climb performance

[sky7]

example:

In a CRJ200 (horrible climb perf)
standard day, clear and a million
normal lapse rates, winds calm, max gross

In the climb cleared to 10,000ft. You level at 10 and accelerate to 330. Then cleared to fl230. do you:

option1: roll the vertical speed up to 3.5ish and bleed the speed quickly down to your 250kt climb profile, therefore gaining altitude quicky up to about 16,000

or,

option2: roll the VS to about 1.5 and bleed off your speed slowly all the way up to fl230.


I know everybody has their own preference and I know atc would like best rate all the time, but what I'm really interested in is a geeky physics-type, factual answer as to which is more efficient. efficiency meaning climb perf, not fuel efficiency.

 

[tgrayson]

I know everybody has their own preference and I know atc would like best rate all the time, but what I'm really interested in is a geeky physics-type, factual answer as to which is more efficient. efficiency meaning climb perf, not fuel efficiency.

There are two ways to climb:

1. Zooming, converting kinetic energy (airspeed) into potential (altitude), and
2. Steady climb, where your fuel is being converted to altitude

The speed at which the most fuel energy is converted into altitude is called Vy. At any other speed, the fuel that could be used to increase altitude is instead being used to produce the power needed to overcome the drag at that particular airspeed.

Therefore, any moment you spend at an airspeed other than Vy produces wasted power and thus wasted climb performance. The altitude produced by the zooming should be pretty much the same in both situation, since the airspeed change is the same.

Based on the above, I'd say it's more efficient to decelerate to your climb airspeed as soon as possible, assuming the 250 knots is actually closer to the best rate of climb.

 

[Minuteman]

Based on the above, I'd say it's more efficient to decelerate to your climb airspeed as soon as possible, assuming the 250 knots is actually closer to the best rate of climb.

:yeahthat:

Convert the kinetic energy (airspeed) into potential (altitude) as efficiently as possible. Just like a continuous descent approach, this will be done with a changing airspeed (that is probably a little higher than the clean-config holding speeds for that altitude/weight). 250 knots might be a decent approximation between 10,000 ft and FL230.

Because the fuel flow rates are lower at high altitude, I would also guess that you're better off getting to a higher altitude quickly.

The (minutes x fuel flow) are going to tick off either way, and you get paid by the hour and are cheap compared to fuel, right?

 

[bhh1128]

The altitude produced by the zooming should be pretty much the same in both situation, since the airspeed change is the same.

It looks like this is the real substance of the question. I may be missing something important here, but it seems like this reasoning would only be valid if excess power were independent of speed, which of course isn't the case.

Yes, the amount of kinetic energy that is exchanged for potential energy is the same in each case, but the rate at which work is done BY THRUST to further increase the potential energy should not be the same in each case.

With the fast zoom, you minimize the time spent at speeds where excess power is small, and you more promptly maximize the excess power.

(I look forward to hearing what I might be missing here. This is an interesting question.)

 

[tgrayson]

(I look forward to hearing what I might be missing here. This is an interesting question.)

You're not missing anything. In fact, you restated my answer.

You said:

With the fast zoom, you minimize the time spent at speeds where excess power is small, and you more promptly maximize the excess power.

and I said:

I'd say it's more efficient to decelerate [fast zoom] to your climb airspeed [maximize excess power] as soon as possible , assuming the 250 knots is actually closer to the best rate of climb.

 

[bhh1128]

Ha. For some reason I completely missed that line of your answer.