Author Topic: Ok, new question, DRAG and WEIGHT.  (Read 1677 times)

Offline Jigster

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Ok, new question, DRAG and WEIGHT.
« Reply #45 on: January 31, 2001, 12:41:00 AM »
Oh btw, given the reason there is no gravitational reaction between, (say a human jumping) because of the Earth's liquid center, and mass.

As the tides rise, the molten core of the earth reacts the same, albeit not as visable because of core pressure. Using the above example, a tiny fraction of the earth actually does move with every object that pulls away from the earth. A very tiny fraction.

This was the reasoning behind the infamous "Chinese Jump Theory", practically a doomsday weapon if Japan ever PO's China.

- Da Bess

funked

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Ok, new question, DRAG and WEIGHT.
« Reply #46 on: January 31, 2001, 12:54:00 AM »
Jig it was Joe Kittinger.  Giant F___ing Cojones. http://www.wpafb.af.mil/museum/afp/afp200.htm
   

[This message has been edited by funked (edited 01-31-2001).]

Offline fscott

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Ok, new question, DRAG and WEIGHT.
« Reply #47 on: January 31, 2001, 01:31:00 AM »
Too al participants, I'm deeply sorry for starting this thread, and the other one too. Our time could be spent much better by blown out of the sky by fscott.

So... My next question: When I rip your wings off with my .50's, what is my awesomeness to weight ratio?  I weigh about 190 lbs.


fscott

Offline Jigster

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Ok, new question, DRAG and WEIGHT.
« Reply #48 on: January 31, 2001, 01:55:00 AM »
Considering Kittinger's right glove de-pressurized while ascending to 100,000ft, and countinuing instead of aborting, I'd say so. Like Yeager breaking those ribs while watching the nudie rodeo a Pancho's Barnes place before taking the X-1 through the transition  

I have the full AvHi write up on Kittinger from F-51 training through the flying circus he's in now. MH and Exc were some cool projects.

- Da Bess

Offline Dingy

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Ok, new question, DRAG and WEIGHT.
« Reply #49 on: January 31, 2001, 08:36:00 AM »
Whereas fscott is apologizing for starting this thread, I salute everyone who participated in here and submitted info.  Its great to know I can still learn things in a game I love so much! <S> All!

-Ding

PakRat

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Ok, new question, DRAG and WEIGHT.
« Reply #50 on: January 31, 2001, 08:45:00 AM »
 
Quote
Originally posted by ljkdern:
Wow is right...sigh... the soap bubble and the bowling ball accelerate at the same rate.... Until one or the other reaches terminal velocity they are accelerating at the same rate. The bowling ball doesnt accelerate faster...it accelerates longer.

Bzzzt - wrong again. All forces are present from the very beginning. They will not accelerate at the same rate because the force speeding up the heavier object will be greater and in the case of the bowling ball / feather argument, the drag force will be much smaller in proportion to the drag force on the feather. The drag on items like the feather will have a greater balancing effect. BUT THESE FORCES ARE PRESENT FROM THE BEGINNING AND THE DRAG WILL GROW AS THE SPEED OF THE OBJECT INCREASES.

What you state would cause a discontinuity in the speed/acceleration of the items. They will not "shift gears" at some point. The speed curves will be smooth - there will be no corner as you imply.

------------------
Rape, pillage, then burn...

Offline jedi

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Ok, new question, DRAG and WEIGHT.
« Reply #51 on: January 31, 2001, 09:35:00 AM »
If the object is to determine which AIRPLANE will dive the best, you probably need to talk about the forces on the AIRPLANE, not soap bubbles  

A diving aircraft has, obviously, gravity working to accelerate it.  But it also has THRUST from its engine, dependent on the engine/prop combo's capability to accelerate it, and drag, dependent on the shape and size of the airframe, working on it.  So you have:

Mass x gravity (different for each plane)
Thrust (different for each plane)
Drag (different for each plane)

Of COURSE they're going to fall at different rates.  The reason is that the sum of the forces on the two planes is NOT equal, otherwise they wouldn't accelerate at ALL.

Sum the forces on each plane.  Divide the resultant force by the plane's mass.  That gives you the resultant acceleration of the plane, which will only be constant until the forces change, which will occur when the drag and thrust change due to atmospheric conditions and airspeed as the altitude decreases.

It's a differential equation, not an example of what Galileo "tested."  It's not a vacuum, it takes place over thousands of feet of constantly changing atmosphere, not 200 feet of "dense" air at the surface, and it's not a "constant acceleration" problem either, so the fact that acceleration due to gravity is independent of mass, while interesting, isn't very relevant  

--jedi

Offline jedi

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Ok, new question, DRAG and WEIGHT.
« Reply #52 on: January 31, 2001, 09:43:00 AM »
P.S. as soon as you attach a powered propeller to your falling body, about 99% of the guys in the sim (including me) are essentially disqualified from having even the slightest clue about what's going on there  

Oh yeah.  Don't forget the lift on the wings in the dive.  It's generating induced drag dependent on the amount of lift.  It varies with speed and density (and thus altitude, which is...er...constantly changing in the dive), unless you're at zero angle of attack, which means it varies with the G-loading.

So add that into the differential equation somewhere  


Offline hitech

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Ok, new question, DRAG and WEIGHT.
« Reply #53 on: January 31, 2001, 11:15:00 AM »
I think I know why the idea that all objects fall at the same rate in a vacuum causes such confusion. It's because it has always been taught as a fundamental Law and not an effect. Most people tend to think of it as an absolute but really it's just and effect of a few simple principles.

If you begin with just the very simple simple laws.
1.
Force = Mass * Acceleration
2.
Force of gravity at a fixed  distance, call this force GF always is a Constant value (Call this GC) times the Mass of the object so we end up with the 2nd equation. (Side Note the GF is what we call Weight)

GF = GC * Mass.

Please keep in mind gravity is just a force, nothing more nothing less.

So very simply combining the 2 equations you get.

GC * Mass = Mass * Acceleration

Now just canceling out the 2 Masses you get

GC = Acceleration

I.E. All objects accelerate/fall at the same rate when no other forces such as drag are at work.

This is really nothing earth shattering, but is really what makes all objects fall at the same rate, and just a different way of understanding why.

HiTech

Offline Badboy

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Ok, new question, DRAG and WEIGHT.
« Reply #54 on: January 31, 2001, 06:10:00 PM »
This type of question often leads to a good deal of confused thinking. Here is an explanation I use with my own students  

What actually happens is by no means obvious, so I'm going to resort to a little math, just to maintain clarity and credibility  

Let's consider two aircraft that are externally identical in every way. However, one aircraft is much heavier than the other. They both have exactly the same thrust and drag. Now we will assume that they both dive from some initial speed and we want to know what happens next. But firstly, let's pause to consider the average layman has to say about it

"I thought Galileo proved that all objects fall with the same acceleration?"

Ok what happened is this... He noticed that he could drop metal balls of different weights, from a building and they would hit the ground at the same time, at least he couldn't measure the difference. This is how the math clocks out. Using Newton's law   f = ma  where f is the force acting on a ball, m is its mass and a is its acceleration towards the ground. We are interested in how the objects accelerates so we write it like this:

                    a = f/m          (1)

That simply says that the acceleration is the force pulling the balls down to the ground divided by their mass. Now if you consider the drag acting on the balls to be negligible, the only force left to consider is the force due to gravity, so we get:

                 f = mg         (2)

Now if you substitute that value for f back into our first expression (1) we get:

                 a  = mg/m

The mass m cancels out leaving:

                 a = g     and g is about 9.8 m/s^2 or 32.2 ft/s^2

This means that if we can ignore drag (which is approximately true for balls that are heavy and smooth) the balls will accelerate at the same rate, regardless of their mass!! Certainly for casual observations that appears to be true.

However, the situation is very different if we consider an aircraft where the drag and thrust can't be ignored, because they are no longer negligible.

So let's consider our aircraft, but  firstly let's just consider the drag, and we can call it d for short. Now we can still write Newton's law just as we did in expression (1) but now when we write down the force acting on the aircraft we need to include the drag too, so we get:

                f = mg - d             (3)

We write a negative sign because drag always opposes the motion. Now look what happens if we substitute the value for the force in expression (3) back into expression (1), we get:

                a = mg/m - d/m

                a = g - d/m           (4)

Now here is the crunch! The mass doesn't cancel out of the drag term, so the mass will now have an influence on the acceleration. But what influence?

Ok,  look at expression (4) again, it is saying that the acceleration is going to be reduced by the drag (yep that makes sense) but as the mass gets larger, the actual value of d/m will get smaller, meaning that for the same amount of drag, heavier objects will accelerate more quickly!!

However, it causes no end of problems because our everyday experience tells us that heavy objects are more difficult to accelerate than lighter ones.

Now take another look at expression (4)... The acceleration is maximum when the d/m term is as small as possible, which means that for the best acceleration in a dive you want a combination of low drag and high weight, which means that the smallest drag/weight ratio possible will give you the best acceleration.

That's the answer to your question... But the situation is not as simple as that, we are still ignoring the thrust involved, and for aircraft that  can produce large amounts of thrust, we can consider what will happen if we call the thrust t and include it. Our expression for the force now becomes:

                   f = mg + t - d

If we now substitute that into expression (1) we get:

                   a = g + t/m - d/m

What happens now depends on how big the thrust and drag are. For example, consider an aircraft that begins to dive from a low speed, at maximum thrust. Now because the speed is low, the drag will be relatively low, so just to clarify this point, lets ignore the drag term just as we previously ignored the thrust. Now we are left with:

                   a = g + t/m

That tells us that the thrust will increase the acceleration (once again, that makes sense) but that the larger mass makes that increase smaller.

So, if our two aircraft have a high thrust to weight ratio and start to dive from low speed, the lighter one will actually accelerate more quickly!! That is contrary to the previous situation. However, because the thrust will get smaller (for a prop aircraft) as the aircraft gets faster, and the drag will increase as the velocity squared, the drag will very quickly become the dominant factor, and so the heavier aircraft will generally very quickly, if not always, exceed the acceleration of the lighter one. Indeed, you often see pilots referring to the better dive acceleration of heavier aircraft, which supports the theory  

But it does open the possibility for exceptions to the rule, a draggy aircraft with low thrust will accelerate poorly no matter how heavy it is  

Now, the argument doesn't stop there, because even though one might suspect that  acceleration is the most important factor in combat, the actual top speed, or terminal velocity that you can reach in a dive, might also be important?

So, if we extend the argument to include terminal velocity, it turns out that if you
are happy to ignore thrust, you can compare the ratio of the terminal velocities of two aircraft as the square root of the ratio of their weights.

So if one of the two aircraft in my previous example was four times heavier than the other, its terminal velocity would be twice as fast.

So, not only does the heavier aircraft accelerate better, it reaches a higher terminal velocity!!

That's more important than you might at first think, because when most folk consider terminal velocity, they imagine a vertical dive... I suspect that few pilots that enter a vertical power dive ever get anywhere close to their terminal velocity, at least not many that live to talk about it  

The point is that there is a terminal velocity for every dive angle, dive at 30 degrees in a heavier aircraft and you will still accelerate better, and reach a higher terminal velocity for that angle of dive.

That is why the heavy US aircraft were always able to extend away from the lighter Japanese aircraft. Even if two aircraft can match speed in level flight, the heavier one could still extend away with better acceleration and higher terminal velocity, for as long as the dive could be maintained.

Hope that helps.


Leon "Badboy" Smith
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Offline ljkdern

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Ok, new question, DRAG and WEIGHT.
« Reply #55 on: January 31, 2001, 11:17:00 PM »
OK....I would like to apologize. It turns I was wrong. Most of the last few messeges are beyond my comprehension but I was able to figure out why a heavier object will accelerate faster than a lighter one in my own simple mind. If only for my own benefit I will explain how I came to this conclusion: (ignoring thrust) Both objects begin at the same acceleration and gradually decrease their acceleration until reaching terminal velocity. Since the heavier object has a higher terminal velocity that can mean only one thing...its decrease in acceleration as it falls happens at a slower rate than the decrease in acceleration of the lighter object. I think it happens this way because of inertia. A given amount of drag force will have more effect on a lighter object than it will on the heavier object...thus it will cause the lighter object to slow more quickly. I may be wrong again but I think im getting closer