- Joined
- Apr 15, 2012
- Messages
- 110
- Reaction score
- 12
My partner and I have long debated the benefit of carrying ballast in our Aztec F. On a couple instances, with just me on board for one leg, then adding about 100# for the next leg, TAS goes up by 4-5 knots. My partner says something else must have changed because the addition of the weight would more than offset any drag reduction.
We both agree that for any given weight, the further back the CG the better. However, I think carrying an extra 100# of dead weight just to push CG to the rear is also beneficial.
This assumes we stay within the allowable limits of the airplane.
My friend wrote to Peter Garrison, who writes for Flying Magazine. His response is attached below. Peter seems to agree with my partner. But experience tells me differently.
First, whatever the answer, the speed difference would be probably erased by random puffs of wind, so the question has no practical importance, as you might suspect from the fact that it is never discussed.
Nevertheless, impractical questions are interesting and fun, so I'll try to tackle this one.
The question boils down to: Is the induced-drag penalty resulting from an additional 100 pounds of dead weight greater or less than the trim drag penalty of the tail downforce with a forward CG?
I'm guessing, but let's say that the stabilator is three times as far from the center of lift of the wing as the baggage compartment is. Therefore, a 100-pound downforce in the baggage compartment is equivalent to a 33 pound downforce at the horizontal stabilizer.
On the Internet I find that the CG range is 15 inches. Let's say the most forward CG is 10 inches ahead of the center of wing lift (ie the 25% of chord point). If the CG is at the forward limit and the airplane weighs 4,000 pounds, the nose-down moment is 40,000 in-lbs.
On the general principle that the stabilator position is around 3 times the wing chord aft, I'll guess that the tail moment arm is around 200 inches and so the stabilator is proving a 200-pound downforce (40,000/200).
So now the question becomes: Is the trim drag saved by reducing that 200 pounds to 167 pounds (that is, 200 - 33) greater than the induced drag increase owing to 100 pounds extra baggage?
With further wild guesses, I'll say that the Aztec cruises at an L/D ratio of 10, and that induced drag is around 30 percent of the total. So the drag is 4,000/10 or 400 pounds, and the induced portion is 120 pounds. Increasing weight by 100 pounds raises the induced drag by 3 pounds.
Now, what is the induced drag of the stabilator, that is, the trim drag? This is tricky, because the stabilator is interrupted in the middle by the large fuselage with its thick boundary layer, so it does not behave like a pure, isolated wing. The lift coefficient of the stabilator, assuming an indicated airspeed of 150 knots and an area of 35 square feet, is 0.075. Using a classical formula for induced drag, CL^2/A*3.14, where A is aspect ratio, and assuming an aspect ratio of 4, we get (roughly) .0005. Let's double that because of the fuselage. So the induced drag is 1/75th of the lift, or 200/75, around 3 pounds.
So, reducing the downforce by adding 100 pounds in the baggage compartment reduced the induced drag of the stabilator by 33/75, or about 1/2 pound. But adding 100 pounds of deadweight increases the induced drag of the airplane by 3 pounds. Thus, the ballast increases total drag by 2.5 pounds. Speed is proportional to the square root of drag. The square root of (402.5/400) is 1.003, so the speed loss is 1/2 knot (150 - 150/1.003).
While all these numbers are approximate, they are conservative (I think) and represent a worst case in terms of trim drag. They suggest to me that ballasting to relieve trim drag is never a good idea. On the other hand, carrying baggage or cargo as far aft as possible (and legal) is beneficial, but only to a very small degree.
If I have made some huge mistake, as frequently happens, I will be glad to be corrected.
We both agree that for any given weight, the further back the CG the better. However, I think carrying an extra 100# of dead weight just to push CG to the rear is also beneficial.
This assumes we stay within the allowable limits of the airplane.
My friend wrote to Peter Garrison, who writes for Flying Magazine. His response is attached below. Peter seems to agree with my partner. But experience tells me differently.
First, whatever the answer, the speed difference would be probably erased by random puffs of wind, so the question has no practical importance, as you might suspect from the fact that it is never discussed.
Nevertheless, impractical questions are interesting and fun, so I'll try to tackle this one.
The question boils down to: Is the induced-drag penalty resulting from an additional 100 pounds of dead weight greater or less than the trim drag penalty of the tail downforce with a forward CG?
I'm guessing, but let's say that the stabilator is three times as far from the center of lift of the wing as the baggage compartment is. Therefore, a 100-pound downforce in the baggage compartment is equivalent to a 33 pound downforce at the horizontal stabilizer.
On the Internet I find that the CG range is 15 inches. Let's say the most forward CG is 10 inches ahead of the center of wing lift (ie the 25% of chord point). If the CG is at the forward limit and the airplane weighs 4,000 pounds, the nose-down moment is 40,000 in-lbs.
On the general principle that the stabilator position is around 3 times the wing chord aft, I'll guess that the tail moment arm is around 200 inches and so the stabilator is proving a 200-pound downforce (40,000/200).
So now the question becomes: Is the trim drag saved by reducing that 200 pounds to 167 pounds (that is, 200 - 33) greater than the induced drag increase owing to 100 pounds extra baggage?
With further wild guesses, I'll say that the Aztec cruises at an L/D ratio of 10, and that induced drag is around 30 percent of the total. So the drag is 4,000/10 or 400 pounds, and the induced portion is 120 pounds. Increasing weight by 100 pounds raises the induced drag by 3 pounds.
Now, what is the induced drag of the stabilator, that is, the trim drag? This is tricky, because the stabilator is interrupted in the middle by the large fuselage with its thick boundary layer, so it does not behave like a pure, isolated wing. The lift coefficient of the stabilator, assuming an indicated airspeed of 150 knots and an area of 35 square feet, is 0.075. Using a classical formula for induced drag, CL^2/A*3.14, where A is aspect ratio, and assuming an aspect ratio of 4, we get (roughly) .0005. Let's double that because of the fuselage. So the induced drag is 1/75th of the lift, or 200/75, around 3 pounds.
So, reducing the downforce by adding 100 pounds in the baggage compartment reduced the induced drag of the stabilator by 33/75, or about 1/2 pound. But adding 100 pounds of deadweight increases the induced drag of the airplane by 3 pounds. Thus, the ballast increases total drag by 2.5 pounds. Speed is proportional to the square root of drag. The square root of (402.5/400) is 1.003, so the speed loss is 1/2 knot (150 - 150/1.003).
While all these numbers are approximate, they are conservative (I think) and represent a worst case in terms of trim drag. They suggest to me that ballasting to relieve trim drag is never a good idea. On the other hand, carrying baggage or cargo as far aft as possible (and legal) is beneficial, but only to a very small degree.
If I have made some huge mistake, as frequently happens, I will be glad to be corrected.
