3
$\begingroup$

I’ve noticed on several graphs in the “Advanced Pilot’s Flight Manual”, that power available is lower for any given speed at 10,000 feet as compared to sea level flight. The power required curves intersect at the sea level best rate-of-climb (BROC) speed. Can someone briefly explain why more power is required at 10,000 feet when below this BROC speed and less power is required at 10,000 feet when above it?

enter image description here

$\endgroup$
5
  • $\begingroup$ "Why is that?" -- why is what? There are at least three factual statements in your post, depending on how one parses it. Is there one in particular you are asking about? Are you asking for a broad discussion of the whole concept of Vy? Please edit your question to clarify what it is your question actually is. $\endgroup$ – Peter Duniho Jul 25 '20 at 2:31
  • 1
    $\begingroup$ Aaron, thanks for your input. I actually just finished CPL and started instructor ratings, and as I go I find there are more and more holes in my knowledge. When i was trained all I got was “more power”, “more nose down”. Not much theory during PPL/CPL. All the questions I have are the ones he sent me to find in the books and many more, most I find in “FTGU” or “Advanced pilot” or online, but some I don’t.. $\endgroup$ – leha007 Jul 25 '20 at 4:13
  • 1
    $\begingroup$ @leha007: Oh, I'm sorry! I thought you had just finished PPL. Even more congratulations are in order! I can totally relate to what you are saying. When I first started helicopter training years ago, I asked my CFI if a certain book would be a good source of information. He told me that he'd seen it before and it was "way too much physics and math". He suggested that I find things with simpler, more functional information... I kept the book and changed the CFI. $\endgroup$ – Aaron Holmes Jul 25 '20 at 4:26
  • 1
    $\begingroup$ Aaron yeah, most CFI’s are like that. I got the one that likes the theory. :) that’s good though, I like learning. $\endgroup$ – leha007 Jul 25 '20 at 4:43
  • 1
    $\begingroup$ You may be interesting about sections about the powercurve of How it flies $\endgroup$ – Manu H Jul 26 '20 at 8:34
3
$\begingroup$

Power required is drag times speed. In essence, you ask why drag at the same lift and in 10,000 ft is growing either side of the best climb speed at sea level.

As @Kolom rightly points out, the coincidence of best climb speed at sea level being equal to lowest drag speed at 10,000 ft is just that, a coincidence. There is no causal relation.

The power required curve is shifted right due to lower density at 10,000 ft which requires the aircraft to fly faster for the same dynamic pressure. The X-axis of the plot is True Airspeed, so all polar points will shift right as altitude increases. Next, you will also note a small upward shift. This is caused by the lower Reynolds number at higher altitude which increases zero-lift drag at the same indicated speed.

The best rate of climb speed is where the local slopes of the power required and the power available curves match. Then the distance between both power curves is greatest and leaves the highest specific excess power after the power required for steady flight has been subtracted from the available power. If the curves would be plotted over Indicated Airspeed, they would simply shift slightly up for the power required and down for the power available as density drops with increasing altitude. The author of your book chose to plot them over True Airspeed, so it might seem that the best ROC at sea level has some significance for the drag at 10,000 ft. It hasn't.

Please make sure you read the linked answers, too, because they should help to give more background. I have kept the answer short, so some things are not fully explained. For example, why there is a minimum drag point at all is explained at the target of the first link.

$\endgroup$
1
$\begingroup$

There is less drag on the aircraft at 10,000 feet. This means that the prop can drive the aircraft along at 120kts while moving less air than it would require at sea level. The trade-off is, when the plane is flying slowly at 10,000 feet, with a lot of drag due to the high angle of attack, the prop struggles to move enough air due to the low density of air molecules at altitude.

$\endgroup$
1
  • 1
    $\begingroup$ Thank you Aaron. $\endgroup$ – leha007 Jul 25 '20 at 4:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.