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Cyber 3 Question 5

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jukzizy created the topic: Cyber 3 Question 5

The centre of gravity range for an echo aeroplane at grossweight of 2785 Kg is.

a) 31.1 mm
b) 2680 mm
c) 2515 mm
d) 165 mm

how do we solve this question if index is not given?

if i use fwd limit formula:

PW-2360×.27+2400
=2785-2360×.27+2400
=2515

2680-2515
= 165 mm

need clarification on this question.
#1

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bobtait replied the topic: Cyber 3 Question 5

The centre of gravity range is simply the distance from the forward limit to the aft limit.

Your working is correct. You don't need to be given any moment index. Simply subtract the forward limit at that weight from the aft limit. The aft limit does not change with weight.
#2
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  • Carello

Carello replied the topic: Cyber 3 Question 5

PW-2360×.27+2400
=2785-2360×.27+2400
=2515


This in incorrect
2785-2360×0.27+2400 = 2515

What you should have written was
Forward Limit = (2785-2360)×0.27+2400 = 2514.75 mm
= 2515 mm (rounded)

Brackets are important. If you remove them, you will get the wrong answer.
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  • John.Heddles
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John.Heddles replied the topic: Cyber 3 Question 5

Some thoughts ...

(a) This in incorrect Not quite the case. While Jukzizy has been a tad casual with his (her?) typing he (she?) has run the calculation following the normal algebraic rules and ended up with the expected answer

(b) there is not much point in rounding off the slope value to 0.27 and then taking the calculated answer to multiple decimals. In this case, carrying the full values would give a CG of 2515.3 to a precision of one decimal place

(c) depending on one's preferences the linear equation for the upper forward limit line can be simplified further to

CG = 160/590 x WT + 1760

= 0.271186441 (etc) x WT + 1760

which we would round off further to something like

= 0.27 x WT + 1760

to get rid of the silly precision of the slope coefficient in the previous line. This round off is the reason for the discrepancy referred to at (b)

Engineering specialist in aircraft performance and weight control.
#4

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jukzizy replied the topic: Cyber 3 Question 5

Thank you very much now it is clear..how to go about it.
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  • Carello

Carello replied the topic: Cyber 3 Question 5

(a) This in incorrect Not quite the case. While Jukzizy has been a tad casual with his (her?) typing he (she?) has run the calculation following the normal algebraic rules and ended up with the expected answer

My point was, the calculation was not run using algebraic rules. Yes, jukzizy got the correct answer, but the working and the answer were inconsistent. The algebra requires brackets.

If the working is set out incorrectly, chances are, the answer will be incorrect. It all comes down to correct technique and attention to detail.

2785-2360×0.27+2400 = 4547.8 not 2515

b) there is not much point in rounding off the slope value to 0.27 and then taking the calculated answer to multiple decimals. In this case, carrying the full values would give a CG of 2515.3 to a precision of one decimal place

Agreed. Using the normal rules of significant figures, the answer cannot be more precise than the inputs.

(c) depending on one's preferences the linear equation for the upper forward limit line can be simplified further to
CG = 160/590 x WT + 1760


For the echo, the forward CG limit is not linear, it changes at 2360 kg.

The equation is correct for all weights between 2360 and 2950 kg inclusive. It falls apart for weights less than 2360 kg. I'm guessing that this is what you meant by "one's preferences"?.

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  • John.Heddles
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John.Heddles replied the topic: Cyber 3 Question 5

I think we are dancing around the same thoughts ? Some further thoughts, if I may ?

My point was, the calculation was not run using algebraic rules. Yes, jukzizy got the correct answer, but the working and the answer were inconsistent. The algebra requires brackets.

Indeed. However, while Jukzizy has been a little lax in that he omitted the parentheses in his post, he has run the calculations with them. I think, perhaps, you are being a tad hard on the post ?

For the echo, the forward CG limit is not linear, it changes at 2360 kg.

As is the case for the very great majority of light aircraft envelopes.

The equation is correct for all weights between 2360 and 2950 kg inclusive. It falls apart for weights less than 2360 kg.

Which is why I specifically specified that I was addressing the "upper forward limit line" rather than the forward limit, in toto. Indeed, the two equations are identical, if presented a little differently. Your comment applies with equal validity to the use of your own equation, as cited.

I'm guessing that this is what you meant by "one's preferences"?.

Not at all.

My comment related to the observation that Bob's equation is presented in a form which provides a link to the derivation. I haven't discussed this with him but I presume the reasoning is for student understanding and, if so, I certainly have no problem with that. However, it is conventional to represent a linear equation in the form y = mx + b and a simple further step with Bob's equation results in the standard form. From my perspective, the standard form offers the benefit of being simpler for the student to commit to memory.

The reference to "preferences" simply relates to selection of one equation form or the other.

Seeing we are discussing this point, it is worthwhile considering the fact that the use of equations for this purpose is quite unjustified (even inappropriate).

The derivation accuracy and rational precision of the envelope limitations do not warrant going to such lengths. If the examiner so requires for examination purposes, so be it, but students should be aware that it is quite inappropriate for routine use. Indeed, were we to do this in the real world, quite apart from its being a tad silly, one would need a new equation for each successive Type which we might operate. In reality, the accuracy obtained by careful reading of the graphs should provide more than sufficient accuracy for the examination. There is one caveat, however - the engineering world favours rational considerations in which regard, the limits should be viewed with a bit of haziness. The legal world, however, is more concerned with the black and white of the written word. It remains essential that pilots ensure that their steeds remain within the AFM/POH prescribed limits which, in turn, derive from the Type Certificate for the aircraft Type.

As a sideline comment, it is important that students are aware that, while the upper forward limit line for the CG vs WT graph usually is linear, this isn't the case for the alternative presentation as moment (or IU) vs WT. For the latter case, the line segment is a curve (quadratic) although this is not always readily obvious from visual inspection unless a more suitable datum is used than that derived from the GAMA Spec 1 POH protocols. Not difficult to figure the equation but, again, quite silly for general usage.

Engineering specialist in aircraft performance and weight control.
#7

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  • Carello

Carello replied the topic: Cyber 3 Question 5

Your comment applies with equal validity to the use of your own equation, as cited.

In fact, there is no linear equation that will correctly model the forward CoG limit over the full range of weights for the Echo. More on this later.

However, it is conventional to represent a linear equation in the form y = mx + b and a simple further step with Bob's equation results in the standard form.

Absolutely true. If you are going to use a linear equation, the slope-intercept form is generally the way to go. That being said, the simple slope-intercept form does lose some important information about what happens at 2360 kg. If the equation is used blindly, the output can be misleading, not to mention wrong.

For this reason I think the original form is more informative; it is still in slope-intercept form. However, the equation below reminds us that something happens when the Wt falls below 2360 kg.

Forward CoG = (160/590)(Wt-2360)+2400

As stated above, neither linear equation is ideal. Each equation has its strengthens and both suffer from problems when the Wt falls below 2360 kt.

The problem is that the forward CoG limit for the Echo is not linear. For that reason, we need a nonlinear function to model the forward CoG limit over the entire weight range [2000-2950 kg].

Luckily, we can model the forward CoG limit with the "Piecewise Function" below:



John, I would have to agree with your closing comments.
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