Author Topic: Thrust to Weight Ratios  (Read 6789 times)

Offline FLS

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Re: Thrust to Weight Ratios
« Reply #120 on: November 08, 2011, 08:09:56 PM »
0 G zoom climb. Weightlessness begins while climbing.

Offline Lusche

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Re: Thrust to Weight Ratios
« Reply #121 on: November 08, 2011, 08:15:35 PM »
In that graphic, the flight path is not illustrated well. Plane does not climb when astronauts experience 0G.

It does.

Here's another one:



Or this one, coming from NASA 

Plane is still gaining several thousand feet during the inital part of the 0G phase.
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Offline 2bighorn

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Re: Thrust to Weight Ratios
« Reply #122 on: November 08, 2011, 08:43:17 PM »
It does.

It's ascending due to being on highly parabolic inertial fall path (orbit), it does not climb in sense of using thrust or lift, otherwise astronauts would end up slamming the walls.








Offline PuppetZ

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Re: Thrust to Weight Ratios
« Reply #123 on: November 08, 2011, 08:49:42 PM »
in the zero G planes, when the zero G phase begin, the pilot push the nose over slowly in a zero G pitch down. If so the graphics are a bit missleading in that the plane will have a more parabolic trajectory instead of this clear cut pitch down at the top and the trajectory will begin the curve when the zero g phase begin. In the graphics, it look like pitch is constant through the first part of this phase.

Taken from wikipedia(I know, I know but it's not ALL BS) : Initially the aircraft climbs with a pitch angle of 45 degrees. The sensation of weightlessness is achieved by reducing thrust and lowering the nose to maintain a zero-lift angle of attack. Weightlessness begins while ascending and lasts all the way "up-and-over the hump", until the craft reaches a declined angle of 30 degrees.
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Offline Lusche

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Re: Thrust to Weight Ratios
« Reply #124 on: November 08, 2011, 08:55:15 PM »
It's ascending due to being on highly parabolic inertial fall path (orbit), it does not climb in sense of using thrust or lift, otherwise astronauts would end up slamming the walls.

It's ascending due to being on highly parabolic inertial fall path (orbit), it does not climb in sense of using thrust or lift, otherwise astronauts would end up slamming the walls.

Neither I nor the picture made any claims on that matter in any way. We both said it still climbs when the occupants start to experience 0G (which it does), and you said that it does not (which is wrong).  :P
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Offline FLS

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Re: Thrust to Weight Ratios
« Reply #125 on: November 08, 2011, 09:12:49 PM »
It's ascending due to being on highly parabolic inertial fall path (orbit), it does not climb in sense of using thrust or lift, otherwise astronauts would end up slamming the walls.


It's climbing in the sense that it's gaining altitude.

Offline 2bighorn

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Re: Thrust to Weight Ratios
« Reply #126 on: November 09, 2011, 12:04:52 AM »
Remember we are not speaking what total physics are happening, but rather what the common term "g loading" refers to.

Sorry, completely missed this one. Yes, I didn't notice you guys were talking about load factors and not physics in general. I shouldn't have interrupted and derail a discussion.

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Offline nrshida

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Re: Thrust to Weight Ratios
« Reply #127 on: November 09, 2011, 12:09:20 AM »
FLS I meant that if you were flying only with the y-component of a parabolic curve (over time) without any movement on the x-axis, then surely you'd need to balance your thrust very carefully against your weight on the way up, or wait until the thrust began to be overtaken by weight to achieve the same effect. If you had a second accelerometer, fore / aft then it would also have to read zero to give you the feeling of weightlessness, right?
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Offline bozon

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Re: Thrust to Weight Ratios
« Reply #128 on: November 09, 2011, 03:03:57 AM »
OK, definitions:

"G", the big capital G in the context of aviation is the acceleration in a direction perpendicular to the movement excluding gravity. In other words, it is the force over mass, where the force include the sum of all forces on the plane except gravity and then taking only the component perpendicular to the velocity.
Since drag is by definition along the velocity direction and thrust is usually close to be aligned with the velocity and we exclude gravity, the only thing that is left is the lift which by definition is perpendicular to the velocity. Hence G=L/M, but expressed in units of earth gravitational acceleration, usually denoted with a small g. So if you use SI or CGS units G=L/(M*g)=L/W where "W" is the weight.

Some finer points:
------------------------
* Generally, G is NOT the actual acceleration of the plane in any axis or total since G does not include gravity.

* a stone being thrown is at 0G from the moment it leaves the hand till it lands, even while going up - the only forces that act on it are drag, which is aligned with the velocity and hence irrelevant to G and gravity which is explicitly excluded from G. Dignify all of us by not arguing that the stone is not spherical or is rotating.

* The pilot feels G by how hard the seat is pressed against hit butt, not his back. In a 90 degree climb (maintaining the angle), the seat is only pushing against the pilots back and not his tuches. The vertical acceleration can be whatever, but since it is aligned with the velocity it is still G=0.
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