Angle v Rate of climb
Thread Starter
Angle v Rate of climb
Hey guys,
I'm doing a practice exam for ATPL P&L and I've come accross a couple of question that have me confused. The answers oppose how I understand the concepts. The offending questions are (with the books answers in bold
A turojet aircraft L/D max speed is at x maximum rate of climb speed is:
X
greater than X
Less than X
A turojet aircraft L/D max speed is at x maximum angle of climb speed is:
X
greater than X
Less than X
My understanding, from my AFT notes (not the same book the questions come from) is that ANGLE is at min drag and rate would be faster than min drag. Am I back to front or are these questions???
I'm quite confused! Can anyone help me out?
Cheers.
I'm doing a practice exam for ATPL P&L and I've come accross a couple of question that have me confused. The answers oppose how I understand the concepts. The offending questions are (with the books answers in bold
A turojet aircraft L/D max speed is at x maximum rate of climb speed is:
X
greater than X
Less than X
A turojet aircraft L/D max speed is at x maximum angle of climb speed is:
X
greater than X
Less than X
My understanding, from my AFT notes (not the same book the questions come from) is that ANGLE is at min drag and rate would be faster than min drag. Am I back to front or are these questions???
I'm quite confused! Can anyone help me out?
Cheers.
Last edited by Username here; 9th May 2013 at 11:37.
Max Angle = max excess thrust over thrust required for level flight
Max Rate = Max excess power over power required for level flight.
Min thrust = min drag
"Power = Thrust * velocity
so whenever you know the thrust of an airplane, you can figure out how much power it's developing by multiplying that number by how fast the airplane is going. Thrust is the more fundamental concept, because without thrust, you have no power.
One way I think about it is that power incorporates not only the concept of thrust, but also the velocity that the thrust produces. This is useful in certain contexts, because some aircraft performance parameters depend on that velocity.
For instance, your angle of climb depends on how much excess thrust you have. The velocity of the aircraft along that flight path is irrelevant. If my climb angle is 45 degrees, it remains 45 degrees no matter how fast I'm flying.
However, rate of climb depends on power. If my climb angle is 45 degrees, the faster I fly along that flight path, the greater my rate of climb is".
Max Rate = Max excess power over power required for level flight.
Min thrust = min drag
"Power = Thrust * velocity
so whenever you know the thrust of an airplane, you can figure out how much power it's developing by multiplying that number by how fast the airplane is going. Thrust is the more fundamental concept, because without thrust, you have no power.
One way I think about it is that power incorporates not only the concept of thrust, but also the velocity that the thrust produces. This is useful in certain contexts, because some aircraft performance parameters depend on that velocity.
For instance, your angle of climb depends on how much excess thrust you have. The velocity of the aircraft along that flight path is irrelevant. If my climb angle is 45 degrees, it remains 45 degrees no matter how fast I'm flying.
However, rate of climb depends on power. If my climb angle is 45 degrees, the faster I fly along that flight path, the greater my rate of climb is".
Last edited by RENURPP; 10th May 2013 at 00:55.
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I don't understand the sample questions, as they don't make grammatical sense as written. However, your understanding is correct.
Taking RENURPP's remark and advancing it one small step:
Turbojet thrust is almost constant with speed. Therefore between thrust & drag, the only variable is drag. Therefore best angle, max excess thrust, occurs at minimum drag speed, which is the same as speed for best L/D ratio.
Turbojet power increases with speed (power is thrust, an almost-constant, multiplied by speed). Therefore best rate, max excess power, occurs at a speed substantially greater than minimum drag speed suitable for best angle.
Taking RENURPP's remark and advancing it one small step:
Turbojet thrust is almost constant with speed. Therefore between thrust & drag, the only variable is drag. Therefore best angle, max excess thrust, occurs at minimum drag speed, which is the same as speed for best L/D ratio.
Turbojet power increases with speed (power is thrust, an almost-constant, multiplied by speed). Therefore best rate, max excess power, occurs at a speed substantially greater than minimum drag speed suitable for best angle.
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My understanding, from my AFT notes (not the same book the questions come from) is that ANGLE is at min drag and rate would be faster than min drag. Am I back to front or are these questions???
I'm quite confused! Can anyone help me out?
I'm quite confused! Can anyone help me out?
You are right about the Best AOC speed. Yes, it is the speed corresponding to the lowest point on the drag curve aka the best L/D speed or the VMD.
The best ROC speed depends on the excess power available and varies with altitude. Theoretically, it is generally higher than the best AOC speed at sea level and at height, as the thrust reduces, it might drop below the best AOC speed.
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as the thrust reduces, it might drop below the best AOC speed.