CG problem..

[ranklein]

I'm looking for a fast/ easy way to answer this question.

Gross weight 285,000lbs.How much weight must be moved from station 1030.0 to station 600.0 in order to move the CG 2.3 inches forward?

I know I might be missing some info but in case there's an answer as it is or just to review those CG shifting problems.

[hawk37]

I'm not familiar with the stations on your aircraft, but assuming sta 600 is ahead of sta 1030, and all in inches, then I'd think the answer is 285,000*2.3/(1030-600) = 1524 lbs
Mind you, I've had a few to drink tonight already
Hawk

[ranklein]

man, with drinks you were right on.So, what's the secret? what's a good way to remember where to put the numbers? What's the formula?

Thanks

[hawk37]

I just draw it out on paper. mark 600 inches, and 1030 inches forward of the nose, or whatever the datum (zero point) is. Put the C of G wherever you want, say at 700 inches. Move 285,000 lbs for 2.3 inches, is the same as 1524 lbs for 430 inches (285,000 * 2.3 = 1524 * 430)
Course, I could be wrong. I'd feel better if someone chimed in and agreed

[PickyPerkins]

Very good, hawk37, I agree with your numbers.
Now please explain the approximate position of the CG on the photo below, and in fact, how the whole combination stays in the air at all. :)
Bonus points are also available for explaining how it got into the air, and for a proposed method of landing.
http://pickyperkins.home.infionline.net/daredevil.jpg
Cheers, http://pickyperkins.home.infionline.net/pi.gif

[ranklein]

Yeah Hawk, I guess it works and make sense too!!

Thanks again

[broadreach]

Picky, you incendiary,

Without a loadplan for the Jenny, do you really expect an answer? Just hipshooting, though, the wire stays are an obvious giveaway; they worked it out so that the cg's limited to x% aft and, from the elevator setting, I'd suggest less than 5% aft. If the says weren't there the plane would have been a bucking bronco.

As to how they took off, doh. And as to landing, well, hope the guy on the pilot ladder hasn't been fooling around with the pilot's sister/wife/etc.

[PickyPerkins]

broad reach and hawk37

Incendiary maybe, but I don’t seem to have started a fire.:)

And yes, I did think I might get some sort of answer.

I think that the photo shows enough information to say at least in which direction the CG moved.

For my money I would say that it probably moved FORWARD rather than rearward.

Cheers, http://pickyperkins.home.infionline.net/pi.gif

[arismount]

This is a CG shift equation that you should commit to memory and become very comfortable working. Practice is the key.
Here we go:
Equation 1.: W / GW = CG Shift / Station Shift
That is: Amount of Weight to be moved is to the Gross Weight, as the Shift of CG is to Shift of Station.
Knowing any three of the variables will allow you to solve for the remaining unknown, viz.:
Equation 2.: W = (GW)(CG Shift) / Station Shift
Equation 3.: CG Shift = (W)(Station Shift) / GW
Equation 4.: Station Shift = (CG Shift)(GW) / W
In your specific problem, we are interested in the amount of Weight to be moved between two given stations so as to produce a given CG Shift; therefore, employ Equation 2. Substituting the given quantitities, this becomes:
W = (285,000)(2.3) / (1030 - 600)
Note that the term in the denominator is the difference, in inches, between the two stations given in the problem, and of course simplifies to 430.
Therefore the equation becomes:
W = (285,000)(2.3) / 430. Performing the math, W = 1524.4186 pounds (I will leave it to you to round this figure accordingly; by the laws of significant figures, it would be 1520 pounds; by ordinary rounding rules, it shoud be 1524 pounds; and as a practical matter, I would say 1525 pounds, since 1524 pounds would not be sufficient to move the CG a minimum of 2.3 inches; the latter consideration is probably splitting hairs, I grant you.)
You can either accept the explanation that the proportion given in Equation 1. is true, or go to any available study guide (ASA, Gleim, etc.) for an explanation as to how the formula is derived.
Good luck to you.

[ranklein]

arismount,

Thanks for a wonderful explanation!!!

[PickyPerkins]

ranklein
At the risk of flogging a dead horse ………………

If you are looking to understand the answer to your CG question rather than be buried in formulae, you might like to try the following:

Fig. 1 below shows an aircraft balanced on a fulcrum positioned under its CG. The fulcrum is on a load cell which can measure the weight of the aircraft. The balance is unstable, so that if the aircraft tips slightly forward it will tend to fall over forward more, and similarly if it tips slightly clockwise it will tend to fall back onto its tail. But for the moment its in perfect balance. However, this is not (yet) the aircraft in the problem.

http://pickyperkins.home.infionline.net/CG_Discussion_Fig1.gifFig. 1

Inside the aircraft is a level floor as shown below, and on this floor are two heavy steel cylinders, “A” and “B“, each of mass m and therefore weight mg. “A” is placed at Station 1030 and “B” is an equal distance on the far side of the fulcrum. The cylinder “A” produces an anti-clockwise turning moment tending to cause the aircraft to fall on its nose, while “B” produces an equal and opposite clockwise turning moment tending to cause the aircraft to fall on its tail. These two moments are equal and opposite, so the aircraft is still in balance. The magnitude of each of these moments is numerically equal to mg times (the distance of the cylinder from the fulcrum). The load cell shows that the aircraft including the cylinders weighs a total of 285,000 lbs. The aircraft including the cylinders represents the aircraft in the problem.

http://pickyperkins.home.infionline.net/CG_Discussion_Fig2.gifFig. 2

If we now enlist the aid of a ghostly FA (who is naturally weightless) to roll “A” from Station 1030 to Station 600, the anti-clockwise moment produced by “A” will INCREASE to mg times (the new distance of “A” from the fulcrum), and the aircraft will tend to fall over forwards (cylinder “B“ is not moved). The CHANGE in the moment produced by “A” will be equal to the DIFFERENCE between the initial and final moments due to “A”, which turns out to be mg times (the distance from Station 1030 to Station 600), or mg x 430 inch-lbs.

http://pickyperkins.home.infionline.net/CG_Discussion_Fig3.gifFig. 3

Now, if you were outside the aircraft, with no knowledge of what was going on inside, you could observe the following sequence:
Initially the aircraft would be in balance and the load cell would show the weight to be 285,000 lbs as in Fig. 1 above.
At a later time the aircraft would be seen to have an unchanged weight of 285,000 lbs but it would be trying to fall over forwards as shown below, and the obvious explanation would be that something had happened internally to move the CG forward. In the “external” view, the anti-clockwise moment causing this would be equal to 285,000 times (the distance that the CG had moved forward) in inch.lbs. We are told that this distance is 2.3 inches.

http://pickyperkins.home.infionline.net/CG_Discussion_Fig4.gifFig. 4

Thus, we have two different descriptions of why the aircraft is tending to fall over forwards:

(a) the “internal” view where we know that the tipping moment is due to moving the cylinder “A” from Station 1030 to Station 600, and is equal to mg x 430, and
(b) the “external” view where we know only that the CG has moved 2.3 inches involving a tipping moment of 285,000 x 2.3 inch.lbs.

Since these are two descriptions of the SAME THING, these two moments must be equal.

That is, the forward-tipping moment = mg x 430, and is also = 285,000 x 2.3 inch.lbs
Therefore, mg = (285,000 x 2.3)/430 = 1,524 lbs.

Cheers, http://pickyperkins.home.infionline.net/pi.gif

[ranklein]

PickyPerkins,

Very nicely described!!!!!!!!!!!!!!!

Excellent explanation!

Cheers