PPRuNe Forums - View Single Post - Why does an aircraft descend quicker when it is lighter?
Old 19th Apr 2010, 14:18
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Keith.Williams.
 
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Jimmygill,

I have no problem with anything that you have just said.

But you appear to be determined to ignore the relevance of power required

As you have said:

At any speed ROD = TAS / (L/D)

Strictly speaking the lift in this equation should be the weight

So we can restate the equation as

ROD = TAS / (W/D)

Rearranging this gives

ROD = TAS x (D/W)

ROD = (TAS x D ) / W

But TAS x Drag = Power required

So ROD = Power required / Weight

Which is what pretty well every text book on this matter states.

Instead of all of the drag curves and polar diagrams, all we needed to do was look at the power required curves for the two aircraft.

My reason for feeling uneasy about excessive emphasis on the L/D ratio is that it ignores the vital part that the TAS has to play in this matter.

As we accelearte towards Vmp the rate of reduction in drag is greater than the rate of increase in TAS, so the power required decreases.

At Vmp the rate of decrease in drag exactly balances the rate of increase in TAS so the power curve is flat.

Between Vmp and Vmd the rate of decrease in drag is less than the rate of increase in TAS so power required increases.

Above Vmd both drag and TAS are increasing, so power required increases.

At each stage along the speed range it is not just the L/D ratio that is determining the shape of the power required curve (and hence the way ROD changes). It is both the TAS and the L/D ratio.


In considering a speed limited descend the case to case variation in TAS is not permitted. Hence for cases where TAS is kept constant RoD is purely a function of the ratio L/D
This is true, but it ignores the fact that the chosen TAS has a bearing on the outcome. Pick a different TAS, and you may well get the opposite outcome.
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