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 mawingho From Hong Kong, joined Jul 2012, 41 posts, RR: 0Posted Sun Sep 2 2012 13:58:29 UTC (3 years 9 months 2 days 8 hours ago) and read 6471 times:

I am not quite understand the Load System Charlie calculation.

Given
Aircraft empty weight = 689 kg
Empty aircraft index units = 19,522
Oil = 7 kg
Row 1: pilot and passenger = 110 kg
Baggage (baggage compartment) = 75 kg
Fuel = 140 litres

What is the maximum passenger weight, in kg, that may be carried in row 2 at take-off?

GW (kg) x distance to move CG (mm) / new STA - current STA

GW = 980
Distance to move = AFT limit - current STA = 3,004 - 2,930 = 74 mm
New STA = 3600
Current STA = 2930
Therefore, ballast to add = 980 x 74 / (3,600 - 2,930) = 108 kg

So, 108 kg is max. passenger weight for row 2 with ZFW and take-off limits checked.
Finally, the key to loading an aircraft is not to exceed any limit. You are advised to check-plot weight moment or weight arm at both ZFW and TOW to ensure compliance.

Reference: Robson, David. "Chapter 10: Loading." Basic Aeronautical Knowledge (BAK). Darra: Aviation Theory Centre Pty, 2009. 329. Print.

Why it's related to AFT limit and moving the STA?

But finally we got the max. passenger weight rather than the distance that the CG need to be moved.

If the AFT limit is 3,004, then why the New STA can be 3,600?

I am just a bit confusing why this formula need to be used and why we always consider the Arm (mm)?

Hope someone can help me out. Thanks!

[Edited 2012-09-02 14:03:44]

 tdscanuck From Canada, joined Jan 2006, 12710 posts, RR: 78 Reply 1, posted Sun Sep 2 2012 17:29:59 UTC (3 years 9 months 2 days 4 hours ago) and read 6402 times:

 Quoting mawingho (Thread starter):I am just a bit confusing why this formula need to be used and why we always consider the Arm (mm)?

I can't answer the specific question about the station vs. aft limit because I agree it's not clear in the calculation, but the reason you always consider the moment arm is because the change in the CG depends on *both* the weight and where you put it (the moment arm). 100kg placed near the CG may be no problem while that same 100kg placed at the extreme ends may put you way outside the CG limits.

Tom.

 zeke From Hong Kong, joined Dec 2006, 10843 posts, RR: 76 Reply 2, posted Sun Sep 2 2012 18:43:30 UTC (3 years 9 months 2 days 3 hours ago) and read 6373 times:

 We are addicted to our thoughts. We cannot change anything if we cannot change our thinking – Santosh Kalwar
 mawingho From Hong Kong, joined Jul 2012, 41 posts, RR: 0 Reply 3, posted Mon Sep 3 2012 08:40:06 UTC (3 years 9 months 1 day 13 hours ago) and read 6209 times:

 Quoting zeke (Reply 2):Adding 135 kg at 3600 mm would add 4860 IU to the existing TOW, the TOW would then be 1115 kg, and 28711.1 + 4860 = 33571.1 IU. 33571.1 *100/115 = 3011 mm (i.e. 7 mm aft of the limit). So you cannot carry 135 kg in row 2.

Thank you for your reply. Hmm..however, I do not quite understand why taking off 135kg from MTOW and then add 135kg at 3600 mm would add 4860 IU to TOW?

I think I am lost... would you mind to explain a bit here? Coz I am new to this stuff..

 zeke From Hong Kong, joined Dec 2006, 10843 posts, RR: 76 Reply 4, posted Mon Sep 3 2012 18:17:41 UTC (3 years 9 months 1 day 3 hours ago) and read 6117 times:

 Quoting mawingho (Reply 3):Hmm..however, I do not quite understand why taking off 135kg from MTOW and then add 135kg at 3600 mm would add 4860 IU to TOW?

It is just the IU for adding that weight

The arms for each station I listed above
Oil 1230
Row 1 2750
Fuel 2950
Row 2 3600
Baggage 4210

The weight at each station is

Oil - 7
Row1 - 110
Row2 - 0
Baggage 75
Fuel - 99

To work out the IU for each station, multiply the weight by the arm
Oil = 1230x 7 = 8610 kg.mm = 86.1 IU (divide by 100 to get IU)
Row 1 = 2750 x 110 = 302500 kg.mm = 3025 IU

So for an empty row 2 the IU is
3600 x 0 = 0 kg.kmm = 0 IU
With 135 kg
3600 x 135 = 486000 kg.mm = 4860 IU.
With 121 kg
3600 x 121 = 435600 kg.mm = 4356 IU.

We see the influence of all three of these on the TOW and IU

With no weight in row 2, we have no change to the TOW or IU

With 135 kg, TOW goes to 980+135 = 1115 kg (i.e. MTOW), the change in IU is 28710 + 4860 = 33570 IU, to work out the CG, first convert the IU to the moment, 33570 x 100 = 3357000 kg.mm.

The CG is then 3357000/1115 = 3011 mm (i.e. aft of the rear limit)

With 121 kg, TOW goes to 980+121 = 1101 kg, the change in IU is 28710 + 4356 = 33066 IU, convert the IU to the moment, 33066 x 100 = 3306600 kg.mm.

The CG is then 3306600/1101 = 3004 mm (i.e. the rear limit)

 We are addicted to our thoughts. We cannot change anything if we cannot change our thinking – Santosh Kalwar
 mawingho From Hong Kong, joined Jul 2012, 41 posts, RR: 0 Reply 5, posted Tue Sep 4 2012 06:51:14 UTC (3 years 9 months 15 hours ago) and read 6050 times:

 Quoting zeke (Reply 4):With 135 kg, TOW goes to 980+135 = 1115 kg (i.e. MTOW), the change in IU is 28710 + 4860 = 33570 IU, to work out the CG, first convert the IU to the moment, 33570 x 100 = 3357000 kg.mm.

Thank you for your reply. I just do not know how u get 28710 and 33570 in the above sentence?

Did you use some mapping tables to get it?

Thanks!

 zeke From Hong Kong, joined Dec 2006, 10843 posts, RR: 76 Reply 6, posted Tue Sep 4 2012 12:23:40 UTC (3 years 9 months 9 hours ago) and read 5984 times:

 Quoting mawingho (Reply 5): Thank you for your reply. I just do not know how u get 28710 and 33570 in the above sentence?

The 28710 comes from your OP
TOW
980 kg
CG arm - 2930 mm
IU - 28710

The 33570 comes about by adding the weight in row 2 where you have it blank
New TOW = 980 + 135 = 1115 kg
New TOW IU = 28710 + 4860 = 33570 IU
New CG arm = IUx100/mass = (33570 * 100) / 1115 = 3011 mm.

[Edited 2012-09-04 12:45:20]

 We are addicted to our thoughts. We cannot change anything if we cannot change our thinking – Santosh Kalwar
 mawingho From Hong Kong, joined Jul 2012, 41 posts, RR: 0 Reply 7, posted Thu Sep 6 2012 06:47:47 UTC (3 years 8 months 4 weeks 15 hours ago) and read 5807 times:

 Quoting zeke (Reply 2):Load to add = (GW x change of CG)/(distance between new CG and where you add the weight) GW=980 kg Change of CG - Aft limit - current CG = 3004 - 2930 = 74 mm Adding weight to row 2, that is 3600 mm, and the new Cg is the aft limit = 3004 mm Load to add = (GW x change of CG)/(distance between new CG and where you add the weight) =980x74/(3600-3004)=121 kg

But the book used (GW x disance to move) / (new STA - current STA)
which is 980 x 74 / (3600 - 2930) = 108kg.
2930 is the curernt STA which is derived from current TOW arm(mm)

Why the formula used in the book is wrong?

 zeke From Hong Kong, joined Dec 2006, 10843 posts, RR: 76 Reply 8, posted Fri Sep 7 2012 20:00:02 UTC (3 years 8 months 3 weeks 6 days 2 hours ago) and read 5700 times:

 Quoting mawingho (Reply 7): Why the formula used in the book is wrong?

I do not know, it is basically the same as the one I used, maybe you have used the wrong arms.

I can derive the formula for you without any problems, it is a simply the sum of moments, we are using 3 stations, the current CG arm (CGA), Row 2 arm (R2A), and the Aft Limit Arm (ALA). We are trying to work out the load to add (LTA) at the RTA to the gross weight mass (GWM) so that the (LTA+GWM)xALA is still in balance.

GWMxCGA=[(GWM+LTA)xALA]-(LTAxR2A)

in words this is the the moment of initial gross weight at CG must be equal to the GW plus the load added in row 2 at the aft limit, minus the load added in row 2 by the row 2 arm.

rearrange this as

(LTAxR2A) = [(GWM+LTA)xALA] - GWMxCGA
(LTAxR2A) = GWMxALA +LTAxALA - GWMxCGA
(LTAxR2A) = GWMx(ALA -CGA) + LTAxALA

ALA-CGA is the "distance to move" or "change of CG"

(LTAxR2A) = GWMx(ALA -CGA) + LTAxALA

rearrange this

GWMx(ALA -CGA) = (LTAxR2A) - (LTAxALA)
GWMx(ALA -CGA) = LTA (R2A - ALA)

R2A-ALA is the distance between where you are adding weight and the aft limit

LTA = GWMx(ALA -CGA) / (R2A - ALA)

Load to add = (GW x change of CG)/(distance between new CG and where you add the weight)

 We are addicted to our thoughts. We cannot change anything if we cannot change our thinking – Santosh Kalwar
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