Dense People!!!!

[Jmmccutc]

i'm putting this in the general section for a reason, because the situation is aviation realted.

senario: my one roomate a CFI for 2 weeks now, has a new favorite line, "I'm a F*ing CFI dude, you're wrong!"

can anyone see why this pisses me off? or is it just me cause i realize the kid is stupid as **CENSORED****CENSORED****CENSORED****CENSORED**, can't do his BASIC physics or math homework (i mostly do it for him when he comes beggin for help). the other things that irratates the piss out of me is he basis everything on Kershner (which i'm not denying that there is alot of good stuff in his books) but even if you can show him in another text a differing statement, he's still right because he's a "CFI"...

i needed to vent so there i did it, if you guys got any suggestions on how i might handle this let me know, maybe you've all flown with someone that fits this mold...


oh one other question...if you're in a constant airspeed decent (sticking to the basic 4 forces, lift, weight, thrust, drag) what causes you to decend? an excess of weight? or a negitive angle of attack? thanks

 

[FN FAL]

oh one other question...if you're in a constant airspeed decent (sticking to the basic 4 forces, lift, weight, thrust, drag) what causes you to decend? an excess of weight? or a negative angle of attack? thanks

Lift opposes weight...so wouldn't the answer be a deficit of lift? For example at a REDUCED angle of attack, you'd be generating less lift, so you'd descend.

A negative angle of attack would cause you to have negative G's, wouldn't it?

 

[Jmmccutc]

i don't know...he's right so what's it matter anyway?

 

[FN FAL]

i don't know...he's right so what's it matter anyway?

Hahahaha...here's a question to ask him.

Ask him if it's legal for a current and qualified instrument pilot to fly an airworthy insturment aircraft in IMC and in uncontrolled airspace, while not on an IFR flight plan.

 

[sqwkvfr]

i don't know...he's right so what's it matter anyway?

SEEK MARRIAGE COUNSELING IMMEDIATELY!!

Good god, it sound's like you two are suffering from an excess of wedded bliss!

 

[JetBlast2000]

You should stop bit$hing at him... he's a fuc&ing CFI...

If he's right ask him to show you where he found the info. If you don't believe him, prove him wrong. My students used to challenge me even when they were blatantly wrong. I would just have to find another way to show them what I was talking about. You could always beat him over the head with the book as a last resort...

 

[minitour]

Hahahaha...here's a question to ask him.

Ask him if it's legal for a current and qualified instrument pilot to fly an airworthy insturment aircraft in IMC and in uncontrolled airspace, while not on an IFR flight plan.

Wait wait...I wanna try...

Without checkin the biblel...Imagonnasay.....Yes...


Am I right?

-mini

 

[FN FAL]

Wait wait...I wanna try...

Without checkin the biblel...Imagonnasay.....Yes...


Am I right?

-mini

Yes...but I bet Jmmcctummsnnnssn's roomate doesn't have a clue about it.

 

[OhDannyBoy]

Hahahaha...here's a question to ask him.

Ask him if it's legal for a current and qualified instrument pilot to fly an airworthy insturment aircraft in IMC and in uncontrolled airspace, while not on an IFR flight plan.

and then ask him what to squawk.

 

[mattpilot]

oh one other question...if you're in a constant airspeed decent (sticking to the basic 4 forces, lift, weight, thrust, drag) what causes you to decend? an excess of weight? or a negative angle of attack? thanks

Lift opposes weight...so wouldn't the answer be a deficit of lift? For example at a REDUCED angle of attack, you'd be generating less lift, so you'd descend.

A negative angle of attack would cause you to have negative G's, wouldn't it?

hmm... The aircraft would be in equilibrium thus Lift = weight. Therefore what pulls you downward would be gravity (and/or the prop).

In an unaccelerated climb same applies. Lift = weight. What makes the airplane climb is excess power/thrust.

basically, you point in the direction you want to go (up or down) by rotating around the CG and your excess power (either gravity induced or from the engine) makes you change altitude.

So if everything stays constant, then the AoA would remain the same for a constant weight. Therefore, Lift will also remain the same. As for your negative AoA - uhmm... i suppose that would make you descent - but not quite what your looking for .

As for excess of weight - weight & lift stays constant, assuming all other factors besides altitude stays constant.





source: 'Flight Theory for Pilots' 4th ed. by Charles E. Dole

 

[smokinhole]

Never thought about that!

What the heck would you squawk?

 

[mattpilot]

What the heck would you squawk?

nothing - your in uncontrolled airspace. Who gives a fook? Either 1200 or turn the thing off.

 

[smokinhole]

DOOOOOHHHHHH! Good point

 

[FN FAL]

hmm... The aircraft would be in equilibrium thus Lift = weight. Therefore what pulls you downward would be gravity (and/or the prop).

In an unaccelerated climb same applies. Lift = weight. What makes the airplane climb is excess power/thrust.

basically, you point in the direction you want to go (up or down) by rotating around the CG and your excess power (either gravity induced or from the engine) makes you change altitude.

So if everything stays constant, then the AoA would remain the same for a constant weight. Therefore, Lift will also remain the same. As for your negative AoA - uhmm... i suppose that would make you descent - but not quite what your looking for .

As for excess of weight - weight & lift stays constant, assuming all other factors besides altitude stays constant.





source: 'Flight Theory for Pilots' 4th ed. by Charles E. Dole

I suppose it would be a deficit of thrust then?

 

[mattpilot]

I suppose it would be a deficit of thrust then?

Well, if your in a nose low situation - gravity would accelerate you if you don't change your power setting. So, to keep a constant airspeed you would need to reduce power/thrust. So yes



Assuming things dont stay constant and the pilot does not change power setting and lets the aircraft accelerate during the descent, then the AoA would get reduced. Why? More airflow over wing = more lift. Thus, to equalize lift & weight, AoA would have to get reduced (again, assuming all other things like Rate of descent, etc.. stay constant).

 

[OhDannyBoy]

nothing - your in uncontrolled airspace. Who gives a fook? Either 1200 or turn the thing off.

yeah, but it will shut the guy's roommate up for a little while.


In an unaccelerated descent a portion of thrust is provided by weight/gravity. at constant a/s, constant aoa and with the same surface area of wing (no flaps etc) the lift must remain constant. This, of course, is for the constant a/s descent.

 

[gfvalvo]

Unaccelerated Flight

oh one other question...if you're in a constant airspeed decent (sticking to the basic 4 forces, lift, weight, thrust, drag) what causes you to decend? an excess of weight? or a negitive angle of attack? thanks

In any regime of unaccelerated flight, all forces are in equilibrium. If the plane is in a nose down attitude during the decent, the Lift, Drag, and Thrust vectors will all have both horizontal and vertical components. Gravity only acts in the vertical component. So:



Lift(vertical) + Drag(vertical) = Gravity + Thrust(vertical)



Lift(horizontal) + Thrust(horizontal) = Drag(horizontal)



With all forces in equilibrium airspeed and vertical speed are constant. The plane descends because:



1. Newton's 1st law



2. Gravity is doing Work on it. This is shown by the change in potential energy:

Mgh(1) - Mgh(2)

Where M is mass of plane, g is gravitational constant, h(1) is initial altitude, h(2) is final altitude.



If the Work is done in a specific amount of time (rate of decent), that specifies the Power. As was stated previously, gravity is supplying this Power.



Same situation for a constant airspeed / rate climb, excess Power from the engine is doing the work to increase the plane's potential energy in the gravitational field.

 

[flyifrvfr]

In any regime of unaccelerated flight, all forces are in equilibrium. If the plane is in a nose down attitude during the decent, the Lift, Drag, and Thrust vectors will all have both horizontal and vertical components. Gravity only acts in the vertical component. So:



Lift(vertical) + Drag(vertical) = Gravity + Thrust(vertical)



Lift(horizontal) + Thrust(horizontal) = Drag(horizontal)



With all forces in equilibrium airspeed and vertical speed are constant. The plane descends because:



1. Newton's 1st law



2. Gravity is doing Work on it. This is shown by the change in potential energy:

Mgh(1) - Mgh(2)

Where M is mass of plane, g is gravitational constant, h(1) is initial altitude, h(2) is final altitude.



If the Work is done in a specific amount of time (rate of decent), that specifies the Power. As was stated previously, gravity is supplying this Power.



Same situation for a constant airspeed / rate climb, excess Power from the engine is doing the work to increase the plane's potential energy in the gravitational field.

I'm getting a fu<king headache

 

[uwochris]

In any situation where the airspeed is constant, you must be in equilibrium (ie. NO unbalanced forces). This is clearly true because if there was an unbalanced force there would be an acceleration (i.e. a change in velocity). Again, we are talking about the horizontal components balancing the horizontal components, while the vertical comp. balance the other vertical comps.

In a descent:

horizontal thrust component + weight component along flight axis= drag component along flight axis.

vertical lift component + vertical thrust component= vertical weight component.

Note: Lift (which acts perpendicular to relative wind) is less than weight (which acts from the CG straight down to the earth) in a climb and descent. This occurs because a component of thrust helps to balance them out. However, ALL vertical components balance out, and ALL horizontal components balance out. Lift= Weight and Thrust = Drag are only approximations.

So what causes the descent? It is simply a deficit of power (or thrust). Tell your roomie to look up Kershner Chapt 3 page 34, which states:

Deficit THP * 33000/ weight = rate of sink.

To be technical, for a prop a/c, the rate of sink will be maximized where there is the greatest deficit of power. On the other hand, the angle of descent will be maximized where there is the greatest deficit of thrust. This occurs because power and thrust are different; they are related, but they are NOT the same.

An a/c requires a certain amount of power to remain S&L at any given speed (or AOA). If there is a deficiency of power, the a/c cannot support itself, and a rate of sink (or angle of descent) will result. If you have excess power, the opposite occurs (a positive rate of climb occurs). Excess thrust causes a positive angle of climb. The descent simply occurs in order to balance the forces out.


Chris.

 

[scangadah]

Who cares, I think you two should box and then tell us who wins.

 

[FN FAL]

To be technical, for a prop a/c, the rate of sink will be maximized where there is the greatest deficit of power. On the other hand, the angle of descent will be maximized where there is the greatest deficit of thrust. This occurs because power and thrust are different; they are related, but they are NOT the same.


Chris.

So I can assume that if I am in an un-accelerated steady state descent...and I feather the prop so that I am generating zero thrust and then push up the power lever to obtain maximum horsepower...

Angle of descent will be maximized because of deficit of thrust and rate of sink will be minimized because a lack of a deficit of power?

 

[Cat Driver]

Friction produces heat, so ask him how long it would take a dog to hump a pail of water to a boil.

 

[FN FAL]

Friction produces heat, so ask him how long it would take a dog to hump a pail of water to a boil.

I don't think a dog would hump a pail of water, but a CFI would!

 

[Dubya]

Who cares, I think you two should box and then tell us who wins.

Aaaaahhh screw that...........squeeze one off in his shampoo bottle...then the next day comment over and over how shiney his hair is.

W

 

[flyboy]

Friction produces heat, so ask him how long it would take a dog to hump a pail of water to a boil.

That is absolutely great!!!!!!!

 

[Jmmccutc]

Friction produces heat, so ask him how long it would take a dog to hump a pail of water to a boil.

dude that is priceless...i was laughing so hard i had Natty coming out of my nose...

 

[uwochris]

FN FAL,

I think that that would be true given the condition and configuration of the a/c. There are many variables to consider however: weight, density altitude, flaps, gear, etc.

 

[Mr. Cole]

As someone with a physics background I've spent some time trying to reconcile how the different concepts of excess power, longitudinal stability, and equilibrium come together in answering this question. As I learned as a physics major, there is more than one way to solve a problem, but the different methods must be consistent and give the same answer. One example is using either Newton's Laws to determine the dynamics of a problem or using the energy method. After reading an aerodynamics book or two I think I've been able to piece it together, but I'm putting my idea out for there a sanity check.

As has been mentioned by other posters a plane is in equilibrium when not only the forces balance, but also all moments. For a plane that has positive long stability, a plot of the moment coefficient versus the coef of lift is negative. In other words an increase in CL produces a nose down moment and a reduction in CL produces a nose up moment. This is how the airplane seeks to obtain equilibrium of the moments about the CG. When the forces are in equalibrium then L = W and T = D. We can also assume this in climbs or descents with small angles, but becomes more complicated the steeper the angle of climb or descent. There are two cases I like to consider below, and since the original question involved a descent I addressed that condition.

The first case is when you push the nose over and don't retrim, but simply take your hand off the yoke. Also assume that the angle is not too steep and there is no reduction in power. As the nose is pushed over the CL is reduced, but the velocity doesn't change immediately and is less than that required for level flight at the new CL, so we have a condition where weight is greater than lift and we begin a descent. In order to restore L = W, the velocity will increase to a value that supports L = W at the new CL. If nothing else occurred the plane would descend with a constant rate of descent. However, you also have the long stability that want to restore the former value of CL, and in order to achieve L = W, the original velocity as well. As we ultimately end up where we started, probably after a few oscillations, the power curve isn't really important.

In the second case we either retrim the plane downward or keep holding forward pressure. If we do either of those the plane will not seek to return to old CL and velocity, but will seek the new velocity that corresponds to the new CL and produces L = W. If the plane remains in a descent at that point it will be constant rate. Unlike the first case the plane doesn't return to it's original state because we have either retrimmed or holding the yoke, so now power will matter. If we operate in the region of normal command of the power curve, the increased airspeed will result in a power deficiency and produce a constant rate of descent corresponding to the deficiency. In the region of reverse command the increased velocity will require less power, creating a power surplus. While the transient condition will initially dictate a descent, as the velocity approaches it's new value the lower power required at that velocity will ultimately produce a climb or reduction in the rate of descent.

Dave

 

[mattpilot]

I just got a headache, dave


I agree with you in principle (or i think i do ).... with the exception of this:

As the nose is pushed over the CL is reduced, but the velocity doesn't change immediately and is less than that required for level flight at the new CL, so we have a condition where weight is greater than lift and we begin a descent

CL being the Coefficent of lift?

Why would CL be reduced if you push the nose down with a constant velocity? Assuming no abrupt control inputs, it should stay the same. Remember, The relative wind also changes since your direction of travel changes by pushing the nose down. Thus Lift would still = Weight.

Once weight EXCEEDS Lift, the airplane would stall, wouldn't it? When approaching stall in slow flight, you continouesly increase AoA to obtain enough Lift to counter weight as you slow down. Same in a high speed stall. When you pull to many G's where the Weight factor exceeds the maximum lift produced by the maximum AoA for steady flight, you're wings stall.

Again, as far as i know, Climbs/Descents are due to acceleration in either direction. Either produced by the engine or by gravity. Of course this is for normal flight. You could, afterall, stall a airplane and let it 'drop like a rock' and 'descend' that way



Man.. this is getting confusing.

(please don't smack me to hard if i got myself mixed up )

 

[SYXDude]

Heck, keep it simple...

• Pull back on the yoke and the houses get smaller
• Push forward on the yoke and the houses get bigger

Of course, if you keep pulling back eventually the houses get bigger.

(The Surgeon General has advised this thread is the leading cause of headaches on flightinfo.com)


Later