before making the quarter turn.
But beyond which, I have no idea.

 

[BvU]

Good. Maybe, because ambiguous. Turning can be done many ways with a jet plane. With a side view and a top view, which kind of turn do you mean ?

 

[negation]

Good. Maybe, because ambiguous. Turning can be done many ways with a jet plane. With a side view and a top view, which kind of turn do you mean ?

The side view.

 

[BvU]

I see no turn, but you made a good choice. It was more or less stipulated in the word "pull" in the OP. Which way do you want the guy to turn ?

 

[BvU]

If he does his thing right, what g does he feel after this scary emergency measure ?

 

[HallsofIvy]

I presume you know that in making a circular turn of radius r, at speed v, the inward acceleration has magnitude \frac{v^2}{r} so "centrifugal force" on the pilot will be \frac{v^2m}{r} where m is the mass of the pilot. I'm not clear whether the "5gs" acceleration the pilot can stand is to include the pilots weight or not. If it is not, then set \frac{v^2m}{r}= 5mg and solve for r. If it is, set \frac{v^2m}{r}= 4mg (subtracting the pilots weight, mg) and solve for r. Of course, if he is making a vertical circle of radius r, he had better be at least "r" above the ground to avoid hitting the ground! We can write a circle of radius r, as parametric equations x= r cos(\omega t), y= r sin(\omega t). differentiating, v_x= -r\omega sin(\omega t), v_y= r\omega cos(\omega t). The acceleration is given by, differentiating again, a_x= -r\omega^2 cos(\omega t), a_y= -r\omega^2 sin(\omega t).

The speed at each instant is given by \sqrt{v_x^2+ v_y^2}= \sqrt{r^2\omega^2}= r\omega= v so we must have \omega= v/r. Then the acceleration is given by a_x= -(v^2/r) cos(\omega t), a_y= -(v

 

[negation]

I see no turn, but you made a good choice. It was more or less stipulated in the word "pull" in the OP. Which way do you want the guy to turn ?

Let's suppose he turns in the -x direction

 

[HallsofIvy]

I presume you know that making a circle of radius r, at speed v, there must be a centripetal force of magnitude mv^2/r. I don't know whether the pilots own weight is included in the "5 gs" or not. If it is not then we must have mv^2/r= 5mg. If it is, then subtract 1 mg for his weight to get mv^2/r= 4mg.

Solve one of those for r. Of course, his initial height, before making that loop must be at least "r" to avoid hitting the ground!

 

[BvU]

Which way is that ? Or was there a coordinate system present already ?

 

[BvU]

@HallsofIvy, don't spoil this! With all due respect, of course...

 

[BvU]

If I were at the controls, I'd prefer the one on the right. Flying upside down at height zero is riskier than head up. Agree?

 

[HallsofIvy]

@HallsofIvy, don't spoil this! With all due respect, of course...

I'm sorry, I foolishly thought we were being serious. I clearly did not read the responses carefully enough.

 

[BvU]

Negation, are you comfortable with the equations for circular motion at constant speed?
All you asked for was a leg-up and now we already have 17 posts!

 

[negation]

Negation, are you comfortable with the equations for circular motion at constant speed?
All you asked for was a leg-up and now we already have 17 posts!

To some degree, I am.

I conceive of a circle with the origin (0,0) and of radius, r. The plane begins from (r,0) and takes the path of an arc length in quadrant 4. At point (-r,0), it begins moving horizontally.
In moving along the arc length, the acceleration changes and I right? And we want the g value to be lesser to or equal to 5g along the arc length.

a = g = v^2/ r

5(9.8) = 333^2/r
r = 2263m

r represents the height at which the pilot must make the turn along the arc length cojoining (-r,0) and (0-r).
Is this it?

 

[negation]

If I were at the controls, I'd prefer the one on the right. Flying upside down at height zero is riskier than head up. Agree?

Yes, this was what I have in mind.
Although, the left one was what I had exactly in mind. But both diagrams are essentially parallel model of one another.

 

[BvU]

You are absolutely right: concerning the physics exercise they are equivalent. But pilot's g-resistance applies to downward g forces: you survive/stay conscious longer with those than with the blood running to your head. (I think -- will ask my son who is in the business)

My battle plan for you was to focus on the worst case: the end of the quarter circle. Then centripetal force and upward force from the gravity pushing the pilot down into the ejector seat align and therefore the magnitude of the sum is at a maximum. This would leave 4g for the centripetal force to execute the 'turn', and the expressions etc. were already posted.

You will do fine on this one, I am sure. Look forward to the next encounter.

 

[negation]

You are absolutely right: concerning the physics exercise they are equivalent. But pilot's g-resistance applies to downward g forces: you survive/stay conscious longer with those than with the blood running to your head. (I think -- will ask my son who is in the business)

My battle plan for you was to focus on the worst case: the end of the quarter circle. Then centripetal force and upward force from the gravity pushing the pilot down into the ejector seat align and therefore the magnitude of the sum is at a maximum. This would leave 4g for the centripetal force to execute the 'turn', and the expressions etc. were already posted.

You will do fine on this one, I am sure. Look forward to the next encounter.

So my answer is right?

 

[BvU]

Actually, I missed your answer at first: it was at the bottom of page 1 of 2 and I picked up at post #19.
Did you read my post #20 ? After passing point Q the pilot experiences 1 g. That is also there just before point Q, so the total in the case of your answer would be 6 g. The guy might pass out!

There is a point of discussion possible here: is the 5g absolute or in excess of normal on-the-ground gravitational acceleration. I agree with rcgldr that it is absolute. That way you can 'easily' calculate that he can make a horizontal circle of radius 2300 m, but for a looping he has to be more careful, especially in the lower half.

So

So my answer is right?

I would say most likely not. But you know what to do.

 

[negation]

Actually, I missed your answer at first: it was at the bottom of page 1 of 2 and I picked up at post #19.
Did you read my post #20 ? After passing point Q the pilot experiences 1 g. That is also there just before point Q, so the total in the case of your answer would be 6 g. The guy might pass out!

There is a point of discussion possible here: is the 5g absolute or in excess of normal on-the-ground gravitational acceleration. I agree with rcgldr that it is absolute. That way you can 'easily' calculate that he can make a horizontal circle of radius 2300 m, but for a looping he has to be more careful, especially in the lower half.

So I would say most likely not. But you know what to do.

I'm a little confused. At point g, the pilot experiences 1g if he travels horizontally
from point Q (0,-r)onward. Between point (-r,0) and (0,-r), is the pilot experiencing 5g or 6g? I'm still quite hazy. And what do we have to take into account the 1g of the pilot?

 

[negation]

You are absolutely right: concerning the physics exercise they are equivalent. But pilot's g-resistance applies to downward g forces: you survive/stay conscious longer with those than with the blood running to your head. (I think -- will ask my son who is in the business)

My battle plan for you was to focus on the worst case: the end of the quarter circle. Then centripetal force and upward force from the gravity pushing the pilot down into the ejector seat align and therefore the magnitude of the sum is at a maximum. This would leave 4g for the centripetal force to execute the 'turn', and the expressions etc. were already posted.

You will do fine on this one, I am sure. Look forward to the next encounter.

If the centripetal force is 4g, then would the mathematical reasoning 4g = v^2/r work?

 

[BvU]

If the centripetal force is 4g, then would the mathematical reasoning 4g = v^2/r work?

For the record: the force needed for the circle trajectory is 4mg towards the centre. 4g is an acceleration. And yes, the acceleration towards the centre of a circular trajectory is v²/r

Between point (-r,0) and (0,-r), is the pilot experiencing 5g or 6g? I'm still quite hazy. And what do we have to take into account the 1g of the pilot?

While going straight down at constant speed, the pilot experiences 1g downward. This is a vector.
As soon as he pulls the wheel (or what do they call it) towards him, a uniform circular motion sets in that requires 4g of acceleration towards the centre. His seat provides the force required for that acceleration. Also a vector, initially pointing to the right. So initially he feels an acceleration of 1 g down, 4g to the right. √17 magnitude total. Gradually going towards 1 g down and 4 g down as well just before point q. Magnitude now 5 g (same direction). He can stand that as per given in the OP. If he's smart he pulls up a little longer, haha. The moment he is going straight again, he experiences 1 g towards the center of the earth.

 

[negation]

For the record: the force needed for the circle trajectory is 4mg towards the centre. 4g is an acceleration. And yes, the acceleration towards the centre of a circular trajectory is v²/r

While going straight down at constant speed, the pilot experiences 1g downward. This is a vector.
As soon as he pulls the wheel (or what do they call it) towards him, a uniform circular motion sets in that requires 4g of acceleration towards the centre. His seat provides the force required for that acceleration. Also a vector, initially pointing to the right. So initially he feels an acceleration of 1 g down, 4g to the right. √17 magnitude total. Gradually going towards 1 g down and 4 g down as well just before point q. Magnitude now 5 g (same direction). He can stand that as per given in the OP. If he's smart he pulls up a little longer, haha. The moment he is going straight again, he experiences 1 g towards the center of the earth.

I can follow your reasoning in this post. But two questions. In moving vertically down(just before making the pull up), why is there a 4g acceleration towards the center? Wouldn't it make more sense to say that there exists a 4g acceleration to the center at the instantaneous moment when the "pull-up" was made?

Gradually going towards 1 g down and 4 g down as well just before point q. Magnitude now 5 g (same direction). He can stand that as per given in the OP. If he's smart he pulls up a little longer, haha. The moment he is going straight again, he experiences 1 g towards the center of the earth.

If the pilot is along the arc length conjoining (-r,0) and (0,-r), How is the magnitude now 5g? Can you demonstrate the mathematical reasoning behind how 5g is achieved? (4g+1g?) The pilot experiences 1g towards the center of the Earth and 4g to the center of the circular path, isn't it? Wouldn't that be a SQRT(17) as you have aforementioned?

 

[rcgldr]

The pilot experiences 1g towards the center of the Earth and 4g to the center of the circular path, isn't it? Wouldn't that be a SQRT(17) as you have aforementioned?

What the pilot experiences is how hard the seat is pushing the pilot "upwards". At the bottom of the climb out, the seat is generating 4 m g of centripetal force and 1 m g of force opposing gravity.

 

[negation]

What the pilot experiences is how hard the seat is pushing the pilot "upwards". At the bottom of the climb out, the seat is generating 4 m g of centripetal force and 1 m g of force opposing gravity.

What's the variable, 'm', by the way?

Edit: I see, mass. mg = w

 

[negation]

What the pilot experiences is how hard the seat is pushing the pilot "upwards". At the bottom of the climb out, the seat is generating 4 m g of centripetal force and 1 m g of force opposing gravity.

EDIT: So in totality, the pilot experiences only net 3g acceleration? At point q(0,-r), there is a 4g acceleration up and 1g acceleration down.

 

[negation]

SQRT(24g^2) = 48ms^-2

g = v^2/r

r = 2310m

 

[rcgldr]

EDIT: So in totality, the pilot experiences only net 3g acceleration? At point q(0,-r), there is a 4g acceleration up and 1g acceleration down.

The pilot's acceleration is 4g's upwards, there's no downwards acceleration because it's opposed by the aircraft and the seat that the pilot sits in. The pilot experiences a force from the seat equal to 5 times the pilots weight, 4x due to the acceleration, 1x due to opposition of gravity.

 

[negation]

The pilot's acceleration is 4g's upwards, there's no downwards acceleration because it's opposed by the aircraft and the seat that the pilot sits in. The pilot experiences a force from the seat equal to 5 times the pilots weight, 4x due to the acceleration, 1x due to opposition of gravity.

Is the radius then 2310?

 

[Lok]

Is the radius then 2310?

Well negation, the true problem is what happens with the plane as it is close to the speed of sound.

This thread was a heap of fun reading from start to "finish", seeing as you are mainly ignored yet somewhat helped :P

The "g force" (the calculated one) at the start of the turn can be 5g, as the gravitational 1g is 90 deg from the centrifugal force. So the two do not mix in the down direction which is actually the horizontal at the beginning of the turn. And the story is not finished yet. The 5g should end up being a 4g at the end of the turn where the gravitational and centrifugal mix at 0 deg one from the other so add up.

If this problem was in an aeronautical setting the 5g varies to 4g during the turn and turn height is bigger. If only a entry level physics problem do ignore the variation. So the ~2300 should be close enough.

 

[negation]

Well negation, the true problem is what happens with the plane as it is close to the speed of sound.

This thread was a heap of fun reading from start to "finish", seeing as you are mainly ignored yet somewhat helped :P

The "g force" (the calculated one) at the start of the turn can be 5g, as the gravitational 1g is 90 deg from the centrifugal force. So the two do not mix in the down direction which is actually the horizontal at the beginning of the turn. And the story is not finished yet. The 5g should end up being a 4g at the end of the turn where the gravitational and centrifugal mix at 0 deg one from the other so add up.

If this problem was in an aeronautical setting the 5g varies to 4g during the turn and turn height is bigger. If only a entry level physics problem do ignore the variation. So the ~2300 should be close enough.

So you are suggesting that the magnitude of the horizontal acceleration before the "pull-up" to be 5g?

If I do take the horizontal acceleration to be 5g and the down ward acceleration, 1g, to be normal to the horizontal, then the magnitude is 50ms^-2

g = v^2/r and r works out to be 2200.

 

[BvU]

In post #26:

I can follow your reasoning in this post. But two questions. In moving vertically down(just before making the pull up), why is there a 4g acceleration towards the center? Wouldn't it make more sense to say that there exists a 4g acceleration to the center at the instantaneous moment when the "pull-up" was made?

I feel misread! Please read it again:
While going straight down at constant speed, the pilot experiences 1g downward. This is a vector.
As soon as he pulls the wheel (or what do they call it -- A stick, I realized since then) towards him, a uniform circular motion sets in that requires 4g of acceleration towards the centre. His seat provides the force required for that acceleration. Also a vector, initially pointing to the right. So initially he feels an acceleration of 1 g down, 4g to the right. √17 magnitude total. Gradually going towards 1 g down and 4 g down as well just before point q. Magnitude now 5 g (same direction). He can stand that as per given in the OP. If he's smart he pulls up a little longer, haha. The moment he is going straight again, he experiences 1 g towards the center of the earth.

In post#29:

EDIT: So in totality, the pilot experiences only net 3g acceleration? At point q(0,-r), there is a 4g acceleration up and 1g acceleration down.

The 1 g acceleration down does not lead to a free fall. His seat pushes him up to oppose the gravitational force and it also pushes him up with 4g to keep him in a circular motion. These seat forces are in the same direction, hence the 5g.

I'm glad rcgldr helps too. Lok is amused, which is nice. But I resent him saying you are being ignored. I try to help you patiently as best I can to deal with the exercise. Where a circular motion is not exactly a given, but already complicated enough for us, right?

 

[negation]

In post #26:
I feel misread! Please read it again:
While going straight down at constant speed, the pilot experiences 1g downward. This is a vector.
As soon as he pulls the wheel (or what do they call it -- A stick, I realized since then) towards him, a uniform circular motion sets in that requires 4g of acceleration towards the centre. His seat provides the force required for that acceleration. Also a vector, initially pointing to the right. So initially he feels an acceleration of 1 g down, 4g to the right. √17 magnitude total. Gradually going towards 1 g down and 4 g down as well just before point q. Magnitude now 5 g (same direction). He can stand that as per given in the OP. If he's smart he pulls up a little longer, haha. The moment he is going straight again, he experiences 1 g towards the center of the earth.

In post#29:

The 1 g acceleration down does not lead to a free fall. His seat pushes him up to oppose the gravitational force and it also pushes him up with 4g to keep him in a circular motion. These seat forces are in the same direction, hence the 5g.

I'm glad rcgldr helps too. Lok is amused, which is nice. But I resent him saying you are being ignored. I try to help you patiently as best I can to deal with the exercise. Where a circular motion is not exactly a given, but already complicated enough for us, right?

Without a detailed sequential diagram, it can be hard to conceive of a mental modelling. Why is it 4g down? Acceleration is always in the direction towards the center, isn't it? Let's just say that when the pilot is at the point (0,-r), he experiences a 4g acceleration towards the center. At the same time, there is a 1g acceleration in the direction towards the ground. But, why, as you'e put it, would the 4g acceleration be in the direction towards the ground? How did this 4g acceleration, initially towards the center changes vector suddenly. It doesn't make sense.

View attachment 66192

 

[lewando]

Mental models are important to establish. Maybe think about it this way:

Scenario 1: You sit in a chair attached securely to the Earth. Let this be a zero-velocity reference frame. The chair is applying a force on you in the upward direction. Earth's gravity, and the fact that you have mass, counters this force exactly, so you don't accelerate.

Scenario 2: Now take away the Earth and attach a rocket under your chair that applies the same equivalent "upward" force as in Scenario 1. You are now accelerating "upwards" at 1g. The amount of blood in your brain remains the same as before and if you close your eyes, it feels exactly like Scenario 1.

Scenario 3: You are sitting in the chair now resting on the curved floor of a rotating centrifuge of an interstellar spacecraft . The centrifuge is a giant "hamster-wheel" that spins with a certain angular velocity that can creates artificial Earthlike gravity. When you close you eyes, it feels exactly like you are sitting in the chair under conditions of either Scenarios 1 or 2. The force of the chair upon your body feels just like it did when you were on the Earth-- "upwards", which is actually towards the center of the centrifuge. As a result of this force, you are accelerating in the direction of "upwards" from your perspective or towards the center of the centrifuge, in reality--"upwards" is relative to you (the direction above your head).

The "g force" (the calculated one) at the start of the turn can be 5g, as the gravitational 1g is 90 deg from the centrifugal force. So the two do not mix in the down direction which is actually the horizontal at the beginning of the turn. And the story is not finished yet. The 5g should end up being a 4g at the end of the turn where the gravitational and centrifugal mix at 0 deg one from the other so add up.

If this problem was in an aeronautical setting the 5g varies to 4g during the turn and turn height is bigger. If only a entry level physics problem do ignore the variation. So the ~2300 should be close enough.

Interesting point, Lok (variable centripetal acceleration). But I think the minimum height at which you begin your turn would still be the same as it would if the turn trajectory was circular (constant centripetal acceleration)

 

[Lok]

Without a detailed sequential diagram, it can be hard to conceive of a mental modelling. Why is it 4g down? Acceleration is always in the direction towards the center, isn't it? Let's just say that when the pilot is at the point (0,-r), he experiences a 4g acceleration towards the center. At the same time, there is a 1g acceleration in the direction towards the ground. But, why, as you'e put it, would the 4g acceleration be in the direction towards the ground? How did this 4g acceleration, initially towards the center changes vector suddenly. It doesn't make sense.
View attachment 66192

I've made a small picture for this.
cg=centrifugal g
gg=gravitational g
As a g force for a human only the downward g(from a human orientation reference) is significant as it is the one that increases or decreases the pressure in the head.
All I am saying is that you would have to have a varying radius in order to keep 5g's. Otherwise you would get more at the end of the turn.

I'm glad rcgldr helps too. Lok is amused, which is nice. But I resent him saying you are being ignored. I try to help you patiently as best I can to deal with the exercise. Where a circular motion is not exactly a given, but already complicated enough for us, right?

@BvU, I did not mean it in a bad way. If you can imagine.

 

[negation]

Mental models are important to establish. Maybe think about it this way:

Scenario 1: You sit in a chair attached securely to the Earth. Let this be a zero-velocity reference frame. The chair is applying a force on you in the upward direction. Earth's gravity, and the fact that you have mass, counters this force exactly, so you don't accelerate.

Scenario 2: Now take away the Earth and attach a rocket under your chair that applies the same equivalent "upward" force as in Scenario 1. You are now accelerating "upwards" at 1g. The amount of blood in your brain remains the same as before and if you close your eyes, it feels exactly like Scenario 1.


I haven't got to the next chapter on forces-will soon- but I shall try to build a vague understanding for the bigger picture to come. I love thinking the big-picture way.

And I suppose, in scenario 1, this occurs arises from Newton's third law: succinctly, that every action has an equal and opposing reaction?
An entity sitting on a chair applies a down ward force on the chair equivalent to the product of the mass and acceleration from the gravity-this translate mathematically to weight since w = mg.
In turn, the chair counters this with an opposing force.

In scenario 2, the net acceleration is 1g. So this implies that the upward acceleration must be 1g greater than the down ward acceleration from gravity. Am I right?

Scenario 3: You are sitting in the chair now resting on the curved floor of a rotating centrifuge of an interstellar spacecraft . The centrifuge is a giant "hamster-wheel" that spins with a certain angular velocity that can creates artificial Earthlike gravity. When you close you eyes, it feels exactly like you are sitting in the chair under conditions of either Scenarios 1 or 2. The force of the chair upon your body feels just like it did when you were on the Earth-- "upwards", which is actually towards the center of the centrifuge. As a result of this force, you are accelerating in the direction of "upwards" from your perspective or towards the center of the centrifuge, in reality--"upwards" is relative to you (the direction above your head).

 

[negation]

I've made a small picture for this.
cg=centrifugal g
gg=gravitational g
As a g force for a human only the downward g(from a human orientation reference) is significant as it is the one that increases or decreases the pressure in the head.
All I am saying is that you would have to have a varying radius in order to keep 5g's. Otherwise you would get more at the end of the turn.


@BvU, I did not mean it in a bad way. If you can imagine.

My modelling is this. Is there any parts I ought to correct?

 

[negation]

I've made a small picture for this.
cg=centrifugal g
gg=gravitational g
As a g force for a human only the downward g(from a human orientation reference) is significant as it is the one that increases or decreases the pressure in the head.
All I am saying is that you would have to have a varying radius in order to keep 5g's. Otherwise you would get more at the end of the turn.


@BvU, I did not mean it in a bad way. If you can imagine.

View attachment 66204

My modelling is this. Is there any parts I ought to correct?
By the way, it occurred to me that in moving around a predictable circular path, the entire phenomenon takes on a SHM modelling.

 

[lewando]

And I suppose, in scenario 1, this occurs arises from Newton's third law: succinctly, that every action has an equal and opposing reaction?

Sure, the 3rd law applies. So does the 2nd law: ∑(all forces) = ma. Since a is zero, ∑(all forces) must be zero, which is the case when adding equal magnitude and opposite direction forces.

In scenario 2, the net acceleration is 1g. So this implies that the upward acceleration must be 1g greater than the down ward acceleration from gravity. Am I right?

In scenario 2, there is no gravity because we disappeared the Earth. Acceleration is a reuslt of the force applied by the rocket: a = F_rocket/m

 

[negation]

Sure, the 3rd law applies. So does the 2nd law: ∑(all forces) = ma. Since a is zero, ∑(all forces) must be zero, which is the case when adding equal magnitude and opposite direction forces.



In scenario 2, there is no gravity because we disappeared the Earth. Acceleration is a reuslt of the force applied by the rocket: a = F_rocket/m

That explains. It's much clearer.

 

[BvU]

Post #26:

1.Without a detailed sequential diagram, it can be hard to conceive of a mental modelling.
2.Why is it 4g down?
3.Acceleration is always in the direction towards the center, isn't it?
4.Let's just say that when the pilot is at the point (0,-r), he experiences a 4g acceleration towards the center.
5.At the same time, there is a 1g acceleration in the direction towards the ground.
6.But, why, as you'e put it, would the 4g acceleration be in the direction towards the ground? How did this 4g acceleration, initially towards the center changes vector suddenly. It doesn't make sense.

1. I agree. Here's another picture, a continuation from the post#22 picture, with the origin shifted to where you have it too.

A. Coming down along a straight line with constant speed. 1mg downward force from the gravitational field of the Earth is counteracted by the pilot's restraining belts. The pilot experiences 1g downwards (for him: forward, towards the nose of the plane, towards the earth) but doesn't accelerate because the forces executed by the belts hold him at the speed of the plane. This is what we call the normal force in static cases: eg. pilot's weight on chair in office: force mg down, equal but opposite normal force up: zero sum, no motion. What you feel in such situation is called 1g. In the picture: 'felt' 1g down, 'felt' acceleration: red Arrow, size g, up. This is present constantly throughout the entire flight of the plane.

B. The moment he pulls the stick and the plane executes the command, he starts to describe a circle. The OP wording and our state of knowledge about kinematics kind of dictate that the plane's trajectory is a quarter part of a circle. Required centripetal force: 4 mg (value to be justified later) towards center, so at point B that is in positive x direction. Required acceleration: 4g in positive x direction. In addition to that, the 1g from before still applies.(What is an SHM model?). Acceleration 4g in positive x direction plus 1g in positive y direction add up to the purple vector sum. Dotted lines: this is how you add vectors, with a parallellogram. Magnitude of the actual acceleration: √17 g. The pilot experiences this √ 17 g in exactly the opposite direction: he was already leaning in his belts (1g), now he's also pushed into his chair (4g).

C. Centripetal force has turned a bit and aligns more with the normal force. Less strain in the belts, more push from the chair.

D. even less belts, more chair

E. Lowest point, just before pilot pushes back the stick to go to level flight. Acceleration to describe circle still 4g, straight up. Acceleration to counteract gravity and NOT fall towards the Earth any further 1g, straight up. Two independent effects, combined to add up to the pilot experiencing 5g pushing him down in his seat. Just what he can stand while remaining conscious. Hence the 4g at B.

Still E. Plane is now horizontal and pilot pushes back the stick to go to level flight. Plane obeys instantly. No more circle. Acceleration to counteract gravity and NOT fall towards the Earth 1g, straight up.

F. Level flight. Force to counteract gravity 1 mg, delivered by the seat. But the pilot feels 1g down.
​

2. It is not 4g down. I used

So initially he feels an acceleration of 1 g down, 4g to the right

Which was wrong and confusing. My apologies.(*) He feels 1 down, 4 to the left. He is accelerated by an acceleration 1g up (belts) plus 1 g down (gravity) and 4g to the right. That way the net force is towards the center of the circle .

This centripetal force rotates $+\pi/2$ until it is 4mg up. Gravity is still there and needs to be counteracted by another 1mg up. Pilot feels 5g pushing him down in his chair.

(*)I realize I repeated this error by copying to post #35 without checking. Sorry once more. Lot has predictive gifts!​

3. For uniform circular motion in the absence of other forces: yes. In this exercise, there is another (constant) force present: Earth gravity.

4. Yes.

5. There is a 1mg force down. But we were told the plane moves at 1200 km/h, a constant speed. Not unreasonable: the air resistance at such speeds is hefty. I would vote for counteracting this 1mg so the trajectory becomes a circle and we can apply our ^v<sup>2/r

6. See the apology under 2. I confused you and didn't realize I had done so. Then I made it worse by repeating.

So much on the quote. I'm running behind. What a dilemma: too quick and you confuse people, too slow and a thread becomes a plate of spaghetti. Fortunately we are having good fun.

Lewando throws in the scenarios, very instructive.

Lok wants a more complicated trajectory by using the full 5g as much as he can. I wonder if he would stay conscious when braking at 5g while simultaneously turning up at 5g. My pilot son tells me the braking is worse because you have the sensation of falling over forward. Nauseating for sure; don't know about influence on consciousness. Will interrogate him some more on this. Any aeronaut listening in?
Lot also turns to the left. What is it with these guys ;-) Good thing none of us means it in a bad way ;-) (I should brush up my smiley knowledge).

Time for me to turn in. Look forward to negation's next exercise! Good thing he likes big-picture thinking (rockets, space stations, no eart nearby and all that)

This surely has become some kind of knotted cable instead of a didactically responsible thread!
Anyway, the ensuing complications are highly undesirable for negation.</sup>

 

[negation]

Post #26:


1. I agree. Here's another picture, a continuation from the post#22 picture, with the origin shifted to where you have it too.

A. Coming down along a straight line with constant speed. 1mg downward force from the gravitational field of the Earth is counteracted by the pilot's restraining belts. The pilot experiences 1g downwards (for him: forward, towards the nose of the plane, towards the earth) but doesn't accelerate because the forces executed by the belts hold him at the speed of the plane. This is what we call the normal force in static cases: eg. pilot's weight on chair in office: force mg down, equal but opposite normal force up: zero sum, no motion. What you feel in such situation is called 1g. In the picture: 'felt' 1g down, 'felt' acceleration: red Arrow, size g, up. This is present constantly throughout the entire flight of the plane.

B. The moment he pulls the stick and the plane executes the command, he starts to describe a circle. The OP wording and our state of knowledge about kinematics kind of dictate that the plane's trajectory is a quarter part of a circle. Required centripetal force: 4 mg (value to be justified later) towards center, so at point B that is in positive x direction. Required acceleration: 4g in positive x direction. In addition to that, the 1g from before still applies.(What is an SHM model?). Acceleration 4g in positive x direction plus 1g in positive y direction add up to the purple vector sum. Dotted lines: this is how you add vectors, with a parallellogram. Magnitude of the actual acceleration: √17 g. The pilot experiences this √ 17 g in exactly the opposite direction: he was already leaning in his belts (1g), now he's also pushed into his chair (4g).

C. Centripetal force has turned a bit and aligns more with the normal force. Less strain in the belts, more push from the chair.

D. even less belts, more chair

E. Lowest point, just before pilot pushes back the stick to go to level flight. Acceleration to describe circle still 4g, straight up. Acceleration to counteract gravity and NOT fall towards the Earth any further 1g, straight up. Two independent effects, combined to add up to the pilot experiencing 5g pushing him down in his seat. Just what he can stand while remaining conscious. Hence the 4g at B.

Still E. Plane is now horizontal and pilot pushes back the stick to go to level flight. Plane obeys instantly. No more circle. Acceleration to counteract gravity and NOT fall towards the Earth 1g, straight up.

F. Level flight. Force to counteract gravity 1 mg, delivered by the seat. But the pilot feels 1g down.
​

2. It is not 4g down. I used Which was wrong and confusing. My apologies.(*) He feels 1 down, 4 to the left. He is accelerated by an acceleration 1g up (belts) plus 1 g down (gravity) and 4g to the right. That way the net force is towards the center of the circle .

This centripetal force rotates $+\pi/2$ until it is 4mg up. Gravity is still there and needs to be counteracted by another 1mg up. Pilot feels 5g pushing him down in his chair.

(*)I realize I repeated this error by copying to post #35 without checking. Sorry once more. Lot has predictive gifts!​

3. For uniform circular motion in the absence of other forces: yes. In this exercise, there is another (constant) force present: Earth gravity.

4. Yes.

5. There is a 1mg force down. But we were told the plane moves at 1200 km/h, a constant speed. Not unreasonable: the air resistance at such speeds is hefty. I would vote for counteracting this 1mg so the trajectory becomes a circle and we can apply our ^v<sup>2/r

6. See the apology under 2. I confused you and didn't realize I had done so. Then I made it worse by repeating.

So much on the quote. I'm running behind. What a dilemma: too quick and you confuse people, too slow and a thread becomes a plate of spaghetti. Fortunately we are having good fun.

Lewando throws in the scenarios, very instructive.

Lok wants a more complicated trajectory by using the full 5g as much as he can. I wonder if he would stay conscious when braking at 5g while simultaneously turning up at 5g. My pilot son tells me the braking is worse because you have the sensation of falling over forward. Nauseating for sure; don't know about influence on consciousness. Will interrogate him some more on this. Any aeronaut listening in?
Lot also turns to the left. What is it with these guys ;-) Good thing none of us means it in a bad way ;-) (I should brush up my smiley knowledge).

Time for me to turn in. Look forward to negation's next exercise! Good thing he likes big-picture thinking (rockets, space stations, no eart nearby and all that)

This surely has become some kind of knotted cable instead of a didactically responsible thread!
Anyway, the ensuing complications are highly undesirable for negation.</sup>

<sup>And so much more clearer. I was rather confused at the outset with the whole "pilot feels 1g acceleration down ward but has a 1g acceleration upward"-rather contradictory initially- but it seems the arrow in the diagram indicates the corollary of what the pilot is subject to-that is, it indicates an opposing acceleration.
And so, the diagram now makes sense. It's so much easier to work with a real-time visual model in my head.

To end the problem, all I've to do is apply the g = v^2/r with g = SQRT(17g^2), yes?</sup>

 

[lewando]

A couple comments on your notations: "g = v^2/r" is more meaningful and better written as a_centripetal = v²/r

(use the [Go Advanced] button and use the X² and X₂ tools. Use the [Preview Post] button you see if what you did looks like what you expected)

"g = SQRT(17g^2)" is based on what, again?

The point made clear in BvU's the last post was that at every point in the turning phase of the flight, there is a vector sum of the acceleration due to gravity, a_g, and the acceleration due to the circular motion, a_centripetal. The magnitude of this sum cannot exceed 5m/s².

Can you tell, from inspection, which of the points in BvU's diagram (B, C, D, or E) represents the largest magnitude of the vector sum?

From that point, you get what you are looking for: the maximum a_centriptetal.

From a_centripetal = v²/r, the maximum a_centripetal gives the smallest r, which is the height that you are looking for.

Edit: I meant "5g m/s²"

 

[negation]

A couple comments on your notations: "g = v^2/r" is more meaningful and better written as a_centripetal = v²/r

(use the [Go Advanced] button and use the X² and X₂ tools. Use the [Preview Post] button you see if what you did looks like what you expected)

"g = SQRT(17g^2)" is based on what, again?

The point made clear in BvM's the last post was that at every point in the turning phase of the flight, there is a vector sum of the acceleration due to gravity, a_g, and the acceleration due to the circular motion, a_centripetal. The magnitude of this sum cannot exceed 5m/s².

Can you tell, from inspection, which of the points in BvM's diagram (B, C, D, or E) represents the largest magnitude of the vector sum?

From that point, you get what you are looking for: the maximum a_centriptetal.

From a_centripetal = v²/r, the maximum a_centripetal gives the smallest r, which is the height that you are looking for.

E is the greatest. It has an acceleration of magnitude 5g.

 

[lewando]

Alright, so... wrap it up, man!

 

[negation]

Alright, so... wrap it up, man!

r =( 333ms^-1)^2 / 5g

Thanks!

 

[BvU]

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?