Lower Speed versus Higher Optimum Altitude
 

I always thought that the B747 FMC calculateda higher optimum altitude as weight decreased. How is it that a higher optimum altitude is actually achieved because the decrease in weight leads to a decrease in speed, and the decrease in speed is the reason why we can get a higher optimum altitude? Thank you!

If I understand you right; The lower speed is because it is based on a constant mach. Mach is a function of temperature.

Both optimum speed and optimum altitude change with weight. Optimum speed is not constant mach, but decreases with fuel burnoff. It is closer to constant angle of attack.

The lower weight does not reduce the speed, however it does increase how high you can climb.

As you climb it gets cooler (until you get to the troposphere, where the temperature remains the same as you climb. at the poles it's low, over the equator it's pretty high) Above 25-30,000 feet aircraft start cruising at mach numbers in stead of knots. As the air cools, mach 1 becomes a lower speed. So, you're right that as you get higher you go slower.

However, higher up there's less air to cause drag. So you need less thrust - quid pro quo - less fuel. The fuel saving due to lower drag penalties outweighs the lower speed.

Tropopause starts at 36,000 Feet,
Cruise altitude for B747 can be 29,000-43,000ft.

At a constant Mach number decreasing temperature is only valid until 36,000 ft.

I think (although I may be wrong as I have no real technical knowledge) that you have 3 variables to think about- speed, weight and altitude. Ideally, you would want to keep the speed and altitude at the optimum numbers, as you can't control the weight. That would give a cruise climb, as you burn off fuel, you're optimum level increases, you slowly climb (assuming that there is a single optimum cruise mach), but because you can't climb continuously, you have to stay at a level and burn off fuel. The FMC makes the best of this by steadily reducing mach number. Once you step-climb to a higher level, the mach increases toward the "best" speed (perhaps faster if above the "optimum" level) then reduces as fuel burns off. At least, that's my understanding of it.

http://www.pprune.org/tech-log/39997...ml#post5416515

This was my answer to a similar question.

It explains the two issues involved.

Jimmygill:
In my ongoing quest to split hairs, I have to point out that the trop is not always at 36,000 feet. It varies constantly from place to place. The average height for the trop is much higher at the equator (up to 58,000) and lower at the poles (around 25,000 feet).

Optimum altidue increase due to reduce in weight and speed,as fuel burns off weight decrease and your 1.32 VMD(which ur minimum drag speed) also decrease so in order to get the best SFC(specific Fuel Consumption)u may climb to ur cruise climb which is ur Optimum altitude at this moment either once or using step climb from the FMS.The idea is to get the best speed for drag,consequently as ur fuel burns off while ur weight is deacreasing climbing to ur optimum altitude help u regain ur best Mach speed and make u save much more fuel than needed.

Hmmm, with that many ur's and u's, it reminds me of Ebonics, just as some of the natives in Oakland once insisted the local schools teach....in place of proper English.:ugh:

Ebonics - A sad sign of the times.

My understanding is that it's an airframe/engine balance. Ideally we'd like to keep the aircraft flying at close to optimum angle of attack (best lift to drag) with the engines operating close to their optimum design rpm (best sfc). For a given altitude as weight decreases and thus also required lift, keeping angle of attack constant we would need to slow down and thus have the engines operating slower than optimum, and not as efficiently. By climbing into thinner air the aircraft has to travel faster for the given angle of attack and also the engines have to operate closer to their design speed thus given better track miles for the fuel burnt.

almost.

It's TAS/drag not lift. If we use lift then we are flying for endurance not range.

SAR=(1/SFC)(TAS/Drag)

see my previous post for a full explanation.

Just taking a swing at this...but it sounds like the goal of 'this flight' is a constant MACH trip..and if that is the case, then a hypotheticaly planned trip at .78 mach at intiatally FL280(or something) would be mean that as your MACH increased to say .80...you might request a higher alt, to put the plane back at .78...and somewhere in this process you might end up at a certain alt, and then at this certain alt to keep .78 for the remainder of the trip, as your weight decreases, you reduce thrust....

Well I guess that's one way to plan a trip....

Modern airliners are designed to cruise at an optimum AoA to stay inside the low drag bucket.
As you burn fuel and get lighter, AoA gets shallow. You then
need to climb into thinner air to get back to your optimum AoA.

If ATC doesnt let you, the second best option is to reduce speed to
increase AoA back to optimum.

Or, get a 787 with variable camber..:p

speed of sound 'a' or 'c' is a function of temp but not Mach which is a function of pressure height

at every pressure height there is a specific temperature and a specific speed of sound so that the T's cancel,...a Mach meter is built like and ASI :)

that's why I think the liberals can be just as racist and short sighted as conservatives:yuk:

terrible:*

but not Mach which is a function of pressure height

.. and CAS. Hence we need speed and height to figure out Mach.

M = (5((qc/ps+1)^0.286-1))^0.5

is a typical sort of equation one finds in the textbooks.

I wouldn't fuss too much about the hieroglyphs .. the important bit is that

M = an equation involving CAS and Hp.

... and CAS :O

it's not like I have not been through this one before:}

.. pass, friend ... :)

Thanks for the reply guys! Your indefatigability towards my contact is much commended for. Hsieh-hsieh ( Thanks in chinese)....:p