W&B answ:notes - Practice weight and balance questions with answers PDF

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Practice weight and balance questions with answers...

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Q1. 24.2%MAC, let me know if you need help figuring this out Q2.

Put the two like numbers of distance along the same line. Below, draw the conversion distance line. Since two lines should overlap, find the difference between the two (subtract). Then, divide this number by the leftover distance and multiply by 100 for percentage. W = 5700kg

d = 35cm

D = 350cm

5700 x 35 = w x 350 w = (5700x35)/350 =570 kg Q3.

Along the same line, plot the two distances that are a given. Next, below, plot the two distances along the same line that are what it should be. Find the difference between the two groups of numbers (subtract) Then plot figures into equation Wxd=wxD to give missing number 2500kg x 90cm = 225000 -154kg x 150cm = -23100 2346kg x ARM

= 201900

ARM = 201900/2346 = 86cm -

Do this like a weight and balance sheet eg. Take-off weight – fuel burn – landing Multiply the biggest weight by its arm to get moment Multiply small weight by arm to get moment Add/subtract weights and moments Then divide to find missing arm value

Q4. answer is 504.6cm aft of datum, there are different ways of getting this, I used W=5700, d = +10, w = 243 I put the weight on top of the original cofg to start then use Wxd=wxD to find how far you can move it back (D) this gives you D = 234.6cm aft on where you put the weight to start with (270 cm aft of datum) so you could put the weight 504.6cm aft of datum (234.6+270) and be on the rear limit. the text book may have a slightly different method, if you got the correct number it doesn’t really matter how you got there though. -

Use biggest weight as what is given Find smallest weight by subtracting actual weight from max weight

-

Q5

Smallest distance will be the limits in which you can move/add cargo, eg. From existing point to absolute furthest point Find big distance through Wxd=wxD to find distance that weight can be added/moved. 9385 lbs x +300lbs

x

9685

68"

= 638180

132" = +39600

x ARM = 677780

ARM = 677780/9685 = 69.98" Yes this is within limits Q6

Think of question again as weight and balance sheet. Start with zero fuel weight using original weight and it’s arm to find moment Multiply additional weight by its arm to get its moment Add the two weights and moments together Divide moment by weight to find the arm 6300kg

x

+38cm =

239400

+250kg

x

-55cm

-13750

=

---------------------------------------------------------------6550kg x

ARM

=

225650

ARM = 225650/6550 = 34.45 cm (+'ve means still aft of datum) , NO this is just out of limits -

Think of question again as weight and balance sheet. Start with zero fuel weight using original weight and it’s arm to find moment Multiply additional weight by its arm to get its moment; however this time it is needed in front of datum to arm is to be negative value Add the two weights and moments together Divide moment by weight to find the arm Because of the negative arm, this causes the CoG to be below limits....