how exactly do autos work?

[Linc]

Lift and Weight as aerodynamic forces, are a subset of physics. When lift = weight in aerodynamics, the height of the aircraft doesn't change. When lift is greater than weight, the aircraft climbs and when lift is less than weight, the aircraft descends.

 

As far as F=ma, the acceleration is equal to the rate of climb, since that is the rate of change from the point of reference; the beginning of the climb. (v2=v1+a*t). Your argument seems to be from the perception of the vestibular system as a validation that the acceleration is no longer occurring. However, the difference between 40mph and 50mph is an acceleration of 10mph, even after you've been going 50mph for a while.

 

You could also talk this out in gs. 1g equals the acceleration of the mass of the aircraft caused by the earth's gravity. 1g applied in the opposite direction only cancels out the 1g acceleration of the mass. Since g is a unit of acceleration, you would have to accelerate the aircraft at greater than 1g to lift the aircraft. As long as you maintain that acceleration to greater than 1g, the aircraft will climb.

[Eric Hunt]

Come on, you guys! Did you pass Helicopter Aerodynamics yet??

 

Witch - the twist of -8 degrees is the change in pitch from the root to the tip. It could be +6 at the root, and the "washout" means the pitch slowly decreases to be -2 degrees at the tip. The aim is to try to develop constant lift over the length of the blade. If the whole blade was at a constant pitch, then the tip of the blade would develop hugely more lift than the root, causing large bending forces. Washout attempts to get a bit more lift at the root and a bit less at the tip.

 

Linc - yes, it is all physics, which was my major at university and apparently not yours.

 

Some definitions first.

Speed (or in truth, velocity) is the rate of change of position. (first differential of position)

Acceleration is the rate of change of velocity (second differential)

Jolt is the rate of change of acceleration (third)

 

You say "When lift is greater than weight, the aircraft climbs and when lift is less than weight, the aircraft descends. " See the previous posts for an explanation. When L>W, you will start to accelerate upwards, but will stabilise at a rate of climb - the acceleration has gone, but the speed lives on.

 

You toss in "V2=V1 + at" which is correct. The final speed is equal to the initial speed plus the acceleration times the time for which it acted. So, 50 kt = 40kt + some extra power for some period of time. The difference between 50 and 40 kt is not an acceleration of 10 kt, it is a difference in SPEED of 10 kt, caused by an ACCELERATION THROUGH 10 kt.

 

You say "1g equals the acceleration of the mass of the aircraft caused by the earth's gravity. 1g applied in the opposite direction only cancels out the 1g acceleration of the mass."

 

So, if the earth pushing back on us against the force of gravity is cancelled out, we would be at zero g sitting on the ground? Interesting physics.

 

When you enter a climb by adding power and making L>W, you DO feel a slight increase in the g force, as you initially accelerate upwards. But as the restraining forces increase to match your extra power, the acceleration stops, the vertical speed settles to be constant, and you only feel one g again.

 

Enter a descent - quickly - and you feel light on the seat, less than 1 g. But then the forces stabilise again, and you are back at 1 g, L=W and rate of descent is constant.

 

You have probably never flown in an aircraft fitted with a g meter, but this meter will tell you this truth. The normal position for its needles is the 1 g position. Enter a climb, it momentarily goes up but then returns to 1. Enter a descent, it momentarily decreases, but then returns to 1.

 

The only time you will ever experience a constant acceleration in a helicopter is if you enter a level turn and stay in the turn. Tighten it up, you feel more g. But stop the turn, you are back at 1 g. Trust me.

[Jon]

Wow guys, thanks a bunch for all the replies and explanations, I understand the basics of it now, I judt near to learn more about the aerodynamics of it all, thanks a bunch

[Guest pokey]

i spent my fair share in physics class too, however, the way i see weight & lift acting on an aircraft is that in level flight you must have equilibrium between the 2: L=W if that equilibrium is upset by more weigh OR more lift? ya either go up or down. How can these two forces be in equilibrium and produce a climb or a dive?

 

and yes, i have spent many (hundreds) of hours in my aerobat with a G-meter, thats exactly what it is too a G-meter-for measuring G's (accelleration). Once the initial accelleration is complete, and we are still (say diving) what is pulling us towards earth? gravity (weight) what is keeping us aloft? lift,,,,,add more lift ( pull back) & we feel & see the G's on our meter but? lift has been added to equal weigh & we are leveling out,,,,,

 

Kinda like weighing yourself in a moving elevator, initialy you weigh more (accelleration) but upon reaching a certain rate of ascent, you weigh the same but the lift force of the elevator going to the top floor exceeds the weight of the car.

[Guest pokey]

ok Eric, i see where i am losing it, and our lines are crossed. We are forgetting about the force that actually makes us go up in an aircraft,,,,, yes i agree lift & wight are equal, but add THRUST! and bingo----all adds up now,,,,,

[Linc]

Yep, you got me, high school education. And not even an high school physics class, so I'll admit to getting all kinds of terms and explanations wrong.

 

v1=40mph

v2=50mph

 

v2=v1+a*t

(v2-v1)/t=a*t/t

a=(v2-v1)/t

a=(50mph-40mph)/t

a=10mph/t

 

which in my explanation was 10mph. Granted, the acceleration would have to be turned into something other than miles per hour, which is only a description of velocity. But if you're talking to laymen, then layman's terms should be acceptable, since everyone can grasp that the vehicle has accelerated 10mph faster than its previous speed, and in order to stay faster than 40mph, it will have to maintain that acceleration on the mass to maintain the force that results in a velocity of 50mph.

 

In every basic flight manual, it is stated that when lift exceeds weight the aircraft will climb, when thrust exceeds drag the aircraft will move forward. There is no way of measuring lift that exceeds weight except by the IVSI and the change on the altimeter. 1g on your g-meter (inertia meter or gravity meter would be more accurate) is a constant acceleration downward caused by the earth's gravity. So, if you're aircraft is climbing and your g-meter shows 1g, why aren't you going down? What the 1g is showing you is that the weight is constant and your inertia has been established in the climb, not lift and weight becoming equal. As you initially accelerate upward, the g-meter will read greater than 1g because the inertia of the helicopter is resisting the climb. Even though you're applying an acceleration in the opposite direction of that 1g, the g-meter will initially register the equal and opposite reaction (g1+g2) until that resting inertia is overcome.

 

Lift only acts in the positive of the vertical axis and weight only acts in the negative vertical axis of the aircraft. You can't measure lift by measuring weight. The aircraft can maintain 1g in the negative vertical axis and still maintain a positive rate of climb in the positive vertical axis, because the difference results in a net positive acceleration of the mass upwards.

 

Lift is an aerodynamic force and F=ma, so L=ma.

Weight is also an aerodynamic force and F=ma again, so W=ma.

 

Since the mass of the aircraft doesn't change, the only thing that can change to change the position of the aircraft vertically is to change the acceleration. So, we could simplify the formula to determine what the aircraft is doing by stating that L and W will represent the accelerations of lift and weight respectively. The net difference between the accelerations will determine whether or not the aircraft climbs or descends.

 

If L>W the aircraft will climb and if L<W the aircraft will descend. If L=W the aircraft will neither climb nor descend.

[joker]

Wow,

 

Jon, it seems like you opened a can of worms here! I told you this was difficult to explain.

 

Anyway, as for the current raging debate, I admit I haven't scrutinised every word and formula that has been stated thus far. However, I suggest all the 'Einsteins' or 'Newtons' here go back to their physics books and look up the chapter on 'Frames of Reference'!

 

The argument exists on this thread, because they are being stated from two different Frames of Reference. One from inside the aircraft, and one from the surface of the earth! Understand this, and we'll see that both the arguements are right.'

 

That's why I haven't read it all in detail!

 

Well, rather than internet links (I leave you all to find the stuff yourself) here's a few tidbits for thought!

 

Newtonian relativity: absolute linear motion at a constant velocity cannot be detected, nor can absolute rest. All motion is relative to a frame of reference. It is not possible to distinguish motion with a constant velocity from rest. All constant velocity frames of reference are equivalent (including frames of reference that appear to be at rest -- after all, a prolonged state of rest is motion with a constant speed of zero).

 

All laws of mechanics remain the same in a frame moving at a constant velocity.

 

Why did the chicken cross the road?

Aristotle:

It is the nature of chickens to cross the road.

Newton:

Because no force caused the chicken's state of uniform road crossing to change.

Einstein:

Is the chicken crossing the road or is the road moving under the chicken?

 

Similar concepts apply to those who tell me centrifugal force doens't exist, by the way.

 

OK One link

 

Joker

[Guest pokey]

Wow,

 

Jon, it seems like you opened a can of worms here! I told you this was difficult to explain.

 

Why did the chicken cross the road?

Aristotle:

It is the nature of chickens to cross the road.

Newton:

Because no force caused the chicken's state of uniform road crossing to change.

Einstein:

Is the chicken crossing the road or is the road moving under the chicken?

 

 

OK One link

 

Joker

 

can of worms indeed !

 

as far as the chicken crossing the road? "do we have an eye witness who actually saw this said chicken crossing alleged road?"

 

good link joker,,,

 

 

BTW,,, i remember from my early fixed wing learning days "pitch controls airspeed, power controls altitude" ALOT of fixed wing pilots i fly with when i ask them that question as we are flying & pressure them for a "quick answer" initially dis-agree. (relevant to this debate/thread?--maybe)

[joker]

Pokey,

 

Your old pilot's saying is explained here:

 

The four forces in straight and level, climbs, and descents

 

This is a subpage of:

 

Forces and Reference Frames

 

Yes, I suppose it is all related!

 

Joker

[HelliBoy]

Come on guys!!! these are trade secrets for us CFIs!

I was planning on charging jon $40 an hour to explain all this!

[Eric Hunt]

You are plucking your physics out of your @sses and making statements that are totally wrong. Not really much point in telling you the truth, you don't listen.

 

Simply put, g = L/W

 

At 1g, L=W.

 

Changing altitude at a constant rate of climb is NOT accelerating, so there are NO unbalanced forces. L=W.If L>W, you will accelerate, and ROC will be steadily increasing. If you did that for a 2-minute climb, your ROC would be off the stops at 10,000 feet per minute.

 

Doesn't happen.

 

Anyway, nice to talk to the kindergarten class, I have some real pilots to talk to.

[Witch]

Bunch of freakin' mathemeticians.

 

AAAAAAAAAAAAAA NO MORE MATH I CAN'T TAKE IT ANY MORE

 

(melts into a puddle)

[67november]

I did this 20+ years ago and my brain still hurts

[joker]

OK,

 

Let's get this clear now, before Mr. Hunt loses his head!

 

Eric is totally correct for steady state flight. The forces are balanced. That's obvious.

 

I presumed that the arguement was one of 'reference frames' without reading the thread in detail. If it is not, then I apologise.

 

Joker

[Linc]

Just told the Standardization Pilot today what you've been saying and he about had a conniption. He wanted to know if you were a pilot, then he asked to make sure and asked if you were a real pilot. I told him the truth, I didn't know.

 

You know, while I'm scribbling with crayons and sniffing the markers and eating the paste, I'm describing some of my thought processes and trying to describe where I'm coming from. I told you I work without a net. I am not a physics guy, I have a diploma not a degree. I try. I'm still pretty sure that there is no equilibrium in a climb. Some force is exceeding the opposite force. If it isn't called an acceleration, fine, but hundreds of pilots have been learning and teaching it that way for years. It must be to the eternal consternation of all the BS holders out there who learned physics in the vacuum and frictionless environment of the university classroom.

 

Oh, wait. You went off to discuss things with real pilots. Disregard all after breaking squelch.

 

On the other hand, I just had a blast flying in the season's first snow, under NVGs.

[Eric Hunt]

" but hundreds of pilots have been learning and teaching it that way for years."

 

The blind leading the blind, brand-new pilots trying to teach learner pilots. A sad system, but that's the way it works in your country.

 

And get your standards man to go back to his books.

 

I'm not trying to get all stand-uppity here. The basics of physics and aerodynamics are perzackly as stated, and if you don't want to believe it, it is your loss. If you are an instructor, it is your students' loss too, because they will try to pass on the erroneous knowledge that you have given them, and swear it is true, because you told them so. And you believe it because your 125-hour instructor told you.

 

Now ..... Mmmmmmmwahh! (We just kissed and made up.)

 

Any more fun questions out there? Anyone for a Hookes Joint Effect? Or why the rate of roll is higher with altitude? The Thick Air Theory?

[Linc]

The question remains, are you a pilot?

 

LMAO! Another couple of my fellow pilots that I spoke with today recommended that you go back to your university and ask for your money back. Keep in mind, I'm probably the least educated of the group. Perchance I'm not conveying your message correctly. In a steady climb, L=W, right?

 

I would say, that for having majored in physics, you sure seem to lack the ability to communicate the message of it. I mean, if it was true, like 1+1=2 true, it wouldn't have to come down to whether or not I believe it. This "whether or not you believe it" bs is like trying to force me to believe a religion. If it requires my belief than the "instructor" hasn't done the job right, or more likely, at all. Are you sure that you understand physics?

 

I can only grin, expecting the kind of response that is going to bring.

[Linc]

"In straight-and-level, unaccelerated forward flight, lift equals weight and thrust equals drag (straight-and-level flight is flight with a constant heading and at a constant altitude). If lift exceeds weight, the helicopter climbs; if lift is less than weight, the helicopter descends. If thrust exceeds drag, the helicopter speeds up; if thrust is less than drag, it slows down."

--FAA Rotorcraft Flying Handbook

How is it that even the FAA hasn't heard about your development regarding lift and weight in climbs and descents? And, you know, I didn't learn from the FAA. More blind leading the blind. I think you need to publish a paper and make sure that everyone learns exactly what you know about physics!

[Eric Hunt]

So, is that all the FAA says about it? There is a little bit - just a little - that they have left out.

 

Let's look at a part of it.

 

"If thrust exceeds drag, the helicopter speeds up". OK, I add a poofteenth of thrust. How much will it speed up? They left that bit out. Will the speed keep increasing to Vne or the speed of light? Or will it only accelerate until the drag from the higher speed balances the extra power added? A poofteenth of power only gives the square root of a poofteenth increase in airspeed. Sure, the initial result of extra thrust is an increase in speed, this is the acceleration caused by the unbalanced increase in thrust. Then the drag balances it out, and you have a new steady speed, and you are using more power.

 

"if thrust is less than drag, it slows down". True. But how much? They left that bit out too. Reduce the power by a smidgin (less than a poofteenth - sorry to use these technical terms on you like this) and it screeches to a stop? No, as you slow down, the drag (proportional to the square of the speed) decreases until it matches the power setting. Forces balance, no deceleration, steady speed.

 

"If lift exceeds weight, the helicopter climbs". Of course it does. It will accelerate upwards, until the extra forces generated from the downwards drag of the airflow adds to the weight, and balances the added vertical thrust, along with other aerodynamic effects such as the increased inflow already spoken about days ago. As another poster correctly stated, lift+thrust=weight+drag. You are then in a steady climb at a steady speed. All acceleration has stopped. No unbalanced forces. Your extra power is adding to your altitude, increasing the Potential Energy.

 

And so on.

 

Blah blah. State this stuff till the cows come home. You won't see the light. Save my bandwidth.

 

And yes, been a pilot for 39 years, same pilots course as Pontius. He understood The Balance of Forces too.

[Linc]

Ahh, now I know where you screwed it up. Thank whatever Supreme Being any may believe in that you weren't my instructor and that I would have to live under the burden of such misunderstanding!

 

It's easy for you to prevent me from wasting your bandwidth, stop responding to my obviously goading responses.

[2rst1]

With the collective all the way down the blades have negative pitch

 

Anyone saying that blades are at negative pitch during an auto, should probably stay on the ground until they've gone over the aerodynamics.

[Sebas]

I have one more question, that just baffles me.

 

If no one actually sees the chicken then how do we know that he even crossed the road?

Maybe he was already on the other side.

[Guest pokey]

I have one more question, that just baffles me.

 

If no one actually sees the chicken then how do we know that he even crossed the road?

Maybe he was already on the other side.

 

Here is even ONE more possibility, maybe the chicken has a twin brother & it is HE who is on the other side?

[Sebas]

Here is even ONE more possibility, maybe the chicken has a twin brother & it is HE who is on the other side?

 

Now I'm really confused.

I also just realized maybe the twin is on this side and the original chicken is on the other side.

I guess I will stick to flying, its much easier when we refer to it as East & West.

[me shakes fist]

Disclaimer: I'm a student pilot with only 20+ hours and I don't have a degree in physics or anything to that effect, but I wanted to add to this as it seems to be fairly simple physics and this is how I understand it.

 

Unless I missed it, I didn't see anyone mention Newton's first law. "An object in motion will remain in motion unless acted upon by an external and unbalanced force."

 

Lets say you are in a climb at a steady 500ft/min. You are at a constant vertical speed which means all forces must be balanced. The lift being produced must be equal to the weight of the helicopter. The example of a car going from 40mph to 50mph cannot be compared to a helicopter in a steady climb. The car going from 40mph to 50mph is an example of acceleration, which is the change in velocity (and velocity is the change in distance). The helicopter in a steady climb is a case of velocity, not acceleration. A car moving at a steady 50mph could be compared to a helicopter in a steady climb.

 

A theoretical example that might help to clear things up (or confuse it more, haha): 2 objects in a zero gravity environment (space) - one is stationary and the other is moving away at 100 ft/min. Now lets say you can "turn on" gravity on the stationary object which would give the object in motion a weight of 5lbs. At the same time you give the object in motion 5lbs of thrust in the opposite direction (rockets, if that helps). Those two forces cancel out and the object continues at 100 ft/min - The object is continuing its steady climb while the lifting force is equal to its weight. Am I correct in this example? If I am right, it should mean the same for a helicopter in a steady climb.

 

To start or arrest a climb or decent the lifting force will not be equal to the weight, but in a steady climb they should be the same.