Thrust to Weight Ratios

[Stoney]

I don't think this statement is technically correct Hitech.

The definition of G loading creates a "divide by zero" mathematical situation at 0g. That doesn't mean that there isn't any lift, just that the denominator is zero.

Think of it this way, if I hold an apple in my hand and ask you to divide it by zero, does that make the apple disappear?

This is what I meant when I said the whole lift = weight relationship could only exist at > or < 0 G.  And really, that should be "required" lift = weight.  Anyway...

Nice to see you Baumer!   

[Baumer]

Thanks Stoney, good to see you to. I took a few months off and just started reading the boards again. Perhaps I'll start flying again soon, who knows? 

I understand where you're coming from FLS, it was just that reading some of the earlier posts I had the impression that some people might be thinking zero gee was also zero induced drag.   

[FLS]

Thanks Stoney, good to see you to. I took a few months off and just started reading the boards again. Perhaps I'll start flying again soon, who knows? 

I understand where you're coming from FLS, it was just that reading some of the earlier posts I had the impression that some people might be thinking zero gee was also zero induced drag.   

That's probably just sloppy writing where minimum induced drag was intended instead of 0 induced drag. Thanks for clearing that up.

[hitech]

I don't think this statement is technically correct Hitech.

The definition of G loading creates a "divide by zero" mathematical situation at 0g. That doesn't mean that there isn't any lift, just that the denominator is zero.

Think of it this way, if I hold an apple in my hand and ask you to divide it by zero, does that make the apple disappear?

Say WHATTT!?

Where is the divide by zero? Weight is a constant hence the denominator can NOT be zero. And therefore can not be a divide by zero error.

The drop of the apple creates zero lift (unless it is irregular shape or spinning ). Lift would move it side to side not slow its fall.

DEFINITIONS !!!!!!!

Drag = force opposite the direction of travel.

Lift = force perpendicular to the direction of travel.

By simple definitions there can not be any net lift at zero g. Zero g is defined as a zero load factor.

As per your equations.

n = l/w.

or solving for L
N * W = L.

for a zero load factor I.E. N = 0
0 * W = L

Since zero * ANYTHing is zero

L = ZERO


HiTech

[hitech]

This is what I meant when I said the whole lift = weight relationship could only exist at > or < 0 G.  And really, that should be "required" lift = weight.  Anyway...

Nice to see you Baumer!   

Stoney the lift = weight thing only applies to sustained level flight. In an steady state attitude that is not horizontal to the ground, lift must always be less then  weight or the aircraft will not be traveling in a straight line.

HiTech

[Mace2004]

Effective AoA is the result of upwash, downwash, prop effects, roll/pitch rates, etc.  You guys are really over-thinking this. All that matters to the guy in the cockpit is that minimum drag, max acceleration, min deceleration all happen at zero G.   

[hitech]

On the standard force diagram like this.



I really wish they drew it like this to remove the confusion of the force definitions.

Also the plane is in a climb in the picture below.




HiTech

[bozon]

This is what I meant when I said the whole lift = weight relationship could only exist at > or < 0 G. 

Stoney, pro tip: when in science or engineering you get a mathematically singular point, it almost certain that you have a mistake somewhere. If the limit behaves badly, i.e. in our case (following your logic) when you approach weight=0, G goes to infinity, you can be CERTAIN that you have a mistake.

You are thinking 0G means weightless. That is not true in the frame of reference of an observer looking at the plane from the outside (i.e. not accelerating). To make things simple: "weight" in our case means the gravitational pulling force acting on the plane, given by weight = mass*g. It is a constant. The little "g" is the earth's gravitational acceleration.

[Midway]

Stoney, pro tip: when in science or engineering you get a mathematically singular point, it almost certain that you have a mistake somewhere. If the limit behaves badly, i.e. in our case (following your logic) when you approach weight=0, G goes to infinity, you can be CERTAIN that you have a mistake.

You are thinking 0G means weightless. That is not true in the frame of reference of an observer looking at the plane from the outside (i.e. not accelerating). To make things simple: "weight" in our case means the gravitational pulling force acting on the plane, given by weight = mass*g. It is a constant. The little "g" is the earth's gravitational acceleration.

<enters brainy think tank room, notices apparent reference to Einstein's relativity gravity / acceleration relationship theory, leaves brainy think tank room rather quickly>  

[Baumer]

Mace I agree with you, we are over thinking this and missing the main point that MINIMAL drag occurs at zero G.

Hitech, apples are not symmetrical and will generate lift. This was actually discussed in one of my aerodynamics classes at ASU.

And just to be clear Mass is a constant, Weight is not a constant.

[bozon]

notices apparent reference to Einstein's relativity gravity / acceleration relationship theory

Nope. Barely Newtons gravity.

Here, have an apple and it will all be clear.
If still not, try these mushrooms.

[FLS]

Midway this is all basic stuff or I wouldn't be contributing. If you do a little background reading you'll understand it.

[hitech]

Mace I agree with you, we are over thinking this and missing the main point that MINIMAL drag occurs at zero G.

Hitech, apples are not symmetrical and will generate lift. This was actually discussed in one of my aerodynamics classes at ASU.

And just to be clear Mass is a constant, Weight is not a constant.

And then they are not at zero load factor or zero G.

HiTech

[Stoney]

And then they are not at zero load factor or zero G.

HiTech

So what you're saying is that if there is relative wind over the wing, and the wing is at an AoA that creates lift, the plane cannot be at 0 G, and vice versa?

[FLS]

So what you're saying is that if there is relative wind over the wing, and the wing is at an AoA that creates lift, the plane cannot be at 0 G, and vice versa?

Can I repeat my earlier question?

Stoney could you give an example of 0G flight with an AOA that creates lift?   
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