stalling speed remains same in terms of IAS for the same config, but in terms TAS, stall speed increases as altitude increases

???
 

1) What is Aerodynamic Centre (AC) ?

2) At 0* angle of attack, what will be the pressure differential around the wings i.e. will it be no pressure diff OR will it be lower pressure above the wings??

3) Minimum no. of satellites for GPS position and RAIM ??
Is it 4/5 ?

The Aerodynamic Centre is that point around which the net result of the different pitching moments ( Thrust-Drag moment and Lift-Weight moment) will act.

at zero degrees, the pressure on top of the wing will still be lower than the pressure below on a normal aerofoil(well chambered thick aerofoil).....

The pressure will be the same on the top and the bottom at zero degrees if it is a symetrical aerofoil....

and Min. No of satellites for a GPS reading is 4. U need 5 for RAIM.





and wrt to the question regarding stall speed,


guys....
with regard to the question on stall speed,

the formula for lift is : LIFT= (C.L) * {(0.5) * (rho) * (TAS) * (TAS)}
where {(0.5) * (rho) * (TAS) * (TAS)} is the dynamic pressure, which gives the indicated airspeed.

So, from the above equation, when we equate for {(0.5) * (rho) * (TAS) * (TAS)} , it will be : (dynamic pressure) = LIFT/(C.L)

and stall occurs when the coefficient of lift is maximum.
Hence, during stall, (dynamic pressure) =LIFT / (C.L max)

and since C.L max is constant for an aerofoil and the load factor is directly proportional to the Lift produced during straight and level flight, if the load factor acting on an A/c is the same, it will always stall at the same IAS.

Also, when (C.L) = (C.L max) , that is the AoA at which the A/c will always stall irrespective of the Altitude or the speed. { Please note that whenever most of the books say speed, they mean the TAS (v) }

Now , from the very first equation , if we equate for the speed {i.e TAS (v)}, we get

(TAS) * (TAS) {or (v squared)} = [{LIFT}/ {(C.L max) * (0.5) * (rho)}]



hence TAS {or (v)} will be equal to the square root of the expression on the right hand side of the above equation.

From the derived equations for the TAS {or (v)} and the IAS , we see that density (rho) is involved only in the equation,

So.... Under constant load factor, the IAS at which an A/c will stall will always remain constant, but the TAS will vary inversely with density

i.e , the A/c will stall at a higher TAS at higher altitudes, since density is lower.


Hope this helps .
Best o Luck to all u guys....

@mike_sierra

Any book on principles of aerodynamics will have the answer if you are looking for the answer.

let me try:

1. The torques or moments acting on an airfoil moving through a fluid can be accounted for by the net lift applied at some point on the foil, and a separate net pitching moment about that point whose magnitude varies with the choice of where the lift is chosen to be applied. The aerodynamic center is the point at which the pitching moment coefficient for the airfoil does not vary with lift coefficient i.e. angle of attack, so this choice makes analysis simpler. Aerodynamic center - Wikipedia, the free encyclopedia

Whats more important is where is it located?
AC is located approximately 25% of the chord from the leading edge of the aerofoil.

2. at 0 AOA symmetric aerofoils doesnt produce any lift as there is nil pressure difference. But in non symmetric aerofoils, lift is being produced as the wings are cambered.

3. for 2D - 3 satellites and 3D - 4

hi
 

hi ..just wanted to know that for the Jet Airways exam....can spatial reasoning...cubes view in 3 dimension have multiple answers....or only one answer ?

yes they can have multiple answers.... especially in cubes/folding type reasoning

regarding the stall speed discussion ...

by how much will the stall speed increase if the a/c banks at 45deg??
since the load factor will increase - but how do we calc. this?

Also,

questions regarding 1:60 Rule ...
will they be asked as there are no calculators allowed?
guys with prior IndiGo and Jet exam experience please shed some light?



could someone please explain turning errors of the compass -
Q) the compass will overread as it is turning thru
E W N or S???

Also,
where can i read DGCA standard Basic Navigation online ?
-earth, departure, relative motion, payload, time, 1n60, maps n charts??

thanks

Q) What is the lift-weight relation in an unaccelerated climb ?

weight is greater than lift and the CoG moves forward in a unaccelerated climb

if m not wrong some basic electricity que. were also asked.....

is there any1 who appeared last time and can tell smth abt electricity part??:confused:

also, how much human factors is given importance..!

by how much will the stall speed increase if the a/c banks at 45deg??
-At 45* bank, the load factor is 1.41, so the IAS will be sq root of the load factor X IAS..

questions regarding 1:60 Rule ...
will they be asked as there are no calculators allowed?
-The formulas and calculations are relatively simple since they do not involve much calculator work..

could someone please explain turning errors of the compass -
Turning errors are caused because of the magnetic dip phenomena.They affect the readings on N and S headings. In the NH, while turning through north from the equator, the compass trends to under-read, so you need to roll out early. Similarly in NH, while turning through south, it tends to over-read
Remember:
Undershoot North Overshoot South

At 0 AOA, lift is produced, which is equivalent to the weight of the a/c... else a/c would sink.. since lift is produced, the upper surface will hv lower preassure than the lower..

stall speed variation with bank angle
 

@ azax

The formula for variation of stall speed with bank angle is :
Vs (new) = Vs(old) * root of [ 1/ cos{phi}] ,

where {phi} is the bank angle....
in your case... it will be 45 degrees.

so

Vs(new) = Vs(old) * root of [ 1/ cos (45 dergees)]
=> Vs(new)=Vs(old)* root of [1/ 0.707]

=> Vs(new) = VS(old) * 1.19

=> Vs(new) / Vs(old) = 1.19

so the stall speed increases by 19% for a bank angle of 45 degrees.


And the compass errors, remember UNOS for the northern hemisphere turning error.

So to attain u're required heading in the northern hemisphere, you will have to undershoot any heading on the northern part of the compass, (you can also say the compass will over-read, but i'm not sure if that is correct. People generally refer to it as over-shooting) (i.e 271 to 089) and u will have to undershoot (again... u can call it under-reading) any heading that is on the southern part of the compass (i.e 091 to 269).

but there is no turning on easterly (090) or westerly (270) headings on both hemispheres.

n i hv no idea abt 1:60 rule questions, but since calculators are not allowed, i think u can expect a couple questions from that topic.

Good Luck :ok:

nopes... generally lift produced is equal to weight at small positive angles of attack ( ideally about + 4 dergees)... Thats y most commercial aircrafts cruise at a small positive AoA, which will be the most efficient angle of attack as well. Angles of attack... at 0 degrees AoA, the lift produced will be less than the weight and the A/c will sink.

At steady climb all forces should be same and balanced so lift and weight will act same in an unaccelerated climb.

i agree.... in an un-accelerated climb, the lift will be equal to the weight.

no the lift is marginally less than the weight in a steady climb ...
the flight is definitely in equilibrium but thats because a vertical component of the thrust vector balances out the excess weight component

i'm sorry... i stand corrected... the sum of the lift and the vertical component of thrust put together will be equal to the weight.

Increase in the Stall Speed at high Altitudes
 

Hello..

First of all I must say that, the Explainations done in this thread till now are very well done! Applaud to all of you'll!

Now coming to the question that our friend asked here,
'The Stall Speed at 8000feet is 45kts, what would it be at 20000feet?'

At such high altitudes there is one important effect that acts on the wing, which is the very well known Mach No. compressibility effect. At such altitudes, the actual equivalent airspeed(EAS) increases because this effect disturbs the pressure pattern and increases the 'effective weight' on the wing, resulting in a higher EAS stall speed.
Hence, it is the EAS(which is IAS corrected for compressibility and position /instrument error) which increases.


Cheers!

So in a descent, is lift more than weight?

Also it could be also said that the TOTAL (if you can capitalize, so can I) lifting component (lift and excess thrust vectors acting together) is equal to weight. If you want to get technical about it, then aren't we both correct?

Anyway, if you and your aerodynamicist friends want to look at it that way, that's fine with me But i'll stick to what the book says and let the scientist worry about the accuracy of it. When I'm crossing the FAF, I'm not worried about whether lift is more or less than weight.

Anyway i don't wanna offend you and don't want to go really that deep into the matter but just FYI i think that you're little confused about the components of the force vectors. The magnitude of the vertical component of the lift vector is less than the magnitude of the vertical component of the weight vector, because the weight vector is at an angle to the lift vector. The magnitude of lift and weight are the same.

The magnitude of the vertical component of thrust plus the magnitude of the vertical component of lift is equal to the magnitude of the vertical component of weight plus the magnitude of the vertical component of drag.