mach nr vs weight:

[norwegian pilot]

Studying the "Ace the techical pilot interview" book and there is a statement in there that makes me totally confused:

What happens to Mach nr on a long leg when weight decreases and all other factors remain constant. Temperature, Pressure alt (FL) and thrust is constant...

The book say mach nr will decrease in this case........ ??????

If weight decrease- required lift decrease - induced drag decrease. Less drag with constant thrust will give a higher TAS and HIGHER Mach nr.... in my opinion.

Someone show me the way out of this fog..

[barit1]

What happens to Mach nr on a long leg when weight decreases and all other factors remain constant. Temperature, Pressure alt (FL) and thrust is constant...

First a caveat: Whenever the phrase "all other factors remain constant" appears in a test question, BEWARE - the test writer may not have considered the possibility that all other conditions CANNOT remain constant, in which case you have to crawl inside the mind of the writer to deduce what he probably meant.

But in this case, flying in a constant atmospheric mass with decreasing weight, induced drag goes down, therefore Mn rises. (Of course profile drag then rises, limiting the amount of speed increase, but faster just the same...)

What "book" tells you that speed decreases?? In the real world, you'll probably clear to a higher altitude for economy, and might then cruise slower, but that violates the premise of the quiz.

[norwegian pilot]

the book is called" Ace the technical pilot interview" written by Gary W. Bristow.

I agree, that statement hve to be wrong.. Unless anyone in here can prove other.

[Keith.Williams.]

This question is fairly typical of the type of stuff that is dreamed up by the JAR examiners.
I think that the examiner wants you to think about how decreasing weight affects best range cruise speed VMRC.
VMRC is approximately 1.32 VMD. As weight decreases due to fuel burn during flight VMD decreases, so VMRC also decreases.
Ideally you would climb to remain at optimum altitude. But the question specifies constant pressure altitude.
So if you reduce your CAS to match the reducing VMRC, your Mach number will also decrease.

Unfortunately all of the above ignores the "constant thrust" condition. But that is also fairly typical of JAR questions.

[Brain Potter]

I've been reading the same book recently and here's another question that puzzled me:

If you had an engine failure between V1 and Vr and you had maximum crosswind, which engine would it be best to lose - upwind or downwind?

It give the answer is the upwind side - because the crosswind will oppose the yawing moment of the downwind engine.

However, surely the weathercocking effect of the crosswind on the fin will be yawing the aircraft into wind. The net yawing moment in this direction would be increased with an upwind engine failure. Conversely the weathercocking would oppose the yawing moment of a downwind engine failure.

[extreme P]

The book is notorious for being full of errors so use your discretion...

If by "long leg" the author means LRC then I believe the answer to be correct.

[Spaghetti Monster]

The 'constant thrust' bit is completely incorrect. If you are cruising for best range, assuming that you don't change altitude, then your optimum M no will gradually decrease as you become lighter. To put it simply, this is to maintain your optimal angle of attack. But of course to do this, the thrust required will decrease.

[Dani]

correct. And if you fly with a constant cost index you also decrease your Mach with lower weight.

[Lemper]

TO BRAINPOTTER
Good thinking, however, I guess the author had something else in mind; try this one: during take off roll with cross wind, in order to keep the nose wheel on the center light, rudder pressure is applied DOWNWIND counteracting the weathercocking. Should an engine fail between V1 and Vr, on the downwind side, rudder pressure would have to be reversed whereas an engine failure on the upwind side would only necessitate an increase or the already applied rudder pressure.

[Brain Potter]

Lemper,

Yes that is another way of thinking. But if downwind rudder is already applied on the take-off roll to cancel the crosswind - then there would be less rudder available on that side for the pilot to apply to counteract the yaw of an upwind engine failure. Wheras with a downwind engine failure the pilot actually has more than the normal amount of rudder travel available to oppose the swing. I still think that the book is wrong. And as for the explanation of dutch roll which says that the upgoing wing actually stalls...!!

[john_tullamarine]

.. as for the explanation of dutch roll which says that the upgoing wing actually stalls...!!

I really MUST read this gem of a tome one of these days ..

[misd-agin]

Try maximum crosswind engine failures in the sim. It's MUCH easier if the upwind engine is failed. It can be a 'night and day' difference in the tracking and initial climbout.

[john_tullamarine]

.. it's MUCH easier if the upwind engine is failed ..

Tell you what ... next time you're in the sim get the guy in back to give you a min weight, max aft CG, min speed schedule failure just approaching V1 with a max crosswind .. one each for opposite side engines .. then come back and tell us the results ..... (Caveat - if the sim is not well set up for this corner of the envelope the observation may not be particularly useful .. if, however, it is a 737 and has an FAA tick in the box .....).

We shall await your comments ...

Normal high speed failure is a doddle regardless of conditions ..

[Brain Potter]

I can see that upwind engine failures may be "easier" to fly because the swing is in the same direction as the weathercock. However I can't see that the validity in the book's explanation that a moment caused by the crosswind opposes the asymmetic yaw of an upwind failure. My aircraft type has an engine-out ferry crosswind limit of 15 knots from the deadside - which supports the argument that there is less rudder authority available with an upwind failure.