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Old 9th Feb 2009, 11:05
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Stuck!

Stuck on a question on my flight planning... can someone explain how this graph works given the information etc?

Hope this is the right place to post this....

ImageShack - Image Hosting :: helpsu9.jpg
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Old 9th Feb 2009, 11:22
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I think you are missing some of the input data for that question.
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Old 9th Feb 2009, 11:25
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Based on the information given, it doesn't seem possible.

Dividing the total moment by all up mass gives the arm as 42in.

This is not on the graph.
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Old 9th Feb 2009, 11:36
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Yep. Seems like a duff question.

CofG is 42 inches aft of datum which is way off the chart and my experience tells me that with a graph like that, even with a very unusual loading you cannot even get the CofG that far forward - unless you fill the back with helium balloons or something.

(Oh, and for your conscience: yes, this is the right place to post questions like this. Even better now that I won a Sharp LCD TV...)
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Old 9th Feb 2009, 11:37
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Hi Dave,

The image does not seem to come across legibly, but in general:

The weight and balance "envelope" as depicted is showing you the maximum permissible center of gravity limits relative, and up to, the maximum gross weight. The actual aircraft weight for the proposed flight, once known by calculation, is plotted up the left side of the graph. Using the calculated C of G (from loading charts provided elsewhere for you, you will have calculated the total aircraft C of G position (doing so may require calculating the moment too, and working backward a bit). You then locate the C of G position along the weight line which you have identified on the chart.

If the plotted C of G position lies outside (left or right) the "envelope" entirely, flight is not permitted. (if it's too high, you're too heavy!) If the plotted point lies inside the "normal" envelope, you may fly the aircraft as a normal category aircraft, and only perform the manuevers stated in the limitations for that type of flight. If the plotted point lies within the "utility" portion of the envelope, additional manuevers (probably including spins) will be permitted. This is to be taken to mean that spins would not be permitted if the C of G point is in the normal envelope. Generally, if it is in the "utility" you may also fly as a "normal" category, though you might be having seat occupancy limitations.

Refer to the actual flight manual (if applicable) for the real information, as what I have told you is only the concept.

If that does not answer the question, let us know in greater detail what you need to know.

Pilot DAR
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Old 9th Feb 2009, 11:39
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Something wrong with those figures - I suspect that the 58800 figure is wrong.
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Old 9th Feb 2009, 12:04
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This is exactly what i thought...... total mass of the plane divided by the total moment should give you the moment of the datum, but 42 shows nowhere on the graph... that's why I was confused... thought I was doing something wrong, couldn't even find it in my PPL manuals..
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Old 9th Feb 2009, 12:28
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As I said before, I suspect that you may be missing more elements of the question. Or as has been mentioned the moment data could be wrong.
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Old 9th Feb 2009, 12:29
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That's off a website designed for your ppl. That's also everything that came with the question..
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Old 9th Feb 2009, 12:36
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Is there any sort of forum associated with that website, or contact details for the people who run it?

I would flag this up as a dodgy question, and see what they say.
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Old 9th Feb 2009, 12:56
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I could try, other than that question the site is very good.

www.pplquiz.co.uk
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Old 9th Feb 2009, 13:23
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Dave,I tried the demo and it seems a good site although I found an error with another unrelated question but can't remember what it was now
I used www.airquiz when I was a student and found it to be very useful.
Good luck
Lister
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Old 9th Feb 2009, 14:03
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Do you have a PPL confuser?

I found mine very useful when I did my PPL.
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Old 9th Feb 2009, 14:10
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Yeah, got one of those too.
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Old 9th Feb 2009, 21:32
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It's a dud question: Bin it and move on............

Cusco.
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Old 15th Feb 2009, 09:33
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ImageShack - Image Hosting :: helpja1.jpg

Can anyone tell me the calculations to working this question out?
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Old 15th Feb 2009, 11:14
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1013-992 = 21mb

21mb x 30 = 630ft

630 + 968 = 1598ft
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Old 16th Feb 2009, 09:45
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Edit... PPLQuiz giving me wrong answers again. Im presuming the last caluclation of an addition is correct?

PPLQUIZ said 338 was the right answer

Last edited by DaveD; 16th Feb 2009 at 10:12.
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Old 16th Feb 2009, 10:26
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What is pressure altitude Dave?

Pressure altitude is the height in the standard atmosphere above the 1013.2mb (hpa) pressure level at which the pressure equals that of the aircraft or point under consideration.
If you were on the ground you would find your pressure altitude by setting 1013 on the altimeter subscale. So you need to add any difference between your current setting and 1013 if lower or deduct it if higher.
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Old 16th Feb 2009, 10:31
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Well, bose-x's calculation is wrong ... as are all of the possible answers given!

Pressure altitude is what the altimeter reads if 1013mb is set on the sub-scale.

If the altimeter reads zero when 992mb is set, it will read 630ft with 1013mb set.

The airfield elevation is a distractor in this question. Unfortunately, it also distracted the questioner!
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