Calculating time of zero g (Glider)

[Bradley87]

Homework Statement

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Finding equation allowing to calculate time of zero-g (weightlessness) in a parabolic flight using non-powered glider (no thrust).

Homework Equations

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I have found a few ways to calculate the time (here is one example which is basing on aircraft's velocity only - https://www.nasa.gov/pdf/466762main_AP_ED_WeightlessWonder.pdf). Here it's basically the same equation where U is initial speed and sin0 is a sinus of the initial angle:

The Attempt at a Solution

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I tried to draw the situation using vectors and forces but there is still something wrong. It is obvious that the vertical acceleration of the glider (acting downward?) should be equal to g=9,81m/s^2 (and cancel it). But how to combine it with other forces?

 

[haruspex]

This is rather a strange question. Where does it come from?
It might help to understand what NASA's aircraft has to do. Even if it were just a box, no wings, no engines, it would not provide a zero g environment. Why not? So what does it do to provide one?

 

[rcgldr]

Since the problem states the zero g path is parabolic, you can assume the only force is gravity, and that the idealized glider follows that parabolic path without any aerodynamic forces such as lift or drag.

 

[haruspex]

Since the problem states the zero g path is parabolic, you can assume the only force is gravity, and that the idealized glider follows that parabolic path without any aerodynamic forces such as lift or drag.

Sorry, I do not follow your reasoning there. Yes, a zero g path is parabolic, and it says the glider's path is therefore to be parabolic, but I do not understand how that permits you to ignore forces other than gravity.

 

[rcgldr]

Since the problem states the zero g path is parabolic, you can assume the only force is gravity, and that the idealized glider follows that parabolic path without any aerodynamic forces such as lift or drag.

Sorry, I do not folow your reasoning there. Yes, a zero g path is parabolic, and it says the glider's path is therefore to be parabolic, but I do not understand how that permits you to ignore forces other than gravity.

Consider what occurs at the peak of the path, if there is aerodynamic drag, it's perpendicular to the force of gravity, and a zero g path that includes a moment of horizontal only velocity doesn't seem possible.

 

[haruspex]

Consider what occurs at the peak of the path, if there is aerodynamic drag, it's perpendicular to the force of gravity, and a zero g path that includes a moment of horizontal only velocity doesn't seem possible.

That's not the same argument. What you are saying now is that if it is possible for there to be a nonzero free fall duration then there must be no drag. I agree with that, though it is not instantly rigorous. We'd need to show that drag and lift cannot cancel.
This is why I wrote up front that it is a strange question. If we ignore drag and lift then the problem reduces to asking how HIGH a glider can go (and with no lift, how does it get there?). If we take drag and lift into account, the answer is either zero or, arguably, a matter of how long it takes the occupant to go from one side of the compartment to the other at the drag+lift acceleration.

Edit: I omitted the word HIGH earlier.

 

[Bradley87]

Thank you for your answers.

Well, I am a student of Aerospace Engineering and also a pilot. I was supposed to prepare a project that is related to physics that - in case it will be good enough - will exempt me from writing final physics exam :)

So I came up with an idea of a project divided into two parts - theoretical and practical. I want to find a way to calculate time (as precisely as possible) and after that take a glider with accelerometer / artificial horizon / a few gopro cameras and a dozen of small soft balls (for better visual effect) and check whether calculations are correct for particular flight parameters (initial angle and speed) by measuring the time of weightlessness during each attempt.

The equation I posted before makes in my opinion everything too simplified. In real life - because of the air resistance, the parabolic trajectory would be slightly smaller - so the time of weightlessness will be shorter.

I have found a good tutorial showing how to calculate projectile motion with drag using numerical modeling and excel - https://www.youtube.com/watch?v=OukRTF6Bgcc

The only problem I have now is that I am not sure whether changing the ball with a glider will work out. Of course I know the drag parameters for my glider. What do you think?

 

[haruspex]

Thank you for your answers.

Well, I am a student of Aerospace Engineering and also a pilot. I was supposed to prepare a project that is related to physics that - in case it will be good enough - will exempt me from writing final physics exam :)

So I came up with an idea of a project divided into two parts - theoretical and practical. I want to find a way to calculate time (as precisely as possible) and after that take a glider with accelerometer / artificial horizon / a few gopro cameras and a dozen of small soft balls (for better visual effect) and check whether calculations are correct for particular flight parameters (initial angle and speed) by measuring the time of weightlessness during each attempt.

The equation I posted before makes in my opinion everything too simplified. In real life - because of the air resistance, the parabolic trajectory would be slightly smaller - so the time of weightlessness will be shorter.

I have found a good tutorial showing how to calculate projectile motion with drag using numerical modeling and excel - https://www.youtube.com/watch?v=OukRTF6Bgcc

The only problem I have now is that I am not sure whether changing the ball with a glider will work out. Of course I know the drag parameters for my glider. What do you think?

it's not merely a matter of being parabolic. It would be parabolic just by having a constant acceleration, but for zero g you need the acceleration to match g. The NASA craft uses its engines to counter drag, as precisely as possible, so that the net force on the frame is g. I see no way to do that with a glider. The nearest I can think of is to find a thermal and pull up into a vertical position near the top of the thermal. As the glider loses speed, there might be a brief period when the velocity of the glider matches that of the thermal. Sounds like conditions for a stall, though.

 

[David Lewis]

Is the glider allowed to dive and gain speed before it pulls up and follows a ballistic trajectory?

 

[haruspex]

Is the glider allowed to dive and gain speed before it pulls up and follows a ballistic trajectory?

I don't see how that is going to help. You would still have to arrange that all forces except g cancel.

 

[David Lewis]

For example, when the glider pulls out of the dive, if its velocity vector is 45 degrees above horizontal, and its acceleration vector is 9.81 m/s² straight down then the glider will follow a parabolic trajectory and it will pull 0 g's.

 

[haruspex]

For example, when the glider pulls out of the dive, if its velocity vector is 45 degrees above horizontal, and its acceleration vector is 9.81 m/s² straight down then the glider will follow a parabolic trajectory and it will pull 0 g's.

If there is no lift and no drag, its acceleration will be g straight down regardless of its speed and angle. With drag, it would have a horizontal acceleration and its vertically downward acceleration would exceed g.

 

[Bradley87]

It is possible to do in a glider. I did it many times by myself, here is also a proof - (starts around 2:15)

The only difference between a glider and a powered plane is that the effect will last shorter - about 5 - 6 seconds only.

 

[haruspex]

It is possible to do in a glider. I did it many times by myself, here is also a proof - (starts around 2:15)

The only difference between a glider and a powered plane is that the effect will last shorter - about 5 - 6 seconds only.

No, that's not zero g. Mostly what the video shows is negative g. That can be achieved going almost vertically upwards. The drag will push the glider down, adding to the gravitational force.

 

[rcgldr]

You can get near zero g, but the aerodynamic drag means objects floating in the cockpit will accelerate forwards instead of hovering.

 

[Bradley87]

But isn't it the matter of proper piloting only (proper pushing the stick forward)? Of course that would be hard but in theory? To get negative G you need to pass 0 before.

When we go up the parabola, drag acts downwards adding to the gravitational force. But we have also (do we?) a little lift acting upwards. This would not compensate the drag but perhaps we can add there something more - a centrifugal force. Does it make any sense? If that is correct I am not sure what would happen on the other leg of the parabola where drag will act upwards..

 

[Bradley87]

I think I know now what is going on. Please check this website: http://rotorlab.tamu.edu/me489_SP11/lectures/Example_writing_MATH_parabolic%20flight.pdf

The only mystery for now is that the lift is "simultaneously removed". Does it happen because of the angle?

 

[haruspex]

I think I know now what is going on. Please check this website: http://rotorlab.tamu.edu/me489_SP11/lectures/Example_writing_MATH_parabolic%20flight.pdf

The only mystery for now is that the lift is "simultaneously removed". Does it happen because of the angle?

My guess is it means spoilers are deployed.

 

[David Lewis]

The glider will not develop any lift when its angle of attack is zero (measured from the zero-lift line). If the glider is carrying ballast, it will possesses more kinetic energy coming out of the dive, and drag will be less of a factor.

 

[rcgldr]

To be in free fall, the only force acting on the glider has to be gravity, all of the aerodynamic forces need to net out to zero force, and as posted previously, lift can't cancel out drag, because it's perpendicular to drag. In powered aircraft, the throttle is adjusted so that thrust is just enough to cancel drag, and the elevator is adjusted to maintain zero lift, with the combination resulting in zero net aerodynamic force (other than a torque that's constantly changing the pitch of the aircraft).