# ATPL theory questions

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Regarding this 480kt vs 450kt wierdness:

https://i.gyazo.com/0401a1f0cda2db36...7f0ffe4008.png

I put the question's numbers in the system, and TAS is indeed given as 480kt.

And, if you compute EAS/CAS, you will find 0.94.

So, the best I can come up with to explain the whole situation is that they made a mistake in thinking the compressibility correction should be applied to the TAS whereas it should be applied to the CAS only.

A situation where the correct answer would be 450kt is the following :

Mach number is measured from (Pt-Ps)/Ps.

So, say you don't have any device to correct your machmeter for compressibility (which is a problem in itself.. but let's assume that).

The machmeter will overread, due to compressibility.

Let's see how it will overread in details :

You are flying at 450 KTAS (but don't know it yet)

You measure Pt, and find you're flying at 265 KCAS (and 252KEAS in reality)

However, since you're measuring a Pt that's higher than the real one, you're actually not accounting for compressibility, and whereas your EAS is 252kt, you mistakenly believe you have an EAS of 265kt

When you correct for density to get TAS, you compute TAS from an EAS that's higher than the real one, so TAS is higher than your real TAS, close to 480kt.

Then you compute your mach number and find a number close to 0.8.

So you believe you're flying at M0.8

Your machmeter measures M0.8, but it actually overread due to non accounting of compressibility.

At this precise moment, you're in your imaginary aircraft with your wierdo-non-corrected-machmeter and an EASA civil servant steps into the cockpit and asserts the compressibility factor is 0.94. Darn, your non-corrected-machmeter overestimated speed by a factor of 6% !

So you're flying at M.75, which is exactly 450kt.

I reckon it is very very far fetched and I don't know what I would answer if this question did come up.

I don't even know if a non corrected machmeter exists...

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https://i.gyazo.com/0401a1f0cda2db36...7f0ffe4008.png

I put the question's numbers in the system, and TAS is indeed given as 480kt.

And, if you compute EAS/CAS, you will find 0.94.

So, the best I can come up with to explain the whole situation is that they made a mistake in thinking the compressibility correction should be applied to the TAS whereas it should be applied to the CAS only.

A situation where the correct answer would be 450kt is the following :

Mach number is measured from (Pt-Ps)/Ps.

So, say you don't have any device to correct your machmeter for compressibility (which is a problem in itself.. but let's assume that).

The machmeter will overread, due to compressibility.

Let's see how it will overread in details :

You are flying at 450 KTAS (but don't know it yet)

You measure Pt, and find you're flying at 265 KCAS (and 252KEAS in reality)

However, since you're measuring a Pt that's higher than the real one, you're actually not accounting for compressibility, and whereas your EAS is 252kt, you mistakenly believe you have an EAS of 265kt

When you correct for density to get TAS, you compute TAS from an EAS that's higher than the real one, so TAS is higher than your real TAS, close to 480kt.

Then you compute your mach number and find a number close to 0.8.

So you believe you're flying at M0.8

Your machmeter measures M0.8, but it actually overread due to non accounting of compressibility.

At this precise moment, you're in your imaginary aircraft with your wierdo-non-corrected-machmeter and an EASA civil servant steps into the cockpit and asserts the compressibility factor is 0.94. Darn, your non-corrected-machmeter overestimated speed by a factor of 6% !

So you're flying at M.75, which is exactly 450kt.

I reckon it is very very far fetched and I don't know what I would answer if this question did come up.

I don't even know if a non corrected machmeter exists...

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It is a mistake because if you use the mach index, you are reading TAS directly off the flight computer, and the compressibilty calculation is to give you TAS anyway.

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And, if you compute EAS/CAS, you will find 0.94.

*f*/

*f_0*using equations 11, 16, and 20, in NACA Report 837 (Aiken, W., 1946). The ratio depends only on pressure altitude and Mach number and is therefore independent of temperature deviation.

Mach number is measured from (Pt-Ps)/Ps.

So, say you don't have any device to correct your machmeter for compressibility (which is a problem in itself.. but let's assume that).

The machmeter will overread, due to compressibility.

So, say you don't have any device to correct your machmeter for compressibility (which is a problem in itself.. but let's assume that).

The machmeter will overread, due to compressibility.

*(Pt - Ps) / Ps*` will result in an error. The Mach number can be related to the normalised pressure rise `

*(Pt - Ps) / Ps*` using a simple algebraic expression without needing to adopt an incompressible flow assumption. The applicable technical standard for Mach meters (SAE AS8018 citing NASA Technical Note D-822) assumes an isentropic flow in relating pressure rise to Mach number. The same is true for airspeed indicators, although in their case the impact pressure

*q_c*–the pressure rise–is normalised by standard sea-level pressure. Impact pressure (for M ≤ 1) is related to Mach number by:

M^2 = 5*(-1+(1+q_c/p)^(2/7)),

and to true airspeed v by:

v^2 = 7*(p/rho)*(-1+(1+q_c/p)^(2/7)), rho is ambient air density.

Because these equations are based on an isentropic model they account for variation in density as air is brought to rest in a pitot probe. The latter was first given in this form by Saint-Venant and Wantzel in Journal de l'École polytechnique, vol 16, 1839 (according to Stanton [link]).

In an incompressible flow the relationship for true airspeed squared is simply,

v^2 = 2*q/rho,

and

*q*is dynamic pressure which is the difference between total and static pressures in an incompressible flow. This incompressible flow model is not used by Mach meters or airspeed indicators. Dynamic pressure may be thought of as a first order approximation of impact pressure - see for example the paragraph following eqn 2, and particularly figure 2, in NACA Report 247 (Zahm, A., 1926). The inverse relationships for impact and dynamic pressures are,

Isentropic flow: q_c = p*(-1+(1+(1/7)*(rho/p)*v^2)^(7/2)), and,

Incompressible flow: q = (1/2)*rho*v^2.

The Maclaurin expansion of p*(-1+(1+(1/7)*(rho/p)*v^2)^(7/2)) about v^2 = 0 is (1/2)*rho*v^2 which is the Euler-Bernoulli solution to Euler's equation for an incompressible flow.

The first use of a compressibility factor, of which I'm aware, in a circular slide rule for obtaining TAS from CAS, as in eqn 9 in NASA Technical Note D-822 or eqn 20 in NACA Report 837, is described in Ezra Kotcher's (of Bell X-1 fame) US patent 2342674 of 29th Feb 1944 for an Air Speed Computer. For on the influence of errors see NACA Technical Note 1605 (Huston, W., 1948),

*Accuracy of Airspeed Measurements and Flight Calibration Procedures*.

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I've lost the will to live already.....

I've checked with a couple of other senior instructors, and they would expect the answer to be 480 knots if it was encountered in an Instruments exam.

I've checked with a couple of other senior instructors, and they would expect the answer to be 480 knots if it was encountered in an Instruments exam.

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I, for one, am fascinated as selfin's explanation goes some way to explaining something that has worried me for some time. I have always found it difficult to reconcile the two statements:

1. On aircraft with Air Data Computers the ASI displays CAS

2. The Air Data Computer calculates CAS using variations of Saint Venant's formulae which assume compressible flow.

Because if #2 were true, the ADC would output what I know as EAS, CAS corrected for compressibility, although it still seems to be called CAS. Is this the case, selfin?

1. On aircraft with Air Data Computers the ASI displays CAS

2. The Air Data Computer calculates CAS using variations of Saint Venant's formulae which assume compressible flow.

Because if #2 were true, the ADC would output what I know as EAS, CAS corrected for compressibility, although it still seems to be called CAS. Is this the case, selfin?

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There is no Mach number in an incompressible flow because the speed of sound is infinite. Neither Mach meters nor airspeed indicators need to be "corrected for compressibility" so it does not necessarily follow that determining Mach number by a quantity such as `

*(Pt - Ps) / Ps*` will result in an error. The Mach number can be related to the normalised pressure rise `

*(Pt - Ps) / Ps*` using a simple algebraic expression without needing to adopt an incompressible flow assumption. The applicable technical standard for Mach meters

Nothing would prevent you from building a machmeter which would would exactly compute the mach from the following formula :

TAS=sqrt(2(Pt-Ps)/rho)

LSS=sqrt(gamma R T)

Rho=Ps/RT

TAS/LSS = sqrt(2(Pt-Ps)/Ps Gamma)

If you measure a higher Pt due to compressibility then yes you're going to need to correct it.

If you measure a CAS of 280kt at FL330, you're probably gonna want to correct something if you want to know your EAS.

In facts, you have to correct for compressibility

**if you use this above mentionned formula**, simply because the formula is not valid for compressible flow.

For compressible flow, where a machmeter becomes useful, there are better suited formulas that you could use, and then you wouldn't need to correct for compressibility, because you avoided the need for it by choosing a better formula.

Then, I could agree that the answer marked as correct was actually wrong, i'm not in the head of the question's writers...

I was simply trying to find an explanation regarding this, without being able to tell whether I would answer 480 or 450 should this question appear on my official test (can we trust aviation exam or this other question bank when they tell us a surprising correct answer like this ?)

And from what you wrote after, I know you'll understand my message because you've written yourself the "better suited formula" I was talking about

Regarding the little technical details of real macheters, I really have no idea wheter early mach meters were able to mechanically reproduce this sort of complicated formula as written in wikipedia :

https://wikimedia.org/api/rest_v1/me...782ff6024667fa

To begin with I don't understand how you could divide a number by another with a system of mechanical gears and hands, so I don't see how you would elevate this ratio to the power 2/7

It is however very easy for modern ADC to use formulas as complicated as required thanks to IT.

1. On aircraft with Air Data Computers the ASI displays CAS

2. The Air Data Computer calculates CAS using variations of Saint Venant's formulae which assume compressible flow.

2. The Air Data Computer calculates CAS using variations of Saint Venant's formulae which assume compressible flow.

Because if #2 were true, the ADC would output what I know as EAS, CAS corrected for compressibility, although it still seems to be called CAS.

KayPam,

Nothing prevents you from calculating a mach number, from speed and temperature, even in an incompressible flow.

*Sur la Vitesse du Son dans l'air et dans l'eau*, Annales des Chimie et de Physique 3, 238–241 (1816)). An excellent summary of the struggle to resolve this problem is given in

*Laplace and the speed of sound*by Bernard Finn who was, like John D Anderson, a curator at the Smithsonian Institution. Anderson provides a straight-forward derivation in Fundamentals of Aerodynamics, section 8.3, which is very similar to this this two-page derivation plucked liberally from a search for "derivation of the speed of sound."

Nothing would prevent you from building a machmeter which would would exactly compute the mach from the following formula

*Multiplcation of Bananas by Umbrellas*.

If you measure a CAS of ...

*dynamic pressure cannot be measured in flight*, and any incompressible flow solutions using a measured isentropic total pressure will be in error (including the fabled Mach meter you've proposed). The proposal amounts to misinterpreting useful information (impact pressure) as an approximation (dynamic pressure), followed by a "correction" to recover the original information. Such a scheme may work at low speed and low altitude but it is inconsistent and unnecessarily restrictive. Indeed it would have been simpler to have used Saint-Venant's solution directly in the first place to obtain CAS and Mach number, which is what modern ASIs and Mach meters do.

It is however very easy for modern ADC to use formulas as complicated as required thanks to IT.

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Sorry for having overcomplicated things in the following messages...

First of all, I don't know why you're saying air would be uncompressible ? It is very much compressible when you squeeze it into a syringe or any kind of air pump or even a balloon. It also suffers waves of compression when sound travels through it. But even if we admit that LSS would be infinite in an incompressible flow :

Mach number is speed/speed of sound (TAS/LSS)

You're saying LSS is infinite in uncompressible flow... And you're probably saying that at low speeds the air is uncompressible.

Well, problem is sound does take time to travel even in still air. Even in water, a fluid that's almost uncompressible, sound takes time to travel.

So basically we have air at low speed : it's almost uncompressible so we can use Bernoulli's theorem. It is a tiny bit compressible so we can use the sqrt(gamma R T) = LSS formula. If speed increases, the LSS formula becomes "more correct" (according to you, you still have to explain why it would less correct at low speeds), but Bernoulli becomes less correct. And that's where the correction factor is needed.

But in any case, nothing prevents the engineer from performing computations that do not make perfect physical sense. One wants to compute a mach number as TAS/sqrt(gamma R T) ? No one can forbid him to do so.. Maybe tell them they're wrong but nothing more.

This is probably the sort of reasons why EASA ends up asking surprising questions like the one we saw.

It is probably the the same sort of reason that could lead to an EASA question like the one we saw yesterday.

Because except by using wrong formulas, I don't see why this 0.94 thingy would be required.

Correct equations allow us to forget about this kind of correction factor.

I was simply explaining the wrong physical reasoning leading to require a 0.94 correction factor.

The equation you would use was derived for a compressible medium and the only value Mach number can take in an incompressible flow is zero.

Mach number is speed/speed of sound (TAS/LSS)

You're saying LSS is infinite in uncompressible flow... And you're probably saying that at low speeds the air is uncompressible.

Well, problem is sound does take time to travel even in still air. Even in water, a fluid that's almost uncompressible, sound takes time to travel.

So basically we have air at low speed : it's almost uncompressible so we can use Bernoulli's theorem. It is a tiny bit compressible so we can use the sqrt(gamma R T) = LSS formula. If speed increases, the LSS formula becomes "more correct" (according to you, you still have to explain why it would less correct at low speeds), but Bernoulli becomes less correct. And that's where the correction factor is needed.

But in any case, nothing prevents the engineer from performing computations that do not make perfect physical sense. One wants to compute a mach number as TAS/sqrt(gamma R T) ? No one can forbid him to do so.. Maybe tell them they're wrong but nothing more.

This is probably the sort of reasons why EASA ends up asking surprising questions like the one we saw.

Such a scheme may work at low speed and low altitude but it is inconsistent and unnecessarily restrictive.

Because except by using wrong formulas, I don't see why this 0.94 thingy would be required.

Correct equations allow us to forget about this kind of correction factor.

I was simply explaining the wrong physical reasoning leading to require a 0.94 correction factor.

Even in water, a fluid that's almost uncompressible, sound takes time to travel.

*K*is finite. Increasing

*K*results in higher speeds. In other words for an acoustic wave to have a finite travel time the volume of the material must change. The more pressure that is required to effect that change, the higher the speed.

In an incompressible flow (and medium)

*K*is infinite which means the propagation speed is infinite, in turn reducing the Mach number to zero for all speeds.

So basically we have air at low speed : it's almost uncompressible so we can use Bernoulli's theorem. It is a tiny bit compressible so we can use the sqrt(gamma R T) = LSS formula. If speed increases, the LSS formula becomes "more correct" (according to you, you still have to explain why it would less correct at low speeds), but Bernoulli becomes less correct.

But in any case, nothing prevents the engineer from performing computations that do not make perfect physical sense. One wants to compute a mach number as TAS/sqrt(gamma R T) ? No one can forbid him to do so.. Maybe tell them they're wrong but nothing more.

Because except by using wrong formulas, I don't see why this 0.94 thingy would be required.

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I pass 33, 81 and 21 this morning.

However, there was an inconsistency in flight planning.

It said that contingency fuel should be 5% of the trip fuel, in the question.

However, we were given the cruise consumption of the aircraft and I was able to calculate that 5% of the trip fuel corresponded to 3 minutes of cruise consumption.

I thought there was a minimum of 5' fuel consumption for contingency ?

In this case, should we believe the question and consider that contingency is 5%, or apply the rules and consider it's 5' ?

Thanks

However, there was an inconsistency in flight planning.

It said that contingency fuel should be 5% of the trip fuel, in the question.

However, we were given the cruise consumption of the aircraft and I was able to calculate that 5% of the trip fuel corresponded to 3 minutes of cruise consumption.

I thought there was a minimum of 5' fuel consumption for contingency ?

In this case, should we believe the question and consider that contingency is 5%, or apply the rules and consider it's 5' ?

Thanks

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Under AMC1 CAT.OP.MPA.150(b) Fuel policy it's the higher of 5% or 5 minutes at holding speed, so 5% would be correct in your case.

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A jet aeroplane is flying long range cruise. How does specific range / fuel flow change?

A Decrease / decrease

D Increase / increase

A Decrease / decrease

**B Decrease / increase [marked wrong]****C Increase / decrease [marked correct]**D Increase / increase

As time passes the aircraft becomes lighter, so the drag decreases. This increases the specific range and decreases the fuel flow.

The greatest surprise to me is that after all these years some of the questions banks do not explain this.